id stringlengths 11 15 | title stringlengths 7 171 | problem stringlengths 9 4.33k | solution stringlengths 6 19k | problem_wikitext stringlengths 9 4.42k | solution_wikitext stringlengths 7 19.1k | proof_title stringlengths 9 171 | theorem_url stringlengths 34 198 | proof_url stringlengths 36 198 | categories listlengths 0 9 | theorem_references listlengths 0 36 | proof_references listlengths 0 253 |
|---|---|---|---|---|---|---|---|---|---|---|---|
proofwiki-19500 | Ordering where every Subclass has Smallest Element is Well-Ordering | Let $A$ be a class.
Let $\RR$ be an ordering on $A$.
Let $\RR$ be such that every subclass of $A$ has a smallest element under $\RR$.
Then $\RR$ is a well-ordering. | We are given that $\RR$ is an ordering.
It will be shown that $\RR$ is a total ordering, from which the fact that it is a well-ordering follows directly {{hypothesis}}.
Let $x$ and $y$ be elements of $A$.
Then $\set {x, y}$ is a subclass of $A$.
Hence {{hypothesis}} $\set {x, y}$ has a smallest element.
That is, either... | Let $A$ be a [[Definition:Class (Class Theory)|class]].
Let $\RR$ be an [[Definition:Ordering (Class Theory)|ordering]] on $A$.
Let $\RR$ be such that every [[Definition:Subclass|subclass]] of $A$ has a [[Definition:Smallest Element (Class Theory)|smallest element]] under $\RR$.
Then $\RR$ is a [[Definition:Well-Or... | We are given that $\RR$ is an [[Definition:Ordering (Class Theory)|ordering]].
It will be shown that $\RR$ is a [[Definition:Total Ordering (Class Theory)|total ordering]], from which the fact that it is a [[Definition:Well-Ordering (Class Theory)|well-ordering]] follows directly {{hypothesis}}.
Let $x$ and $y$ be [... | Ordering where every Subclass has Smallest Element is Well-Ordering | https://proofwiki.org/wiki/Ordering_where_every_Subclass_has_Smallest_Element_is_Well-Ordering | https://proofwiki.org/wiki/Ordering_where_every_Subclass_has_Smallest_Element_is_Well-Ordering | [
"Well-Orderings"
] | [
"Definition:Class (Class Theory)",
"Definition:Ordering/Class Theory",
"Definition:Subclass",
"Definition:Smallest Element/Class Theory",
"Definition:Well-Ordering/Class Theory"
] | [
"Definition:Ordering/Class Theory",
"Definition:Total Ordering/Class Theory",
"Definition:Well-Ordering/Class Theory",
"Definition:Element/Class",
"Definition:Subclass",
"Definition:Smallest Element/Class Theory",
"Definition:Total Ordering/Class Theory",
"Definition:Total Ordering/Class Theory",
"D... |
proofwiki-19501 | Survival Function preserves Inequality | Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $f, g : X \to \overline \R$ be $\Sigma$-measurable functions such that:
:$\size f \le \size g$ $\mu$-almost everywhere.
Let $F_f$ and $F_g$ be the survival functions of $f$ and $g$ respectively.
Then:
:$F_f \le F_g$ | We aim to show that:
:$\map {F_f} \alpha \le \map {F_g} \alpha$ for all $\alpha \in \hointr 0 \infty$.
That is:
:$\map \mu {\set {x \in X : \size {\map f x} \ge \alpha} } \le \map \mu {\set {x \in X : \size {\map g x} \ge \alpha} }$ for all $\alpha \in \hointr 0 \infty$.
Since $f \le g$ $\mu$-almost everywhere, there... | Let $\struct {X, \Sigma, \mu}$ be a [[Definition:Measure Space|measure space]].
Let $f, g : X \to \overline \R$ be [[Definition:Measurable Function|$\Sigma$-measurable functions]] such that:
:$\size f \le \size g$ [[Definition:Almost Everywhere|$\mu$-almost everywhere]].
Let $F_f$ and $F_g$ be the [[Definition:Surv... | We aim to show that:
:$\map {F_f} \alpha \le \map {F_g} \alpha$ for all $\alpha \in \hointr 0 \infty$.
That is:
:$\map \mu {\set {x \in X : \size {\map f x} \ge \alpha} } \le \map \mu {\set {x \in X : \size {\map g x} \ge \alpha} }$ for all $\alpha \in \hointr 0 \infty$.
Since $f \le g$ [[Definition:Almost Everyw... | Survival Function preserves Inequality | https://proofwiki.org/wiki/Survival_Function_preserves_Inequality | https://proofwiki.org/wiki/Survival_Function_preserves_Inequality | [
"Survival Functions"
] | [
"Definition:Measure Space",
"Definition:Measurable Function",
"Definition:Almost Everywhere",
"Definition:Survival Function"
] | [
"Definition:Almost Everywhere",
"Definition:Null Set",
"Measurable Functions Determine Measurable Sets",
"Definition:Null Set",
"Null Sets Closed under Subset",
"Definition:Countably Additive Function",
"Measure is Monotone",
"Measure is Monotone",
"Definition:Subset"
] |
proofwiki-19502 | Bound on Survival Function of Pointwise Sum | Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $f, g : X \to \overline \R$ be $\Sigma$-measurable functions such that the pointwise sum $f + g$ is well-defined.
Let $F_{f + g}$ be the survival function of the pointwise scalar multiple $f + g$.
Let $F_f$ and $F_g$ be the survival functions of $f$ and $g$ respec... | Let $\alpha, \beta \in \hointr 0 \infty$.
We show that:
:$\set {x \in X : \size {\map f x + \map g x} \ge \alpha + \beta} \subseteq \set {x \in X : \size {\map f x} \ge \alpha} \cup \set {x \in X : \size {\map g x} \ge \beta}$
Let $x \in X$ be such that $\size {\map f x + \map g x} \ge \alpha + \beta$.
If $\size {\ma... | Let $\struct {X, \Sigma, \mu}$ be a [[Definition:Measure Space|measure space]].
Let $f, g : X \to \overline \R$ be [[Definition:Measurable Function|$\Sigma$-measurable functions]] such that the [[Definition:Pointwise Addition of Extended Real-Valued Functions|pointwise sum]] $f + g$ is well-defined.
Let $F_{f + g}$ b... | Let $\alpha, \beta \in \hointr 0 \infty$.
We show that:
:$\set {x \in X : \size {\map f x + \map g x} \ge \alpha + \beta} \subseteq \set {x \in X : \size {\map f x} \ge \alpha} \cup \set {x \in X : \size {\map g x} \ge \beta}$
Let $x \in X$ be such that $\size {\map f x + \map g x} \ge \alpha + \beta$.
If $\size ... | Bound on Survival Function of Pointwise Sum | https://proofwiki.org/wiki/Bound_on_Survival_Function_of_Pointwise_Sum | https://proofwiki.org/wiki/Bound_on_Survival_Function_of_Pointwise_Sum | [
"Survival Functions"
] | [
"Definition:Measure Space",
"Definition:Measurable Function",
"Definition:Pointwise Addition of Extended Real-Valued Functions",
"Definition:Survival Function",
"Definition:Pointwise Scalar Multiplication of Extended Real-Valued Functions",
"Definition:Survival Function"
] | [
"Triangle Inequality",
"Definition:Subset",
"Measure is Monotone",
"Measure is Subadditive"
] |
proofwiki-19503 | Survival Function is Decreasing | Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $f : X \to \overline \R$ be a $\Sigma$-measurable function.
Let $F_f$ be the survival function of $f$.
Then $F_f$ is a decreasing function. | Let $\alpha, \beta \in \hointr 0 \infty$ with $\alpha \le \beta$.
We show:
:$\map {F_f} \beta \le \map {F_f} \alpha$
We first show:
:$\set {x \in X : \size {\map f x} \ge \beta} \subseteq \set {x \in X : \size {\map f x} \ge \alpha}$
Let $x \in X$ have $\size {\map f x} \ge \beta$.
Then, since $\beta \ge \alpha$ we h... | Let $\struct {X, \Sigma, \mu}$ be a [[Definition:Measure Space|measure space]].
Let $f : X \to \overline \R$ be a [[Definition:Measurable Function|$\Sigma$-measurable function]].
Let $F_f$ be the [[Definition:Survival Function|survival function]] of $f$.
Then $F_f$ is a [[Definition:Decreasing Function|decreasing f... | Let $\alpha, \beta \in \hointr 0 \infty$ with $\alpha \le \beta$.
We show:
:$\map {F_f} \beta \le \map {F_f} \alpha$
We first show:
:$\set {x \in X : \size {\map f x} \ge \beta} \subseteq \set {x \in X : \size {\map f x} \ge \alpha}$
Let $x \in X$ have $\size {\map f x} \ge \beta$.
Then, since $\beta \ge \alpha... | Survival Function is Decreasing | https://proofwiki.org/wiki/Survival_Function_is_Decreasing | https://proofwiki.org/wiki/Survival_Function_is_Decreasing | [
"Survival Functions"
] | [
"Definition:Measure Space",
"Definition:Measurable Function",
"Definition:Survival Function",
"Definition:Decreasing/Real Function"
] | [
"Definition:Subset",
"Measure is Monotone",
"Definition:Survival Function",
"Definition:Decreasing/Real Function"
] |
proofwiki-19504 | Survival Function is Well-Defined | Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $f: X \to \overline \R$ be a $\Sigma$-measurable function.
Then the survival function $F_f$ is well-defined. | From Absolute Value of Measurable Function is Measurable:
:$\size f$ is $\Sigma$-measurable.
Then from Characterization of Measurable Functions, we have:
:$\set {x \in X : \size {\map f x} \ge \alpha} \in \Sigma$ for all $\alpha \in \hointr 0 \infty$.
So:
:$\map \mu {\set {x \in X : \size {\map f x} \ge \alpha} }$
is ... | Let $\struct {X, \Sigma, \mu}$ be a [[Definition:Measure Space|measure space]].
Let $f: X \to \overline \R$ be a [[Definition:Measurable Function|$\Sigma$-measurable function]].
Then the [[Definition:Survival Function|survival function]] $F_f$ is well-defined. | From [[Absolute Value of Measurable Function is Measurable]]:
:$\size f$ is [[Definition:Measurable Function|$\Sigma$-measurable]].
Then from [[Characterization of Measurable Functions]], we have:
:$\set {x \in X : \size {\map f x} \ge \alpha} \in \Sigma$ for all $\alpha \in \hointr 0 \infty$.
So:
:$\map \mu {\se... | Survival Function is Well-Defined | https://proofwiki.org/wiki/Survival_Function_is_Well-Defined | https://proofwiki.org/wiki/Survival_Function_is_Well-Defined | [
"Survival Functions",
"Examples of Well-Defined Mappings"
] | [
"Definition:Measure Space",
"Definition:Measurable Function",
"Definition:Survival Function"
] | [
"Absolute Value of Measurable Function is Measurable",
"Definition:Measurable Function",
"Characterization of Measurable Functions",
"Definition:Survival Function",
"Category:Survival Functions",
"Category:Examples of Well-Defined Mappings"
] |
proofwiki-19505 | Restriction of Norm on Vector Space to Subspace is Norm | Let $\Bbb F$ be a subfield of $\C$.
Let $X$ be a vector space over $\Bbb F$.
Let $\norm \cdot_X : X \to \hointr 0 \infty$ be a norm on $X$.
Let $Y$ be a vector subspace of $X$.
Let $\norm \cdot_Y$ denote the restriction of $\norm \cdot_X$ to $Y$.
Then $\norm \cdot_Y$ is a norm on $Y$. | Since $\norm x_X \ge 0$ for any $x \in X$, we have $\norm y_Y \ge 0$ for any $y \in Y$.
We verify each of the axioms for a norm on a vector space. | Let $\Bbb F$ be a [[Definition:Subfield|subfield]] of $\C$.
Let $X$ be a [[Definition:Vector Space|vector space]] over $\Bbb F$.
Let $\norm \cdot_X : X \to \hointr 0 \infty$ be a [[Definition:Norm on Vector Space|norm]] on $X$.
Let $Y$ be a [[Definition:Vector Subspace|vector subspace]] of $X$.
Let $\norm \cdot_Y$... | Since $\norm x_X \ge 0$ for any $x \in X$, we have $\norm y_Y \ge 0$ for any $y \in Y$.
We verify each of the [[Axiom:Vector Space Norm Axioms|axioms for a norm on a vector space]]. | Restriction of Norm on Vector Space to Subspace is Norm | https://proofwiki.org/wiki/Restriction_of_Norm_on_Vector_Space_to_Subspace_is_Norm | https://proofwiki.org/wiki/Restriction_of_Norm_on_Vector_Space_to_Subspace_is_Norm | [
"Normed Vector Spaces"
] | [
"Definition:Subfield",
"Definition:Vector Space",
"Definition:Norm/Vector Space",
"Definition:Vector Subspace",
"Definition:Restriction/Mapping",
"Definition:Norm/Vector Space"
] | [
"Axiom:Vector Space Norm Axioms"
] |
proofwiki-19506 | Closed Subspace of Banach Space forms Banach Space | Let $\struct {X, \norm \cdot_X}$ be a Banach space.
Let $Y$ be a closed linear subspace of $X$.
Let $\norm \cdot_Y$ be the restriction of $\norm \cdot_X$ to $Y$.
Then $\struct {Y, \norm \cdot_Y}$ is a Banach space. | From Restriction of Norm on Vector Space to Subspace is Norm, $\struct {Y, \norm \cdot_Y}$ is a normed vector space.
We now show that $\struct {Y, \norm \cdot_Y}$ is complete.
Let $\sequence {y_n}_{n \mathop \in \N}$ be a Cauchy sequence in $\struct {Y, \norm \cdot_Y}$.
Then, for each $\epsilon > 0$ there exists $N \i... | Let $\struct {X, \norm \cdot_X}$ be a [[Definition:Banach Space|Banach space]].
Let $Y$ be a [[Definition:Closed Linear Subspace|closed linear subspace]] of $X$.
Let $\norm \cdot_Y$ be the [[Definition:Restriction of Mapping|restriction]] of $\norm \cdot_X$ to $Y$.
Then $\struct {Y, \norm \cdot_Y}$ is a [[Definiti... | From [[Restriction of Norm on Vector Space to Subspace is Norm]], $\struct {Y, \norm \cdot_Y}$ is a [[Definition:Normed Vector Space|normed vector space]].
We now show that $\struct {Y, \norm \cdot_Y}$ is [[Definition:Complete Normed Vector Space|complete]].
Let $\sequence {y_n}_{n \mathop \in \N}$ be a [[Definition:... | Closed Subspace of Banach Space forms Banach Space | https://proofwiki.org/wiki/Closed_Subspace_of_Banach_Space_forms_Banach_Space | https://proofwiki.org/wiki/Closed_Subspace_of_Banach_Space_forms_Banach_Space | [
"Banach Spaces"
] | [
"Definition:Banach Space",
"Definition:Closed Linear Subspace",
"Definition:Restriction/Mapping",
"Definition:Banach Space"
] | [
"Restriction of Norm on Vector Space to Subspace is Norm",
"Definition:Normed Vector Space",
"Definition:Banach Space",
"Definition:Cauchy Sequence/Normed Vector Space",
"Definition:Restriction/Mapping",
"Definition:Cauchy Sequence/Normed Vector Space",
"Definition:Banach Space",
"Definition:Convergen... |
proofwiki-19507 | Condition for Equivalence of Norms that induce Banach Spaces | Let $\Bbb F \in \set {\R, \C}$.
Let $X$ be a vector space over $\Bbb F$.
Let $\norm \cdot_1$ and $\norm \cdot_2$ be norms on $X$ such that $\struct {X, \norm \cdot_1}$ and $\struct {X, \norm \cdot_2}$ are Banach spaces.
Suppose that, for some real number $C > 0$:
:$\norm x_2 \le C \norm x_1$ for all $x \in X$.
Then $... | Consider $I : \struct {X, \norm \cdot_1} \to \struct {X, \norm \cdot_2}$ defined by:
:$I x = x$ for each $x \in X$.
Then for $x, y \in X$ and $\lambda \in \Bbb F$.
Then we have:
:$\map I {\lambda x + y} = \lambda x + y = \lambda I x + I y$
so $I$ is linear.
Similarly, we have:
:$\norm {I x}_2 = \norm x_2 \le C \no... | Let $\Bbb F \in \set {\R, \C}$.
Let $X$ be a [[Definition:Vector Space|vector space]] over $\Bbb F$.
Let $\norm \cdot_1$ and $\norm \cdot_2$ be [[Definition:Norm on Vector Space|norms]] on $X$ such that $\struct {X, \norm \cdot_1}$ and $\struct {X, \norm \cdot_2}$ are [[Definition:Banach Space|Banach spaces]].
Suppo... | Consider $I : \struct {X, \norm \cdot_1} \to \struct {X, \norm \cdot_2}$ defined by:
:$I x = x$ for each $x \in X$.
Then for $x, y \in X$ and $\lambda \in \Bbb F$.
Then we have:
:$\map I {\lambda x + y} = \lambda x + y = \lambda I x + I y$
so $I$ is [[Definition:Linear Transformation|linear]].
Similarly, we ... | Condition for Equivalence of Norms that induce Banach Spaces | https://proofwiki.org/wiki/Condition_for_Equivalence_of_Norms_that_induce_Banach_Spaces | https://proofwiki.org/wiki/Condition_for_Equivalence_of_Norms_that_induce_Banach_Spaces | [
"Banach Spaces"
] | [
"Definition:Vector Space",
"Definition:Norm/Vector Space",
"Definition:Banach Space",
"Definition:Real Number",
"Definition:Equivalence of Norms"
] | [
"Definition:Linear Transformation",
"Definition:Bounded Linear Transformation",
"Definition:Bijection",
"Banach Isomorphism Theorem",
"Definition:Inverse Mapping",
"Definition:Bounded Linear Transformation",
"Definition:Real Number",
"Definition:Equivalence of Norms",
"Category:Banach Spaces"
] |
proofwiki-19508 | Condition for Total Ordering to be Well-Ordering | Let $A$ be a non-empty class under a total ordering $\preccurlyeq$.
Then $\preccurlyeq$ is a well-ordering on $A$ {{iff}} the following $3$ conditions are fulfilled:
:$(1): \quad A$ has a smallest element $x_0$
:$(2): \quad$ Every element of $A$ except the greatest (if there is one) has an immediate successor
:$(3): \q... | === Sufficient Condition ===
Let $\preccurlyeq$ be a well-ordering on $A$.
Then by definition of well-ordering, there exists a smallest element of $A$ which we may call $x_0$.
Hence condition $(1)$ is fulfilled.
Let there exist an element $x$ of $A$ such that $x$ has no immediate successor.
Let $S$ be the subclass of $... | Let $A$ be a [[Definition:Non-Empty Class|non-empty]] [[Definition:Class (Class Theory)|class]] under a [[Definition:Total Ordering|total ordering]] $\preccurlyeq$.
Then $\preccurlyeq$ is a [[Definition:Well-Ordering|well-ordering]] on $A$ {{iff}} the following $3$ conditions are fulfilled:
:$(1): \quad A$ has a [[D... | === Sufficient Condition ===
Let $\preccurlyeq$ be a [[Definition:Well-Ordering|well-ordering]] on $A$.
Then by definition of [[Definition:Well-Ordering|well-ordering]], there exists a [[Definition:Smallest Element (Class Theory)|smallest element]] of $A$ which we may call $x_0$.
Hence condition $(1)$ is fulfilled.
... | Condition for Total Ordering to be Well-Ordering | https://proofwiki.org/wiki/Condition_for_Total_Ordering_to_be_Well-Ordering | https://proofwiki.org/wiki/Condition_for_Total_Ordering_to_be_Well-Ordering | [
"Well-Orderings",
"Total Orderings"
] | [
"Definition:Non-Empty Set/Class Theory",
"Definition:Class (Class Theory)",
"Definition:Total Ordering",
"Definition:Well-Ordering",
"Definition:Smallest Element/Class Theory",
"Definition:Element/Class",
"Definition:Greatest Element/Class Theory",
"Definition:Immediate Successor Element/Class Theory"... | [
"Definition:Well-Ordering",
"Definition:Well-Ordering",
"Definition:Smallest Element/Class Theory",
"Definition:Element/Class",
"Definition:Immediate Successor Element/Class Theory",
"Definition:Subclass",
"Definition:Well-Ordering",
"Definition:Smallest Element/Class Theory",
"Definition:Element/Cl... |
proofwiki-19509 | Backward Path of Well-Ordering is Finite | Let $A$ be a class under a well-ordering $\preccurlyeq$.
Let $S$ be a backward path of $\preccurlyeq$ in $A$.
Then $S$ is finite. | First suppose that $S$ has no terms.
Then $S$ is finite by default.
Hence let it be assumed that $S$ has at least $1$ term.
Let $T = \set {x: x \in \sequence {x_n} }$ be the range of $S$.
Then by definition $T$ is a subclass of $A$.
{{AimForCont}} $S$ is infinite.
By definition of well-ordering, $T$ has a smallest elem... | Let $A$ be a [[Definition:Class (Class Theory)|class]] under a [[Definition:Well-Ordering|well-ordering]] $\preccurlyeq$.
Let $S$ be a [[Definition:Backward Path (Class Theory)|backward path]] of $\preccurlyeq$ in $A$.
Then $S$ is [[Definition:Finite Sequence|finite]]. | First suppose that $S$ has no [[Definition:Term of Sequence|terms]].
Then $S$ is [[Definition:Finite Sequence|finite]] by default.
Hence let it be assumed that $S$ has at least $1$ [[Definition:Term of Sequence|term]].
Let $T = \set {x: x \in \sequence {x_n} }$ be the [[Definition:Range of Sequence|range]] of $S$.
... | Backward Path of Well-Ordering is Finite | https://proofwiki.org/wiki/Backward_Path_of_Well-Ordering_is_Finite | https://proofwiki.org/wiki/Backward_Path_of_Well-Ordering_is_Finite | [
"Well-Orderings"
] | [
"Definition:Class (Class Theory)",
"Definition:Well-Ordering",
"Definition:Backward Path (Class Theory)",
"Definition:Finite Sequence"
] | [
"Definition:Term of Sequence",
"Definition:Finite Sequence",
"Definition:Term of Sequence",
"Definition:Range of Sequence",
"Definition:Subclass",
"Definition:Sequence/Infinite Sequence",
"Definition:Well-Ordering",
"Definition:Smallest Element/Class Theory",
"Definition:Backward Path (Class Theory)... |
proofwiki-19510 | P-adic Norm forms Non-Archimedean Valued Field/Rational Numbers | The $p$-adic norm $\norm{\,\cdot\,}_p$ forms a non-Archimedean norm on the rational numbers $\Q$.
The rational numbers $\struct{\Q, \norm{\,\cdot\,}_p}$ with the $p$-adic norm is a valued field with a non-Archimedean norm. | First we note that the $p$-adic norm is a norm.
Let $\nu_p$ denote the $p$-adic valuation on the rational numbers.
Recall the definition of the $p$-adic norm:
:<nowiki>$\forall q \in \Q: \norm q_p := \begin{cases}
0 & : q = 0 \\
p^{-\map {\nu_p} q} & : q \ne 0
\end{cases}$
</nowiki>
We must show the following holds... | The [[Definition:P-adic Norm on Rational Numbers|$p$-adic norm]] $\norm{\,\cdot\,}_p$ forms a [[Definition:Non-Archimedean Division Ring Norm|non-Archimedean norm]] on the [[Definition:Rational Number|rational numbers]] $\Q$.
The [[Definition:Rational Number|rational numbers]] $\struct{\Q, \norm{\,\cdot\,}_p}$ with t... | First we note that the [[P-adic Norm is Norm|$p$-adic norm is a norm]].
Let $\nu_p$ denote the [[Definition:P-adic Valuation|$p$-adic valuation]] on the [[Definition:Rational Number|rational numbers]].
Recall the definition of the [[Definition:P-adic Norm|$p$-adic norm]]:
:<nowiki>$\forall q \in \Q: \norm q_p := \b... | P-adic Norm forms Non-Archimedean Valued Field/Rational Numbers | https://proofwiki.org/wiki/P-adic_Norm_forms_Non-Archimedean_Valued_Field/Rational_Numbers | https://proofwiki.org/wiki/P-adic_Norm_forms_Non-Archimedean_Valued_Field/Rational_Numbers | [
"P-adic Norm forms Non-Archimedean Valued Field"
] | [
"Definition:P-adic Norm/Rational Numbers",
"Definition:Non-Archimedean/Norm (Division Ring)",
"Definition:Rational Number",
"Definition:Rational Number",
"Definition:P-adic Norm/Rational Numbers",
"Definition:Valued Field",
"Definition:Non-Archimedean/Norm (Division Ring)"
] | [
"P-adic Norm is Norm",
"Definition:P-adic Valuation",
"Definition:Rational Number",
"Definition:P-adic Norm",
"P-adic Valuation is Valuation"
] |
proofwiki-19511 | P-adic Norm forms Non-Archimedean Valued Field/P-adic Numbers | Let $p$ be a prime number.
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers.
Then:
:$\struct{\Q_p, \norm {\,\cdot\,}_p}$ is a valued field
:$\norm {\,\cdot\,}_p$ is a non-Archimedean norm
That is, the $p$-adic numbers $\struct {\Q_p, \norm {\,\cdot\,}_p}$ form a valued field with a non-Archimedean norm... | Let $\norm {\,\cdot\,}^\Q_p$ be the $p$-adic norm on the rationals $\Q$.
From $p$-adic Norm on Rational Numbers is Non-Archimedean Norm:
:$\struct{Q, \norm {\,\cdot\,}^\Q_p}$ is a valued field with non-Archimedean norm $\norm {\,\cdot\,}_p$
By definition of the $p$-adic numbers:
:$\Q_p$ is the quotient ring $\CC \, \b... | Let $p$ be a [[Definition:Prime Number|prime number]].
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the [[Definition:Valued Field of P-adic Numbers|$p$-adic numbers]].
Then:
:$\struct{\Q_p, \norm {\,\cdot\,}_p}$ is a [[Definition:Valued Field|valued field]]
:$\norm {\,\cdot\,}_p$ is a [[Definition:Non-Archimedean Di... | Let $\norm {\,\cdot\,}^\Q_p$ be the [[Definition:P-adic Norm|$p$-adic norm]] on the [[Definition:Rational Numbers|rationals $\Q$]].
From [[P-adic Norm on Rational Numbers is Non-Archimedean Norm|$p$-adic Norm on Rational Numbers is Non-Archimedean Norm]]:
:$\struct{Q, \norm {\,\cdot\,}^\Q_p}$ is a [[Definition:Valued ... | P-adic Norm forms Non-Archimedean Valued Field/P-adic Numbers | https://proofwiki.org/wiki/P-adic_Norm_forms_Non-Archimedean_Valued_Field/P-adic_Numbers | https://proofwiki.org/wiki/P-adic_Norm_forms_Non-Archimedean_Valued_Field/P-adic_Numbers | [
"P-adic Norm forms Non-Archimedean Valued Field"
] | [
"Definition:Prime Number",
"Definition:Valued Field of P-adic Numbers",
"Definition:Valued Field",
"Definition:Non-Archimedean/Norm (Division Ring)",
"Definition:Valued Field of P-adic Numbers",
"Definition:Valued Field",
"Definition:Non-Archimedean/Norm (Division Ring)"
] | [
"Definition:P-adic Norm",
"Definition:Rational Number",
"P-adic Norm forms Non-Archimedean Valued Field/Rational Numbers",
"Definition:Valued Field",
"Definition:Non-Archimedean/Norm (Division Ring)",
"Definition:Field of P-adic Numbers",
"Definition:Quotient Ring",
"Definition:Ring of Cauchy Sequence... |
proofwiki-19512 | Composition of Linear Transformations is Linear Transformation | Let $K$ be a field.
Let $X, Y, Z$ be vector spaces over $K$.
Let $T_1 : X \to Y$ and $T_2 : Y \to Z$ be linear transformations.
Then the composition $T_2 \circ T_1 : X \to Z$ is a linear transformation. | Let $\lambda \in K$ and $u, v \in X$.
Then, we have:
{{begin-eqn}}
{{eqn | l = \map {\paren {T_2 \circ T_1} } {\lambda u + v}
| r = \map {T_2} {\map {T_1} {\lambda u + v} }
}}
{{eqn | r = \map {T_2} {\lambda T_1 u + T_1 v}
| c = {{Defof|Linear Transformation}}
}}
{{eqn | r = \lambda \paren {T_2 T_1} u + \paren {T... | Let $K$ be a [[Definition:Field (Abstract Algebra)|field]].
Let $X, Y, Z$ be [[Definition:Vector Space|vector spaces]] over $K$.
Let $T_1 : X \to Y$ and $T_2 : Y \to Z$ be [[Definition:Linear Transformation|linear transformations]].
Then the [[Definition:Composition of Mappings|composition]] $T_2 \circ T_1 : X \to ... | Let $\lambda \in K$ and $u, v \in X$.
Then, we have:
{{begin-eqn}}
{{eqn | l = \map {\paren {T_2 \circ T_1} } {\lambda u + v}
| r = \map {T_2} {\map {T_1} {\lambda u + v} }
}}
{{eqn | r = \map {T_2} {\lambda T_1 u + T_1 v}
| c = {{Defof|Linear Transformation}}
}}
{{eqn | r = \lambda \paren {T_2 T_1} u + \paren ... | Composition of Linear Transformations is Linear Transformation | https://proofwiki.org/wiki/Composition_of_Linear_Transformations_is_Linear_Transformation | https://proofwiki.org/wiki/Composition_of_Linear_Transformations_is_Linear_Transformation | [
"Linear Transformations"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Vector Space",
"Definition:Linear Transformation",
"Definition:Composition of Mappings",
"Definition:Linear Transformation"
] | [
"Definition:Linear Transformation",
"Category:Linear Transformations"
] |
proofwiki-19513 | Image of Vector Subspace under Linear Transformation is Vector Subspace | Let $K$ be a field.
Let $X$ and $Y$ be vector spaces over $K$.
Let $U$ be a vector subspace of $X$.
Let $T : X \to Y$ be a linear transformation.
Then $T \sqbrk U$ is a vector subspace of $Y$. | Since $U$ is a non-empty set, we can apply the One-Step Vector Subspace Test to $T \sqbrk U$.
Let $\lambda \in K$ and $u, v \in T \sqbrk U$.
Then there exists $x, y \in U$ such that:
:$u = T x$
and:
:$v = T y$
Then:
{{begin-eqn}}
{{eqn | l = \lambda u + v
| r = \lambda T x + T y
}}
{{eqn | r = \map T {\lambda x + ... | Let $K$ be a [[Definition:Field (Abstract Algebra)|field]].
Let $X$ and $Y$ be [[Definition:Vector Space|vector spaces]] over $K$.
Let $U$ be a [[Definition:Vector Subspace|vector subspace]] of $X$.
Let $T : X \to Y$ be a [[Definition:Linear Transformation|linear transformation]].
Then $T \sqbrk U$ is a [[Definit... | Since $U$ is a [[Definition:Non-Empty Set|non-empty set]], we can apply the [[One-Step Vector Subspace Test]] to $T \sqbrk U$.
Let $\lambda \in K$ and $u, v \in T \sqbrk U$.
Then there exists $x, y \in U$ such that:
:$u = T x$
and:
:$v = T y$
Then:
{{begin-eqn}}
{{eqn | l = \lambda u + v
| r = \lambda T x +... | Image of Vector Subspace under Linear Transformation is Vector Subspace | https://proofwiki.org/wiki/Image_of_Vector_Subspace_under_Linear_Transformation_is_Vector_Subspace | https://proofwiki.org/wiki/Image_of_Vector_Subspace_under_Linear_Transformation_is_Vector_Subspace | [
"Vector Subspaces",
"Linear Transformations"
] | [
"Definition:Field (Abstract Algebra)",
"Definition:Vector Space",
"Definition:Vector Subspace",
"Definition:Linear Transformation",
"Definition:Vector Subspace"
] | [
"Definition:Non-Empty Set",
"One-Step Vector Subspace Test",
"Definition:Vector Subspace",
"Definition:Vector Subspace",
"Category:Vector Subspaces",
"Category:Linear Transformations"
] |
proofwiki-19514 | Equivalence of Formulations of Axiom of Choice/Formulation 1 iff Formulation 4 | The following formulation of the Axiom of Choice: | We note from Set equals Union of Power Set that:
:$x = \ds \map \bigcup {\powerset x}$
Setting $\powerset A =: s$, we see that from Formulation $1$:
:$\ds \paren {\O \notin \powerset A \implies \exists \paren {f: \powerset A \to \bigcup \powerset A}: \forall x \in \powerset A: \map f x \in x}$
That is:
:for every non-... | The following formulation of the [[Axiom:Axiom of Choice|Axiom of Choice]]: | We note from [[Set equals Union of Power Set]] that:
:$x = \ds \map \bigcup {\powerset x}$
Setting $\powerset A =: s$, we see that from [[Axiom:Axiom of Choice/Formulation 1|Formulation $1$]]:
:$\ds \paren {\O \notin \powerset A \implies \exists \paren {f: \powerset A \to \bigcup \powerset A}: \forall x \in \powerset... | Equivalence of Formulations of Axiom of Choice/Formulation 1 iff Formulation 4 | https://proofwiki.org/wiki/Equivalence_of_Formulations_of_Axiom_of_Choice/Formulation_1_iff_Formulation_4 | https://proofwiki.org/wiki/Equivalence_of_Formulations_of_Axiom_of_Choice/Formulation_1_iff_Formulation_4 | [
"Equivalence of Formulations of Axiom of Choice"
] | [
"Axiom:Axiom of Choice",
"Axiom:Axiom of Choice"
] | [
"Set equals Union of Power Set",
"Axiom:Axiom of Choice/Formulation 1",
"Definition:Non-Empty Set",
"Definition:Proper Subset",
"Definition:Mapping",
"Definition:Non-Empty Set",
"Definition:Proper Subset",
"Axiom:Axiom of Choice/Formulation 4"
] |
proofwiki-19515 | Equivalence of Definitions of Limit Point of Set/Definition from Open Neighborhood iff Definition from Relative Complement | Let $T = \struct {S, \tau}$ be a topological space.
Let $A \subseteq S$.
{{TFAE|def = Limit Point of Set|view = limit point}}
=== Definition from Open Neighborhood===
{{:Definition:Limit Point of Set/Definition from Open Neighborhood}}
=== Definition from Relative Complement===
{{:Definition:Limit Point of Set/Definiti... | The following equivalence holds:
{{begin-eqn}}
{{eqn | o =
| c = There exists an open neighborhood $U$ of $x$ such that $A \cap \paren {U \setminus \set x} = \O$
}}
{{eqn | o = \leadstoandfrom
| c = There exists an open neighborhood $U$ of $x$ such that $U \subseteq \paren{S \setminus A} \cup \set x$
... | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $A \subseteq S$.
{{TFAE|def = Limit Point of Set|view = limit point}}
=== [[Definition:Limit Point of Set/Definition from Open Neighborhood|Definition from Open Neighborhood]]===
{{:Definition:Limit Point of Set/Definition from... | The following [[Definition:Logical Equivalence|equivalence]] holds:
{{begin-eqn}}
{{eqn | o =
| c = There exists an [[Definition:Open Neighborhood of Point|open neighborhood]] $U$ of $x$ such that $A \cap \paren {U \setminus \set x} = \O$
}}
{{eqn | o = \leadstoandfrom
| c = There exists an [[Definition:Ope... | Equivalence of Definitions of Limit Point of Set/Definition from Open Neighborhood iff Definition from Relative Complement/Proof 1 | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Limit_Point_of_Set/Definition_from_Open_Neighborhood_iff_Definition_from_Relative_Complement | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Limit_Point_of_Set/Definition_from_Open_Neighborhood_iff_Definition_from_Relative_Complement/Proof_1 | [
"Equivalence of Definitions of Limit Point of Set"
] | [
"Definition:Topological Space",
"Definition:Limit Point of Set/Definition from Open Neighborhood",
"Definition:Limit Point of Set/Definition from Relative Complement"
] | [
"Definition:Logical Equivalence",
"Definition:Open Neighborhood/Point",
"Definition:Open Neighborhood/Point",
"Modus Ponendo Tollens",
"Definition:Neighborhood (Topology)/Point",
"Rule of Transposition"
] |
proofwiki-19516 | Equivalence of Definitions of Limit Point of Set/Definition from Open Neighborhood iff Definition from Relative Complement | Let $T = \struct {S, \tau}$ be a topological space.
Let $A \subseteq S$.
{{TFAE|def = Limit Point of Set|view = limit point}}
=== Definition from Open Neighborhood===
{{:Definition:Limit Point of Set/Definition from Open Neighborhood}}
=== Definition from Relative Complement===
{{:Definition:Limit Point of Set/Definiti... | The following equivalence holds:
There exists an open neighborhood $U$ of $x$ such that $A \cap \paren {U \setminus \set x} = \O$
{{begin-eqn}}
{{eqn | l = \O
| r = A \cap \paren {U \setminus \set x}
}}
{{eqn | ll = \leadstoandfrom
| l = \O
| r = \paren {U \cap A} \setminus \set x
| cc = Inter... | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $A \subseteq S$.
{{TFAE|def = Limit Point of Set|view = limit point}}
=== [[Definition:Limit Point of Set/Definition from Open Neighborhood|Definition from Open Neighborhood]]===
{{:Definition:Limit Point of Set/Definition from... | The following [[Definition:Logical Equivalence|equivalence]] holds:
There exists an [[Definition:Open Neighborhood of Point|open neighborhood]] $U$ of $x$ such that $A \cap \paren {U \setminus \set x} = \O$
{{begin-eqn}}
{{eqn | l = \O
| r = A \cap \paren {U \setminus \set x}
}}
{{eqn | ll = \leadstoandfrom
... | Equivalence of Definitions of Limit Point of Set/Definition from Open Neighborhood iff Definition from Relative Complement/Proof 2 | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Limit_Point_of_Set/Definition_from_Open_Neighborhood_iff_Definition_from_Relative_Complement | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Limit_Point_of_Set/Definition_from_Open_Neighborhood_iff_Definition_from_Relative_Complement/Proof_2 | [
"Equivalence of Definitions of Limit Point of Set"
] | [
"Definition:Topological Space",
"Definition:Limit Point of Set/Definition from Open Neighborhood",
"Definition:Limit Point of Set/Definition from Relative Complement"
] | [
"Definition:Logical Equivalence",
"Definition:Open Neighborhood/Point",
"Intersection with Set Difference is Set Difference with Intersection",
"Intersection is Commutative",
"Intersection with Set Difference is Set Difference with Intersection",
"Complement of Complement",
"Intersection with Complement... |
proofwiki-19517 | Equivalence of Definitions of Limit Point in Metric Space | Let $M = \struct {S, d}$ be a metric space.
Let $\tau$ be the topology induced by the metric $d$.
Let $A \subseteq S$ be a subset of $S$.
Let $\alpha \in S$.
{{TFAE|def = Limit Point/Metric Space|view = limit point in metric space}}
=== Definition 1 ===
{{:Definition:Limit Point/Metric Space/Definition 1}}
=== Definit... | === $(1) \implies (2)$ ===
Let:
:$\forall \epsilon \in \R_{>0}: \paren {\map {B_\epsilon} \alpha \setminus \set \alpha} \cap A \ne \O$
{{:Equivalence of Definitions of Limit Point in Metric Space/Definition 1 implies Definition 2}}{{qed|lemma}} | Let $M = \struct {S, d}$ be a [[Definition:Metric Space|metric space]].
Let $\tau$ be the [[Definition:Topology Induced by Metric|topology induced]] by the metric $d$.
Let $A \subseteq S$ be a [[Definition:Subset|subset]] of $S$.
Let $\alpha \in S$.
{{TFAE|def = Limit Point/Metric Space|view = limit point in metr... | === [[Equivalence of Definitions of Limit Point in Metric Space/Definition 1 implies Definition 2|$(1) \implies (2)$]] ===
Let:
:$\forall \epsilon \in \R_{>0}: \paren {\map {B_\epsilon} \alpha \setminus \set \alpha} \cap A \ne \O$
{{:Equivalence of Definitions of Limit Point in Metric Space/Definition 1 implies Defin... | Equivalence of Definitions of Limit Point in Metric Space | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Limit_Point_in_Metric_Space | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Limit_Point_in_Metric_Space | [
"Limit Points in Metric Spaces",
"Equivalence of Definitions of Limit Point in Metric Space"
] | [
"Definition:Metric Space",
"Definition:Topology Induced by Metric",
"Definition:Subset",
"Definition:Limit Point/Metric Space/Definition 1",
"Definition:Limit Point/Metric Space/Definition 2",
"Definition:Limit Point/Metric Space/Definition 3"
] | [
"Equivalence of Definitions of Limit Point in Metric Space/Definition 1 implies Definition 2"
] |
proofwiki-19518 | Equivalence of Definitions of Limit Point in Metric Space/Definition 1 implies Definition 2 | Let $M = \struct {S, d}$ be a metric space.
Let $\tau$ be the topology induced by the metric $d$.
Let $A \subseteq S$ be a subset of $S$.
Let $\alpha \in S$.
Let:
:$\forall \epsilon \in \R_{>0}: \paren {\map {B_\epsilon} \alpha \setminus \set \alpha} \cap A \ne \O$
Then:
:there is a sequence $\sequence{\alpha_n}$ in $... | A sequence $\sequence {\alpha_n}$ is constructed using finite recursion.
Let $\alpha_0$ be some arbitrary point in $A \cap \paren{\map {B_1} \alpha \setminus \set \alpha}$.
$\alpha_{n + 1}$ is some arbitrary point in $A \cap \paren{\map {B_{\frac {\map d {\alpha, \alpha_n} } 2} } \alpha \setminus \set \alpha}$ where $\... | Let $M = \struct {S, d}$ be a [[Definition:Metric Space|metric space]].
Let $\tau$ be the [[Definition:Topology Induced by Metric|topology induced]] by the metric $d$.
Let $A \subseteq S$ be a [[Definition:Subset|subset]] of $S$.
Let $\alpha \in S$.
Let:
:$\forall \epsilon \in \R_{>0}: \paren {\map {B_\epsilon} \... | A [[Definition:Sequence|sequence]] $\sequence {\alpha_n}$ is constructed using finite recursion.
Let $\alpha_0$ be some arbitrary point in $A \cap \paren{\map {B_1} \alpha \setminus \set \alpha}$.
$\alpha_{n + 1}$ is some arbitrary point in $A \cap \paren{\map {B_{\frac {\map d {\alpha, \alpha_n} } 2} } \alpha \setmi... | Equivalence of Definitions of Limit Point in Metric Space/Definition 1 implies Definition 2 | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Limit_Point_in_Metric_Space/Definition_1_implies_Definition_2 | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Limit_Point_in_Metric_Space/Definition_1_implies_Definition_2 | [
"Equivalence of Definitions of Limit Point in Metric Space"
] | [
"Definition:Metric Space",
"Definition:Topology Induced by Metric",
"Definition:Subset",
"Definition:Sequence",
"Definition:Limit of Sequence/Topological Space",
"Definition:Sequence",
"Definition:Sequence"
] | [
"Definition:Sequence",
"Definition:Open Ball",
"Definition:Natural Numbers",
"Axiom of Archimedes",
"Definition:Limit of Sequence/Metric Space"
] |
proofwiki-19519 | Equivalence of Definitions of Extreme Point of Convex Set | {{finish|The definitions have changed}}
{{TFAE|def = Extreme Point of Convex Set}}
Let $X$ be a vector space over $\R$.
Let $K$ be a convex subset of $X$.
=== Definition 1 ===
{{:Definition:Extreme Point of Convex Set/Definition 1}}
=== Definition 2 ===
{{:Definition:Extreme Point of Convex Set/Definition 2}} | === Definition 1 implies Definition 2 ===
Suppose that:
:whenever $a = t x + \paren {1 - t} y$ for $t \in \openint 0 1$, we have $x = y = a$.
Then, if $x, y \in K \setminus \set a$ and $t \in \openint 0 1$, we have:
:$t x + \paren {1 - t} y \ne a$
If $t = 0$, then we have:
:$t x + \paren {1 - t} y = y \ne a$
and if... | {{finish|The definitions have changed}}
{{TFAE|def = Extreme Point of Convex Set}}
Let $X$ be a [[Definition:Vector Space|vector space]] over $\R$.
Let $K$ be a [[Definition:Convex Set (Vector Space)|convex subset]] of $X$.
=== [[Definition:Extreme Point of Convex Set/Definition 1|Definition 1]] ===
{{:Definition:... | === [[Definition:Extreme Point of Convex Set/Definition 1|Definition 1]] implies [[Definition:Extreme Point of Convex Set/Definition 2|Definition 2]] ===
Suppose that:
:whenever $a = t x + \paren {1 - t} y$ for $t \in \openint 0 1$, we have $x = y = a$.
Then, if $x, y \in K \setminus \set a$ and $t \in \openint 0 1... | Equivalence of Definitions of Extreme Point of Convex Set | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Extreme_Point_of_Convex_Set | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Extreme_Point_of_Convex_Set | [
"Convex Sets (Vector Spaces)",
"Extreme Points of Convex Sets"
] | [
"Definition:Vector Space",
"Definition:Convex Set (Vector Space)",
"Definition:Extreme Point of Convex Set/Definition 1",
"Definition:Extreme Point of Convex Set/Definition 2"
] | [
"Definition:Extreme Point of Convex Set/Definition 1",
"Definition:Extreme Point of Convex Set/Definition 2",
"Definition:Convex Set (Vector Space)",
"Definition:Convex Set (Vector Space)",
"Definition:Extreme Point of Convex Set/Definition 2",
"Definition:Extreme Point of Convex Set/Definition 1",
"Def... |
proofwiki-19520 | Axioms of Hilbert Proof System Instance 1 for Predicate Logic are Tautologies | Let $\mathscr H$ be Instance 1 of a Hilbert proof system for predicate logic.
Then the axioms of $\mathscr H$ are tautologies. | {{improve|For each axiom this should link to a theorem page instead, but I had difficulties coming up with names. Maybe subpages?}}
=== Axiom 1 ===
This is precisely the statement of Propositional Tautology is Tautology in Predicate Logic.
{{qed|lemma}}
{{proofread|below}}
{{improve|The theorem cited below is actually ... | Let $\mathscr H$ be [[Definition:Hilbert Proof System Instance 1 for Predicate Logic|Instance 1 of a Hilbert proof system for predicate logic]].
Then the [[Definition:Axiom (Formal Systems)|axioms]] of $\mathscr H$ are [[Definition:Tautology (Predicate Logic)|tautologies]]. | {{improve|For each axiom this should link to a theorem page instead, but I had difficulties coming up with names. Maybe subpages?}}
=== Axiom 1 ===
This is precisely the statement of [[Propositional Tautology is Tautology in Predicate Logic]].
{{qed|lemma}}
{{proofread|below}}
{{improve|The theorem cited below is act... | Axioms of Hilbert Proof System Instance 1 for Predicate Logic are Tautologies | https://proofwiki.org/wiki/Axioms_of_Hilbert_Proof_System_Instance_1_for_Predicate_Logic_are_Tautologies | https://proofwiki.org/wiki/Axioms_of_Hilbert_Proof_System_Instance_1_for_Predicate_Logic_are_Tautologies | [
"Proof Systems",
"Predicate Logic"
] | [
"Definition:Hilbert Proof System/Predicate Logic/Instance 1",
"Definition:Axiom/Formal Systems",
"Definition:Tautology/Formal Semantics/Predicate Logic"
] | [
"Propositional Tautology is Tautology in Predicate Logic",
"Universal Generalisation",
"Well-Formed Formula is Tautology iff Universal Closure is Tautology",
"Definition:Language of Predicate Logic/Formal Grammar",
"Definition:Universal Closure of Well-Formed Formula",
"Definition:Assignment for Structure... |
proofwiki-19521 | Equivalence of Definitions of Limit Point in Metric Space/Definition 2 implies Definition 3 | Let $M = \struct {S, d}$ be a metric space.
Let $\tau$ be the topology induced by the metric $d$.
Let $A \subseteq S$ be a subset of $S$.
Let $\alpha \in S$.
Let there exist a sequence $\sequence{\alpha_n}$ in $A \setminus \set \alpha$ such that $\sequence{\alpha_n}$ converges to $\alpha$ in $S$.
Then:
:$\alpha$ is a ... | Let $U \in \tau$ be an arbitrary open set:
:$\alpha \in U$
By {{Defof|Topology Induced by Metric}}:
:$\exists \epsilon \in \R_{>0} : \map {B_\epsilon} \alpha \subseteq U$
By {{Defof|Convergent Sequence (Metric Space)}}:
:$\exists N \in \N : \forall n \ge N : d(\alpha, \alpha_n) < \epsilon$
By {{Defof|Open Ball}}:
:$\al... | Let $M = \struct {S, d}$ be a [[Definition:Metric Space|metric space]].
Let $\tau$ be the [[Definition:Topology Induced by Metric|topology induced]] by the metric $d$.
Let $A \subseteq S$ be a [[Definition:Subset|subset]] of $S$.
Let $\alpha \in S$.
Let there exist a [[Definition:Sequence|sequence]] $\sequence{\a... | Let $U \in \tau$ be an arbitrary [[Definition:Open Set (Topology)|open set]]:
:$\alpha \in U$
By {{Defof|Topology Induced by Metric}}:
:$\exists \epsilon \in \R_{>0} : \map {B_\epsilon} \alpha \subseteq U$
By {{Defof|Convergent Sequence (Metric Space)}}:
:$\exists N \in \N : \forall n \ge N : d(\alpha, \alpha_n) < \e... | Equivalence of Definitions of Limit Point in Metric Space/Definition 2 implies Definition 3 | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Limit_Point_in_Metric_Space/Definition_2_implies_Definition_3 | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Limit_Point_in_Metric_Space/Definition_2_implies_Definition_3 | [
"Equivalence of Definitions of Limit Point in Metric Space"
] | [
"Definition:Metric Space",
"Definition:Topology Induced by Metric",
"Definition:Subset",
"Definition:Sequence",
"Definition:Convergent Sequence/Metric Space",
"Definition:Limit Point/Topology/Set",
"Definition:Topological Space"
] | [
"Definition:Open Set/Topology",
"Definition:Hypothesis",
"Intersection with Set Difference is Set Difference with Intersection",
"Set Intersection Preserves Subsets",
"Set Difference over Subset",
"Intersection with Set Difference is Set Difference with Intersection",
"Definition:Limit Point/Topology/Se... |
proofwiki-19522 | Equivalence of Definitions of Limit Point in Metric Space/Definition 3 implies Definition 1 | Let $M = \struct {S, d}$ be a metric space.
Let $\tau$ be the topology induced by the metric $d$.
Let $A \subseteq S$ be a subset of $S$.
Let $\alpha \in S$.
Let $\alpha$ be a limit point in the topological space $\struct{S, \tau}$.
Then:
:$\forall \epsilon \in \R_{>0}: \paren {\map {B_\epsilon} \alpha \setminus \set ... | From Open Ball is Open Set:
:$\forall \epsilon \in \R_{>0}: \map {B_\epsilon} \alpha$ is open set in $M$
By {{Defof|Topology Induced by Metric}}:
:$\forall \epsilon \in \R_{>0}: \map {B_\epsilon} \alpha$ is open set in $\struct{S,\tau}$
By {{Defof|Limit Point of Set}}:
:$\forall \epsilon \in \R_{>0}: \paren {\map {B_\e... | Let $M = \struct {S, d}$ be a [[Definition:Metric Space|metric space]].
Let $\tau$ be the [[Definition:Topology Induced by Metric|topology induced]] by the metric $d$.
Let $A \subseteq S$ be a [[Definition:Subset|subset]] of $S$.
Let $\alpha \in S$.
Let $\alpha$ be a [[Definition:Limit Point of Set|limit point]] ... | From [[Open Ball is Open Set]]:
:$\forall \epsilon \in \R_{>0}: \map {B_\epsilon} \alpha$ is [[Definition:Open Set (Metric Space)|open set]] in $M$
By {{Defof|Topology Induced by Metric}}:
:$\forall \epsilon \in \R_{>0}: \map {B_\epsilon} \alpha$ is [[Definition:Open Set (Topology)|open set]] in $\struct{S,\tau}$
By ... | Equivalence of Definitions of Limit Point in Metric Space/Definition 3 implies Definition 1 | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Limit_Point_in_Metric_Space/Definition_3_implies_Definition_1 | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Limit_Point_in_Metric_Space/Definition_3_implies_Definition_1 | [
"Equivalence of Definitions of Limit Point in Metric Space"
] | [
"Definition:Metric Space",
"Definition:Topology Induced by Metric",
"Definition:Subset",
"Definition:Limit Point/Topology/Set",
"Definition:Topological Space"
] | [
"Open Ball is Open Set",
"Definition:Open Set/Metric Space",
"Definition:Open Set/Topology"
] |
proofwiki-19523 | Union minus Symmetric Difference equals Intersection | :$\paren {A \cup B} \setminus \paren {A \symdif B} = A \cap B$ | {{begin-eqn}}
{{eqn | l = \paren {A \cup B} \setminus \paren {A \symdif B}
| r = \paren {A \cup B} \setminus \paren {\paren {A \cup B} \setminus \paren {A \cap B} }
| c = {{Defof|Symmetric Difference|index = 2}}
}}
{{eqn | r = \paren {A \cup B} \cap \paren {A \cap B}
| c = Set Difference with Set Diff... | :$\paren {A \cup B} \setminus \paren {A \symdif B} = A \cap B$ | {{begin-eqn}}
{{eqn | l = \paren {A \cup B} \setminus \paren {A \symdif B}
| r = \paren {A \cup B} \setminus \paren {\paren {A \cup B} \setminus \paren {A \cap B} }
| c = {{Defof|Symmetric Difference|index = 2}}
}}
{{eqn | r = \paren {A \cup B} \cap \paren {A \cap B}
| c = [[Set Difference with Set Di... | Union minus Symmetric Difference equals Intersection | https://proofwiki.org/wiki/Union_minus_Symmetric_Difference_equals_Intersection | https://proofwiki.org/wiki/Union_minus_Symmetric_Difference_equals_Intersection | [
"Set Union",
"Set Intersection",
"Symmetric Difference"
] | [] | [
"Set Difference with Set Difference",
"Intersection of Union with Intersection",
"Category:Set Union",
"Category:Set Intersection",
"Category:Symmetric Difference"
] |
proofwiki-19524 | Locally Convex Space is Hausdorff iff induces Hausdorff Topology | Let $\struct {X, \mathcal P}$ be a locally convex space.
Let $\tau$ be the standard topology on $\struct {X, \mathcal P}$.
Then $\struct {X, \mathcal P}$ is Hausdorff {{iff}} $\struct {X, \tau}$ is a Hausdorff topological space. | === Sufficient Condition ===
Suppose that $\struct {X, \tau}$ is a Hausdorff space.
Let $x \in X$ have $x \ne {\mathbf 0}_X$.
If there exists no such $x$, then we have the claim by vacuous truth.
Otherwise, since $\struct {X, \tau}$ is Hausdorff, there exists $U, V \in \tau$ with $x \in U$ and ${\mathbf 0}_X \in V$ and... | Let $\struct {X, \mathcal P}$ be a [[Definition:Locally Convex Space|locally convex space]].
Let $\tau$ be the [[Definition:Locally Convex Space/Standard Topology|standard topology on $\struct {X, \mathcal P}$]].
Then $\struct {X, \mathcal P}$ is [[Definition:Locally Convex Space/Hausdorff|Hausdorff]] {{iff}} $\stru... | === Sufficient Condition ===
Suppose that $\struct {X, \tau}$ is a [[Definition:Hausdorff Space|Hausdorff space]].
Let $x \in X$ have $x \ne {\mathbf 0}_X$.
If there exists no such $x$, then we have the claim by [[Definition:Vacuous Truth|vacuous truth]].
Otherwise, since $\struct {X, \tau}$ is [[Definition:Hausdor... | Locally Convex Space is Hausdorff iff induces Hausdorff Topology | https://proofwiki.org/wiki/Locally_Convex_Space_is_Hausdorff_iff_induces_Hausdorff_Topology | https://proofwiki.org/wiki/Locally_Convex_Space_is_Hausdorff_iff_induces_Hausdorff_Topology | [
"Locally Convex Spaces",
"Hausdorff Spaces"
] | [
"Definition:Locally Convex Space",
"Definition:Locally Convex Space/Standard Topology",
"Definition:Locally Convex Space/Hausdorff",
"Definition:T2 Space"
] | [
"Definition:T2 Space",
"Definition:Vacuous Truth",
"Definition:T2 Space",
"Open Sets in Standard Topology of Locally Convex Space",
"Definition:Seminorm",
"Definition:Locally Convex Space/Hausdorff",
"Definition:Locally Convex Space/Hausdorff",
"Definition:Locally Convex Space/Hausdorff",
"Definitio... |
proofwiki-19525 | Seminorm Maps Zero Vector to Zero | Let $\struct {K, +, \circ}$ be a division ring with norm $\norm {\,\cdot\,}_K$.
Let $X$ be a vector space over $\struct {K, \norm {\,\cdot\,}_K}$.
Let $\mathbf 0_X$ be the zero vector of $X$.
Let $p$ be a seminorm on $X$.
Then $\map p {\mathbf 0_X} = 0$. | We have:
{{begin-eqn}}
{{eqn | l = \map p {\mathbf 0_X}
| r = \map p {0 \circ \mathbf 0_X}
| c = Zero Vector Scaled is Zero Vector
}}
{{eqn | r = \norm 0_K \map p {\mathbf 0_X}
| c = $(\text N 2)$ in {{Defof|Seminorm}}
}}
{{eqn | r = 0 \; \map p {\mathbf 0_X}
| c = $(\text N 1)$ in {{Defof|Norm ... | Let $\struct {K, +, \circ}$ be a [[Definition:Division Ring|division ring]] with [[Definition:Norm on Division Ring|norm]] $\norm {\,\cdot\,}_K$.
Let $X$ be a [[Definition:Vector Space|vector space]] over $\struct {K, \norm {\,\cdot\,}_K}$.
Let $\mathbf 0_X$ be the [[Definition:Zero Vector|zero vector]] of $X$.
Let ... | We have:
{{begin-eqn}}
{{eqn | l = \map p {\mathbf 0_X}
| r = \map p {0 \circ \mathbf 0_X}
| c = [[Zero Vector Scaled is Zero Vector]]
}}
{{eqn | r = \norm 0_K \map p {\mathbf 0_X}
| c = $(\text N 2)$ in {{Defof|Seminorm}}
}}
{{eqn | r = 0 \; \map p {\mathbf 0_X}
| c = $(\text N 1)$ in {{Defof|... | Seminorm Maps Zero Vector to Zero | https://proofwiki.org/wiki/Seminorm_Maps_Zero_Vector_to_Zero | https://proofwiki.org/wiki/Seminorm_Maps_Zero_Vector_to_Zero | [
"Seminorms",
"Zero Vectors"
] | [
"Definition:Division Ring",
"Definition:Norm/Division Ring",
"Definition:Vector Space",
"Definition:Zero Vector",
"Definition:Seminorm"
] | [
"Zero Vector Scaled is Zero Vector",
"Category:Seminorms",
"Category:Zero Vectors"
] |
proofwiki-19526 | Principle of Superinduction | Let $A$ be a class.
Let $g: A \to A$ be a mapping on $A$.
Let $A$ be minimally superinductive under $g$.
Let $P: A \to \set {\T, \F}$ be a propositional function on $A$.
Suppose that:
{{begin-axiom}}
{{axiom | n = 1
| m = \map P \O = \T
}}
{{axiom | n = 2
| q = \forall x \in A
| m = \map P x = \... | We are given that $A$ is a minimally superinductive class under $g$.
That is, $A$ is a superinductive class under $g$ with the extra property that for $A$ has no proper class which is also superinductive under $g$.
Let $P$ be a propositional function on $A$ which has the properties specified:
{{begin-axiom}}
{{axiom | ... | Let $A$ be a [[Definition:Class (Class Theory)|class]].
Let $g: A \to A$ be a [[Definition:Class Mapping|mapping]] on $A$.
Let $A$ be [[Definition:Minimally Superinductive Class|minimally superinductive under $g$]].
Let $P: A \to \set {\T, \F}$ be a [[Definition:Propositional Function|propositional function]] on $A... | We are given that $A$ is a [[Definition:Minimally Superinductive Class|minimally superinductive class under $g$]].
That is, $A$ is a [[Definition:Superinductive Class|superinductive class under $g$]] with the extra property that for $A$ has no [[Definition:Proper Class|proper class]] which is also [[Definition:Superin... | Principle of Superinduction | https://proofwiki.org/wiki/Principle_of_Superinduction | https://proofwiki.org/wiki/Principle_of_Superinduction | [
"Mathematical Induction",
"Proof Techniques",
"Proof by Superinduction",
"Closure under Chain Unions"
] | [
"Definition:Class (Class Theory)",
"Definition:Mapping/Class Theory",
"Definition:Minimally Superinductive Class",
"Definition:Propositional Function",
"Definition:Chain (Order Theory)/Subset Relation",
"Definition:Element/Class"
] | [
"Definition:Minimally Superinductive Class",
"Definition:Superinductive Class",
"Definition:Class (Class Theory)/Proper Class",
"Definition:Superinductive Class",
"Definition:Propositional Function",
"Definition:Chain (Order Theory)/Subset Relation",
"Definition:Element/Class",
"Definition:Class (Clas... |
proofwiki-19527 | Double Superinduction Principle | Let $M$ be a class.
Let $g: M \to M$ be a mapping on $M$.
Let $M$ be a minimally superinductive class under $g$.
Let $\RR$ be a relation on $M$ which satisfies:
{{begin-axiom}}
{{axiom | n = \text D_1
| q = \forall x \in M
| m = \map \RR {x, \O}
}}
{{axiom | n = \text D_2
| q = \forall x, y \in ... | The proof proceeds by Proof by Superinduction.
Let an element $x$ of $M$ be defined as:
:'''left normal''' with respect to $\RR$ {{iff}} $\map \RR {x, y}$ for all $y \in M$
:'''right normal''' with respect to $\RR$ {{iff}} $\map \RR {y, x}$ for all $y \in M$.
Let the hypothesis be assumed.
First we demonstrate a lemma: | Let $M$ be a [[Definition:Class (Class Theory)|class]].
Let $g: M \to M$ be a [[Definition:Class Mapping|mapping]] on $M$.
Let $M$ be a [[Definition:Minimally Superinductive Class|minimally superinductive class]] under $g$.
Let $\RR$ be a [[Definition:Relation (Class Theory)|relation]] on $M$ which satisfies:
{{beg... | The proof proceeds by [[Proof by Superinduction]].
Let an [[Definition:Element of Class|element]] $x$ of $M$ be defined as:
:'''[[Definition:Left Normal Element of Relation|left normal]]''' with respect to $\RR$ {{iff}} $\map \RR {x, y}$ for all $y \in M$
:'''[[Definition:Right Normal Element of Relation|right normal]... | Double Superinduction Principle | https://proofwiki.org/wiki/Double_Superinduction_Principle | https://proofwiki.org/wiki/Double_Superinduction_Principle | [
"Mathematical Induction",
"Named Theorems",
"Proof Techniques",
"Double Superinduction Principle",
"Closure under Chain Unions"
] | [
"Definition:Class (Class Theory)",
"Definition:Mapping/Class Theory",
"Definition:Minimally Superinductive Class",
"Definition:Relation/Class Theory",
"Definition:Chain (Order Theory)/Subset Relation",
"Definition:Element/Class"
] | [
"Principle of Superinduction",
"Definition:Element/Class",
"Definition:Left Normal Element of Relation",
"Definition:Right Normal Element of Relation",
"Definition:Lemma",
"Principle of Superinduction",
"Definition:Right Normal Element of Relation",
"Definition:Right Normal Element of Relation",
"De... |
proofwiki-19528 | Pasting Lemma/Union of Open Sets | Let $T = \struct {X, \tau}$ and $S = \struct {Y, \sigma}$ be topological spaces.
Let $I$ be an indexing set.
Let $\family {C_i}_{i \mathop \in I}$ be a family of open sets of $T$.
Let $f: X \to Y$ be a mapping such that the restriction $f \restriction_{C_i}$ is continuous for all $i \in I$.
Then $f$ is continuous on $C... | Let $V$ be an open set of $S$.
By assumption, we have that, for all $i \in I$, $U_i = \paren {f \restriction_{C_i} }^{-1} \sqbrk V$ is also open in $T$.
By {{Defof|Restriction of Mapping}}, we have that $U_i = C_i \cap f^{-1} \sqbrk V$.
Therefore, we can compute:
{{begin-eqn}}
{{eqn | l = \paren {f \restriction_C}^{-1}... | Let $T = \struct {X, \tau}$ and $S = \struct {Y, \sigma}$ be [[Definition:Topological Space|topological spaces]].
Let $I$ be an [[Definition:Indexing Set|indexing set]].
Let $\family {C_i}_{i \mathop \in I}$ be a [[Definition:Indexed Family|family]] of [[Definition:Open Set (Topology)|open sets]] of $T$.
Let $f: X \... | Let $V$ be an [[Definition:Open Set (Topology)|open set]] of $S$.
By assumption, we have that, for all $i \in I$, $U_i = \paren {f \restriction_{C_i} }^{-1} \sqbrk V$ is also open in $T$.
By {{Defof|Restriction of Mapping}}, we have that $U_i = C_i \cap f^{-1} \sqbrk V$.
Therefore, we can compute:
{{begin-eqn}}
{{e... | Pasting Lemma/Union of Open Sets | https://proofwiki.org/wiki/Pasting_Lemma/Union_of_Open_Sets | https://proofwiki.org/wiki/Pasting_Lemma/Union_of_Open_Sets | [
"Pasting Lemma",
"Open Sets"
] | [
"Definition:Topological Space",
"Definition:Indexing Set",
"Definition:Indexing Set/Family",
"Definition:Open Set/Topology",
"Definition:Mapping",
"Definition:Restriction/Mapping",
"Definition:Continuous Mapping (Topology)",
"Definition:Continuous Mapping (Topology)",
"Definition:Continuous Mapping ... | [
"Definition:Open Set/Topology",
"Intersection Distributes over Union",
"Definition:Set Union",
"Definition:Open Set/Topology",
"Definition:Open Set/Topology",
"Definition:Topology",
"Definition:Continuous Mapping (Topology)"
] |
proofwiki-19529 | Pasting Lemma/Finite Union of Closed Sets | Let $T = \struct {X, \tau}$ and $S = \struct {Y, \sigma}$ be topological spaces.
Let $I$ be a finite indexing set.
Let $\family {C_i}_{i \mathop \in I}$ be a finite family of closed sets of $T$.
Let $f: X \to Y$ be a mapping such that the restriction $f \restriction_{C_i}$ is continuous for all $i \in I$.
Then $f$ is c... | Let $V \subset S$ be a closed set.
By Continuity Defined from Closed Sets, we have that $U_i = \paren {f \restriction_{C_i} }^{-1} \sqbrk V$ is also closed.
From the definition of a restriction, we have that $U_i = C_i \cap f^{-1} \sqbrk V$.
Therefore, we can compute:
{{begin-eqn}}
{{eqn | l = \paren {f \restriction_{C... | Let $T = \struct {X, \tau}$ and $S = \struct {Y, \sigma}$ be [[Definition:Topological Space|topological spaces]].
Let $I$ be a [[Definition:Finite Set|finite]] [[Definition:Indexing Set|indexing set]].
Let $\family {C_i}_{i \mathop \in I}$ be a [[Definition:Finite Set|finite]] [[Definition:Indexed Family|family]] of ... | Let $V \subset S$ be a [[Definition:Closed Set (Topology)|closed set]].
By [[Continuity Defined from Closed Sets]], we have that $U_i = \paren {f \restriction_{C_i} }^{-1} \sqbrk V$ is also [[Definition:Closed Set (Topology)|closed]].
From the definition of a [[Definition:Restriction of Mapping|restriction]], we have... | Pasting Lemma/Finite Union of Closed Sets | https://proofwiki.org/wiki/Pasting_Lemma/Finite_Union_of_Closed_Sets | https://proofwiki.org/wiki/Pasting_Lemma/Finite_Union_of_Closed_Sets | [
"Pasting Lemma",
"Closed Sets"
] | [
"Definition:Topological Space",
"Definition:Finite Set",
"Definition:Indexing Set",
"Definition:Finite Set",
"Definition:Indexing Set/Family",
"Definition:Closed Set/Topology",
"Definition:Mapping",
"Definition:Restriction/Mapping",
"Definition:Continuous Mapping (Topology)",
"Definition:Continuou... | [
"Definition:Closed Set/Topology",
"Continuity Defined from Closed Sets",
"Definition:Closed Set/Topology",
"Definition:Restriction/Mapping",
"Intersection Distributes over Union",
"Definition:Set Union",
"Definition:Closed Set/Topology",
"Definition:Closed Set/Topology",
"Definition:Topology",
"Co... |
proofwiki-19530 | Pasting Lemma/Counterexample of Infinite Union of Closed Sets | Let $T = \struct {X, \tau}$ and $S = \struct {Y, \sigma}$ be topological spaces.
Let $I$ be an infinite indexing set.
For all $i \in I$, let $C_i$ be closed in $T$.
Let $f: X \to Y$ be a mapping such that the restriction $f \restriction_{C_i}$ is continuous for all $i$.
Then it does not necessarily hold that $f$ is con... | Let $f : \closedint {-1} 1 \to \R$ be the function on the closed interval $\closedint {-1} 1$ defined by:
:<nowiki>$\map f x = \begin{cases}
1 & : x \ge 0 \\
-1 & : x < 0
\end{cases}$</nowiki>
The function $f$ is discontinuous at $0$.
The function $f$ restricted to the closed intervals:
:$\closedint {-1} {-\dfrac 1 2... | Let $T = \struct {X, \tau}$ and $S = \struct {Y, \sigma}$ be [[Definition:Topological Space|topological spaces]].
Let $I$ be an [[Definition:Infinite Set|infinite]] [[Definition:Indexing Set|indexing set]].
For all $i \in I$, let $C_i$ be [[Definition:Closed Set (Topology)|closed]] in $T$.
Let $f: X \to Y$ be a [[De... | Let $f : \closedint {-1} 1 \to \R$ be the [[Definition:Real-Valued Function|function]] on the [[Definition:Closed Interval|closed interval]] $\closedint {-1} 1$ defined by:
:<nowiki>$\map f x = \begin{cases}
1 & : x \ge 0 \\
-1 & : x < 0
\end{cases}$</nowiki>
The function $f$ is [[Definition:Discontinuous Real Func... | Pasting Lemma/Counterexample of Infinite Union of Closed Sets | https://proofwiki.org/wiki/Pasting_Lemma/Counterexample_of_Infinite_Union_of_Closed_Sets | https://proofwiki.org/wiki/Pasting_Lemma/Counterexample_of_Infinite_Union_of_Closed_Sets | [
"Pasting Lemma",
"Closed Sets"
] | [
"Definition:Topological Space",
"Definition:Infinite Set",
"Definition:Indexing Set",
"Definition:Closed Set/Topology",
"Definition:Mapping",
"Definition:Restriction/Mapping",
"Definition:Continuous Mapping (Topology)",
"Definition:Continuous Mapping (Topology)",
"Definition:Continuous Mapping (Topo... | [
"Definition:Real-Valued Function",
"Definition:Interval/Ordered Set/Closed",
"Definition:Discontinuous Mapping/Real Function/Point",
"Definition:Restriction/Mapping",
"Definition:Interval/Ordered Set/Closed",
"Definition:Constant Mapping",
"Constant Function is Continuous",
"Definition:Restriction/Map... |
proofwiki-19531 | Soundness Theorem for Hilbert Proof System for Predicate Logic | Let $\mathscr H$ be instance 1 of a Hilbert proof system for predicate logic.
Let $\mathrm{PL}$ be the formal semantics of structures for predicate logic.
Then $\mathscr H$ is a strongly sound proof system for $\mathrm{PL}$:
:Every $\mathscr H$-provable consequence is a $\mathrm{PL}$-semantic consequence. | By Modus Ponendo Ponens for Semantic Consequence in Predicate Logic, the sole rule of inference of $\mathscr H$ preserves $\mathrm{PL}$-semantic consequence.
By definition a $\mathscr H$-provable consequence $\FF \vdash_{\mathscr H} \mathbf A$ means that $\mathbf A$ is a theorem of $\map {\mathscr H} \FF$.
It is to be ... | Let $\mathscr H$ be [[Definition:Hilbert Proof System Instance 1 for Predicate Logic|instance 1 of a Hilbert proof system for predicate logic]].
Let $\mathrm{PL}$ be the [[Definition:Formal Semantics of Structures for Predicate Logic|formal semantics of structures for predicate logic]].
Then $\mathscr H$ is a [[Defi... | By [[Modus Ponendo Ponens for Semantic Consequence in Predicate Logic]], the sole [[Definition:Rule of Inference|rule of inference]] of $\mathscr H$ preserves [[Definition:Semantic Consequence|$\mathrm{PL}$-semantic consequence]].
By definition a [[Definition:Provable Consequence|$\mathscr H$-provable consequence]] $... | Soundness Theorem for Hilbert Proof System for Predicate Logic | https://proofwiki.org/wiki/Soundness_Theorem_for_Hilbert_Proof_System_for_Predicate_Logic | https://proofwiki.org/wiki/Soundness_Theorem_for_Hilbert_Proof_System_for_Predicate_Logic | [
"Soundness Theorem",
"Hilbert Proof System Instance 1 for Predicate Logic"
] | [
"Definition:Hilbert Proof System/Predicate Logic/Instance 1",
"Definition:Structure for Predicate Logic/Formal Semantics",
"Definition:Sound Proof System/Strongly Sound",
"Definition:Provable Consequence",
"Definition:Semantic Consequence"
] | [
"Modus Ponendo Ponens for Semantic Consequence in Predicate Logic",
"Definition:Rule of Inference",
"Definition:Semantic Consequence",
"Definition:Provable Consequence",
"Definition:Theorem/Formal System",
"Definition:Semantic Consequence",
"Definition:Structure for Predicate Logic",
"Axioms of Hilber... |
proofwiki-19532 | Metric on Shift of Finite Type is Metric | Let $\struct {X _\mathbf A, \sigma_\mathbf A}$ be a shift of finite type.
Let $\theta \in \openint 0 1$.
Then the metric $d_\theta$ is indeed a metric on $X_\mathbf A$.
That is, $\struct {X _\mathbf A, d _\theta} $ is a metric space. | === M1 ===
Let $x\in X _\mathbf A$
Obviously, $x_i = x_i$ for all $i \in \openint {-\infty} \infty$.
Therefore:
:$\map {d _\theta} {x,x} = \theta ^\infty = 0$.
{{qed|lemma}} | Let $\struct {X _\mathbf A, \sigma_\mathbf A}$ be a [[Definition:Shift of Finite Type|shift of finite type]].
Let $\theta \in \openint 0 1$.
Then the [[Definition:Metric on Shift of Finite Type|metric $d_\theta$]] is indeed a [[Definition:Metric|metric]] on $X_\mathbf A$.
That is, $\struct {X _\mathbf A, d _\theta}... | === M1 ===
Let $x\in X _\mathbf A$
Obviously, $x_i = x_i$ for all $i \in \openint {-\infty} \infty$.
Therefore:
:$\map {d _\theta} {x,x} = \theta ^\infty = 0$.
{{qed|lemma}} | Metric on Shift of Finite Type is Metric | https://proofwiki.org/wiki/Metric_on_Shift_of_Finite_Type_is_Metric | https://proofwiki.org/wiki/Metric_on_Shift_of_Finite_Type_is_Metric | [
"Metric Spaces"
] | [
"Definition:Shift of Finite Type",
"Definition:Metric/Shift of Finite Type",
"Definition:Metric Space/Metric",
"Definition:Metric Space"
] | [] |
proofwiki-19533 | G-Tower is Nest | Let $M$ be a class.
Let $g: M \to M$ be a progressing mapping on $M$.
Let $M$ be a $g$-tower.
Then $M$ is a nest. | We need to show that:
:$\forall x, y \in M: x \subseteq y$ or $y \subseteq x$
First some lemmata: | Let $M$ be a [[Definition:Class (Class Theory)|class]].
Let $g: M \to M$ be a [[Definition:Progressing Mapping|progressing mapping]] on $M$.
Let $M$ be a [[Definition:G-Tower|$g$-tower]].
Then $M$ is a [[Definition:Nest (Class Theory)|nest]]. | We need to show that:
:$\forall x, y \in M: x \subseteq y$ or $y \subseteq x$
First some [[Definition:Lemma|lemmata]]: | G-Tower is Nest | https://proofwiki.org/wiki/G-Tower_is_Nest | https://proofwiki.org/wiki/G-Tower_is_Nest | [
"G-Tower is Nest",
"G-Towers",
"Nests"
] | [
"Definition:Class (Class Theory)",
"Definition:Progressing Mapping",
"Definition:G-Tower",
"Definition:Nest/Class Theory"
] | [
"Definition:Lemma"
] |
proofwiki-19534 | Sandwich Principle for G-Towers | Let $M$ be a class.
Let $g: M \to M$ be a progressing mapping on $M$.
Let $M$ be a $g$-tower.
Let $x, y \in M$ be arbitrary.
Then it cannot be the case that:
:$x \subsetneqq y \subsetneqq \map g x$ | From Lemma $2$ of $g$-Tower is Nest we have that:
:$\forall x, y \in M: \map g x \subseteq y \lor y \subseteq x$
From the Sandwich Principle:
:$\forall x, y \in M: x \subseteq y \subseteq \map g x \implies x = y \lor y = \map g x$
That is, there is no element $y$ of $M$ such that:
:$x \subsetneqq y \subsetneqq \map g x... | Let $M$ be a [[Definition:Class (Class Theory)|class]].
Let $g: M \to M$ be a [[Definition:Progressing Mapping|progressing mapping]] on $M$.
Let $M$ be a [[Definition:G-Tower|$g$-tower]].
Let $x, y \in M$ be arbitrary.
Then it cannot be the case that:
:$x \subsetneqq y \subsetneqq \map g x$ | From [[G-Tower is Nest/Lemma 2|Lemma $2$]] of [[G-Tower is Nest|$g$-Tower is Nest]] we have that:
:$\forall x, y \in M: \map g x \subseteq y \lor y \subseteq x$
From the [[Sandwich Principle]]:
:$\forall x, y \in M: x \subseteq y \subseteq \map g x \implies x = y \lor y = \map g x$
That is, there is no [[Definiti... | Sandwich Principle for G-Towers | https://proofwiki.org/wiki/Sandwich_Principle_for_G-Towers | https://proofwiki.org/wiki/Sandwich_Principle_for_G-Towers | [
"G-Towers",
"Sandwich Principle"
] | [
"Definition:Class (Class Theory)",
"Definition:Progressing Mapping",
"Definition:G-Tower"
] | [
"G-Tower is Nest/Lemma 2",
"G-Tower is Nest",
"Sandwich Principle",
"Definition:Element/Class"
] |
proofwiki-19535 | Sandwich Principle for G-Towers/Corollary 1 | Let $M$ be a class.
Let $g: M \to M$ be a progressing mapping on $M$.
Let $M$ be a $g$-tower.
Let $x, y \in M$ be arbitrary.
Then:
:$x \subsetneqq y \implies \map g x \subseteq y$ | From {{Lemma|G-Tower is Nest|2|disp = $g$-Tower is Nest|proof = yes}}:
:$\forall x, y \in M: \map g x \subseteq y \lor y \subseteq x$
From {{Corollary|Sandwich Principle|1}}:
:$\forall x, y \in M: x \subseteq y \subseteq \map g x \implies x = y \lor y = \map g x$
That is, if:
:$x \subsetneqq y$
then:
:$\map g x \subset... | Let $M$ be a [[Definition:Class (Class Theory)|class]].
Let $g: M \to M$ be a [[Definition:Progressing Mapping|progressing mapping]] on $M$.
Let $M$ be a [[Definition:G-Tower|$g$-tower]].
Let $x, y \in M$ be arbitrary.
Then:
:$x \subsetneqq y \implies \map g x \subseteq y$ | From {{Lemma|G-Tower is Nest|2|disp = $g$-Tower is Nest|proof = yes}}:
:$\forall x, y \in M: \map g x \subseteq y \lor y \subseteq x$
From {{Corollary|Sandwich Principle|1}}:
:$\forall x, y \in M: x \subseteq y \subseteq \map g x \implies x = y \lor y = \map g x$
That is, if:
:$x \subsetneqq y$
then:
:$\map g x ... | Sandwich Principle for G-Towers/Corollary 1 | https://proofwiki.org/wiki/Sandwich_Principle_for_G-Towers/Corollary_1 | https://proofwiki.org/wiki/Sandwich_Principle_for_G-Towers/Corollary_1 | [
"G-Towers",
"Sandwich Principle"
] | [
"Definition:Class (Class Theory)",
"Definition:Progressing Mapping",
"Definition:G-Tower"
] | [] |
proofwiki-19536 | Sandwich Principle for G-Towers/Corollary 2 | Let $M$ be a class.
Let $g: M \to M$ be a progressing mapping on $M$.
Let $M$ be a $g$-tower.
Let $x, y \in M$ be arbitrary such that $x \subseteq y$.
Then:
:$\map g x \subseteq \map g y$ | From {{Lemma|G-Tower is Nest|2|disp = $g$-Tower is Nest|proof = yes}}:
:$\forall x, y \in M: \map g x \subseteq y \lor y \subseteq x$
From {{Corollary|Sandwich Principle|2}}:
:$\forall x, y \in M: x \subseteq y\implies \map g x \subseteq \map g y$
{{qed}} | Let $M$ be a [[Definition:Class (Class Theory)|class]].
Let $g: M \to M$ be a [[Definition:Progressing Mapping|progressing mapping]] on $M$.
Let $M$ be a [[Definition:G-Tower|$g$-tower]].
Let $x, y \in M$ be arbitrary such that $x \subseteq y$.
Then:
:$\map g x \subseteq \map g y$ | From {{Lemma|G-Tower is Nest|2|disp = $g$-Tower is Nest|proof = yes}}:
:$\forall x, y \in M: \map g x \subseteq y \lor y \subseteq x$
From {{Corollary|Sandwich Principle|2}}:
:$\forall x, y \in M: x \subseteq y\implies \map g x \subseteq \map g y$
{{qed}} | Sandwich Principle for G-Towers/Corollary 2 | https://proofwiki.org/wiki/Sandwich_Principle_for_G-Towers/Corollary_2 | https://proofwiki.org/wiki/Sandwich_Principle_for_G-Towers/Corollary_2 | [
"G-Towers",
"Sandwich Principle"
] | [
"Definition:Class (Class Theory)",
"Definition:Progressing Mapping",
"Definition:G-Tower"
] | [] |
proofwiki-19537 | Union of Mappings which Agree is Mapping/Family of Mappings | Let $Y$ be a set.
Let $\family {A_i}_{i \mathop \in I}$ be a family of sets indexed by $I$.
Let $\family {f_i: A_i \to Y}$ be a family of mappings indexed by $I$.
Let $X = \ds \bigcup_{i \mathop \in I} A_i$.
Let $f = \ds \bigcup_{i \mathop \in I} f_i : X \to Y$ where $\ds \bigcup_{i \mathop \in I} f_i$ is the union of ... | By Union of Relations is Relation, $f = \ds \bigcup_{i \mathop \in I} f_i$ is a relation whose domain is $X = \ds \bigcup_{i \mathop \in I} A_i$.
Let $\tuple {x, y_1}, \tuple {x, y_2} \in f$.
Then there exists $i, j \in I$ such that:
:$\tuple {x, y_1} \in f_i$ and $\tuple {x, y_2} \in f_j$ | Let $Y$ be a [[Definition:Set|set]].
Let $\family {A_i}_{i \mathop \in I}$ be a [[Definition:Indexed Family of Sets|family of sets indexed by $I$]].
Let $\family {f_i: A_i \to Y}$ be a [[Definition:Indexed Family of Sets|family of mappings indexed by $I$]].
Let $X = \ds \bigcup_{i \mathop \in I} A_i$.
Let $f = \ds... | By [[Union of Relations is Relation]], $f = \ds \bigcup_{i \mathop \in I} f_i$ is a [[Definition:Relation|relation]] whose [[Definition:Domain of Relation|domain]] is $X = \ds \bigcup_{i \mathop \in I} A_i$.
Let $\tuple {x, y_1}, \tuple {x, y_2} \in f$.
Then there exists $i, j \in I$ such that:
:$\tuple {x, y_1} \in ... | Union of Mappings which Agree is Mapping/Family of Mappings | https://proofwiki.org/wiki/Union_of_Mappings_which_Agree_is_Mapping/Family_of_Mappings | https://proofwiki.org/wiki/Union_of_Mappings_which_Agree_is_Mapping/Family_of_Mappings | [
"Mapping Theory",
"Set Union",
"Indexed Families"
] | [
"Definition:Set",
"Definition:Indexing Set/Family of Sets",
"Definition:Indexing Set/Family of Sets",
"Definition:Set Union",
"Definition:Relation",
"Definition:Agreement/Mappings",
"Definition:Mapping"
] | [
"Union of Relations is Relation",
"Definition:Relation",
"Definition:Domain (Set Theory)/Relation"
] |
proofwiki-19538 | Pasting Lemma/Continuous Mappings on Open Sets | Let $T = \struct {X, \tau}$ and $S = \struct {Y, \sigma}$ be topological spaces.
Let $I$ be an indexing set.
Let $\family {C_i}_{i \mathop \in I}$ be a family of open sets of $T$.
Let $\family {f_i : C_i \to Y}_{i \mathop \in I}$ be a family of continuous mappings.
Let $C = \ds \bigcup_{i \mathop \in I} C_i$.
Let $f = ... | From Union of Family of Mappings which Agree is Mapping:
:$f$ is a mapping from $C$ to $Y$.
From Restriction of Union of Mappings which Agree Equals Mapping:
:$\forall i \in I : f \restriction_{C_i} = f_i$
Hence:
:$\forall i \in I : f \restriction_{C_i}$ is continuous
From Pasting Lemma for Union of Open Sets
:$f$ is c... | Let $T = \struct {X, \tau}$ and $S = \struct {Y, \sigma}$ be [[Definition:Topological Space|topological spaces]].
Let $I$ be an [[Definition:Indexing Set|indexing set]].
Let $\family {C_i}_{i \mathop \in I}$ be a [[Definition:Indexed Family|family]] of [[Definition:Open Set (Topology)|open sets]] of $T$.
Let $\famil... | From [[Union of Family of Mappings which Agree is Mapping]]:
:$f$ is a [[Definition:Mapping|mapping]] from $C$ to $Y$.
From [[Restriction of Union of Mappings which Agree Equals Mapping]]:
:$\forall i \in I : f \restriction_{C_i} = f_i$
Hence:
:$\forall i \in I : f \restriction_{C_i}$ is [[Definition:Continuous Mappi... | Pasting Lemma/Continuous Mappings on Open Sets | https://proofwiki.org/wiki/Pasting_Lemma/Continuous_Mappings_on_Open_Sets | https://proofwiki.org/wiki/Pasting_Lemma/Continuous_Mappings_on_Open_Sets | [
"Pasting Lemma",
"Open Sets"
] | [
"Definition:Topological Space",
"Definition:Indexing Set",
"Definition:Indexing Set/Family",
"Definition:Open Set/Topology",
"Definition:Indexing Set/Family",
"Definition:Continuous Mapping (Topology)",
"Definition:Set Union",
"Definition:Relation",
"Definition:Agreement/Mappings",
"Definition:Con... | [
"Union of Mappings which Agree is Mapping/Family of Mappings",
"Definition:Mapping",
"Restriction of Union of Mappings which Agree Equals Mapping",
"Definition:Continuous Mapping (Topology)",
"Pasting Lemma/Union of Open Sets",
"Definition:Continuous Mapping (Topology)"
] |
proofwiki-19539 | Fixed Point of g-Tower is Greatest Element | Let $M$ be a class.
Let $g: M \to M$ be a progressing mapping on $M$.
Let $M$ be a $g$-tower.
Let $x = \map g x$.
Then $x$ is the greatest element (under the subset relation) of $M$. | Let $x$ be an element of $M$ such that $\map g x = x$.
We will demonstrate by the Principle of Superinduction on $y$ that:
:$\forall y \in M: y \subseteq x$
The proof proceeds by superinduction.
For all $y \in M$, let $\map P y$ be the proposition:
:$y \subseteq x$ | Let $M$ be a [[Definition:Class (Class Theory)|class]].
Let $g: M \to M$ be a [[Definition:Progressing Mapping|progressing mapping]] on $M$.
Let $M$ be a [[Definition:G-Tower|$g$-tower]].
Let $x = \map g x$.
Then $x$ is the [[Definition:Greatest Set by Set Inclusion (Class Theory)|greatest element]] (under the [[D... | Let $x$ be an [[Definition:Element of Class|element]] of $M$ such that $\map g x = x$.
We will demonstrate by the [[Principle of Superinduction]] on $y$ that:
:$\forall y \in M: y \subseteq x$
The proof proceeds by [[Principle of Superinduction|superinduction]].
For all $y \in M$, let $\map P y$ be the [[Definition... | Fixed Point of g-Tower is Greatest Element | https://proofwiki.org/wiki/Fixed_Point_of_g-Tower_is_Greatest_Element | https://proofwiki.org/wiki/Fixed_Point_of_g-Tower_is_Greatest_Element | [
"G-Towers"
] | [
"Definition:Class (Class Theory)",
"Definition:Progressing Mapping",
"Definition:G-Tower",
"Definition:Greatest Set by Set Inclusion/Class Theory",
"Definition:Subset Relation"
] | [
"Definition:Element/Class",
"Principle of Superinduction",
"Principle of Superinduction",
"Definition:Proposition",
"Definition:Element/Class",
"Principle of Superinduction"
] |
proofwiki-19540 | Restriction of Union of Mappings which Agree Equals Mapping | Let $Y$ be a set.
Let $\family {A_i}_{i \mathop \in I}$ be a family of sets indexed by $I$.
Let $\family {f_i: A_i \to Y}$ be a family of mappings indexed by $I$.
Let $X = \ds \bigcup_{i \mathop \in I} A_i$.
Let $f = \ds \bigcup_{i \mathop \in I} f_i : X \to Y$ where $\ds \bigcup_{i \mathop \in I} f_i$ is the union of ... | From Union of Family of Mappings which Agree is Mapping:
:$f$ is a mapping from $X$ to $Y$.
Let $i \in I$ be an arbitrary index of $I$.
Consider the restriction of $f$ to $A_i$:
{{begin-eqn}}
{{eqn | l = f \restriction _ { A_i }
| r = f \cap \paren {A_i \times Y}
| c = {{Defof|Restriction of Mapping}}
}}
{{... | Let $Y$ be a [[Definition:Set|set]].
Let $\family {A_i}_{i \mathop \in I}$ be a [[Definition:Indexed Family of Sets|family of sets indexed by $I$]].
Let $\family {f_i: A_i \to Y}$ be a [[Definition:Indexed Family of Sets|family of mappings indexed by $I$]].
Let $X = \ds \bigcup_{i \mathop \in I} A_i$.
Let $f = \ds... | From [[Union of Family of Mappings which Agree is Mapping]]:
:$f$ is a [[Definition:Mapping|mapping]] from $X$ to $Y$.
Let $i \in I$ be an arbitrary [[Definition:Indexing Set|index]] of $I$.
Consider the [[Definition:Restriction of Mapping|restriction]] of $f$ to $A_i$:
{{begin-eqn}}
{{eqn | l = f \restriction _ { A... | Restriction of Union of Mappings which Agree Equals Mapping | https://proofwiki.org/wiki/Restriction_of_Union_of_Mappings_which_Agree_Equals_Mapping | https://proofwiki.org/wiki/Restriction_of_Union_of_Mappings_which_Agree_Equals_Mapping | [
"Mapping Theory",
"Set Union",
"Indexed Families",
"Restrictions"
] | [
"Definition:Set",
"Definition:Indexing Set/Family of Sets",
"Definition:Indexing Set/Family of Sets",
"Definition:Set Union",
"Definition:Relation",
"Definition:Agreement/Mappings",
"Definition:Restriction/Mapping"
] | [
"Union of Mappings which Agree is Mapping/Family of Mappings",
"Definition:Mapping",
"Definition:Indexing Set",
"Definition:Restriction/Mapping",
"Intersection Distributes over Union",
"Definition:Indexing Set",
"Definition:Mapping"
] |
proofwiki-19541 | Metric on Shift of Finite Type is Non-Archimedean | Let $\struct {X_\mathbf A, \sigma_\mathbf A}$ be a shift of finite type.
Let $\theta \in \openint 0 1$.
Then the metric $d_\theta$ on $X_\mathbf A$ is non-archimedean metric.
That is, $\struct {X_\mathbf A, d _\theta} $ is an ultrametric space. | Let $x, y, z \in X_\mathbf A$.
Let:
:$N_1 := \sup \set {n \in \N \cup \set \infty : x_i = y_i \text { for all } i \in \openint {-n} n}$
and:
:$N_2 := \sup \set {n \in \N \cup \set \infty : y_i = z_i \text { for all } i \in \openint {-n} n}$
so that:
:$\map {d_\theta} {x, y} = \theta^{N_1}$
and:
:$\map {d_\theta} {y, z}... | Let $\struct {X_\mathbf A, \sigma_\mathbf A}$ be a [[Definition:Shift of Finite Type|shift of finite type]].
Let $\theta \in \openint 0 1$.
Then the [[Definition:Metric on Shift of Finite Type|metric $d_\theta$]] on $X_\mathbf A$ is [[Definition:Non-Archimedean Metric|non-archimedean metric]].
That is, $\struct {X_... | Let $x, y, z \in X_\mathbf A$.
Let:
:$N_1 := \sup \set {n \in \N \cup \set \infty : x_i = y_i \text { for all } i \in \openint {-n} n}$
and:
:$N_2 := \sup \set {n \in \N \cup \set \infty : y_i = z_i \text { for all } i \in \openint {-n} n}$
so that:
:$\map {d_\theta} {x, y} = \theta^{N_1}$
and:
:$\map {d_\theta} {y, z... | Metric on Shift of Finite Type is Non-Archimedean | https://proofwiki.org/wiki/Metric_on_Shift_of_Finite_Type_is_Non-Archimedean | https://proofwiki.org/wiki/Metric_on_Shift_of_Finite_Type_is_Non-Archimedean | [
"Metric Spaces"
] | [
"Definition:Shift of Finite Type",
"Definition:Metric/Shift of Finite Type",
"Definition:Non-Archimedean/Metric",
"Definition:Ultrametric Space"
] | [
"Category:Metric Spaces"
] |
proofwiki-19542 | Completely Metrizable Space is Metrizable Space | Let $\struct {S, \tau}$ be a completely metrizable topological space.
Then, $\struct {S, \tau}$ is a metrizable space. | Since $\struct {S, \tau}$ is completely metrizable, $\tau$ is induced by a (complete) metric $d$.
In particular, $\struct {S, d}$ is a metric space.
As $\struct {S, \tau}$ is homeomorphic to itself by the identity mapping, it is metrizable.
{{qed}}
Category:Completely Metrizable Spaces
Category:Metrizable Spaces
pcwkc8... | Let $\struct {S, \tau}$ be a [[Definition:Completely Metrizable Space|completely metrizable topological space]].
Then, $\struct {S, \tau}$ is a [[Definition:Metrizable Space|metrizable space]]. | Since $\struct {S, \tau}$ is [[Definition:Completely Metrizable Space|completely metrizable]], $\tau$ is [[Definition:Topology Induced by Metric|induced]] by a ([[Definition:Complete Metric Space|complete]]) [[Definition:Metric|metric]] $d$.
In particular, $\struct {S, d}$ is a [[Definition:Metric Space|metric space]]... | Completely Metrizable Space is Metrizable Space | https://proofwiki.org/wiki/Completely_Metrizable_Space_is_Metrizable_Space | https://proofwiki.org/wiki/Completely_Metrizable_Space_is_Metrizable_Space | [
"Completely Metrizable Spaces",
"Metrizable Spaces"
] | [
"Definition:Completely Metrizable Space",
"Definition:Metrizable Space"
] | [
"Definition:Completely Metrizable Space",
"Definition:Topology Induced by Metric",
"Definition:Complete Metric Space",
"Definition:Metric Space/Metric",
"Definition:Metric Space",
"Definition:Homeomorphism/Topological Spaces",
"Identity Mapping is Homeomorphism",
"Definition:Metrizable Space",
"Cate... |
proofwiki-19543 | Pasting Lemma/Continuous Mappings on Closed Sets | Let $T = \struct {X, \tau}$ and $S = \struct {Y, \sigma}$ be topological spaces.
Let $I$ be a finite indexing set.
Let $\family {C_i}_{i \mathop \in I}$ be a finite family of closed sets of $T$.
Let $\family {f_i : C_i \to Y}_{i \mathop \in I}$ be a finite family of continuous mappings.
Let $C = \ds \bigcup_{i \mathop ... | From Union of Family of Mappings which Agree is Mapping:
:$f$ is a mapping from $C$ to $Y$.
From Restriction of Union of Mappings which Agree Equals Mapping:
:$\forall i \in I : f \restriction_{C_i} = f_i$
Hence:
:$\forall i \in I : f \restriction_{C_i}$ is continuous
From Pasting Lemma for Finite Union of Closed Sets
... | Let $T = \struct {X, \tau}$ and $S = \struct {Y, \sigma}$ be [[Definition:Topological Space|topological spaces]].
Let $I$ be a [[Definition:Finite Set|finite]] [[Definition:Indexing Set|indexing set]].
Let $\family {C_i}_{i \mathop \in I}$ be a [[Definition:Finite Set|finite]] [[Definition:Indexed Family|family]] of ... | From [[Union of Family of Mappings which Agree is Mapping]]:
:$f$ is a [[Definition:Mapping|mapping]] from $C$ to $Y$.
From [[Restriction of Union of Mappings which Agree Equals Mapping]]:
:$\forall i \in I : f \restriction_{C_i} = f_i$
Hence:
:$\forall i \in I : f \restriction_{C_i}$ is [[Definition:Continuous Mappi... | Pasting Lemma/Continuous Mappings on Closed Sets | https://proofwiki.org/wiki/Pasting_Lemma/Continuous_Mappings_on_Closed_Sets | https://proofwiki.org/wiki/Pasting_Lemma/Continuous_Mappings_on_Closed_Sets | [
"Pasting Lemma",
"Closed Sets"
] | [
"Definition:Topological Space",
"Definition:Finite Set",
"Definition:Indexing Set",
"Definition:Finite Set",
"Definition:Indexing Set/Family",
"Definition:Closed Set/Topology",
"Definition:Finite Set",
"Definition:Indexing Set/Family",
"Definition:Continuous Mapping (Topology)",
"Definition:Set Un... | [
"Union of Mappings which Agree is Mapping/Family of Mappings",
"Definition:Mapping",
"Restriction of Union of Mappings which Agree Equals Mapping",
"Definition:Continuous Mapping (Topology)",
"Pasting Lemma/Finite Union of Closed Sets",
"Definition:Continuous Mapping (Topology)"
] |
proofwiki-19544 | Pasting Lemma/Pair of Continuous Mappings on Open Sets | Let $X$ and $Y$ be topological spaces.
Let $A$ and $B$ be open in $X$.
Let $f: A \to Y$ and $g: B \to Y$ be continuous mappings that agree on $A \cap B$.
Let $f \cup g$ be the union of the mappings $f$ and $g$:
:$\forall x \in A \cup B: \map {f \cup g} x = \begin {cases} \map f x & : x \in A \\ \map g x & : x \in B \en... | Follows directly from Pasting Lemma for Continuous Mappings on Open Sets.
{{qed}} | Let $X$ and $Y$ be [[Definition:Topological Space|topological spaces]].
Let $A$ and $B$ be [[Definition:Open Set (Topology)|open]] in $X$.
Let $f: A \to Y$ and $g: B \to Y$ be [[Definition:Continuous Mapping on Set|continuous mappings]] that [[Definition:Agreement of Mappings|agree]] on $A \cap B$.
Let $f \cup g$ be... | Follows directly from [[Pasting Lemma for Continuous Mappings on Open Sets]].
{{qed}} | Pasting Lemma/Pair of Continuous Mappings on Open Sets/Proof 1 | https://proofwiki.org/wiki/Pasting_Lemma/Pair_of_Continuous_Mappings_on_Open_Sets | https://proofwiki.org/wiki/Pasting_Lemma/Pair_of_Continuous_Mappings_on_Open_Sets/Proof_1 | [
"Pasting Lemma",
"Pasting Lemma for Pair of Continuous Mappings on Open Sets"
] | [
"Definition:Topological Space",
"Definition:Open Set/Topology",
"Definition:Continuous Mapping (Topology)/Set",
"Definition:Agreement/Mappings",
"Definition:Set Union",
"Definition:Mapping",
"Definition:Continuous Mapping (Topology)/Set"
] | [
"Pasting Lemma/Continuous Mappings on Open Sets"
] |
proofwiki-19545 | Pasting Lemma/Pair of Continuous Mappings on Open Sets | Let $X$ and $Y$ be topological spaces.
Let $A$ and $B$ be open in $X$.
Let $f: A \to Y$ and $g: B \to Y$ be continuous mappings that agree on $A \cap B$.
Let $f \cup g$ be the union of the mappings $f$ and $g$:
:$\forall x \in A \cup B: \map {f \cup g} x = \begin {cases} \map f x & : x \in A \\ \map g x & : x \in B \en... | From Union of Mappings which Agree is Mapping
:$f \cup g$ is well-defined.
By {{Defof|Continuous Mapping}}:
:$f \cup g$ is continuous {{iff}} $\paren {f \cup g}^{-1} \sqbrk U$ is open in $A \cup B$ for every open $U$ in $Y$.
Let $U$ be an arbitrary open subset in $Y$.
From Preimage of Union Mapping is Union of Preimage... | Let $X$ and $Y$ be [[Definition:Topological Space|topological spaces]].
Let $A$ and $B$ be [[Definition:Open Set (Topology)|open]] in $X$.
Let $f: A \to Y$ and $g: B \to Y$ be [[Definition:Continuous Mapping on Set|continuous mappings]] that [[Definition:Agreement of Mappings|agree]] on $A \cap B$.
Let $f \cup g$ be... | From [[Union of Mappings which Agree is Mapping]]
:$f \cup g$ is [[Definition:Well-Defined Mapping|well-defined]].
By {{Defof|Continuous Mapping}}:
:$f \cup g$ is [[Definition:Continuous Mapping on Set|continuous]] {{iff}} $\paren {f \cup g}^{-1} \sqbrk U$ is [[Definition:Open Set (Topology)|open]] in $A \cup B$ for ... | Pasting Lemma/Pair of Continuous Mappings on Open Sets/Proof 2 | https://proofwiki.org/wiki/Pasting_Lemma/Pair_of_Continuous_Mappings_on_Open_Sets | https://proofwiki.org/wiki/Pasting_Lemma/Pair_of_Continuous_Mappings_on_Open_Sets/Proof_2 | [
"Pasting Lemma",
"Pasting Lemma for Pair of Continuous Mappings on Open Sets"
] | [
"Definition:Topological Space",
"Definition:Open Set/Topology",
"Definition:Continuous Mapping (Topology)/Set",
"Definition:Agreement/Mappings",
"Definition:Set Union",
"Definition:Mapping",
"Definition:Continuous Mapping (Topology)/Set"
] | [
"Union of Mappings which Agree is Mapping",
"Definition:Well-Defined/Mapping",
"Definition:Continuous Mapping (Topology)/Set",
"Definition:Open Set/Topology",
"Definition:Open Set/Topology",
"Definition:Open Set/Topology",
"Preimage of Union Mapping is Union of Preimages",
"Definition:Open Set/Topolog... |
proofwiki-19546 | Pasting Lemma/Pair of Continuous Mappings on Closed Sets | Let $X$ and $Y$ be topological spaces.
Let $A$ and $B$ be closed in $X$.
Let $f: A \to Y$ and $g: B \to Y$ be continuous mappings that agree on $A \cap B$.
Let $f \cup g$ be the union of the mappings $f$ and $g$:
:$\forall x \in A \cup B: \map {f \cup g} x = \begin {cases} \map f x & : x \in A \\ \map g x & : x \in B \... | Follows directly from Pasting Lemma for Continuous Mappings on Closed Sets
{{qed}} | Let $X$ and $Y$ be [[Definition:Topological Space|topological spaces]].
Let $A$ and $B$ be [[Definition:Closed Set (Topology)|closed]] in $X$.
Let $f: A \to Y$ and $g: B \to Y$ be [[Definition:Continuous Mapping on Set|continuous mappings]] that [[Definition:Agreement of Mappings|agree]] on $A \cap B$.
Let $f \cup g... | Follows directly from [[Pasting Lemma for Continuous Mappings on Closed Sets]]
{{qed}} | Pasting Lemma/Pair of Continuous Mappings on Closed Sets | https://proofwiki.org/wiki/Pasting_Lemma/Pair_of_Continuous_Mappings_on_Closed_Sets | https://proofwiki.org/wiki/Pasting_Lemma/Pair_of_Continuous_Mappings_on_Closed_Sets | [
"Pasting Lemma"
] | [
"Definition:Topological Space",
"Definition:Closed Set/Topology",
"Definition:Continuous Mapping (Topology)/Set",
"Definition:Agreement/Mappings",
"Definition:Set Union",
"Definition:Mapping",
"Definition:Continuous Mapping (Topology)/Set"
] | [
"Pasting Lemma/Continuous Mappings on Closed Sets"
] |
proofwiki-19547 | Shift of Finite Type is Compact | Let $\struct {X_\mathbf A, \sigma_\mathbf A}$ be a shift of finite type.
Then $X_\mathbf A $ is compact. | {{ProofWanted}}
Category:Compact Topological Spaces
Category:Topology
5khghc3au403lqdd3srns5yx0pcq2ez | Let $\struct {X_\mathbf A, \sigma_\mathbf A}$ be a [[Definition:Shift of Finite Type|shift of finite type]].
Then $X_\mathbf A $ is [[Definition:Compact Topological Space|compact]]. | {{ProofWanted}}
[[Category:Compact Topological Spaces]]
[[Category:Topology]]
5khghc3au403lqdd3srns5yx0pcq2ez | Shift of Finite Type is Compact | https://proofwiki.org/wiki/Shift_of_Finite_Type_is_Compact | https://proofwiki.org/wiki/Shift_of_Finite_Type_is_Compact | [
"Compact Topological Spaces",
"Topology"
] | [
"Definition:Shift of Finite Type",
"Definition:Compact Topological Space"
] | [
"Category:Compact Topological Spaces",
"Category:Topology"
] |
proofwiki-19548 | Union of g-Tower is Greatest Element and Unique Fixed Point | Let $M$ be a set.
Let $g: M \to M$ be a progressing mapping on $M$.
Let $M$ be a $g$-tower.
Then:
:$\ds \bigcup M \in M$
:$\ds \bigcup M$ is the greatest element (under the subset relation) of $M$
:$\ds \bigcup M$ is the unique fixed point of $M$. | Let the hypothesis be assumed.
By G-Tower is Nest, $M$ is a nest. By assumption, $M$ is a set.
Hence by definition $M$ is a chain.
By $g$-Tower is Closed under Chain Unions:
:$\ds \bigcup M \in M$
It follows by Draft:Union of Set of Sets is Greatest Element under Subset Relation that $\ds \bigcup M$ is the greatest ele... | Let $M$ be a [[Definition:Set|set]].
Let $g: M \to M$ be a [[Definition:Progressing Mapping|progressing mapping]] on $M$.
Let $M$ be a [[Definition:G-Tower|$g$-tower]].
Then:
:$\ds \bigcup M \in M$
:$\ds \bigcup M$ is the [[Definition:Greatest Set by Set Inclusion (Class Theory)|greatest element]] (under the [[Defi... | Let the hypothesis be assumed.
By [[G-Tower is Nest]], $M$ is a [[Definition:Nest (Class Theory)|nest]]. By assumption, $M$ is a [[Definition:Set|set]].
Hence by definition $M$ is a [[Definition:Chain of Sets|chain]].
By [[G-Tower is Closed under Chain Unions|$g$-Tower is Closed under Chain Unions]]:
:$\ds \bigcup M... | Union of g-Tower is Greatest Element and Unique Fixed Point | https://proofwiki.org/wiki/Union_of_g-Tower_is_Greatest_Element_and_Unique_Fixed_Point | https://proofwiki.org/wiki/Union_of_g-Tower_is_Greatest_Element_and_Unique_Fixed_Point | [
"G-Towers"
] | [
"Definition:Set",
"Definition:Progressing Mapping",
"Definition:G-Tower",
"Definition:Greatest Set by Set Inclusion/Class Theory",
"Definition:Subset Relation",
"Definition:Unique",
"Definition:Fixed Point"
] | [
"G-Tower is Nest",
"Definition:Nest/Class Theory",
"Definition:Set",
"Definition:Chain (Order Theory)/Subset Relation",
"G-Tower is Closed under Chain Unions",
"Draft:Union of Set of Sets is Greatest Element under Subset Relation",
"Definition:Greatest Set by Set Inclusion/Class Theory",
"Definition:S... |
proofwiki-19549 | G-Tower is Closed under Chain Unions | Let $M$ be a class.
Let $g: M \to M$ be a progressing mapping on $M$.
Let $M$ be a $g$-tower.
Then $M$ is closed under chain unions. | By definition:
:a $g$-tower is a class which is minimally superinductive under $g$
:a class which is minimally superinductive under $g$ is superinductive under $g$
:a superinductive class is closed under chain unions.
Hence the result.
{{qed}}
Category:G-Towers
Category:Closure under Chain Unions
reymphwfm5emep94xq63af... | Let $M$ be a [[Definition:Class (Class Theory)|class]].
Let $g: M \to M$ be a [[Definition:Progressing Mapping|progressing mapping]] on $M$.
Let $M$ be a [[Definition:G-Tower|$g$-tower]].
Then $M$ is [[Definition:Closure under Chain Unions|closed under chain unions]]. | By definition:
:a [[Definition:G-Tower|$g$-tower]] is a [[Definition:Class (Class Theory)|class]] which is [[Definition:Minimally Superinductive Class|minimally superinductive]] under $g$
:a [[Definition:Class (Class Theory)|class]] which is [[Definition:Minimally Superinductive Class|minimally superinductive]] under $... | G-Tower is Closed under Chain Unions | https://proofwiki.org/wiki/G-Tower_is_Closed_under_Chain_Unions | https://proofwiki.org/wiki/G-Tower_is_Closed_under_Chain_Unions | [
"G-Towers",
"Closure under Chain Unions"
] | [
"Definition:Class (Class Theory)",
"Definition:Progressing Mapping",
"Definition:G-Tower",
"Definition:Closure under Chain Unions"
] | [
"Definition:G-Tower",
"Definition:Class (Class Theory)",
"Definition:Minimally Superinductive Class",
"Definition:Class (Class Theory)",
"Definition:Minimally Superinductive Class",
"Definition:Superinductive Class",
"Definition:Superinductive Class",
"Definition:Closure under Chain Unions",
"Catego... |
proofwiki-19550 | Intersection of Set whose Every Element is Closed under Mapping is also Closed under Mapping | Let $S$ be a set of sets.
Let $g$ be a mapping such that:
:for every $x \in S$, $x$ is closed under $g$.
Then the intersection $\bigcap S$ of $S$ is also closed under $g$. | The domain of $g$ is not made clear, but the assumption is that:
:$\forall x \in S: \forall y \in x: y \in \Dom g$
First we note that by definition of intersection of $S$:
:$\bigcap S := \set {y: \forall x \in S: y \in x}$
Then:
{{begin-eqn}}
{{eqn | l = y
| o = \in
| r = \bigcap S
| c =
}}
{{eqn | l... | Let $S$ be a [[Definition:Set of Sets|set of sets]].
Let $g$ be a [[Definition:Mapping|mapping]] such that:
:for every $x \in S$, $x$ is [[Definition:Closed under Mapping|closed under $g$]].
Then the [[Definition:Intersection of Set of Sets|intersection]] $\bigcap S$ of $S$ is also [[Definition:Closed under Mapping|... | The [[Definition:Domain of Mapping|domain]] of $g$ is not made clear, but the assumption is that:
:$\forall x \in S: \forall y \in x: y \in \Dom g$
First we note that by definition of [[Definition:Intersection of Set of Sets|intersection of $S$]]:
:$\bigcap S := \set {y: \forall x \in S: y \in x}$
Then:
{{begin-eqn}... | Intersection of Set whose Every Element is Closed under Mapping is also Closed under Mapping/Proof | https://proofwiki.org/wiki/Intersection_of_Set_whose_Every_Element_is_Closed_under_Mapping_is_also_Closed_under_Mapping | https://proofwiki.org/wiki/Intersection_of_Set_whose_Every_Element_is_Closed_under_Mapping_is_also_Closed_under_Mapping/Proof | [
"Closedness under Mappings",
"Set Intersection",
"Intersection of Set whose Every Element is Closed under Mapping is also Closed under Mapping"
] | [
"Definition:Set of Sets",
"Definition:Mapping",
"Definition:Closed under Mapping",
"Definition:Set Intersection/Set of Sets",
"Definition:Closed under Mapping"
] | [
"Definition:Domain (Set Theory)/Mapping",
"Definition:Set Intersection/Set of Sets",
"Definition:Closed under Mapping",
"Definition:Closed under Mapping"
] |
proofwiki-19551 | Intersection of Set whose Every Element is Closed under Chain Unions is also Closed under Chain Unions | Let $S$ be a set of sets.
Let $S$ be such that:
:$\forall x \in S: x$ is closed under chain unions.
Then the intersection $\ds \bigcap S$ of $S$ is also closed under chain unions. | First we note that by definition of intersection of $S$:
:$\ds \bigcap S := \set {y: \forall x \in S: y \in x}$
Recall the definition of closed under chain unions:
$S$ is '''closed under chain unions''' {{iff}}:
:for every chain $C$ of elements of $S$, $\ds \bigcup C$ is also in $S$.
Let $C_\cap$ be a chain in $\ds \bi... | Let $S$ be a [[Definition:Set of Sets|set of sets]].
Let $S$ be such that:
:$\forall x \in S: x$ is [[Definition:Closure under Chain Unions|closed under chain unions]].
Then the [[Definition:Intersection of Set of Sets|intersection]] $\ds \bigcap S$ of $S$ is also [[Definition:Closure under Chain Unions|closed under... | First we note that by definition of [[Definition:Intersection of Set of Sets|intersection of $S$]]:
:$\ds \bigcap S := \set {y: \forall x \in S: y \in x}$
Recall the definition of [[Definition:Closure under Chain Unions|closed under chain unions]]:
$S$ is '''closed under chain unions''' {{iff}}:
:for every [[Defini... | Intersection of Set whose Every Element is Closed under Chain Unions is also Closed under Chain Unions/Proof | https://proofwiki.org/wiki/Intersection_of_Set_whose_Every_Element_is_Closed_under_Chain_Unions_is_also_Closed_under_Chain_Unions | https://proofwiki.org/wiki/Intersection_of_Set_whose_Every_Element_is_Closed_under_Chain_Unions_is_also_Closed_under_Chain_Unions/Proof | [
"Closure under Chain Unions",
"Set Intersection",
"Intersection of Set whose Every Element is Closed under Chain Unions is also Closed under Chain Unions"
] | [
"Definition:Set of Sets",
"Definition:Closure under Chain Unions",
"Definition:Set Intersection/Set of Sets",
"Definition:Closure under Chain Unions"
] | [
"Definition:Set Intersection/Set of Sets",
"Definition:Closure under Chain Unions",
"Definition:Chain (Order Theory)/Subset Relation",
"Definition:Element/Class",
"Definition:Chain (Order Theory)/Subset Relation",
"Definition:Chain (Order Theory)/Subset Relation",
"Definition:Closure under Chain Unions"... |
proofwiki-19552 | Inverse of Composition of Subsets of Topological Group | Let $\struct {G, \odot, \tau}$ be a topological group.
Let $A, B \subseteq G$.
Then:
:$\paren {A \odot B}^{-1} = B^{-1} \odot A^{-1}$ | The proof is identical to Inverse of Product of Subsets of Group.
{{qed}} | Let $\struct {G, \odot, \tau}$ be a [[Definition:Topological Group|topological group]].
Let $A, B \subseteq G$.
Then:
:$\paren {A \odot B}^{-1} = B^{-1} \odot A^{-1}$ | The proof is identical to [[Inverse of Product of Subsets of Group]].
{{qed}} | Inverse of Composition of Subsets of Topological Group | https://proofwiki.org/wiki/Inverse_of_Composition_of_Subsets_of_Topological_Group | https://proofwiki.org/wiki/Inverse_of_Composition_of_Subsets_of_Topological_Group | [
"Topological Groups"
] | [
"Definition:Topological Group"
] | [
"Inverse of Product of Subsets of Group"
] |
proofwiki-19553 | Set which is Superinductive under Progressing Mapping has Fixed Point | Let $S$ be a set.
Let $g: S \to S$ be a progressing mapping on $S$.
Let $S$ be superinductive under $g$.
Then there exists $x \in S$ such that $x$ is a fixed point of $g$. | Let us assume the hypothesis.
Let $M$ be the intersection of all subsets of $S$ that are superinductive under $g$.
From:
:Intersection of Set whose Every Element is Closed under Mapping is also Closed under Mapping
:Intersection of Set whose Every Element is Closed under Chain Unions is also Closed under Chain Unions
i... | Let $S$ be a [[Definition:Set|set]].
Let $g: S \to S$ be a [[Definition:Progressing Mapping|progressing mapping]] on $S$.
Let $S$ be [[Definition:Superinductive Class|superinductive under $g$]].
Then there exists $x \in S$ such that $x$ is a [[Definition:Fixed Point|fixed point]] of $g$. | Let us assume the hypothesis.
Let $M$ be the [[Definition:Set Intersection|intersection]] of all [[Definition:Subset|subsets]] of $S$ that are [[Definition:Superinductive Class|superinductive under $g$]].
From:
:[[Intersection of Set whose Every Element is Closed under Mapping is also Closed under Mapping]]
:[[Inters... | Set which is Superinductive under Progressing Mapping has Fixed Point | https://proofwiki.org/wiki/Set_which_is_Superinductive_under_Progressing_Mapping_has_Fixed_Point | https://proofwiki.org/wiki/Set_which_is_Superinductive_under_Progressing_Mapping_has_Fixed_Point | [
"Superinductive Classes",
"Progressing Mappings"
] | [
"Definition:Set",
"Definition:Progressing Mapping",
"Definition:Superinductive Class",
"Definition:Fixed Point"
] | [
"Definition:Set Intersection",
"Definition:Subset",
"Definition:Superinductive Class",
"Intersection of Set whose Every Element is Closed under Mapping is also Closed under Mapping",
"Intersection of Set whose Every Element is Closed under Chain Unions is also Closed under Chain Unions",
"Definition:Minim... |
proofwiki-19554 | Equivalence of Definitions of Strictly Progressing Mappings | Let $g$ be a mapping.
{{TFAE|def = Strictly Progressing Mapping}} | === $(1)$ implies $(2)$ ===
Let $g$ be a strictly progressing mapping by definition $1$.
Then by definition:
:$\forall x \in \Dom g: x \subsetneqq \map g x$
Hence {{afortiori}}:
:$\forall x \in \Dom g: x \subseteq \map g x$
and so $g$ is a progressing mapping.
Also {{afortiori}}:
:$\forall x \in \Dom g: x \ne \map g x$... | Let $g$ be a [[Definition:Class Mapping|mapping]].
{{TFAE|def = Strictly Progressing Mapping}} | === $(1)$ implies $(2)$ ===
Let $g$ be a [[Definition:Strictly Progressing Mapping/Definition 1|strictly progressing mapping by definition $1$]].
Then by definition:
:$\forall x \in \Dom g: x \subsetneqq \map g x$
Hence {{afortiori}}:
:$\forall x \in \Dom g: x \subseteq \map g x$
and so $g$ is a [[Definition:Progres... | Equivalence of Definitions of Strictly Progressing Mappings | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Strictly_Progressing_Mappings | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Strictly_Progressing_Mappings | [
"Strictly Progressing Mappings"
] | [
"Definition:Mapping/Class Theory"
] | [
"Definition:Strictly Progressing Mapping/Definition 1",
"Definition:Progressing Mapping",
"Definition:Fixed Point",
"Definition:Strictly Progressing Mapping/Definition 2",
"Definition:Strictly Progressing Mapping/Definition 2",
"Definition:Strictly Progressing Mapping/Definition 1"
] |
proofwiki-19555 | Superinductive Class under Strictly Progressing Mapping is Proper Class | Let $A$ be a class.
Let $g: A \to A$ be a strictly progressing mapping on $A$.
Let $A$ be superinductive under $g$.
Then $A$ cannot be a set, and thus is a proper class. | {{AimForCont}} $A$ is a set.
Then from Set which is Superinductive under Progressing Mapping has Fixed Point, $A$ has a fixed point.
However, we have that $g$ is a strictly progressing mapping on $A$.
Hence {{afortiori}} $g$ has no fixed point in $A$.
Hence the result by Proof by Contradiction.
{{Qed}} | Let $A$ be a [[Definition:Class (Class Theory)|class]].
Let $g: A \to A$ be a [[Definition:Strictly Progressing Mapping|strictly progressing mapping]] on $A$.
Let $A$ be [[Definition:Superinductive Class|superinductive]] under $g$.
Then $A$ cannot be a [[Definition:Set|set]], and thus is a [[Definition:Proper Class... | {{AimForCont}} $A$ is a [[Definition:Set|set]].
Then from [[Set which is Superinductive under Progressing Mapping has Fixed Point]], $A$ has a [[Definition:Fixed Point|fixed point]].
However, we have that $g$ is a [[Definition:Strictly Progressing Mapping|strictly progressing mapping]] on $A$.
Hence {{afortiori}} $g... | Superinductive Class under Strictly Progressing Mapping is Proper Class | https://proofwiki.org/wiki/Superinductive_Class_under_Strictly_Progressing_Mapping_is_Proper_Class | https://proofwiki.org/wiki/Superinductive_Class_under_Strictly_Progressing_Mapping_is_Proper_Class | [
"Superinductive Classes",
"Strictly Progressing Mappings"
] | [
"Definition:Class (Class Theory)",
"Definition:Strictly Progressing Mapping",
"Definition:Superinductive Class",
"Definition:Set",
"Definition:Class (Class Theory)/Proper Class"
] | [
"Definition:Set",
"Set which is Superinductive under Progressing Mapping has Fixed Point",
"Definition:Fixed Point",
"Definition:Strictly Progressing Mapping",
"Definition:Fixed Point",
"Proof by Contradiction"
] |
proofwiki-19556 | G-Tower is Well-Ordered under Subset Relation | Let $M$ be a class.
Let $g: M \to M$ be a progressing mapping on $M$.
Let $M$ be a $g$-tower.
Then $M$ is well-ordered under the subset relation such that:
{{begin-axiom}}
{{axiom | n = 1
| lc= Smallest Element:
| t = $\O$ is the smallest element of $M$
}}
{{axiom | n = 2
| lc= Immediate Success... | Let $M$ be a $g$-tower.
By $g$-Tower is Nest, $M$ is a nest.
Hence {{afortiori}} $\subseteq$ is a total ordering on $M$.
It remains to be shown that $\subseteq$ is a well-ordering, by showing the following:
Let $A \subseteq M$ be an arbitrary non-empty subclass of $M$.
Let $L$ be the set of all elements $x$ which are p... | Let $M$ be a [[Definition:Class (Class Theory)|class]].
Let $g: M \to M$ be a [[Definition:Progressing Mapping|progressing mapping]] on $M$.
Let $M$ be a [[Definition:G-Tower|$g$-tower]].
Then $M$ is [[Definition:Well-Ordered Class|well-ordered]] under the [[Definition:Subset Relation|subset relation]] such that:
... | Let $M$ be a [[Definition:G-Tower|$g$-tower]].
By [[G-Tower is Nest|$g$-Tower is Nest]], $M$ is a [[Definition:Nest (Class Theory)|nest]].
Hence {{afortiori}} $\subseteq$ is a [[Definition:Total Ordering|total ordering]] on $M$.
It remains to be shown that $\subseteq$ is a [[Definition:Well-Ordering (Class Theory)|... | G-Tower is Well-Ordered under Subset Relation | https://proofwiki.org/wiki/G-Tower_is_Well-Ordered_under_Subset_Relation | https://proofwiki.org/wiki/G-Tower_is_Well-Ordered_under_Subset_Relation | [
"G-Towers",
"Well-Orderings",
"Subset Relation",
"G-Tower is Well-Ordered under Subset Relation"
] | [
"Definition:Class (Class Theory)",
"Definition:Progressing Mapping",
"Definition:G-Tower",
"Definition:Well-Ordered Class",
"Definition:Subset Relation",
"Definition:Smallest Element/Class Theory",
"Definition:Smallest Element/Class Theory",
"Definition:Immediate Successor Element/Class Theory",
"De... | [
"Definition:G-Tower",
"G-Tower is Nest",
"Definition:Nest/Class Theory",
"Definition:Total Ordering",
"Definition:Well-Ordering/Class Theory",
"Definition:Non-Empty Set/Class Theory",
"Definition:Subclass",
"Definition:Set",
"Definition:Element/Class",
"Definition:Proper Subset",
"Definition:Ele... |
proofwiki-19557 | Modus Ponendo Ponens for Semantic Consequence in Predicate Logic | Let $\mathrm{PL}$ be the formal semantics of structures for predicate logic.
Denote with $\models_{\mathrm{PL}}$ $\mathrm{PL}$-semantic consequence.
Let $\mathbf A$ and $\mathbf B$ be sentences.
Let $\FF$ be a set of sentences.
Suppose that:
:$\FF \models_{\mathrm{PL}} \mathbf A$
:$\FF \models_{\mathrm{PL}} \mathbf A \... | To establish $\FF \models_{\mathrm{PL}} \mathbf B$, we need to show that for all structures $\AA$:
:$\AA \models_{\mathrm{PL}} \FF$ implies $\AA \models_{\mathrm{PL}} \mathbf B$
where $\models_{\mathrm{PL}}$ denotes the models relation.
Then since $\AA \models_{\mathrm{PL}} \FF$, it follows that:
:$\AA \models_{\mathrm... | Let $\mathrm{PL}$ be the [[Definition:Formal Semantics of Structures for Predicate Logic|formal semantics of structures for predicate logic]].
Denote with $\models_{\mathrm{PL}}$ [[Definition:Semantic Consequence (Predicate Logic)|$\mathrm{PL}$-semantic consequence]].
Let $\mathbf A$ and $\mathbf B$ be [[Definition:S... | To establish $\FF \models_{\mathrm{PL}} \mathbf B$, we need to show that for all [[Definition:Structure for Predicate Logic|structures]] $\AA$:
:$\AA \models_{\mathrm{PL}} \FF$ implies $\AA \models_{\mathrm{PL}} \mathbf B$
where $\models_{\mathrm{PL}}$ denotes the [[Definition:Model (Predicate Logic)|models relation]... | Modus Ponendo Ponens for Semantic Consequence in Predicate Logic | https://proofwiki.org/wiki/Modus_Ponendo_Ponens_for_Semantic_Consequence_in_Predicate_Logic | https://proofwiki.org/wiki/Modus_Ponendo_Ponens_for_Semantic_Consequence_in_Predicate_Logic | [
"Model Theory for Predicate Logic",
"Modus Ponendo Ponens"
] | [
"Definition:Structure for Predicate Logic/Formal Semantics",
"Definition:Semantic Consequence/Predicate Logic",
"Definition:Classes of WFFs/Sentence",
"Definition:Set",
"Definition:Classes of WFFs/Sentence",
"Modus Ponendo Ponens"
] | [
"Definition:Structure for Predicate Logic",
"Definition:Model (Predicate Logic)",
"Definition:Conditional/Truth Function",
"Definition:Value of Formula under Assignment/Sentence",
"Definition:Conditional/Truth Function",
"Category:Model Theory for Predicate Logic",
"Category:Modus Ponendo Ponens"
] |
proofwiki-19558 | Short Exact Sequence Condition of Noetherian Modules | Let $A$ be a commutative ring with unity.
Let:
:$0 \longrightarrow M' \stackrel {\alpha} {\longrightarrow} M \stackrel {\beta} {\longrightarrow} M' ' \longrightarrow 0$
be a short exact sequence of $A$-modules.
Then:
:$M$ is Noetherian
{{iff}}:
:$M'$ and $M' '$ are Noetherian | === Necessary condition ===
Assume $M$ is Noetherian. | Let $A$ be a [[Definition:Commutative and Unitary Ring|commutative ring with unity]].
Let:
:$0 \longrightarrow M' \stackrel {\alpha} {\longrightarrow} M \stackrel {\beta} {\longrightarrow} M' ' \longrightarrow 0$
be a [[Definition:Short Exact Sequence of Modules|short exact sequence]] of [[Definition:Module over Ring|... | === Necessary condition ===
Assume $M$ is [[Definition:Noetherian Module|Noetherian]]. | Short Exact Sequence Condition of Noetherian Modules | https://proofwiki.org/wiki/Short_Exact_Sequence_Condition_of_Noetherian_Modules | https://proofwiki.org/wiki/Short_Exact_Sequence_Condition_of_Noetherian_Modules | [
"Noetherian Modules"
] | [
"Definition:Commutative and Unitary Ring",
"Definition:Short Exact Sequence of Modules",
"Definition:Module over Ring",
"Definition:Noetherian Module",
"Definition:Noetherian Module"
] | [
"Definition:Noetherian Module",
"Definition:Noetherian Module",
"Definition:Noetherian Module",
"Definition:Noetherian Module"
] |
proofwiki-19559 | Subspace of Euclidean Space is Closed | Let $\R^n$ be a Euclidean space.
Every subspace of $\R^n$ is closed in $\R^n$. | First we note that from Euclidean Space is Normed Vector Space, $\R^n$ has a norm $\norm {\, \cdot \,}$.
Let $V$ be a subspace of $\R^n$.
Let $\set {\mathbf v_1, \ldots, \mathbf v_k}$ be a basis for $V$.
Extend this to basis $\set {\mathbf v_1, \ldots, \mathbf v_k, \mathbf v_{k + 1}, \ldots, \mathbf v_n}$ for $\R^n$.
B... | Let $\R^n$ be a [[Definition:Euclidean Space|Euclidean space]].
Every [[Definition:Vector Subspace|subspace]] of $\R^n$ is [[Definition:Closed Set of Normed Vector Space|closed]] in $\R^n$. | First we note that from [[Euclidean Space is Normed Vector Space]], $\R^n$ has a [[Definition:Norm on Vector Space|norm]] $\norm {\, \cdot \,}$.
Let $V$ be a [[Definition:Vector Subspace|subspace]] of $\R^n$.
Let $\set {\mathbf v_1, \ldots, \mathbf v_k}$ be a [[Definition:Basis of Vector Space|basis]] for $V$.
Exten... | Subspace of Euclidean Space is Closed | https://proofwiki.org/wiki/Subspace_of_Euclidean_Space_is_Closed | https://proofwiki.org/wiki/Subspace_of_Euclidean_Space_is_Closed | [
"Real Euclidean Spaces"
] | [
"Definition:Euclidean Space",
"Definition:Vector Subspace",
"Definition:Closed Set/Normed Vector Space"
] | [
"Euclidean Space is Normed Vector Space",
"Definition:Norm/Vector Space",
"Definition:Vector Subspace",
"Definition:Basis of Vector Space",
"Definition:Basis of Vector Space",
"Gram-Schmidt Orthogonalization",
"Definition:Orthonormal Basis of Vector Space",
"Definition:Generated Submodule/Linear Span"... |
proofwiki-19560 | Characterization of Lipschitz Continuity on Shift of Finite Type by Variations | Let $\struct {X_\mathbf A, \sigma_\mathbf A}$ be a shift of finite type.
Let $f : X_\mathbf A \to \C$ be a continuous mapping.
Let $\theta \in \openint 0 1$.
Let $C > 0$.
Then:
:$\forall x, y \in X_\mathbf A : \size {\map f x - \map f y} \le C \map {d_\theta} {x, y}$
{{iff}}:
:$\forall n \in \N : \map {\mathrm {var}_n}... | === Sufficient Condition ===
Suppose:
:$\forall x, y \in X_\mathbf A : \size {\map f x - \map f y} \le C \map {d_\theta} {x, y}$
Let $n\in\N$.
By definition of the metric:
:$\forall x, y \in X_\mathbf A:$
::$\forall i \in \openint {-n} n : x_i = y_i \implies \map {d _\theta} {x, y} \le \theta ^n$
Thus, by the assumptio... | Let $\struct {X_\mathbf A, \sigma_\mathbf A}$ be a [[Definition:Shift of Finite Type|shift of finite type]].
Let $f : X_\mathbf A \to \C$ be a [[Definition:Continuous Mapping (Metric Space)|continuous]] [[Definition:Mapping|mapping]].
Let $\theta \in \openint 0 1$.
Let $C > 0$.
Then:
:$\forall x, y \in X_\mathbf A... | === Sufficient Condition ===
Suppose:
:$\forall x, y \in X_\mathbf A : \size {\map f x - \map f y} \le C \map {d_\theta} {x, y}$
Let $n\in\N$.
By definition of the [[Definition:Metric on Shift of Finite Type|metric]]:
:$\forall x, y \in X_\mathbf A:$
::$\forall i \in \openint {-n} n : x_i = y_i \implies \map {d _\th... | Characterization of Lipschitz Continuity on Shift of Finite Type by Variations | https://proofwiki.org/wiki/Characterization_of_Lipschitz_Continuity_on_Shift_of_Finite_Type_by_Variations | https://proofwiki.org/wiki/Characterization_of_Lipschitz_Continuity_on_Shift_of_Finite_Type_by_Variations | [
"Lipschitz Continuity",
"Metric Spaces"
] | [
"Definition:Shift of Finite Type",
"Definition:Continuous Mapping (Metric Space)",
"Definition:Mapping",
"Definition:Metric/Shift of Finite Type",
"Definition:Variation of Function/Shift of Finite Type",
"Definition:Lipschitz Seminorm"
] | [
"Definition:Metric/Shift of Finite Type"
] |
proofwiki-19561 | G-Tower is Closed under Mapping | Let $M$ be a class.
Let $g: M \to M$ be a progressing mapping on $M$.
Let $M$ be a $g$-tower.
Then $M$ is closed under $g$:
:$\forall x: \paren {x \in M \implies \map g x \in M}$ | By definition:
:a $g$-tower is a class which is minimally superinductive under $g$
:a class which is minimally superinductive under $g$ is superinductive under $g$
:a superinductive class is inductive under $g$
:an inductive class under $g$ is closed under $g$.
Hence the result.
{{qed}}
Category:G-Towers
Category:Close... | Let $M$ be a [[Definition:Class (Class Theory)|class]].
Let $g: M \to M$ be a [[Definition:Progressing Mapping|progressing mapping]] on $M$.
Let $M$ be a [[Definition:G-Tower|$g$-tower]].
Then $M$ is [[Definition:Closed Class under Mapping|closed under $g$]]:
:$\forall x: \paren {x \in M \implies \map g x \in M}$ | By definition:
:a [[Definition:G-Tower|$g$-tower]] is a [[Definition:Class (Class Theory)|class]] which is [[Definition:Minimally Superinductive Class|minimally superinductive]] under $g$
:a [[Definition:Class (Class Theory)|class]] which is [[Definition:Minimally Superinductive Class|minimally superinductive]] under $... | G-Tower is Closed under Mapping | https://proofwiki.org/wiki/G-Tower_is_Closed_under_Mapping | https://proofwiki.org/wiki/G-Tower_is_Closed_under_Mapping | [
"G-Towers",
"Closedness under Mappings"
] | [
"Definition:Class (Class Theory)",
"Definition:Progressing Mapping",
"Definition:G-Tower",
"Definition:Closed under Mapping/Class Theory"
] | [
"Definition:G-Tower",
"Definition:Class (Class Theory)",
"Definition:Minimally Superinductive Class",
"Definition:Class (Class Theory)",
"Definition:Minimally Superinductive Class",
"Definition:Superinductive Class",
"Definition:Superinductive Class",
"Definition:Inductive Class/General",
"Definitio... |
proofwiki-19562 | Strict Lower Closure of G-Tower is Set of Elements which are Proper Subsets | Let $M$ be a class.
Let $g: M \to M$ be a progressing mapping on $M$.
Let $M$ be a $g$-tower.
Let $x \in M$.
Then the strict lower closure of $x$ is the set of all elements of $M$ that are proper subsets of $M$. | Follows directly from the definitions.
{{qed}} | Let $M$ be a [[Definition:Class (Class Theory)|class]].
Let $g: M \to M$ be a [[Definition:Progressing Mapping|progressing mapping]] on $M$.
Let $M$ be a [[Definition:G-Tower|$g$-tower]].
Let $x \in M$.
Then the [[Definition:Strict Lower Closure of Element (Class Theory)|strict lower closure]] of $x$ is the [[Defi... | Follows directly from the definitions.
{{qed}} | Strict Lower Closure of G-Tower is Set of Elements which are Proper Subsets | https://proofwiki.org/wiki/Strict_Lower_Closure_of_G-Tower_is_Set_of_Elements_which_are_Proper_Subsets | https://proofwiki.org/wiki/Strict_Lower_Closure_of_G-Tower_is_Set_of_Elements_which_are_Proper_Subsets | [
"Lower Closures",
"G-Towers"
] | [
"Definition:Class (Class Theory)",
"Definition:Progressing Mapping",
"Definition:G-Tower",
"Definition:Strict Lower Closure/Element/Class Theory",
"Definition:Set",
"Definition:Element/Class",
"Definition:Proper Subset"
] | [] |
proofwiki-19563 | P-Seminorm is Seminorm | Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $p \in \hointr 1 \infty$.
Let $\map {\LL^p} {X, \Sigma, \mu}$ be the Lebesgue $p$-space on $\struct {X, \Sigma, \mu}$.
Let $\norm {\, \cdot \,}_p$ be the $p$-seminorm on $\map {\LL^p} {X, \Sigma, \mu}$.
Then $\norm {\, \cdot \,}_p$ is a seminorm on $\map {\LL^p} {... | Let $f \in \map {\LL^p} {X, \Sigma, \mu}$.
From the construction of the integral of a positive $\Sigma$-measurable function, we have:
:$\ds \paren {\int \size f^p \rd \mu}^{1/p} \ge 0$
so:
:$\norm f_p \ge 0$
We also have:
:$\ds \int \size f^p \rd \mu < \infty$
So $\norm {\, \cdot \,}$ maps from $\map {\LL^p} {X, \Si... | Let $\struct {X, \Sigma, \mu}$ be a [[Definition:Measure Space|measure space]].
Let $p \in \hointr 1 \infty$.
Let $\map {\LL^p} {X, \Sigma, \mu}$ be the [[Definition:Lebesgue Space|Lebesgue $p$-space]] on $\struct {X, \Sigma, \mu}$.
Let $\norm {\, \cdot \,}_p$ be the [[Definition:P-Seminorm|$p$-seminorm]] on $\map ... | Let $f \in \map {\LL^p} {X, \Sigma, \mu}$.
From the construction of the [[Definition:Integral of Positive Measurable Function|integral of a positive $\Sigma$-measurable function]], we have:
:$\ds \paren {\int \size f^p \rd \mu}^{1/p} \ge 0$
so:
:$\norm f_p \ge 0$
We also have:
:$\ds \int \size f^p \rd \mu < \i... | P-Seminorm is Seminorm | https://proofwiki.org/wiki/P-Seminorm_is_Seminorm | https://proofwiki.org/wiki/P-Seminorm_is_Seminorm | [
"Lebesgue Spaces",
"Seminorms"
] | [
"Definition:Measure Space",
"Definition:Lebesgue Space",
"Definition:P-Seminorm",
"Definition:Seminorm"
] | [
"Definition:Integral of Positive Measurable Function",
"Definition:Mapping",
"Definition:Positive/Real Number"
] |
proofwiki-19564 | Lp Space is Subset of Space of Real-Valued Measurable Functions Identified by A.E. Equality | Let $\struct {X, \Sigma, \mu}$ be a measure space, and let $p \in \closedint 1 \infty$.
Let $\map {\mathcal M} {X, \Sigma, \R}/\sim_\mu$ be the space of real-valued measurable functions identified by $\mu$-A.E. equality.
Let $\map {L^p} {X, \Sigma, \mu}$ be the $L^p$ space of $\struct {X, \Sigma, \mu}$.
Then:
:$\map {... | Let $f \in \map {\LL^p} {X, \Sigma, \mu}$.
Denote by $\eqclass f {\sim_\mu}$ the $\sim_\mu$-equivalence class of $f$ in $\map {\LL^p} {X, \Sigma, \mu}$.
Let $\eqclass f {\sim_\mu}^\ast$ be the $\sim_\mu$-equivalence class of $f$ in $\map {\mathcal M} {X, \Sigma, \R}$.
We aim to show that:
:$\eqclass f {\sim_\mu} = \eq... | Let $\struct {X, \Sigma, \mu}$ be a [[Definition:Measure Space|measure space]], and let $p \in \closedint 1 \infty$.
Let $\map {\mathcal M} {X, \Sigma, \R}/\sim_\mu$ be the [[Definition:Space of Measurable Functions Identified by A.E. Equality|space of real-valued measurable functions identified by $\mu$-A.E. equality... | Let $f \in \map {\LL^p} {X, \Sigma, \mu}$.
Denote by $\eqclass f {\sim_\mu}$ the [[Definition:Equivalence Class|$\sim_\mu$-equivalence class]] of $f$ in $\map {\LL^p} {X, \Sigma, \mu}$.
Let $\eqclass f {\sim_\mu}^\ast$ be the [[Definition:Equivalence Class|$\sim_\mu$-equivalence class]] of $f$ in $\map {\mathcal M} {... | Lp Space is Subset of Space of Real-Valued Measurable Functions Identified by A.E. Equality | https://proofwiki.org/wiki/Lp_Space_is_Subset_of_Space_of_Real-Valued_Measurable_Functions_Identified_by_A.E._Equality | https://proofwiki.org/wiki/Lp_Space_is_Subset_of_Space_of_Real-Valued_Measurable_Functions_Identified_by_A.E._Equality | [
"Lp Spaces",
"Space of Real-Valued Measurable Functions Identified by A.E. Equality"
] | [
"Definition:Measure Space",
"Definition:Space of Measurable Functions Identified by A.E. Equality",
"Definition:Lp Space"
] | [
"Definition:Equivalence Class",
"Definition:Equivalence Class",
"Pointwise Exponentiation preserves A.E. Equality",
"Category:Lp Spaces",
"Category:Space of Real-Valued Measurable Functions Identified by A.E. Equality"
] |
proofwiki-19565 | Space of Real-Valued Measurable Functions Identified by A.E. Equality is Vector Space | Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $\map \MM {X, \Sigma, \R}$ be the set of real-valued $\Sigma$-measurable functions on $X$.
Let $\sim_\mu$ be the almost-everywhere equality relation on $\map \MM {X, \Sigma, \R}$ with respect to $\mu$.
Let $+$ denote pointwise addition on $\map \MM {X, \Sigma, \R}... | We verify each of the vector space axioms. | Let $\struct {X, \Sigma, \mu}$ be a [[Definition:Measure Space|measure space]].
Let $\map \MM {X, \Sigma, \R}$ be the set of [[Definition:Measurable Real-Valued Function|real-valued $\Sigma$-measurable functions]] on $X$.
Let $\sim_\mu$ be the [[Definition:Almost-Everywhere Equality Relation|almost-everywhere equali... | We verify each of the [[Axiom:Vector Space Axioms|vector space axioms]]. | Space of Real-Valued Measurable Functions Identified by A.E. Equality is Vector Space | https://proofwiki.org/wiki/Space_of_Real-Valued_Measurable_Functions_Identified_by_A.E._Equality_is_Vector_Space | https://proofwiki.org/wiki/Space_of_Real-Valued_Measurable_Functions_Identified_by_A.E._Equality_is_Vector_Space | [
"Vector Spaces",
"Space of Real-Valued Measurable Functions Identified by A.E. Equality"
] | [
"Definition:Measure Space",
"Definition:Measurable Function/Real-Valued Function",
"Definition:Almost-Everywhere Equality Relation",
"Definition:Pointwise Addition on Space of Real-Valued Measurable Functions Identified by A.E. Equality",
"Definition:Pointwise Scalar Multiplication on Space of Real-Valued M... | [
"Axiom:Vector Space Axioms"
] |
proofwiki-19566 | G-Tower is Well-Ordered under Subset Relation/Empty Set | $\O$ is the smallest element of $M$. | Follows directly from $g$-Tower is Well-Ordered under Subset Relation.
{{qed}} | $\O$ is the [[Definition:Smallest Element (Class Theory)|smallest element]] of $M$. | Follows directly from [[G-Tower is Well-Ordered under Subset Relation|$g$-Tower is Well-Ordered under Subset Relation]].
{{qed}} | G-Tower is Well-Ordered under Subset Relation/Empty Set | https://proofwiki.org/wiki/G-Tower_is_Well-Ordered_under_Subset_Relation/Empty_Set | https://proofwiki.org/wiki/G-Tower_is_Well-Ordered_under_Subset_Relation/Empty_Set | [
"G-Tower is Well-Ordered under Subset Relation"
] | [
"Definition:Smallest Element/Class Theory"
] | [
"G-Tower is Well-Ordered under Subset Relation"
] |
proofwiki-19567 | G-Tower is Well-Ordered under Subset Relation/Successor of Non-Greatest Element | Let $x \in M$ such that $x$ is not the greatest element of $M$.
Then the immediate successor of $x$ is $\map g x$. | We have that $g$-Tower is Well-Ordered under Subset Relation.
Let $x \in M$ such that $x$ is not the greatest element of $M$.
Then from Fixed Point of $g$-Tower is Greatest Element:
:$x \ne \map g x$
Hence:
:$x \subsetneqq \map g x$
Hence by the Sandwich Principle for $g$-Towers, there is no $y \in M$ such that:
:$x \s... | Let $x \in M$ such that $x$ is not the [[Definition:Greatest Element (Class Theory)|greatest element]] of $M$.
Then the [[Definition:Immediate Successor Element (Class Theory)|immediate successor]] of $x$ is $\map g x$. | We have that [[G-Tower is Well-Ordered under Subset Relation|$g$-Tower is Well-Ordered under Subset Relation]].
Let $x \in M$ such that $x$ is not the [[Definition:Greatest Element (Class Theory)|greatest element]] of $M$.
Then from [[Fixed Point of g-Tower is Greatest Element|Fixed Point of $g$-Tower is Greatest Ele... | G-Tower is Well-Ordered under Subset Relation/Successor of Non-Greatest Element | https://proofwiki.org/wiki/G-Tower_is_Well-Ordered_under_Subset_Relation/Successor_of_Non-Greatest_Element | https://proofwiki.org/wiki/G-Tower_is_Well-Ordered_under_Subset_Relation/Successor_of_Non-Greatest_Element | [
"G-Tower is Well-Ordered under Subset Relation"
] | [
"Definition:Greatest Element/Class Theory",
"Definition:Immediate Successor Element/Class Theory"
] | [
"G-Tower is Well-Ordered under Subset Relation",
"Definition:Greatest Element/Class Theory",
"Fixed Point of g-Tower is Greatest Element",
"Sandwich Principle for G-Towers",
"Definition:Immediate Successor Element/Class Theory"
] |
proofwiki-19568 | G-Tower is Well-Ordered under Subset Relation/Union of Limit Elements | Let $x \in M$ be a limit element of $M$.
Then:
:$x = \bigcup x^\subset$
where $\bigcup x^\subset$ denotes the union of the lower section of $x$. | We have that $g$-Tower is Well-Ordered under Subset Relation.
Let $x \in M$ be a limit element of $M$.
Then by definition $x$ is not an immediate successor element.
Hence by $g$-Tower is Well-Ordered under Subset Relation: Successor of Non-Greatest Element:
:there exists no $y \in M$ such that $x = \map g y$
otherwise ... | Let $x \in M$ be a [[Definition:Limit Element under Well-Ordering|limit element]] of $M$.
Then:
:$x = \bigcup x^\subset$
where $\bigcup x^\subset$ denotes the [[Definition:Union of Set of Sets|union]] of the [[Definition:Lower Section|lower section]] of $x$. | We have that [[G-Tower is Well-Ordered under Subset Relation|$g$-Tower is Well-Ordered under Subset Relation]].
Let $x \in M$ be a [[Definition:Limit Element under Well-Ordering|limit element]] of $M$.
Then by definition $x$ is not an [[Definition:Immediate Successor Element (Class Theory)|immediate successor element... | G-Tower is Well-Ordered under Subset Relation/Union of Limit Elements | https://proofwiki.org/wiki/G-Tower_is_Well-Ordered_under_Subset_Relation/Union_of_Limit_Elements | https://proofwiki.org/wiki/G-Tower_is_Well-Ordered_under_Subset_Relation/Union_of_Limit_Elements | [
"G-Tower is Well-Ordered under Subset Relation"
] | [
"Definition:Limit Element under Well-Ordering",
"Definition:Set Union/Set of Sets",
"Definition:Lower Section"
] | [
"G-Tower is Well-Ordered under Subset Relation",
"Definition:Limit Element under Well-Ordering",
"Definition:Immediate Successor Element/Class Theory",
"G-Tower is Well-Ordered under Subset Relation/Successor of Non-Greatest Element",
"Definition:Immediate Successor Element/Class Theory",
"G-Tower is Well... |
proofwiki-19569 | Slow g-Tower is Slowly Well-Ordered under Subset Relation | Let $M$ be a class.
Let $g: M \to M$ be a slowly progressing mapping on $M$.
Let $M$ be a slow $g$-tower.
Then $M$ is slowly well-ordered under the subset relation. | Let $M$ be a slow $g$-tower.
By $g$-Tower is Well-Ordered under Subset Relation: Empty Set:
:$\O$ is the smallest element of $M$.
This is condition $(\text S_1)$ of the definition of slowly well-ordered class under subset relation.
By $g$-Tower is Well-Ordered under Subset Relation: Union of Limit Elements:
:Each limit... | Let $M$ be a [[Definition:Class (Class Theory)|class]].
Let $g: M \to M$ be a [[Definition:Slowly Progressing Mapping|slowly progressing mapping]] on $M$.
Let $M$ be a [[Definition:Slow g-Tower|slow $g$-tower]].
Then $M$ is [[Definition:Slowly Well-Ordered Class under Subset Relation|slowly well-ordered under the s... | Let $M$ be a [[Definition:Slow g-Tower|slow $g$-tower]].
By [[G-Tower is Well-Ordered under Subset Relation/Empty Set|$g$-Tower is Well-Ordered under Subset Relation: Empty Set]]:
:$\O$ is the [[Definition:Smallest Element (Class Theory)|smallest element]] of $M$.
This is condition $(\text S_1)$ of the definition of ... | Slow g-Tower is Slowly Well-Ordered under Subset Relation | https://proofwiki.org/wiki/Slow_g-Tower_is_Slowly_Well-Ordered_under_Subset_Relation | https://proofwiki.org/wiki/Slow_g-Tower_is_Slowly_Well-Ordered_under_Subset_Relation | [
"G-Towers",
"Well-Orderings",
"Subset Relation"
] | [
"Definition:Class (Class Theory)",
"Definition:Slowly Progressing Mapping",
"Definition:Slow g-Tower",
"Definition:Slowly Well-Ordered Class under Subset Relation"
] | [
"Definition:Slow g-Tower",
"G-Tower is Well-Ordered under Subset Relation/Empty Set",
"Definition:Smallest Element/Class Theory",
"Definition:Slowly Well-Ordered Class under Subset Relation",
"G-Tower is Well-Ordered under Subset Relation/Union of Limit Elements",
"Definition:Limit Element under Well-Orde... |
proofwiki-19570 | Union of Slowly Well-Ordered Class under Subset Relation is Well-Orderable | Let $N$ be a class.
Let $N$ be slowly well-ordered under the subset relation.
Then $\bigcup N$ is well-orderable. | Let $N$ be slowly well-ordered under $\subseteq$.
Let $A = \bigcup N$.
For $a \in A$, let $\map F a$ denote the smallest element of $N$ that contains $a$.
Then for $a, b \in A$, we define $a \preccurlyeq b \iff \map F a \subseteq \map F b$.
It will be shown that $\preccurlyeq$ is a well-ordering on $A$.
By Class which ... | Let $N$ be a [[Definition:Class (Class Theory)|class]].
Let $N$ be [[Definition:Slowly Well-Ordered Class under Subset Relation|slowly well-ordered under the subset relation]].
Then $\bigcup N$ is [[Definition:Well-Orderable Class|well-orderable]]. | Let $N$ be [[Definition:Slowly Well-Ordered Class under Subset Relation|slowly well-ordered under $\subseteq$]].
Let $A = \bigcup N$.
For $a \in A$, let $\map F a$ denote the [[Definition:Smallest Element (Class Theory)|smallest element]] of $N$ that contains $a$.
Then for $a, b \in A$, we define $a \preccurlyeq b ... | Union of Slowly Well-Ordered Class under Subset Relation is Well-Orderable | https://proofwiki.org/wiki/Union_of_Slowly_Well-Ordered_Class_under_Subset_Relation_is_Well-Orderable | https://proofwiki.org/wiki/Union_of_Slowly_Well-Ordered_Class_under_Subset_Relation_is_Well-Orderable | [
"Well-Orderings",
"Subset Relation",
"Class Union",
"Union of Slowly Well-Ordered Class under Subset Relation is Well-Orderable"
] | [
"Definition:Class (Class Theory)",
"Definition:Slowly Well-Ordered Class under Subset Relation",
"Definition:Well-Orderable Set/Class Theory"
] | [
"Definition:Slowly Well-Ordered Class under Subset Relation",
"Definition:Smallest Element/Class Theory",
"Definition:Well-Ordering",
"Class which has Injection to Subclass of Well-Orderable Class is Well-Orderable",
"Definition:Injection",
"Definition:Slowly Well-Ordered Class under Subset Relation",
"... |
proofwiki-19571 | Deduction Theorem for Hilbert Proof System for Predicate Logic | Let $\mathscr H$ be instance 1 of a Hilbert proof system for predicate logic.
Then the deduction rule:
::$\dfrac{U,\mathbf A \vdash \mathbf B}{U \vdash \mathbf A \implies \mathbf B}$
is a derived rule for provable consequences in $\mathscr H$. | For any proof of $U, \mathbf A \vdash \mathbf B$, we indicate how to transform it into a proof of $U \vdash \mathbf A \implies \mathbf B$ without using the deduction rule.
This is done by applying the Second Principle of Mathematical Induction to the length $n$ of the proof of $U,\mathbf A \vdash \mathbf B$.
By definit... | Let $\mathscr H$ be [[Definition:Hilbert Proof System Instance 1 for Predicate Logic|instance 1 of a Hilbert proof system for predicate logic]].
Then the [[Definition:Deduction Rule|deduction rule]]:
::$\dfrac{U,\mathbf A \vdash \mathbf B}{U \vdash \mathbf A \implies \mathbf B}$
is a [[Definition:Derived Rule|deriv... | For any [[Definition:Formal Proof|proof]] of $U, \mathbf A \vdash \mathbf B$, we indicate how to transform it into a proof of $U \vdash \mathbf A \implies \mathbf B$ without using the [[Definition:Deduction Rule|deduction rule]].
This is done by applying the [[Second Principle of Mathematical Induction]] to the length... | Deduction Theorem for Hilbert Proof System for Predicate Logic | https://proofwiki.org/wiki/Deduction_Theorem_for_Hilbert_Proof_System_for_Predicate_Logic | https://proofwiki.org/wiki/Deduction_Theorem_for_Hilbert_Proof_System_for_Predicate_Logic | [
"Deduction Theorem",
"Hilbert Proof System Instance 1 for Predicate Logic"
] | [
"Definition:Hilbert Proof System/Predicate Logic/Instance 1",
"Definition:Deduction Rule",
"Definition:Derived Rule",
"Definition:Provable Consequence"
] | [
"Definition:Proof System/Formal Proof",
"Definition:Deduction Rule",
"Second Principle of Mathematical Induction",
"Definition:Proof System/Formal Proof",
"Definition:Hilbert Proof System/Predicate Logic/Instance 1",
"Definition:Axiom/Formal Systems",
"Modus Ponendo Ponens",
"Definition:Proof System/F... |
proofwiki-19572 | Union of Slow g-Tower is Well-Orderable | Let $M$ be a class.
Let $g: M \to M$ be a slowly progressing mapping on $M$.
Let $M$ be a slow $g$-tower.
Then $\ds \bigcup M$ is well-orderable. | Let $M$ be a slow $g$-tower.
From Slow $g$-Tower is Slowly Well-Ordered under Subset Relation we have that $M$ is slowly well-ordered under $\subseteq$.
The result follows from Union of Slowly Well-Ordered Class under Subset Relation is Well-Orderable.
{{qed}} | Let $M$ be a [[Definition:Class (Class Theory)|class]].
Let $g: M \to M$ be a [[Definition:Slowly Progressing Mapping|slowly progressing mapping]] on $M$.
Let $M$ be a [[Definition:Slow g-Tower|slow $g$-tower]].
Then $\ds \bigcup M$ is [[Definition:Well-Orderable Class|well-orderable]]. | Let $M$ be a [[Definition:Slow g-Tower|slow $g$-tower]].
From [[Slow g-Tower is Slowly Well-Ordered under Subset Relation|Slow $g$-Tower is Slowly Well-Ordered under Subset Relation]] we have that $M$ is [[Definition:Slowly Well-Ordered Class under Subset Relation|slowly well-ordered under $\subseteq$]].
The result f... | Union of Slow g-Tower is Well-Orderable | https://proofwiki.org/wiki/Union_of_Slow_g-Tower_is_Well-Orderable | https://proofwiki.org/wiki/Union_of_Slow_g-Tower_is_Well-Orderable | [
"G-Towers",
"Well-Orderings",
"Class Union"
] | [
"Definition:Class (Class Theory)",
"Definition:Slowly Progressing Mapping",
"Definition:Slow g-Tower",
"Definition:Well-Orderable Set/Class Theory"
] | [
"Definition:Slow g-Tower",
"Slow g-Tower is Slowly Well-Ordered under Subset Relation",
"Definition:Slowly Well-Ordered Class under Subset Relation",
"Union of Slowly Well-Ordered Class under Subset Relation is Well-Orderable"
] |
proofwiki-19573 | Measure is countably subadditive | Let $\left({X, \Sigma, \mu}\right)$ be a measure space.
Then $\mu$ is $\sigma$-subadditive, that is:
:$\ds\forall \sequence {A_n} _{n \in \N} \subseteq \Sigma: \map \mu {\bigcup _{n \in \N} A_n} \le \sum _{n \in \N} \map \mu {A_n}$ | For each $n\in\N$, let:
:$\ds A'_n := A_n \setminus \bigcup _{i=0} ^{n-1} A_i$
Then, by (SA3) and (SA2'):
:$\forall n\in\N : A'_n \in \Sigma$
Furthermore, by construction:
:$\bigcup _{n \in \N} A'_n = \bigcup _{n \in \N} A_n$
:$\sequence {A'_n} _{n \in \N}$ are pairwise disjoint
:$\forall n\in\N : A' _n\subseteq A_n$
T... | Let $\left({X, \Sigma, \mu}\right)$ be a [[Definition:Measure Space|measure space]].
Then $\mu$ is [[Definition:Countably Additive Function|$\sigma$-subadditive]], that is:
:$\ds\forall \sequence {A_n} _{n \in \N} \subseteq \Sigma: \map \mu {\bigcup _{n \in \N} A_n} \le \sum _{n \in \N} \map \mu {A_n}$ | For each $n\in\N$, let:
:$\ds A'_n := A_n \setminus \bigcup _{i=0} ^{n-1} A_i$
Then, by [[Definition:Sigma-Algebra/Definition 1|(SA3)]] and [[Definition:Sigma-Algebra/Definition 2|(SA2')]]:
:$\forall n\in\N : A'_n \in \Sigma$
Furthermore, by construction:
:$\bigcup _{n \in \N} A'_n = \bigcup _{n \in \N} A_n$
:$\seque... | Measure is countably subadditive | https://proofwiki.org/wiki/Measure_is_countably_subadditive | https://proofwiki.org/wiki/Measure_is_countably_subadditive | [
"Measures"
] | [
"Definition:Measure Space",
"Definition:Countably Additive Function"
] | [
"Definition:Sigma-Algebra/Definition 1",
"Definition:Sigma-Algebra/Definition 2",
"Definition:Pairwise Disjoint/Family",
"Definition:Measure (Measure Theory)",
"Measure is Monotone",
"Category:Measures"
] |
proofwiki-19574 | Set with Slowly Progressing Mapping on Power Set with Self as Fixed Point is Well-Orderable | Let $S$ be a set of sets.
Let there exist a slowly progressing mapping $g: \powerset S \to \powerset S$, where $\powerset S$ denotes the power set of $S$.
Let $g$ have exactly one fixed point $S$.
Then $S$ is well-orderable. | Let the hypothesis be assumed.
We have that:
:$\O \in \powerset S$
:$\powerset S$ is closed under $g$
:$\powerset S$ is closed under chain unions.
Hence $\powerset S$ is superinductive under $g$.
Let $M$ be the intersection of all subsets of $S$ that are superinductive under $g$.
From:
:Intersection of Set whose Every ... | Let $S$ be a [[Definition:Set of Sets|set of sets]].
Let there exist a [[Definition:Slowly Progressing Mapping|slowly progressing mapping]] $g: \powerset S \to \powerset S$, where $\powerset S$ denotes the [[Definition:Power Set|power set]] of $S$.
Let $g$ have exactly one [[Definition:Fixed Point|fixed point]] $S$.
... | Let the hypothesis be assumed.
We have that:
:$\O \in \powerset S$
:$\powerset S$ is [[Definition:Closed under Mapping|closed]] under $g$
:$\powerset S$ is [[Definition:Closure under Chain Unions|closed under chain unions]].
Hence $\powerset S$ is [[Definition:Superinductive Class|superinductive]] under $g$.
Let $M... | Set with Slowly Progressing Mapping on Power Set with Self as Fixed Point is Well-Orderable | https://proofwiki.org/wiki/Set_with_Slowly_Progressing_Mapping_on_Power_Set_with_Self_as_Fixed_Point_is_Well-Orderable | https://proofwiki.org/wiki/Set_with_Slowly_Progressing_Mapping_on_Power_Set_with_Self_as_Fixed_Point_is_Well-Orderable | [
"Slowly Progressing Mappings",
"Well-Orderings"
] | [
"Definition:Set of Sets",
"Definition:Slowly Progressing Mapping",
"Definition:Power Set",
"Definition:Fixed Point",
"Definition:Well-Orderable Set"
] | [
"Definition:Closed under Mapping",
"Definition:Closure under Chain Unions",
"Definition:Superinductive Class",
"Definition:Set Intersection",
"Definition:Subset",
"Definition:Superinductive Class",
"Intersection of Set whose Every Element is Closed under Mapping is also Closed under Mapping",
"Interse... |
proofwiki-19575 | L-Infinity Norm is Well-Defined | Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $\map {\LL^\infty} {X, \Sigma, \mu}$ be the Lebesgue $\infty$-space for $\struct {X, \Sigma, \mu}$.
Let $\sim$ be the $\mu$-almost-everywhere equality relation on $\map {\LL^\infty} {X, \Sigma, \mu}$.
Let $\map {L^\infty} {X, \Sigma, \mu}$ be the $L^\infty$ space o... | We show that for $E \in \map {L^\infty} {X, \Sigma, \mu}$, $\norm E_\infty$ is independent of the representative chosen for $E$.
Let:
:$E = \eqclass f \sim = \eqclass g \sim$
for $\eqclass f \sim, \eqclass g \sim \in \map {L^\infty} {X, \Sigma, \mu}$.
We show that:
:$\norm f_\infty = \norm g_\infty$
From Equivalence C... | Let $\struct {X, \Sigma, \mu}$ be a [[Definition:Measure Space|measure space]].
Let $\map {\LL^\infty} {X, \Sigma, \mu}$ be the [[Definition:Lebesgue Infinity-Space|Lebesgue $\infty$-space]] for $\struct {X, \Sigma, \mu}$.
Let $\sim$ be the [[Definition:Almost-Everywhere Equality Relation|$\mu$-almost-everywhere equa... | We show that for $E \in \map {L^\infty} {X, \Sigma, \mu}$, $\norm E_\infty$ is independent of the [[Definition:Representative of Equivalence Class|representative]] chosen for $E$.
Let:
:$E = \eqclass f \sim = \eqclass g \sim$
for $\eqclass f \sim, \eqclass g \sim \in \map {L^\infty} {X, \Sigma, \mu}$.
We show that:... | L-Infinity Norm is Well-Defined | https://proofwiki.org/wiki/L-Infinity_Norm_is_Well-Defined | https://proofwiki.org/wiki/L-Infinity_Norm_is_Well-Defined | [
"L-Infinity Norm"
] | [
"Definition:Measure Space",
"Definition:Lebesgue Space/L-Infinity",
"Definition:Almost-Everywhere Equality Relation",
"Definition:Lp Space",
"Definition:Lp Norm/L-Infinity Norm"
] | [
"Definition:Equivalence Class/Representative",
"Equivalence Class Equivalent Statements",
"Definition:Almost-Everywhere Equality Relation",
"Definition:Almost Everywhere",
"Definition:Null Set",
"Null Sets Closed under Subset",
"Definition:Disjoint Sets",
"Definition:Real Number",
"Definition:Counta... |
proofwiki-19576 | L-Infinity Norm is Norm | Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $\map {L^\infty} {X, \Sigma, \mu}$ be the $L^\infty$ vector space on $\struct {X, \Sigma, \mu}$.
Let $\norm \cdot_\infty$ be the $L^\infty$ norm.
Then $\norm \cdot_\infty$ is a norm on $\map {L^\infty} {X, \Sigma, \mu}$. | {{MissingLinks}}
Let $\map {\LL^\infty} {X, \Sigma, \mu}$ be the Lebesgue $\infty$-space.
Let $\sim$ be the $\mu$-almost everywhere equality relation on $\map {\LL^\infty} {X, \Sigma, \mu}$.
Let $\eqclass f \sim \in \map {L^\infty} {X, \Sigma, \mu}$.
Then, we have by the definition of the $L^\infty$ norm we have:
:$... | Let $\struct {X, \Sigma, \mu}$ be a [[Definition:Measure Space|measure space]].
Let $\map {L^\infty} {X, \Sigma, \mu}$ be the [[Definition:Lp Vector Space|$L^\infty$ vector space]] on $\struct {X, \Sigma, \mu}$.
Let $\norm \cdot_\infty$ be the [[Definition:L-Infinity Norm|$L^\infty$ norm]].
Then $\norm \cdot_\infty... | {{MissingLinks}}
Let $\map {\LL^\infty} {X, \Sigma, \mu}$ be the [[Definition:Lebesgue Infinity-Space|Lebesgue $\infty$-space]].
Let $\sim$ be the [[Definition:Almost-Everywhere Equality Relation on Lebesgue Space|$\mu$-almost everywhere equality relation on $\map {\LL^\infty} {X, \Sigma, \mu}$]].
Let $\eqclass f \... | L-Infinity Norm is Norm | https://proofwiki.org/wiki/L-Infinity_Norm_is_Norm | https://proofwiki.org/wiki/L-Infinity_Norm_is_Norm | [
"L-Infinity Norm",
"Normed Vector Spaces"
] | [
"Definition:Measure Space",
"Definition:Lp Space/Vector Space",
"Definition:Lp Norm/L-Infinity Norm",
"Definition:Norm/Vector Space"
] | [
"Definition:Lebesgue Space/L-Infinity",
"Definition:Almost-Everywhere Equality Relation/Lebesgue Space",
"Definition:Lp Norm/L-Infinity Norm",
"P-Seminorm is Seminorm",
"Definition:Mapping",
"Definition:Positive/Real Number",
"Axiom:Vector Space Norm Axioms",
"Definition:Lp Norm/L-Infinity Norm",
"P... |
proofwiki-19577 | Set with Choice Function is Well-Orderable | Let $S$ be a set.
Let there exist a choice function for $S$.
Then $S$ is well-orderable. | Let us define the notation:
:$\map {\PP_{\ne \O} } S := \powerset S \setminus \set \O$
for the power set of $S$ without the empty set.
Let $C: \map {\PP_{\ne \O} } S \to S$ be a choice function for $S$.
Let $x \subseteq S$ be a subset of $S$.
From Subset is Element of Power Set we have:
:$x \subseteq S \iff x \in \powe... | Let $S$ be a [[Definition:Set|set]].
Let there exist a [[Definition:Choice Function on Power Set|choice function for $S$]].
Then $S$ is [[Definition:Well-Orderable Set|well-orderable]]. | Let us define the notation:
:$\map {\PP_{\ne \O} } S := \powerset S \setminus \set \O$
for the [[Definition:Power Set|power set]] of $S$ without the [[Definition:Empty Set|empty set]].
Let $C: \map {\PP_{\ne \O} } S \to S$ be a [[Definition:Choice Function on Power Set|choice function for $S$]].
Let $x \subseteq S$ ... | Set with Choice Function is Well-Orderable | https://proofwiki.org/wiki/Set_with_Choice_Function_is_Well-Orderable | https://proofwiki.org/wiki/Set_with_Choice_Function_is_Well-Orderable | [
"Choice Functions",
"Well-Orderings"
] | [
"Definition:Set",
"Definition:Choice Function/Power Set",
"Definition:Well-Orderable Set"
] | [
"Definition:Power Set",
"Definition:Empty Set",
"Definition:Choice Function/Power Set",
"Definition:Subset",
"Subset is Element of Power Set",
"Definition:Set Difference",
"Definition:Proper Subset",
"Definition:Element",
"Definition:Slowly Progressing Mapping",
"Definition:Unique",
"Definition:... |
proofwiki-19578 | P-Seminorm of Function Zero iff A.E. Zero | Let $\struct {X, \Sigma, \mu}$ be a measure space and let $p \in \closedint 1 \infty$.
Let $\map {\LL^p} {X, \Sigma, \mu}$ be the Lebesgue $p$-space.
Let $f \in \map {\LL^p} {X, \Sigma, \mu}$.
Then:
:$\norm f_p = 0$ {{iff}} $f = 0$ $\mu$-almost everywhere
where $\norm \cdot_p$ is the $p$-seminorm. | === Case 1: $1 \le p < \infty$ ===
We have that:
:$\norm f_p = 0$ {{iff}} $\norm f_p^p = 0$.
That is, $\norm f_p = 0$ {{iff}}:
:$\ds \int \size f^p \rd \mu = 0$
From Measurable Function Zero A.E. iff Absolute Value has Zero Integral, this is equivalent to:
:$\size f^p = 0$ almost everywhere.
From Pointwise Exponenti... | Let $\struct {X, \Sigma, \mu}$ be a [[Definition:Measure Space|measure space]] and let $p \in \closedint 1 \infty$.
Let $\map {\LL^p} {X, \Sigma, \mu}$ be the [[Definition:Lebesgue Space|Lebesgue $p$-space]].
Let $f \in \map {\LL^p} {X, \Sigma, \mu}$.
Then:
:$\norm f_p = 0$ {{iff}} $f = 0$ [[Definition:Almost Eve... | === Case 1: $1 \le p < \infty$ ===
We have that:
:$\norm f_p = 0$ {{iff}} $\norm f_p^p = 0$.
That is, $\norm f_p = 0$ {{iff}}:
:$\ds \int \size f^p \rd \mu = 0$
From [[Measurable Function Zero A.E. iff Absolute Value has Zero Integral]], this is equivalent to:
:$\size f^p = 0$ [[Definition:Almost Everywhere|al... | P-Seminorm of Function Zero iff A.E. Zero | https://proofwiki.org/wiki/P-Seminorm_of_Function_Zero_iff_A.E._Zero | https://proofwiki.org/wiki/P-Seminorm_of_Function_Zero_iff_A.E._Zero | [
"Lebesgue Spaces"
] | [
"Definition:Measure Space",
"Definition:Lebesgue Space",
"Definition:Almost Everywhere",
"Definition:P-Seminorm"
] | [
"Measurable Function Zero A.E. iff Absolute Value has Zero Integral",
"Definition:Almost Everywhere",
"Pointwise Exponentiation preserves A.E. Equality",
"Definition:Almost Everywhere",
"Definition:Almost Everywhere",
"Definition:Almost Everywhere"
] |
proofwiki-19579 | Pointwise Exponentiation preserves A.E. Equality | Let $\struct {X, \Sigma, \mu}$ be a measure space and let $p \in \hointr 0 \infty$.
Let $f, g : X \to \R$ be real-valued $\Sigma$-measurable functions such that:
:$\size f = \size g$ $\mu$-almost everywhere.
Then:
:$\size f^p = \size g^p$ $\mu$-almost everywhere. | Since:
:$\size f = \size g$ $\mu$-almost everywhere
there exists a $\mu$-null set $N$ such that:
:if $\size {\map f x} \ne \size {\map g x}$ then $x \in N$.
If $\size {\map f x} = \size {\map g x}$, then $\size {\map f x}^p = \size {\map g x}^p$.
So by the Rule of Transposition, we have if $\size {\map f x}^p \ne \s... | Let $\struct {X, \Sigma, \mu}$ be a [[Definition:Measure Space|measure space]] and let $p \in \hointr 0 \infty$.
Let $f, g : X \to \R$ be [[Definition:Measurable Real-Valued Function|real-valued $\Sigma$-measurable functions]] such that:
:$\size f = \size g$ [[Definition:Almost Everywhere|$\mu$-almost everywhere]].... | Since:
:$\size f = \size g$ [[Definition:Almost Everywhere|$\mu$-almost everywhere]]
there exists a [[Definition:Null Set|$\mu$-null set]] $N$ such that:
:if $\size {\map f x} \ne \size {\map g x}$ then $x \in N$.
If $\size {\map f x} = \size {\map g x}$, then $\size {\map f x}^p = \size {\map g x}^p$.
So by th... | Pointwise Exponentiation preserves A.E. Equality | https://proofwiki.org/wiki/Pointwise_Exponentiation_preserves_A.E._Equality | https://proofwiki.org/wiki/Pointwise_Exponentiation_preserves_A.E._Equality | [
"Almost-Everywhere Equality Relation"
] | [
"Definition:Measure Space",
"Definition:Measurable Function/Real-Valued Function",
"Definition:Almost Everywhere",
"Definition:Almost Everywhere"
] | [
"Definition:Almost Everywhere",
"Definition:Null Set",
"Rule of Transposition",
"Definition:Almost Everywhere",
"Category:Almost-Everywhere Equality Relation"
] |
proofwiki-19580 | Integral on L-1 Space is Well-Defined | Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $\map {L^1} {X, \Sigma, \mu}$ be the $L^1$ space of $\struct {X, \Sigma, \mu}$.
Let $E \in \map {L^1} {X, \Sigma, \mu}$.
Then the $\mu$-integral of $E$ is well-defined. | Let $E = \eqclass f \sim$.
We first establish that:
:$\ds \int f \rd \mu$
is a well-understood real number.
We then will show that the $\mu$-integral of $E$ is independent of the choice of representative for $E$.
Since $E \in \map {L^1} {X, \Sigma, \mu}$, we have that $f \in \map {\LL^1} {X, \Sigma, \mu}$.
Then:
:$\d... | Let $\struct {X, \Sigma, \mu}$ be a [[Definition:Measure Space|measure space]].
Let $\map {L^1} {X, \Sigma, \mu}$ be the [[Definition:Lp Space|$L^1$ space]] of $\struct {X, \Sigma, \mu}$.
Let $E \in \map {L^1} {X, \Sigma, \mu}$.
Then the [[Definition:Integral on L-1 Space|$\mu$-integral]] of $E$ is well-defined. | Let $E = \eqclass f \sim$.
We first establish that:
:$\ds \int f \rd \mu$
is a well-understood [[Definition:Real Number|real number]].
We then will show that the [[Definition:Integral on L-1 Space|$\mu$-integral]] of $E$ is independent of the choice of [[Definition:Representative of Equivalence Class|representative... | Integral on L-1 Space is Well-Defined | https://proofwiki.org/wiki/Integral_on_L-1_Space_is_Well-Defined | https://proofwiki.org/wiki/Integral_on_L-1_Space_is_Well-Defined | [
"Lp Spaces"
] | [
"Definition:Measure Space",
"Definition:Lp Space",
"Definition:Integral on L-1 Space"
] | [
"Definition:Real Number",
"Definition:Integral on L-1 Space",
"Definition:Equivalence Class/Representative",
"Characterization of Integrable Functions",
"Equivalence Class Equivalent Statements",
"Definition:Almost Everywhere",
"Definition:Almost-Everywhere Equality Relation",
"A.E. Equal Positive Mea... |
proofwiki-19581 | Riemann Integral Operator is Continuous Linear Transformation | Let $\struct {C \closedint a b, \norm {\, \cdot \,}_\infty}$ be the normed vector space of real-valued functions continuous on $\closedint a b \subseteq \R$ equipped with the supremum norm.
Let $T : C \closedint a b \to \R$ be the Riemann integral operator:
:$\ds \forall \mathbf x \in C \closedint a b : \map T {\mathbf... | We have that Integral Operator is Linear.
Furthermore:
{{begin-eqn}}
{{eqn | l = \size {T \mathbf x}
| r = \size {\int_a^b \map {\mathbf x} t \rd t}
}}
{{eqn | o = \le
| r = \int_a^b \size {\map {\mathbf x} t} \rd t
}}
{{eqn | o = \le
| r = \int_a^b \norm {\map {\mathbf x} t}_\infty \rd t
| c =... | Let $\struct {C \closedint a b, \norm {\, \cdot \,}_\infty}$ be the [[Space of Continuous on Closed Interval Real-Valued Functions with Supremum Norm forms Normed Vector Space|normed vector space]] of [[Definition:Space of Real-Valued Functions Continuous on Closed Interval|real-valued functions continuous on $\closedi... | We have that [[Integral Operator is Linear]].
Furthermore:
{{begin-eqn}}
{{eqn | l = \size {T \mathbf x}
| r = \size {\int_a^b \map {\mathbf x} t \rd t}
}}
{{eqn | o = \le
| r = \int_a^b \size {\map {\mathbf x} t} \rd t
}}
{{eqn | o = \le
| r = \int_a^b \norm {\map {\mathbf x} t}_\infty \rd t
... | Riemann Integral Operator is Continuous Linear Transformation | https://proofwiki.org/wiki/Riemann_Integral_Operator_is_Continuous_Linear_Transformation | https://proofwiki.org/wiki/Riemann_Integral_Operator_is_Continuous_Linear_Transformation | [
"Operator Theory",
"Definite Integrals",
"Linear Transformations",
"Continuous Transformations"
] | [
"Space of Continuous on Closed Interval Real-Valued Functions with Supremum Norm forms Normed Vector Space",
"Definition:Space of Real-Valued Functions Continuous on Closed Interval",
"Definition:Supremum Norm",
"Definition:Definite Integral/Riemann/Integral Operator",
"Definition:Continuous Mapping (Normed... | [
"Integral Operator is Linear",
"Definition:Supremum Norm/Continuous on Closed Interval Real-Valued Function",
"Continuity of Linear Transformation/Normed Vector Space",
"Definition:Continuous Mapping (Normed Vector Space)/Space/Definition 1"
] |
proofwiki-19582 | Zermelo's Well-Ordering Theorem/Converse | Let it be supposed that every set is well-orderable.
Then the Axiom of Choice holds. | Let $S$ be an arbitrary set.
By assumption $S$ is well-orderable.
From Well-Orderable Set has Choice Function, $S$ has a choice function.
As $S$ is arbitrary, the result follows.
{{qed}} | Let it be supposed that every [[Definition:Set|set]] is [[Definition:Well-Orderable Set|well-orderable]].
Then the [[Axiom:Axiom of Choice|Axiom of Choice]] holds. | Let $S$ be an arbitrary [[Definition:Set|set]].
By assumption $S$ is [[Definition:Well-Orderable Set|well-orderable]].
From [[Well-Orderable Set has Choice Function]], $S$ has a [[Definition:Choice Function|choice function]].
As $S$ is arbitrary, the result follows.
{{qed}} | Zermelo's Well-Ordering Theorem/Converse/Proof 1 | https://proofwiki.org/wiki/Zermelo's_Well-Ordering_Theorem/Converse | https://proofwiki.org/wiki/Zermelo's_Well-Ordering_Theorem/Converse/Proof_1 | [
"Zermelo's Well-Ordering Theorem"
] | [
"Definition:Set",
"Definition:Well-Orderable Set",
"Axiom:Axiom of Choice"
] | [
"Definition:Set",
"Definition:Well-Orderable Set",
"Well-Orderable Set has Choice Function",
"Definition:Choice Function"
] |
proofwiki-19583 | Zermelo's Well-Ordering Theorem/Converse | Let it be supposed that every set is well-orderable.
Then the Axiom of Choice holds. | Let $\FF$ be an arbitrary collection of sets.
By assumption all sets can be well-ordered.
Hence the set $\bigcup \FF$ of all elements of sets contained in $\FF$ is well-ordered by some ordering $\preceq$.
By definition then, every subset of $\ds \bigcup \FF$ has a smallest element under $\preceq$.
Also, note that each ... | Let it be supposed that every [[Definition:Set|set]] is [[Definition:Well-Orderable Set|well-orderable]].
Then the [[Axiom:Axiom of Choice|Axiom of Choice]] holds. | Let $\FF$ be an arbitrary [[Definition:Set of Sets|collection of sets]].
By assumption all [[Definition:Set|sets]] can be [[Definition:Well-Ordered Set|well-ordered]].
Hence the [[Definition:Set|set]] $\bigcup \FF$ of all [[Definition:Element|elements]] of [[Definition:Set|sets]] contained in $\FF$ is [[Definition:W... | Zermelo's Well-Ordering Theorem/Converse/Proof 2 | https://proofwiki.org/wiki/Zermelo's_Well-Ordering_Theorem/Converse | https://proofwiki.org/wiki/Zermelo's_Well-Ordering_Theorem/Converse/Proof_2 | [
"Zermelo's Well-Ordering Theorem"
] | [
"Definition:Set",
"Definition:Well-Orderable Set",
"Axiom:Axiom of Choice"
] | [
"Definition:Set of Sets",
"Definition:Set",
"Definition:Well-Ordered Set",
"Definition:Set",
"Definition:Element",
"Definition:Set",
"Definition:Well-Ordered Set",
"Definition:Ordering",
"Definition:Subset",
"Definition:Smallest Element",
"Definition:Set",
"Definition:Subset",
"Definition:Ch... |
proofwiki-19584 | L-2 Inner Product is Well-Defined | Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $\map {\LL^2} {X, \Sigma, \mu}$ be the Lebesgue $2$-space of $\struct {X, \Sigma, \mu}$.
Let $\sim$ be the $\mu$-almost-everywhere equality relation on $\map {\LL^2} {X, \Sigma, \mu}$.
Let $\map {L^2} {X, \Sigma, \mu}$ be the $L^p$ space on $\struct {X, \Sigma, \m... | Let $E = \eqclass f \sim$ and $F = \eqclass g \sim$ where $\sim$ is the $\mu$-almost everywhere equality relation.
We aim to show that:
:$\ds \int \paren {f \cdot g} \rd \mu$
exists as a real number and is independent of the choice of representatives $f$ and $g$ of $E$ and $F$.
From the definition of the $L^2$ space,... | Let $\struct {X, \Sigma, \mu}$ be a [[Definition:Measure Space|measure space]].
Let $\map {\LL^2} {X, \Sigma, \mu}$ be the [[Definition:Lebesgue Space|Lebesgue $2$-space]] of $\struct {X, \Sigma, \mu}$.
Let $\sim$ be the [[Definition:Almost-Everywhere Equality Relation|$\mu$-almost-everywhere equality relation]] on ... | Let $E = \eqclass f \sim$ and $F = \eqclass g \sim$ where $\sim$ is the [[Definition:Almost-Everywhere Equality Relation|$\mu$-almost everywhere equality relation]].
We aim to show that:
:$\ds \int \paren {f \cdot g} \rd \mu$
exists as a [[Definition:Real Number|real number]] and is independent of the choice of [[... | L-2 Inner Product is Well-Defined | https://proofwiki.org/wiki/L-2_Inner_Product_is_Well-Defined | https://proofwiki.org/wiki/L-2_Inner_Product_is_Well-Defined | [
"Lp Spaces"
] | [
"Definition:Measure Space",
"Definition:Lebesgue Space",
"Definition:Almost-Everywhere Equality Relation",
"Definition:Lp Space",
"Definition:L-2 Inner Product"
] | [
"Definition:Almost-Everywhere Equality Relation",
"Definition:Real Number",
"Definition:Equivalence Class/Representative",
"Definition:Lp Space",
"Hölder's Inequality for Integrals",
"Definition:Integral of Measure-Integrable Function",
"Definition:Real Number",
"Equivalence Class Equivalent Statement... |
proofwiki-19585 | L-2 Inner Product is Inner Product | Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $\map {\mathcal L^2} {X, \Sigma, \mu}$ be the Lebesgue $2$-space of $\struct {X, \Sigma, \mu}$.
Let $\map {L^2} {X, \Sigma, \mu}$ be the $L^2$ space of $\struct {X, \Sigma, \mu}$.
Let $\innerprod \cdot \cdot$ be the $L^2$ inner product.
Then $\innerprod \cdot \cdot... | === Proof of Symmetry ===
Let $E, F \in \map {L^2} {X, \Sigma, \mu}$.
Let:
:$E = \eqclass f \sim$
and:
:$F = \eqclass g \sim$
where $\eqclass f \sim$ and $\eqclass g \sim$ are the equivalence class of $f, g \in \map {\LL^2} {X, \Sigma, \mu}$ under the $\mu$-almost everywhere equality relation.
Then:
{{begin-eqn}}
{{... | Let $\struct {X, \Sigma, \mu}$ be a [[Definition:Measure Space|measure space]].
Let $\map {\mathcal L^2} {X, \Sigma, \mu}$ be the [[Definition:Lebesgue Space|Lebesgue $2$-space]] of $\struct {X, \Sigma, \mu}$.
Let $\map {L^2} {X, \Sigma, \mu}$ be the [[Definition:Lp Space|$L^2$ space]] of $\struct {X, \Sigma, \mu}$.
... | === Proof of Symmetry ===
Let $E, F \in \map {L^2} {X, \Sigma, \mu}$.
Let:
:$E = \eqclass f \sim$
and:
:$F = \eqclass g \sim$
where $\eqclass f \sim$ and $\eqclass g \sim$ are the [[Definition:Equivalence Class|equivalence class]] of $f, g \in \map {\LL^2} {X, \Sigma, \mu}$ under the [[Definition:Almost-Everywhere ... | L-2 Inner Product is Inner Product | https://proofwiki.org/wiki/L-2_Inner_Product_is_Inner_Product | https://proofwiki.org/wiki/L-2_Inner_Product_is_Inner_Product | [
"Lp Spaces",
"Examples of Inner Products"
] | [
"Definition:Measure Space",
"Definition:Lebesgue Space",
"Definition:Lp Space",
"Definition:L-2 Inner Product",
"Definition:Inner Product"
] | [
"Definition:Equivalence Class",
"Definition:Almost-Everywhere Equality Relation",
"Real Multiplication is Commutative"
] |
proofwiki-19586 | Norm on Vector Space is Quasinorm | Let $\struct {R, +, \circ}$ be a division ring with norm $\norm {\,\cdot\,}_R$.
Let $V$ be a vector space over $R$.
Let $\norm \cdot$ be a norm on $V$.
Then $\norm \cdot$ is a quasinorm on $V$. | {{improve|In order to make it completely clear what is being proven here, especially with the labels identifying the axioms, I have started a style in which those axioms are repeated at the top of the question: "Recall what we are to prove" or however to word it. This is made easier if the axioms are in a separately tr... | Let $\struct {R, +, \circ}$ be a [[Definition:Division Ring|division ring]] with [[Definition:Norm on Division Ring|norm]] $\norm {\,\cdot\,}_R$.
Let $V$ be a [[Definition:Vector Space over Division Ring|vector space]] over $R$.
Let $\norm \cdot$ be a [[Definition:Norm on Vector Space|norm]] on $V$.
Then $\norm \cd... | {{improve|In order to make it completely clear what is being proven here, especially with the labels identifying the axioms, I have started a style in which those axioms are repeated at the top of the question: "Recall what we are to prove" or however to word it. This is made easier if the axioms are in a separately tr... | Norm on Vector Space is Quasinorm | https://proofwiki.org/wiki/Norm_on_Vector_Space_is_Quasinorm | https://proofwiki.org/wiki/Norm_on_Vector_Space_is_Quasinorm | [
"Quasinorms"
] | [
"Definition:Division Ring",
"Definition:Norm/Division Ring",
"Definition:Vector Space/Division Ring",
"Definition:Norm/Vector Space",
"Definition:Quasinorm"
] | [] |
proofwiki-19587 | Well-Orderable Set has Choice Function | Let $S$ be a well-orderable set.
Then $S$ has a choice function. | Let $\preccurlyeq$ be a well-ordering on $S$.
Let $T \subseteq S$ be an arbitrary non-empty subset of $S$.
As $S$ is a well-ordered set, $T$ has a unique smallest element by $\preccurlyeq$.
Thus, we may define the choice function $C: \powerset S \setminus \set \O \to S$ as:
:$\forall T \in \powerset S \setminus \set \O... | Let $S$ be a [[Definition:Well-Orderable Set|well-orderable set]].
Then $S$ has a [[Definition:Choice Function on Power Set|choice function]]. | Let $\preccurlyeq$ be a [[Definition:Well-Ordering|well-ordering]] on $S$.
Let $T \subseteq S$ be an arbitrary [[Definition:Non-Empty Set|non-empty]] [[Definition:Subset|subset]] of $S$.
As $S$ is a [[Definition:Well-Ordered Set|well-ordered set]], $T$ has a [[Definition:Unique|unique]] [[Definition:Smallest Element|... | Well-Orderable Set has Choice Function | https://proofwiki.org/wiki/Well-Orderable_Set_has_Choice_Function | https://proofwiki.org/wiki/Well-Orderable_Set_has_Choice_Function | [
"Choice Functions",
"Well-Orderings"
] | [
"Definition:Well-Orderable Set",
"Definition:Choice Function/Power Set"
] | [
"Definition:Well-Ordering",
"Definition:Non-Empty Set",
"Definition:Subset",
"Definition:Well-Ordered Set",
"Definition:Unique",
"Definition:Smallest Element",
"Definition:Choice Function",
"Definition:Smallest Element",
"Definition:Choice Function"
] |
proofwiki-19588 | Countable Set has Choice Function | Let $S$ be a countable set.
Let $\mathbb S = \powerset S \setminus \set \O$ be the power set of $S$ excluding the empty set $\O$.
Then there exists a choice function $C$ for $S$:
:$\forall x \in \mathbb S: \map C x \in x$ | From Countable Set is Well-Orderable, we have that $S$ is a well-orderable set.
The result follows from Well-Orderable Set has Choice Function.
{{qed}} | Let $S$ be a [[Definition:Countable Set|countable set]].
Let $\mathbb S = \powerset S \setminus \set \O$ be the [[Definition:Power Set|power set]] of $S$ excluding the [[Definition:Empty Set|empty set]] $\O$.
Then there exists a [[Definition:Choice Function|choice function]] $C$ for $S$:
:$\forall x \in \mathbb S: \... | From [[Countable Set is Well-Orderable]], we have that $S$ is a [[Definition:Well-Orderable Set|well-orderable set]].
The result follows from [[Well-Orderable Set has Choice Function]].
{{qed}} | Countable Set has Choice Function | https://proofwiki.org/wiki/Countable_Set_has_Choice_Function | https://proofwiki.org/wiki/Countable_Set_has_Choice_Function | [
"Countable Sets",
"Choice Functions"
] | [
"Definition:Countable Set",
"Definition:Power Set",
"Definition:Empty Set",
"Definition:Choice Function"
] | [
"Countable Set is Well-Orderable",
"Definition:Well-Orderable Set",
"Well-Orderable Set has Choice Function"
] |
proofwiki-19589 | Maximal Principles are Equivalent to Axiom of Choice | The Maximal Principles are equivalent to the Axiom of Choice. | {{ProofWanted|Establish this for each separate one, as and when I reach them in S&F}} | The [[Maximal Principles]] are [[Definition:Logical Equivalence|equivalent]] to the [[Axiom:Axiom of Choice|Axiom of Choice]]. | {{ProofWanted|Establish this for each separate one, as and when I reach them in S&F}} | Maximal Principles are Equivalent to Axiom of Choice | https://proofwiki.org/wiki/Maximal_Principles_are_Equivalent_to_Axiom_of_Choice | https://proofwiki.org/wiki/Maximal_Principles_are_Equivalent_to_Axiom_of_Choice | [
"Maximal Principles",
"Axiom of Choice"
] | [
"Maximal Principles",
"Definition:Logical Equivalence",
"Axiom:Axiom of Choice"
] | [] |
proofwiki-19590 | Maximal Element of Nest is Greatest Element | Let $A$ be a nest.
Let $x$ be a maximal element of $A$.
Then $x$ is the greatest element of $A$. | Let $x$ be a maximal element of $A$.
Let $y \in A$ be arbitrary such that $x \ne y$.
We have by definition of maximal element that:
:$x \not \subset y$
and as $x \ne y$:
:$x \nsubseteq y$
But because $A$ is a nest:
:$x \subseteq y$ or $y \subseteq x$
from which it follows that it must be the case that:
:$y \subseteq x$... | Let $A$ be a [[Definition:Nest (Class Theory)|nest]].
Let $x$ be a [[Definition:Maximal Element (Class Theory)|maximal element]] of $A$.
Then $x$ is the [[Definition:Greatest Element (Class Theory)|greatest element]] of $A$. | Let $x$ be a [[Definition:Maximal Element (Class Theory)|maximal element]] of $A$.
Let $y \in A$ be arbitrary such that $x \ne y$.
We have by definition of [[Definition:Maximal Element (Class Theory)|maximal element]] that:
:$x \not \subset y$
and as $x \ne y$:
:$x \nsubseteq y$
But because $A$ is a [[Definition:Nes... | Maximal Element of Nest is Greatest Element | https://proofwiki.org/wiki/Maximal_Element_of_Nest_is_Greatest_Element | https://proofwiki.org/wiki/Maximal_Element_of_Nest_is_Greatest_Element | [
"Maximal Elements",
"Nests"
] | [
"Definition:Nest/Class Theory",
"Definition:Maximal Element/Class Theory",
"Definition:Greatest Element/Class Theory"
] | [
"Definition:Maximal Element/Class Theory",
"Definition:Maximal Element/Class Theory",
"Definition:Nest/Class Theory",
"Definition:Greatest Element/Class Theory"
] |
proofwiki-19591 | Non-Empty Set of Type M has Maximal Element | Let $S$ be a non-empty set of sets which is of type $M$.
Then $S$ has a maximal element under the subset relation. | Let $S$ be a non-empty set type $M$ set.
Let $x \in S$ be arbitrary.
Then by definition $x$ is a subset of a maximal element of $S$ under the subset relation.
Hence there has to actually ''be'' such a maximal element of $S$ for $x$ to be a subset of.
{{qed}} | Let $S$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Set of Sets|set of sets]] which is of [[Definition:Type M Set|type $M$]].
Then $S$ has a [[Definition:Maximal Element|maximal element]] under the [[Definition:Subset Relation|subset relation]]. | Let $S$ be a [[Definition:Non-Empty Set|non-empty set]] [[Definition:Type M Set|type $M$ set]].
Let $x \in S$ be arbitrary.
Then by definition $x$ is a [[Definition:Subset|subset]] of a [[Definition:Maximal Element|maximal element]] of $S$ under the [[Definition:Subset Relation|subset relation]].
Hence there has to ... | Non-Empty Set of Type M has Maximal Element | https://proofwiki.org/wiki/Non-Empty_Set_of_Type_M_has_Maximal_Element | https://proofwiki.org/wiki/Non-Empty_Set_of_Type_M_has_Maximal_Element | [
"Type M Sets"
] | [
"Definition:Non-Empty Set",
"Definition:Set of Sets",
"Definition:Type M Set",
"Definition:Maximal/Element",
"Definition:Subset Relation"
] | [
"Definition:Non-Empty Set",
"Definition:Type M Set",
"Definition:Subset",
"Definition:Maximal/Element",
"Definition:Subset Relation",
"Definition:Maximal/Element",
"Definition:Subset"
] |
proofwiki-19592 | Axiom of Choice implies Kuratowski's Lemma | Let the Axiom of Choice be accepted.
Then Kuratowski's Lemma holds. | We have:
* Axiom of Choice implies Zorn's Lemma
* Zorn's Lemma implies Kuratowski's Lemma.
{{qed}} | Let the [[Axiom:Axiom of Choice|Axiom of Choice]] be accepted.
Then [[Kuratowski's Lemma]] holds. | We have:
* [[Axiom of Choice implies Zorn's Lemma]]
* [[Zorn's Lemma implies Kuratowski's Lemma]].
{{qed}} | Axiom of Choice implies Kuratowski's Lemma/Proof 1 | https://proofwiki.org/wiki/Axiom_of_Choice_implies_Kuratowski's_Lemma | https://proofwiki.org/wiki/Axiom_of_Choice_implies_Kuratowski's_Lemma/Proof_1 | [
"Kuratowski's Lemma",
"Axiom of Choice",
"Axiom of Choice implies Kuratowski's Lemma"
] | [
"Axiom:Axiom of Choice",
"Kuratowski's Lemma"
] | [
"Axiom of Choice implies Zorn's Lemma",
"Zorn's Lemma implies Kuratowski's Lemma"
] |
proofwiki-19593 | Closed Set under Chain Unions with Choice Function is of Type M | Let $S$ be a set which is closed under chain unions.
Let there exist a choice function for $S$.
Then $S$ is of type $M$, that is:
:every element of $S$ is a subset of a maximal element of $S$ under the subset relation. | Let $S$ be closed under chain unions.
Let $C$ be a choice function for $S$.
For $x \in S$, let $x^*$ be defined as the set of all elements $y$ of $S$ such that $x$ is a proper subset of $y$:
:$\forall x \in S: x^* := \set {y \in S: x \subsetneqq y}$
We have that $x^*$ is empty {{iff}} $x$ is a maximal element of $S$ un... | Let $S$ be a [[Definition:Set|set]] which is [[Definition:Closure under Chain Unions|closed under chain unions]].
Let there exist a [[Definition:Choice Function|choice function]] for $S$.
Then $S$ is of [[Definition:Type M Set|type $M$]], that is:
:every [[Definition:Element|element]] of $S$ is a [[Definition:Subse... | Let $S$ be [[Definition:Closure under Chain Unions|closed under chain unions]].
Let $C$ be a [[Definition:Choice Function|choice function]] for $S$.
For $x \in S$, let $x^*$ be defined as the [[Definition:Set|set]] of all [[Definition:Element|elements]] $y$ of $S$ such that $x$ is a [[Definition:Proper Subset|proper ... | Closed Set under Chain Unions with Choice Function is of Type M | https://proofwiki.org/wiki/Closed_Set_under_Chain_Unions_with_Choice_Function_is_of_Type_M | https://proofwiki.org/wiki/Closed_Set_under_Chain_Unions_with_Choice_Function_is_of_Type_M | [
"Type M Sets",
"Choice Functions",
"Closure under Chain Unions"
] | [
"Definition:Set",
"Definition:Closure under Chain Unions",
"Definition:Choice Function",
"Definition:Type M Set",
"Definition:Element",
"Definition:Subset",
"Definition:Maximal/Element",
"Definition:Subset Relation"
] | [
"Definition:Closure under Chain Unions",
"Definition:Choice Function",
"Definition:Set",
"Definition:Element",
"Definition:Proper Subset",
"Definition:Empty Set",
"Definition:Maximal/Element",
"Definition:Subset Relation",
"Definition:Progressing Mapping",
"Definition:Maximal/Element",
"Definiti... |
proofwiki-19594 | Maximal Element under Subset Relation need not be Greatest Element | Let $A$ be a class.
Let $M \in A$ be a maximal element of $A$ under the subset relation
Then $M$ is not necessarily the greatest element of $A$. | Let $A = \set {x, y}$ such that:
:$x = \set \O$
:$y = \set {\set \O}$
Then:
:$x$ and $y$ are both maximal elements of $A$ by definition.
However:
:$x \not \subseteq y$
and:
:$y \not \subseteq x$
and so neither $x$ nor $y$ are the greatest element of $A$.
{{qed}} | Let $A$ be a [[Definition:Class (Class Theory)|class]].
Let $M \in A$ be a [[Definition:Maximal Element (Class Theory)|maximal element]] of $A$ under the [[Definition:Subset Relation|subset relation]]
Then $M$ is not necessarily the [[Definition:Greatest Element (Class Theory)|greatest element]] of $A$. | Let $A = \set {x, y}$ such that:
:$x = \set \O$
:$y = \set {\set \O}$
Then:
:$x$ and $y$ are both [[Definition:Maximal Element (Class Theory)|maximal elements]] of $A$ by definition.
However:
:$x \not \subseteq y$
and:
:$y \not \subseteq x$
and so neither $x$ nor $y$ are the [[Definition:Greatest Element (Class Theo... | Maximal Element under Subset Relation need not be Greatest Element | https://proofwiki.org/wiki/Maximal_Element_under_Subset_Relation_need_not_be_Greatest_Element | https://proofwiki.org/wiki/Maximal_Element_under_Subset_Relation_need_not_be_Greatest_Element | [
"Maximal Elements",
"Greatest Elements"
] | [
"Definition:Class (Class Theory)",
"Definition:Maximal Element/Class Theory",
"Definition:Subset Relation",
"Definition:Greatest Element/Class Theory"
] | [
"Definition:Maximal Element/Class Theory",
"Definition:Greatest Element/Class Theory"
] |
proofwiki-19595 | Sigma-Algebra Generated by Finite Partition is Finite Sub-Sigma-Algebra | Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $\xi$ be a finite partition of $\Omega$.
Let $\map \sigma \xi$ the generated $\sigma$-algebra by $\xi$.
Then, $\map \sigma \xi$ is a finite sub-$\sigma$-algebra of $\Sigma$.
Furthermore:
:$\ds \map \sigma \xi = \set {\bigcup S: S \subseteq \xi}$ | Let:
:$\ds \Gamma := \set {\bigcup S: S \subseteq \xi}$
By Power Set of Finite Set is Finite, $\Gamma$ is finite.
Therefore, it suffices to show:
:$\map \sigma \xi = \Gamma$
As $\xi$ is finite, from (SA3) follows:
:$\map \sigma \xi \supseteq \Gamma$
To conclude the equality, by definition of $\map\sigma\xi$, we need t... | Let $\struct {\Omega, \Sigma, \Pr}$ be a [[Definition:Probability Space|probability space]].
Let $\xi$ be a [[Definition:Finite Partition (Probability Theory)|finite partition]] of $\Omega$.
Let $\map \sigma \xi$ the [[Definition:Sigma-Algebra Generated by Collection of Subsets|generated $\sigma$-algebra]] by $\xi$.
... | Let:
:$\ds \Gamma := \set {\bigcup S: S \subseteq \xi}$
By [[Power Set of Finite Set is Finite]], $\Gamma$ is [[Definition:Finite Set|finite]].
Therefore, it suffices to show:
:$\map \sigma \xi = \Gamma$
As $\xi$ is [[Definition:Finite Set|finite]], from [[Definition:Sigma-Algebra/Definition 1|(SA3)]] follows:
:$\... | Sigma-Algebra Generated by Finite Partition is Finite Sub-Sigma-Algebra | https://proofwiki.org/wiki/Sigma-Algebra_Generated_by_Finite_Partition_is_Finite_Sub-Sigma-Algebra | https://proofwiki.org/wiki/Sigma-Algebra_Generated_by_Finite_Partition_is_Finite_Sub-Sigma-Algebra | [
"Probability Theory",
"Ergodic Theory"
] | [
"Definition:Probability Space",
"Definition:Finite Partition (Probability Theory)",
"Definition:Sigma-Algebra Generated by Collection of Subsets",
"Definition:Finite Sub-Sigma-Algebra"
] | [
"Power Set of Finite Set is Finite",
"Definition:Finite Set",
"Definition:Finite Set",
"Definition:Sigma-Algebra/Definition 1",
"Definition:Equals",
"Definition:Sigma-Algebra Generated by Collection of Subsets/Definition 1",
"Definition:Sigma-Algebra",
"Definition:Sigma-Algebra/Definition 1",
"Defin... |
proofwiki-19596 | Properties of Join | Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $\eta, \gamma$ be finite partitions of $\Omega$.
Then:
:$\map \sigma {\eta \vee \gamma} = \map \sigma \eta \vee \map \sigma \gamma$
where:
:$\map \sigma \cdot$ denotes the generated $\sigma$-algebra
:$\vee$ on the {{LHS}} denotes the join of finite partiti... | {{proofWanted}} | Let $\struct {\Omega, \Sigma, \Pr}$ be a [[Definition:Probability Space|probability space]].
Let $\eta, \gamma$ be [[Definition:Finite Partition (Probability Theory)|finite partitions]] of $\Omega$.
Then:
:$\map \sigma {\eta \vee \gamma} = \map \sigma \eta \vee \map \sigma \gamma$
where:
:$\map \sigma \cdot$ denotes... | {{proofWanted}} | Properties of Join | https://proofwiki.org/wiki/Properties_of_Join | https://proofwiki.org/wiki/Properties_of_Join | [
"Probability Theory",
"Ergodic Theory"
] | [
"Definition:Probability Space",
"Definition:Finite Partition (Probability Theory)",
"Definition:Sigma-Algebra Generated by Collection of Subsets",
"Definition:Join of Finite Partitions",
"Definition:Join of Finite Sub-Sigma-Algebras"
] | [] |
proofwiki-19597 | Characterization of Almost Everywhere Zero | <onlyinclude>
Let $\struct {X, \Sigma}$ be a measurable space.
Let $\mu$ be a measure on $\struct {X, \Sigma}$.
Let $f : X \to \overline \R$ be a measurable function.
Then:
:$f = 0$ $\mu$-almost everywhere
{{iff}}:
:$\ds \forall A \in \Sigma : \int \paren {\chi_A \cdot f} \rd \mu = 0$
where:
:$\chi_A$ is the characteri... | === Necessary condition ===
First, by Measurable Function Zero A.E. iff Absolute Value has Zero Integral:
:$\ds \int \size f \rd \mu = 0$
Let $A\in\Sigma$.
Let $\paren {\chi_A \cdot f}^+$, $\paren {\chi_A \cdot f}^-$ be the positive and negative parts of $\chi_A \cdot f$, respectively.
Observe:
:$\paren {\chi_A \cdot f... | <onlyinclude>
Let $\struct {X, \Sigma}$ be a [[Definition:Measurable Space|measurable space]].
Let $\mu$ be a [[Definition:Measure (Measure Theory)|measure]] on $\struct {X, \Sigma}$.
Let $f : X \to \overline \R$ be a [[Definition:Measurable Extended Real-Valued Function|measurable function]].
Then:
:$f = 0$ $\mu$-... | === Necessary condition ===
First, by [[Measurable Function Zero A.E. iff Absolute Value has Zero Integral]]:
:$\ds \int \size f \rd \mu = 0$
Let $A\in\Sigma$.
Let $\paren {\chi_A \cdot f}^+$, $\paren {\chi_A \cdot f}^-$ be the [[Definition:Positive Part|positive]] and [[Definition:Negative Part|negative parts]] of... | Characterization of Almost Everywhere Zero | https://proofwiki.org/wiki/Characterization_of_Almost_Everywhere_Zero | https://proofwiki.org/wiki/Characterization_of_Almost_Everywhere_Zero | [
"Measure Theory"
] | [
"Definition:Measurable Space",
"Definition:Measure (Measure Theory)",
"Definition:Measurable Function/Extended Real-Valued Function",
"Definition:Almost Everywhere",
"Definition:Characteristic Function (Set Theory)/Set"
] | [
"Measurable Function Zero A.E. iff Absolute Value has Zero Integral",
"Definition:Positive Part",
"Definition:Negative Part",
"Integral of Positive Measurable Function is Monotone",
"Definition:Integral of Measure-Integrable Function",
"Integral of Positive Measurable Function is Monotone"
] |
proofwiki-19598 | Nesthood has Finite Character | Let $P$ be the property of sets defined as:
:$\forall x: \map P x$ denotes that $x$ is a nest.
Then $P$ is of finite character.
That is:
:$x$ is a nest
{{iff}}:
:every finite subset of $x$ is a nest. | By definition, a nest $N$ is a class on which $\subseteq$ is a total ordering.
Here we are given that $N$ is a set.
The result follows from Property of being Totally Ordered is of Finite Character.
{{qed}} | Let $P$ be the [[Definition:Property|property]] of [[Definition:Set|sets]] defined as:
:$\forall x: \map P x$ denotes that $x$ is a [[Definition:Nest (Class Theory)|nest]].
Then $P$ is of [[Definition:Finite Character (Property of Sets)|finite character]].
That is:
:$x$ is a [[Definition:Nest (Class Theory)|nest]]
{{... | By definition, a [[Definition:Nest (Class Theory)|nest]] $N$ is a [[Definition:Class (Class Theory)|class]] on which $\subseteq$ is a [[Definition:Total Ordering (Class Theory)|total ordering]].
Here we are given that $N$ is a [[Definition:Set|set]].
The result follows from [[Property of being Totally Ordered is of F... | Nesthood has Finite Character/Proof 1 | https://proofwiki.org/wiki/Nesthood_has_Finite_Character | https://proofwiki.org/wiki/Nesthood_has_Finite_Character/Proof_1 | [
"Nests",
"Finite Character",
"Nesthood has Finite Character"
] | [
"Definition:Property",
"Definition:Set",
"Definition:Nest/Class Theory",
"Definition:Finite Character/Property of Sets",
"Definition:Nest/Class Theory",
"Definition:Finite Subset",
"Definition:Nest/Class Theory"
] | [
"Definition:Nest/Class Theory",
"Definition:Class (Class Theory)",
"Definition:Total Ordering/Class Theory",
"Definition:Set",
"Property of being Totally Ordered is of Finite Character"
] |
proofwiki-19599 | Nesthood has Finite Character | Let $P$ be the property of sets defined as:
:$\forall x: \map P x$ denotes that $x$ is a nest.
Then $P$ is of finite character.
That is:
:$x$ is a nest
{{iff}}:
:every finite subset of $x$ is a nest. | By definition, a nest is a class on which $\subseteq$ is a total ordering.
=== Sufficient Condition ===
Let $x$ be a nest.
Let $y \subseteq x$.
From Restriction of Total Ordering is Total Ordering it follows that $y$ is also a nest.
This holds in particular if $y$ is a finite set.
Hence the result.
{{qed|lemma}}
=== Ne... | Let $P$ be the [[Definition:Property|property]] of [[Definition:Set|sets]] defined as:
:$\forall x: \map P x$ denotes that $x$ is a [[Definition:Nest (Class Theory)|nest]].
Then $P$ is of [[Definition:Finite Character (Property of Sets)|finite character]].
That is:
:$x$ is a [[Definition:Nest (Class Theory)|nest]]
{{... | By definition, a [[Definition:Nest (Class Theory)|nest]] is a [[Definition:Class (Class Theory)|class]] on which $\subseteq$ is a [[Definition:Total Ordering (Class Theory)|total ordering]].
=== Sufficient Condition ===
Let $x$ be a [[Definition:Nest (Class Theory)|nest]].
Let $y \subseteq x$.
From [[Restriction o... | Nesthood has Finite Character/Proof 2 | https://proofwiki.org/wiki/Nesthood_has_Finite_Character | https://proofwiki.org/wiki/Nesthood_has_Finite_Character/Proof_2 | [
"Nests",
"Finite Character",
"Nesthood has Finite Character"
] | [
"Definition:Property",
"Definition:Set",
"Definition:Nest/Class Theory",
"Definition:Finite Character/Property of Sets",
"Definition:Nest/Class Theory",
"Definition:Finite Subset",
"Definition:Nest/Class Theory"
] | [
"Definition:Nest/Class Theory",
"Definition:Class (Class Theory)",
"Definition:Total Ordering/Class Theory",
"Definition:Nest/Class Theory",
"Restriction of Total Ordering is Total Ordering",
"Definition:Nest/Class Theory",
"Definition:Finite Set",
"Definition:Finite Subset",
"Definition:Nest/Class ... |
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