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-4800 | Axiom of Pairing from Powers and Replacement | The {{axiom-link|Pairing|Set Theory}} is a consequence of:
:the {{axiom-link|the Empty Set|Set Theory}}
and
:the {{axiom-link|Powers|Set Theory}}
and also
:the {{axiom-link|Replacement}}. | It is to be shown that $2 = \set {\O, \set \O}$ is a set.
By the {{axiom-link|the Empty Set|Set Theory}}, it is seen that $\O$ is a set.
By the {{axiom-link|Powers|Set Theory}}, it is seen that $\powerset \O$ is also a set.
It is to be shown that $\powerset \O = \set \O$.
To determine the elements of $\powerset \O$, we... | The {{axiom-link|Pairing|Set Theory}} is a consequence of:
:the {{axiom-link|the Empty Set|Set Theory}}
and
:the {{axiom-link|Powers|Set Theory}}
and also
:the {{axiom-link|Replacement}}. | It is to be shown that $2 = \set {\O, \set \O}$ is a [[Definition:Set|set]].
By the {{axiom-link|the Empty Set|Set Theory}}, it is seen that $\O$ is a [[Definition:Set|set]].
By the {{axiom-link|Powers|Set Theory}}, it is seen that $\powerset \O$ is also a [[Definition:Set|set]].
It is to be shown that $\powerset ... | Axiom of Pairing from Powers and Replacement | https://proofwiki.org/wiki/Axiom_of_Pairing_from_Powers_and_Replacement | https://proofwiki.org/wiki/Axiom_of_Pairing_from_Powers_and_Replacement | [
"Axiom of Pairing",
"Doubletons"
] | [] | [
"Definition:Set",
"Definition:Set",
"Definition:Set",
"Definition:Element",
"Definition:Set",
"Definition:Empty Set",
"Definition:Logical Implication",
"Rule of Transposition",
"Definition:Empty Set",
"Empty Set is Unique",
"Empty Set is Subset of All Sets",
"Set is Subset of Itself",
"Defin... |
proofwiki-4801 | Axiom of Pairing from Axiom of Specification | Let it be supposed that there exists a set which contains at least two elements.
Then the Axiom of Pairing is a consequence of the Axiom of Specification. | Let $A$ be a set which contains at least two elements.
Let $a$ and $b$ be any two elements of $A$.
Let $\map P x$ be the propositional function:
:$\map P x := \paren {x = a \lor x = b}$
Then we may use the Axiom of Specification to define $B$ as:
:$x \in B \iff \set {x \in A: \map P x}$
Hence we can define:
:$B := \set... | Let it be supposed that there exists a [[Definition:Set|set]] which contains at least two [[Definition:Element|elements]].
Then the [[Axiom:Axiom of Pairing (Set Theory)|Axiom of Pairing]] is a consequence of the [[Axiom:Axiom of Specification (Sets)|Axiom of Specification]]. | Let $A$ be a [[Definition:Set|set]] which contains at least two [[Definition:Element|elements]].
Let $a$ and $b$ be any two [[Definition:Element|elements]] of $A$.
Let $\map P x$ be the [[Definition:Propositional Function|propositional function]]:
:$\map P x := \paren {x = a \lor x = b}$
Then we may use the [[Axiom:... | Axiom of Pairing from Axiom of Specification | https://proofwiki.org/wiki/Axiom_of_Pairing_from_Axiom_of_Specification | https://proofwiki.org/wiki/Axiom_of_Pairing_from_Axiom_of_Specification | [
"Axiom of Pairing",
"Doubletons"
] | [
"Definition:Set",
"Definition:Element",
"Axiom:Axiom of Pairing/Set Theory",
"Axiom:Axiom of Specification/Set Theory"
] | [
"Definition:Set",
"Definition:Element",
"Definition:Element",
"Definition:Propositional Function",
"Axiom:Axiom of Specification/Set Theory"
] |
proofwiki-4802 | Existence of Singleton Set | Let $a$ be a set.
Then the singleton set $\set a$ may be constructed such that:
:$a \in \set a$ | Let $a$ be a set.
From the Axiom of Pairing the set $\set {a, a}$ may be formed.
From the Axiom of Extension it follows that:
:$\set {a, a} = \set a$
{{qed}} | Let $a$ be a [[Definition:Set|set]].
Then the [[Definition:Singleton|singleton set]] $\set a$ may be constructed such that:
:$a \in \set a$ | Let $a$ be a [[Definition:Set|set]].
From the [[Axiom:Axiom of Pairing (Set Theory)|Axiom of Pairing]] the set $\set {a, a}$ may be formed.
From the [[Axiom:Axiom of Extension (Sets)|Axiom of Extension]] it follows that:
:$\set {a, a} = \set a$
{{qed}} | Existence of Singleton Set | https://proofwiki.org/wiki/Existence_of_Singleton_Set | https://proofwiki.org/wiki/Existence_of_Singleton_Set | [
"Axiomatic Set Theory",
"Singletons"
] | [
"Definition:Set",
"Definition:Singleton"
] | [
"Definition:Set",
"Axiom:Axiom of Pairing/Set Theory",
"Axiom:Axiom of Extension/Set Theory"
] |
proofwiki-4803 | Intersection is Subset of Union of Intersections with Complements | Let $R, S, T$ be sets.
Then:
:$S \cap T \subseteq \paren {R \cap S} \cup \paren {\overline R \cap T}$
where $\overline R$ denotes the complement of $R$. | Let $x \in S \cap T$.
Then by definition of set intersection, $x \in S \land x \in T$.
From Conjunction implies Disjunction of Conjunctions with Complements, it follows that:
:$\paren {x \in S \land \psi} \lor \paren {x \in T \land \neg \psi}$
where $\psi$ is any arbitrary statement.
Let $\psi$ be the statement $x \in ... | Let $R, S, T$ be [[Definition:Set|sets]].
Then:
:$S \cap T \subseteq \paren {R \cap S} \cup \paren {\overline R \cap T}$
where $\overline R$ denotes the [[Definition:Set Complement|complement]] of $R$. | Let $x \in S \cap T$.
Then by definition of [[Definition:Set Intersection|set intersection]], $x \in S \land x \in T$.
From [[Conjunction implies Disjunction of Conjunctions with Complements]], it follows that:
:$\paren {x \in S \land \psi} \lor \paren {x \in T \land \neg \psi}$
where $\psi$ is any arbitrary [[Defin... | Intersection is Subset of Union of Intersections with Complements | https://proofwiki.org/wiki/Intersection_is_Subset_of_Union_of_Intersections_with_Complements | https://proofwiki.org/wiki/Intersection_is_Subset_of_Union_of_Intersections_with_Complements | [
"Set Union",
"Set Intersection",
"Set Complement"
] | [
"Definition:Set",
"Definition:Set Complement"
] | [
"Definition:Set Intersection",
"Conjunction implies Disjunction of Conjunctions with Complements",
"Definition:Statement",
"Definition:Statement",
"Definition:Subset",
"Definition:Set Intersection",
"Definition:Set Union",
"Definition:Set Complement"
] |
proofwiki-4804 | Intersection of Unions with Complements is Subset of Union | Let $R, S, T$ be sets.
Then:
:$\left({R \cup S}\right) \cap \left({\overline R \cup T}\right) \subseteq S \cup T$ | Let $x \in \left({R \cup S}\right) \cap \left({\overline R \cup T}\right)$.
Then by definition of set intersection, set union and set complement, we have:
:$\left({x \in R \lor x \in S}\right) \land \left({x \notin R \lor x \in T}\right)$
From Conjunction of Disjunctions with Complements implies Disjunction it follows ... | Let $R, S, T$ be [[Definition:Set|sets]].
Then:
:$\left({R \cup S}\right) \cap \left({\overline R \cup T}\right) \subseteq S \cup T$ | Let $x \in \left({R \cup S}\right) \cap \left({\overline R \cup T}\right)$.
Then by definition of [[Definition:Set Intersection|set intersection]], [[Definition:Set Union|set union]] and [[Definition:Set Complement|set complement]], we have:
:$\left({x \in R \lor x \in S}\right) \land \left({x \notin R \lor x \in T}\... | Intersection of Unions with Complements is Subset of Union | https://proofwiki.org/wiki/Intersection_of_Unions_with_Complements_is_Subset_of_Union | https://proofwiki.org/wiki/Intersection_of_Unions_with_Complements_is_Subset_of_Union | [
"Set Union",
"Set Intersection",
"Set Complement"
] | [
"Definition:Set"
] | [
"Definition:Set Intersection",
"Definition:Set Union",
"Definition:Set Complement",
"Conjunction of Disjunctions with Complements implies Disjunction",
"Definition:Set Union",
"Definition:Subset"
] |
proofwiki-4805 | Conjunction implies Disjunction of Conjunctions with Complements | :$p \land q \vdash \paren {p \land r} \lor \paren {q \land \neg r}$ | {{BeginTableau|p \land q \vdash \paren {p \land r} \lor \paren {q \land \neg r} }}
{{Premise|1|p \land q}}
{{ExcludedMiddle|2|r \lor \neg r}}
{{Simplification|3|1|p|1|1}}
{{Simplification|4|1|q|1|2}}
{{Assumption|5|r}}
{{Conjunction|6|1, 5|p \land r|1|5}}
{{Addition|7|1, 5|\paren {p \land r} \lor \paren {q \land \neg r... | :$p \land q \vdash \paren {p \land r} \lor \paren {q \land \neg r}$ | {{BeginTableau|p \land q \vdash \paren {p \land r} \lor \paren {q \land \neg r} }}
{{Premise|1|p \land q}}
{{ExcludedMiddle|2|r \lor \neg r}}
{{Simplification|3|1|p|1|1}}
{{Simplification|4|1|q|1|2}}
{{Assumption|5|r}}
{{Conjunction|6|1, 5|p \land r|1|5}}
{{Addition|7|1, 5|\paren {p \land r} \lor \paren {q \land \neg r... | Conjunction implies Disjunction of Conjunctions with Complements | https://proofwiki.org/wiki/Conjunction_implies_Disjunction_of_Conjunctions_with_Complements | https://proofwiki.org/wiki/Conjunction_implies_Disjunction_of_Conjunctions_with_Complements | [
"Conjunction",
"Disjunction"
] | [] | [
"Category:Conjunction",
"Category:Disjunction"
] |
proofwiki-4806 | Conjunction of Disjunctions with Complements implies Disjunction | :$\paren {p \lor r} \land \paren {q \lor \neg r} \vdash p \lor q$ | {{BeginTableau|\paren {p \lor r} \land \paren {q \lor \neg r} \vdash p \lor q}}
{{Premise|1|\paren {p \lor r} \land \paren {q \lor \neg r} }}
{{Simplification|2|1|p \lor r|1|1|The aim is to use Proof by Cases on this ...}}
{{Assumption|3|p|Assume the first of the disjuncts ...}}
{{Addition|4|3|p \lor q|3|1|... and demo... | :$\paren {p \lor r} \land \paren {q \lor \neg r} \vdash p \lor q$ | {{BeginTableau|\paren {p \lor r} \land \paren {q \lor \neg r} \vdash p \lor q}}
{{Premise|1|\paren {p \lor r} \land \paren {q \lor \neg r} }}
{{Simplification|2|1|p \lor r|1|1|The aim is to use [[Proof by Cases]] on this ...}}
{{Assumption|3|p|Assume the first of the disjuncts ...}}
{{Addition|4|3|p \lor q|3|1|... and ... | Conjunction of Disjunctions with Complements implies Disjunction | https://proofwiki.org/wiki/Conjunction_of_Disjunctions_with_Complements_implies_Disjunction | https://proofwiki.org/wiki/Conjunction_of_Disjunctions_with_Complements_implies_Disjunction | [
"Conjunction",
"Disjunction"
] | [] | [
"Proof by Cases",
"Rule of Material Implication",
"Category:Conjunction",
"Category:Disjunction"
] |
proofwiki-4807 | Intersection of Elements of Power Set | Let $S$ be a set.
Let:
:$\ds \mathbb S = \bigcap_{X \mathop \in \powerset S} X$
where $\powerset S$ is the power set of $S$.
Then $\mathbb S = \O$. | By Intersection is Subset:
:$\ds \forall X \in \powerset S: \bigcap_{X \mathop \in \powerset S} X \subseteq X$
From Empty Set is Element of Power Set:
:$\O \in \powerset S$
So:
:$\ds \bigcap_{X \mathop \in \powerset S} X \subseteq \O$
From Empty Set is Subset of All Sets:
:$\ds \O \subseteq \bigcap_{X \mathop \in \powe... | Let $S$ be a [[Definition:Set|set]].
Let:
:$\ds \mathbb S = \bigcap_{X \mathop \in \powerset S} X$
where $\powerset S$ is the [[Definition:Power Set|power set]] of $S$.
Then $\mathbb S = \O$. | By [[Intersection is Subset]]:
:$\ds \forall X \in \powerset S: \bigcap_{X \mathop \in \powerset S} X \subseteq X$
From [[Empty Set is Element of Power Set]]:
:$\O \in \powerset S$
So:
:$\ds \bigcap_{X \mathop \in \powerset S} X \subseteq \O$
From [[Empty Set is Subset of All Sets]]:
:$\ds \O \subseteq \bigcap_{X \... | Intersection of Elements of Power Set | https://proofwiki.org/wiki/Intersection_of_Elements_of_Power_Set | https://proofwiki.org/wiki/Intersection_of_Elements_of_Power_Set | [
"Power Set",
"Set Intersection",
"Empty Set"
] | [
"Definition:Set",
"Definition:Power Set"
] | [
"Intersection is Subset",
"Empty Set is Element of Power Set",
"Empty Set is Subset of All Sets",
"Definition:Set Equality/Definition 2"
] |
proofwiki-4808 | Power Set of Subset | Let $S \subseteq T$ where $S$ and $T$ are both sets.
Then:
:$\powerset S \subseteq \powerset T$
where $\powerset S$ denotes the power set of $S$. | {{begin-eqn}}
{{eqn | l = X
| o = \in
| r = \powerset S
| c =
}}
{{eqn | ll= \leadsto
| l = X
| o = \subseteq
| r = S
| c = {{Defof|Power Set}}
}}
{{eqn | ll= \leadsto
| l = X
| o = \subseteq
| r = T
| c = as $S \subseteq T$: Subset Relation is Transitiv... | Let $S \subseteq T$ where $S$ and $T$ are both [[Definition:Set|sets]].
Then:
:$\powerset S \subseteq \powerset T$
where $\powerset S$ denotes the [[Definition:Power Set|power set]] of $S$. | {{begin-eqn}}
{{eqn | l = X
| o = \in
| r = \powerset S
| c =
}}
{{eqn | ll= \leadsto
| l = X
| o = \subseteq
| r = S
| c = {{Defof|Power Set}}
}}
{{eqn | ll= \leadsto
| l = X
| o = \subseteq
| r = T
| c = as $S \subseteq T$: [[Subset Relation is Transit... | Power Set of Subset | https://proofwiki.org/wiki/Power_Set_of_Subset | https://proofwiki.org/wiki/Power_Set_of_Subset | [
"Power Set",
"Subsets"
] | [
"Definition:Set",
"Definition:Power Set"
] | [
"Subset Relation is Transitive"
] |
proofwiki-4809 | Union of Elements of Power Set | Let $S$ be a set.
Then:
:$\ds S = \bigcup_{X \mathop \in \powerset S} X$
where $\powerset S$ denotes the power set of $S$. | By Subset of Union:
:$\ds \forall X \in \powerset S: X \subseteq \bigcup_{X \mathop \in \powerset S} X$
From Set is Subset of Itself, $S \subseteq S$ and so $S \in \powerset S$.
So:
:$\ds S \subseteq \bigcup_{X \mathop \in \powerset S} X$
From Union is Smallest Superset:
:$\ds \paren {\forall X \in \mathbb S: X \subset... | Let $S$ be a [[Definition:Set|set]].
Then:
:$\ds S = \bigcup_{X \mathop \in \powerset S} X$
where $\powerset S$ denotes the [[Definition:Power Set|power set]] of $S$. | By [[Subset of Union]]:
:$\ds \forall X \in \powerset S: X \subseteq \bigcup_{X \mathop \in \powerset S} X$
From [[Set is Subset of Itself]], $S \subseteq S$ and so $S \in \powerset S$.
So:
:$\ds S \subseteq \bigcup_{X \mathop \in \powerset S} X$
From [[Union is Smallest Superset]]:
:$\ds \paren {\forall X \in \ma... | Union of Elements of Power Set | https://proofwiki.org/wiki/Union_of_Elements_of_Power_Set | https://proofwiki.org/wiki/Union_of_Elements_of_Power_Set | [
"Power Set",
"Set Union"
] | [
"Definition:Set",
"Definition:Power Set"
] | [
"Set is Subset of Union",
"Set is Subset of Itself",
"Union is Smallest Superset",
"Set is Subset of Itself",
"Definition:Power Set",
"Definition:Tautology",
"Definition:Set Equality/Definition 2"
] |
proofwiki-4810 | Subset of Cartesian Product | Let $S$ be a set of ordered pairs.
Then $S$ is the subset of the cartesian product of two sets. | Let $S$ be a set of ordered pairs.
Let $x \in S$ such that $x = \set {\set a, \set {a, b} }$ as defined in Kuratowski Formalization of Ordered Pair.
Since the elements of $S$ are sets, we can form the union $\mathbb S = \bigcup S$ of the sets in $S$.
Since $x \in S$ it follows that the elements of $x$ are elements of $... | Let $S$ be a [[Definition:Set|set]] of [[Definition:Ordered Pair|ordered pairs]].
Then $S$ is the [[Definition:Subset|subset]] of the [[Definition:Cartesian Product|cartesian product]] of two sets. | Let $S$ be a [[Definition:Set|set]] of [[Definition:Ordered Pair|ordered pairs]].
Let $x \in S$ such that $x = \set {\set a, \set {a, b} }$ as defined in [[Kuratowski Formalization of Ordered Pair]].
Since the elements of $S$ are [[Definition:Set|sets]], we can form the [[Definition:Set Union|union]] $\mathbb S = \bi... | Subset of Cartesian Product | https://proofwiki.org/wiki/Subset_of_Cartesian_Product | https://proofwiki.org/wiki/Subset_of_Cartesian_Product | [
"Cartesian Product",
"Axiomatic Set Theory"
] | [
"Definition:Set",
"Definition:Ordered Pair",
"Definition:Subset",
"Definition:Cartesian Product"
] | [
"Definition:Set",
"Definition:Ordered Pair",
"Equivalence of Definitions of Ordered Pair",
"Definition:Set",
"Definition:Set Union",
"Definition:Element",
"Definition:Set Union",
"Equivalence of Definitions of Ordered Pair",
"Definition:Subset",
"Axiom:Axiom of Specification/Set Theory",
"Defini... |
proofwiki-4811 | Diagonal Relation is Smallest Equivalence Relation | The diagonal relation $\Delta_S$ on $S$ is the smallest equivalence in $S$, in the sense that:
:$\forall \EE \subseteq S \times S: \Delta_S \subseteq \EE$
where $\EE$ denotes a general equivalence relation. | It is confirmed that, from Diagonal Relation is Equivalence, $\Delta_S$ is an equivalence relation.
Let $\EE$ be an arbitrary equivalence relation.
By definition, $\EE$ is reflexive.
From Relation Contains Diagonal Relation iff Reflexive it follows that as $\Delta_S \subseteq \EE$.
{{qed}} | The [[Definition:Diagonal Relation|diagonal relation]] $\Delta_S$ on $S$ is the [[Definition:Minimal Set|smallest]] [[Definition:Equivalence Relation|equivalence]] in $S$, in the sense that:
:$\forall \EE \subseteq S \times S: \Delta_S \subseteq \EE$
where $\EE$ denotes a general [[Definition:Equivalence Relation|equi... | It is confirmed that, from [[Diagonal Relation is Equivalence]], $\Delta_S$ is an [[Definition:Equivalence Relation|equivalence relation]].
Let $\EE$ be an arbitrary [[Definition:Equivalence Relation|equivalence relation]].
By definition, $\EE$ is [[Definition:Reflexive Relation|reflexive]].
From [[Relation Contains... | Diagonal Relation is Smallest Equivalence Relation | https://proofwiki.org/wiki/Diagonal_Relation_is_Smallest_Equivalence_Relation | https://proofwiki.org/wiki/Diagonal_Relation_is_Smallest_Equivalence_Relation | [
"Equivalence Relations"
] | [
"Definition:Diagonal Relation",
"Definition:Minimal/Set",
"Definition:Equivalence Relation",
"Definition:Equivalence Relation"
] | [
"Diagonal Relation is Equivalence",
"Definition:Equivalence Relation",
"Definition:Equivalence Relation",
"Definition:Reflexive Relation",
"Equivalence of Definitions of Reflexive Relation"
] |
proofwiki-4812 | Trivial Relation is Largest Equivalence Relation | The trivial relation $\TT$ on $S$ is the largest equivalence in $S$, in the sense that:
:$\forall \EE \subseteq S \times S: \EE \subseteq \TT$
where $\EE$ denotes a general equivalence relation. | The trivial relation $\TT$ on $S$ is defined as:
:$\TT = S \times S$
It is confirmed from Trivial Relation is Equivalence that the trivial relation is in fact an equivalence relation.
Let $\EE$ be an arbitrary equivalence relation on $S$.
By definition of relation, $\EE \subseteq S \times S$ and so (trivially) $\EE \su... | The [[Definition:Trivial Relation|trivial relation]] $\TT$ on $S$ is the largest [[Definition:Equivalence Relation|equivalence]] in $S$, in the sense that:
:$\forall \EE \subseteq S \times S: \EE \subseteq \TT$
where $\EE$ denotes a general [[Definition:Equivalence Relation|equivalence relation]]. | The [[Definition:Trivial Relation|trivial relation]] $\TT$ on $S$ is defined as:
:$\TT = S \times S$
It is confirmed from [[Trivial Relation is Equivalence]] that the [[Definition:Trivial Relation|trivial relation]] is in fact an [[Definition:Equivalence Relation|equivalence relation]].
Let $\EE$ be an arbitrary [[De... | Trivial Relation is Largest Equivalence Relation | https://proofwiki.org/wiki/Trivial_Relation_is_Largest_Equivalence_Relation | https://proofwiki.org/wiki/Trivial_Relation_is_Largest_Equivalence_Relation | [
"Equivalence Relations",
"Trivial Relation"
] | [
"Definition:Trivial Relation",
"Definition:Equivalence Relation",
"Definition:Equivalence Relation"
] | [
"Definition:Trivial Relation",
"Trivial Relation is Equivalence",
"Definition:Trivial Relation",
"Definition:Equivalence Relation",
"Definition:Equivalence Relation",
"Definition:Relation"
] |
proofwiki-4813 | Continuously Differentiable Curve has Finite Arc Length | Let $y = \map f x$ be a real function which is continuous on the closed interval $\closedint a b$ and continuously differentiable on the open interval $\openint a b$.
The definite integral:
:$s = \ds \int_{x \mathop = a}^{x \mathop = b} \sqrt {1 + \paren {\frac {\d y} {\d x} }^2} \rd x$
exists, and is called the '''ar... | It intuitively makes sense to define the length of a line segment to be the distance between the two end points, as given by the Distance Formula:
:$\sqrt {\paren {x_1 - x_2}^2 + \paren {y_1 - y_2}^2}$
Similarly, it is reasonable to assume that the actual length of the curve would be approximately equal to the sum of t... | Let $y = \map f x$ be a [[Definition:Real Function|real function]] which is [[Definition:Continuous Real Function on Interval|continuous]] on the [[Definition:Closed Real Interval|closed interval]] $\closedint a b$ and [[Definition:Continuously Differentiable|continuously differentiable]] on the [[Definition:Open Real ... | It intuitively makes sense to define the [[Definition:Linear Measure|length]] of a [[Definition:Line Segment|line segment]] to be the distance between the two end points, as given by the [[Distance Formula]]:
:$\sqrt {\paren {x_1 - x_2}^2 + \paren {y_1 - y_2}^2}$
Similarly, it is reasonable to assume that the actual ... | Continuously Differentiable Curve has Finite Arc Length | https://proofwiki.org/wiki/Continuously_Differentiable_Curve_has_Finite_Arc_Length | https://proofwiki.org/wiki/Continuously_Differentiable_Curve_has_Finite_Arc_Length | [
"Integral Calculus",
"Analytic Geometry"
] | [
"Definition:Real Function",
"Definition:Continuous Real Function/Interval",
"Definition:Real Interval/Closed",
"Definition:Continuously Differentiable",
"Definition:Real Interval/Open",
"Definition:Definite Integral",
"Definition:Arc Length"
] | [
"Definition:Linear Measure",
"Definition:Line/Segment",
"Distance Formula",
"File:ArcLength1.png",
"Definition:Real Interval/Closed",
"Definition:Radicand",
"Definition:Radicand",
"Definition:By Hypothesis",
"Definition:Continuous Real Function/Interval",
"Definition:Differentiable Mapping/Real Fu... |
proofwiki-4814 | Arc Length for Parametric Equations | Let $x = \map f t$ and $y = \map g t$ be real functions of a parameter $t$.
Let these equations describe a curve $\CC$ that is continuous for all $t \in \closedint a b$ and continuously differentiable for all $t \in \openint a b$.
Suppose that the graph of the curve does not intersect itself for any $t \in \openint a ... | {{begin-eqn}}
{{eqn | l = s
| r = \int_a^b \sqrt {1 + \paren {\frac {\d y} {\d x} }^2} \rd x
| c = {{Defof|Arc Length}}
}}
{{eqn | r = \int_a^b \sqrt {\paren {\frac {\frac {\d x} {\d t} } {\frac {\d x} {\d t} } }^2 + \paren {\frac {\frac {\d y}{\d t} } {\frac {\d x} {\d t} } }^2} \rd x
| c = because $... | Let $x = \map f t$ and $y = \map g t$ be [[Definition:Real Function|real functions]] of a parameter $t$.
Let these equations describe a [[Definition:Curve|curve]] $\CC$ that is [[Definition:Continuous on Interval|continuous]] for all $t \in \closedint a b$ and [[Definition:Continuously Differentiable|continuously dif... | {{begin-eqn}}
{{eqn | l = s
| r = \int_a^b \sqrt {1 + \paren {\frac {\d y} {\d x} }^2} \rd x
| c = {{Defof|Arc Length}}
}}
{{eqn | r = \int_a^b \sqrt {\paren {\frac {\frac {\d x} {\d t} } {\frac {\d x} {\d t} } }^2 + \paren {\frac {\frac {\d y}{\d t} } {\frac {\d x} {\d t} } }^2} \rd x
| c = because $... | Arc Length for Parametric Equations | https://proofwiki.org/wiki/Arc_Length_for_Parametric_Equations | https://proofwiki.org/wiki/Arc_Length_for_Parametric_Equations | [
"Arc Length",
"Integral Calculus"
] | [
"Definition:Real Function",
"Definition:Line/Curve",
"Definition:Continuous Real Function/Interval",
"Definition:Continuously Differentiable",
"Definition:Graph of Mapping",
"Definition:Arc Length"
] | [
"Definition:Radicand",
"Definition:Absolute Value",
"Definition:Linear Measure",
"Derivative of Inverse Function",
"Integration by Substitution"
] |
proofwiki-4815 | Inverse Element of Injection | Let $S$ and $T$ be sets.
Let $f: S \to T$ be an injection.
Then:
:$\map {f^{-1} } y = x \iff \map f x = y$ | === Necessary Condition ===
Let $y = \map f x$.
From the definition of the preimage of an element:
:$\map {f^{-1} } y = \set {x \in S: \tuple {y, x} \in f}$
Thus:
:$x \in \map {f^{-1} } y$
By definition of injection, $\map {f^{-1} } y$ is a singleton:
:$\map {f^{-1} } y = \set x$
which can be expressed as:
:$\map {f^{-... | Let $S$ and $T$ be [[Definition:Set|sets]].
Let $f: S \to T$ be an [[Definition:Injection|injection]].
Then:
:$\map {f^{-1} } y = x \iff \map f x = y$ | === Necessary Condition ===
Let $y = \map f x$.
From the definition of the [[Definition:Preimage of Element under Mapping|preimage of an element]]:
:$\map {f^{-1} } y = \set {x \in S: \tuple {y, x} \in f}$
Thus:
:$x \in \map {f^{-1} } y$
By definition of [[Definition:Injection/Definition 4|injection]], $\map {f^{-1... | Inverse Element of Injection | https://proofwiki.org/wiki/Inverse_Element_of_Injection | https://proofwiki.org/wiki/Inverse_Element_of_Injection | [
"Injections",
"Inverse Mappings"
] | [
"Definition:Set",
"Definition:Injection"
] | [
"Definition:Preimage/Mapping/Element",
"Definition:Injection/Definition 4",
"Definition:Singleton"
] |
proofwiki-4816 | Intersection of Inductive Sets | Let $\mathbb S$ be a non-empty indexed family of inductive sets.
Then $\bigcap \mathbb S$ is an inductive set. | From definition, a set $X$ is an inductive set {{iff}} both the following holds:
:$\O \in X$
:$x \in X \implies x^+ \in X$
For all $S \in \mathbb S$, $S$ is an inductive set.
From definition of an inductive set, $\O \in S$.
By definition of set intersection, $\O \in \bigcap \mathbb S$.
Now suppose $x \in \bigcap \mathb... | Let $\mathbb S$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Indexed Family of Sets|indexed family]] of [[Definition:Inductive Set|inductive sets]].
Then $\bigcap \mathbb S$ is an [[Definition:Inductive Set|inductive set]]. | From definition, a [[Definition:Set|set]] $X$ is an [[Definition:Inductive Set|inductive set]] {{iff}} both the following holds:
:$\O \in X$
:$x \in X \implies x^+ \in X$
For all $S \in \mathbb S$, $S$ is an [[Definition:Inductive Set|inductive set]].
From definition of an [[Definition:Inductive Set|inductive set]],... | Intersection of Inductive Sets | https://proofwiki.org/wiki/Intersection_of_Inductive_Sets | https://proofwiki.org/wiki/Intersection_of_Inductive_Sets | [
"Inductive Sets",
"Set Intersection"
] | [
"Definition:Non-Empty Set",
"Definition:Indexing Set/Family of Sets",
"Definition:Inductive Set",
"Definition:Inductive Set"
] | [
"Definition:Set",
"Definition:Inductive Set",
"Definition:Inductive Set",
"Definition:Inductive Set",
"Definition:Set Intersection",
"Definition:Set Intersection",
"Definition:Inductive Set",
"Definition:Set Intersection",
"Definition:Inductive Set"
] |
proofwiki-4817 | Minimally Inductive Set Exists | There exists a minimally inductive set $\omega$ that is a subset of every other inductive set. | From the Axiom of Infinity, there is a set $S$ such that:
:$\O \in S$
:$x \in S \implies x^+ \in S$
where $\O$ denotes the empty set and $x^+$ is the successor set of $x$.
That is, there exists an inductive set.
Next, by the Axiom of Specification, the minimally inductive set $\omega$:
:$\omega := \ds \bigcap \set {S' ... | There exists a [[Definition:Minimally Inductive Set|minimally inductive set]] $\omega$ that is a [[Definition:Subset|subset]] of every other [[Definition:Inductive Set|inductive set]]. | From the [[Axiom:Axiom of Infinity|Axiom of Infinity]], there is a [[Definition:Set|set]] $S$ such that:
:$\O \in S$
:$x \in S \implies x^+ \in S$
where $\O$ denotes the [[Definition:Empty Set|empty set]] and $x^+$ is the [[Definition:Successor Set|successor set]] of $x$.
That is, there exists an [[Definition:Induct... | Minimally Inductive Set Exists | https://proofwiki.org/wiki/Minimally_Inductive_Set_Exists | https://proofwiki.org/wiki/Minimally_Inductive_Set_Exists | [
"Minimally Inductive Set"
] | [
"Definition:Minimally Inductive Set",
"Definition:Subset",
"Definition:Inductive Set"
] | [
"Axiom:Axiom of Infinity",
"Definition:Set",
"Definition:Empty Set",
"Definition:Successor Mapping/Successor Set",
"Definition:Inductive Set",
"Axiom:Axiom of Specification/Set Theory",
"Definition:Minimally Inductive Set",
"Definition:Inductive Set",
"Intersection of Inductive Sets",
"Definition:... |
proofwiki-4818 | Minimally Inductive Set forms Peano Structure | Let $\omega$ be the minimally inductive set.
Let $\cdot^+: \omega \to \omega$ be the mapping assigning to a set its successor set:
:$n^+ := n \cup \set n$
Let $\O \in \omega$ be the empty set.
Then $\struct {\omega, \cdot^+, \O}$ is a Peano structure. | We need to check that all of Peano's axioms hold for $\struct {\omega, \cdot^+, \O}$.
Suppose first that for $m, n \in \omega$, we have $m^+ = n^+$.
Since $n \in n^+$ it follows that $n \in m^+$.
Hence, either $n \in m$ or $n = m$.
Similarly, either $m \in n$ or $m = n$.
Now if $n \ne m$, both $m \in n$ and $n \in m$.
... | Let $\omega$ be the [[Definition:Minimally Inductive Set|minimally inductive set]].
Let $\cdot^+: \omega \to \omega$ be the [[Definition:Mapping|mapping]] assigning to a [[Definition:Set|set]] its [[Definition:Successor Set|successor set]]:
:$n^+ := n \cup \set n$
Let $\O \in \omega$ be the [[Definition:Empty Set|em... | We need to check that all of [[Axiom:Peano's Axioms|Peano's axioms]] hold for $\struct {\omega, \cdot^+, \O}$.
Suppose first that for $m, n \in \omega$, we have $m^+ = n^+$.
Since $n \in n^+$ it follows that $n \in m^+$.
Hence, either $n \in m$ or $n = m$.
Similarly, either $m \in n$ or $m = n$.
Now if $n \ne m$,... | Minimally Inductive Set forms Peano Structure | https://proofwiki.org/wiki/Minimally_Inductive_Set_forms_Peano_Structure | https://proofwiki.org/wiki/Minimally_Inductive_Set_forms_Peano_Structure | [
"Peano's Axioms",
"Minimally Inductive Set"
] | [
"Definition:Minimally Inductive Set",
"Definition:Mapping",
"Definition:Set",
"Definition:Successor Mapping/Successor Set",
"Definition:Empty Set",
"Definition:Peano Structure"
] | [
"Axiom:Peano's Axioms",
"Element of Minimally Inductive Set is Transitive Set",
"Finite Ordinal is not Subset of one of its Elements",
"Definition:Inductive Set",
"Definition:Minimally Inductive Set",
"Definition:Set Equality/Definition 2",
"Definition:Peano Structure"
] |
proofwiki-4819 | Finite Ordinal is not Subset of one of its Elements | Let $n$ be a finite ordinal.
Then:
:$\nexists x \in n: n \subseteq x$
that is, $n$ is not a subset of one of its elements. | Let $S$ be the set of all those finite ordinals $n$ which are not a subset of any of its elements.
That is:
:$n \in S \iff n \in \omega \land \forall x \in n: n \nsubseteq x$
We know that $0 = \O$ is not a subset of any of its elements, as $\O$ by definition has no elements.
So $0 \in S$.
Now suppose $n \in S$.
From Se... | Let $n$ be a [[Definition:Finite Ordinal|finite ordinal]].
Then:
:$\nexists x \in n: n \subseteq x$
that is, $n$ is not a [[Definition:Subset|subset]] of one of its [[Definition:Element|elements]]. | Let $S$ be the [[Definition:Set|set]] of all those [[Definition:Finite Ordinal|finite ordinals]] $n$ which are not a [[Definition:Subset|subset]] of any of its [[Definition:Element|elements]].
That is:
:$n \in S \iff n \in \omega \land \forall x \in n: n \nsubseteq x$
We know that $0 = \O$ is not a [[Definition:Subse... | Finite Ordinal is not Subset of one of its Elements | https://proofwiki.org/wiki/Finite_Ordinal_is_not_Subset_of_one_of_its_Elements | https://proofwiki.org/wiki/Finite_Ordinal_is_not_Subset_of_one_of_its_Elements | [
"Finite Ordinals"
] | [
"Definition:Finite Ordinal",
"Definition:Subset",
"Definition:Element"
] | [
"Definition:Set",
"Definition:Finite Ordinal",
"Definition:Subset",
"Definition:Element",
"Definition:Subset",
"Definition:Element",
"Definition:Empty Set",
"Definition:Element",
"Set is Subset of Itself",
"Definition:Successor Mapping/Successor Set",
"Subset Relation is Transitive",
"Definiti... |
proofwiki-4820 | Element of Minimally Inductive Set is Transitive Set | Let $\omega$ be the minimally inductive set.
Let $n \in \omega$.
Then $x \in n \implies x \subseteq n$.
That is, every element of $n$ is also a subset of it.
In other words, each element of $\omega$ is a transitive set. | Let $S \subseteq$ be the set of all transitive elements of $\omega$.
That is:
:$n \in S \iff n \in \omega \land \forall x \in n: x \subseteq n$
It is vacuously true that $0 \in S$, as there are no $x \in 0$.
Now suppose $n \in S$.
If $x \in n^+$ then either $x \in n$ or $x = n$.
In the first case:
:$x \subseteq n$ as $... | Let $\omega$ be the [[Definition:Minimally Inductive Set|minimally inductive set]].
Let $n \in \omega$.
Then $x \in n \implies x \subseteq n$.
That is, every [[Definition:Element|element]] of $n$ is also a [[Definition:Subset|subset]] of it.
In other words, each element of $\omega$ is a [[Definition:Transitive Se... | Let $S \subseteq$ be the [[Definition:Set|set]] of all [[Definition:Transitive Set|transitive]] [[Definition:Element|elements]] of $\omega$.
That is:
:$n \in S \iff n \in \omega \land \forall x \in n: x \subseteq n$
It is [[Definition:Vacuous Truth|vacuously true]] that $0 \in S$, as there are no $x \in 0$.
Now sup... | Element of Minimally Inductive Set is Transitive Set | https://proofwiki.org/wiki/Element_of_Minimally_Inductive_Set_is_Transitive_Set | https://proofwiki.org/wiki/Element_of_Minimally_Inductive_Set_is_Transitive_Set | [
"Minimally Inductive Set"
] | [
"Definition:Minimally Inductive Set",
"Definition:Element",
"Definition:Subset",
"Definition:Transitive Class"
] | [
"Definition:Set",
"Definition:Transitive Class",
"Definition:Element",
"Definition:Vacuous Truth",
"Definition:Successor Mapping/Successor Set",
"Principle of Mathematical Induction for Minimally Inductive Set"
] |
proofwiki-4821 | Primitive of Exponential Function | :$\ds \int e^x \rd x = e^x + C$
where $C$ is an arbitrary constant. | {{begin-eqn}}
{{eqn | l = \map {\dfrac \d {\d x} } {a^x}
| r = a^x \ln a
| c = Derivative of General Exponential Function
}}
{{eqn | ll= \leadsto
| l = \map {\dfrac \d {\d x} } {\dfrac {a^x} {\ln a} }
| r = a^x
| c = Derivative of Constant Multiple
}}
{{eqn | ll= \leadsto
| l = \int ... | :$\ds \int e^x \rd x = e^x + C$
where $C$ is an [[Definition:Arbitrary Constant (Calculus)|arbitrary constant]]. | {{begin-eqn}}
{{eqn | l = \map {\dfrac \d {\d x} } {a^x}
| r = a^x \ln a
| c = [[Derivative of General Exponential Function]]
}}
{{eqn | ll= \leadsto
| l = \map {\dfrac \d {\d x} } {\dfrac {a^x} {\ln a} }
| r = a^x
| c = [[Derivative of Constant Multiple]]
}}
{{eqn | ll= \leadsto
| l... | Primitive of Exponential Function/General Result/Proof 1 | https://proofwiki.org/wiki/Primitive_of_Exponential_Function | https://proofwiki.org/wiki/Primitive_of_Exponential_Function/General_Result/Proof_1 | [
"Primitive of Exponential Function",
"Primitives involving Exponential Function",
"Exponential Function"
] | [
"Definition:Primitive (Calculus)/Constant of Integration"
] | [
"Derivative of General Exponential Function",
"Derivative of Constant Multiple"
] |
proofwiki-4822 | Primitive of Exponential Function | :$\ds \int e^x \rd x = e^x + C$
where $C$ is an arbitrary constant. | Let $u = x \ln a$.
{{begin-eqn}}
{{eqn | l = \int a^x \rd x
| r = \int \map \exp {x \ln a} \rd x
| c = {{Defof|Power to Real Number}}
}}
{{eqn | r = \frac 1 {\ln a} \int \map \exp u \rd u
| c = Primitive of Function of Constant Multiple
}}
{{eqn | r = \frac {\map \exp u} {\ln a} + C
| c = Primit... | :$\ds \int e^x \rd x = e^x + C$
where $C$ is an [[Definition:Arbitrary Constant (Calculus)|arbitrary constant]]. | Let $u = x \ln a$.
{{begin-eqn}}
{{eqn | l = \int a^x \rd x
| r = \int \map \exp {x \ln a} \rd x
| c = {{Defof|Power to Real Number}}
}}
{{eqn | r = \frac 1 {\ln a} \int \map \exp u \rd u
| c = [[Primitive of Function of Constant Multiple]]
}}
{{eqn | r = \frac {\map \exp u} {\ln a} + C
| c = [... | Primitive of Exponential Function/General Result/Proof 2 | https://proofwiki.org/wiki/Primitive_of_Exponential_Function | https://proofwiki.org/wiki/Primitive_of_Exponential_Function/General_Result/Proof_2 | [
"Primitive of Exponential Function",
"Primitives involving Exponential Function",
"Exponential Function"
] | [
"Definition:Primitive (Calculus)/Constant of Integration"
] | [
"Primitive of Function of Constant Multiple",
"Primitive of Exponential Function"
] |
proofwiki-4823 | Natural Number Addition Commutes with Zero | Let $\N$ be the natural numbers.
Then:
:$\forall n \in \N: 0 + n = n = n + 0$ | Proof by induction:
From definition of addition:
{{begin-eqn}}
{{eqn | q = \forall m, n \in \N
| l = m + 0
| r = m
| c =
}}
{{eqn | l = m + n^+
| r = \paren {m + n}^+
| c =
}}
{{end-eqn}}
For all $n \in \N$, let $\map P n$ be the proposition:
:$0 + n = n = n + 0$ | Let $\N$ be the [[Definition:Natural Numbers|natural numbers]].
Then:
:$\forall n \in \N: 0 + n = n = n + 0$ | Proof by [[Principle of Mathematical Induction|induction]]:
From definition of [[Definition:Addition in Minimally Inductive Set|addition]]:
{{begin-eqn}}
{{eqn | q = \forall m, n \in \N
| l = m + 0
| r = m
| c =
}}
{{eqn | l = m + n^+
| r = \paren {m + n}^+
| c =
}}
{{end-eqn}}
For ... | Natural Number Addition Commutes with Zero | https://proofwiki.org/wiki/Natural_Number_Addition_Commutes_with_Zero | https://proofwiki.org/wiki/Natural_Number_Addition_Commutes_with_Zero | [
"Natural Number Addition",
"Examples of Commutative Operations"
] | [
"Definition:Natural Numbers"
] | [
"Principle of Mathematical Induction",
"Definition:Addition in Minimally Inductive Set",
"Definition:Proposition",
"Principle of Mathematical Induction",
"Principle of Mathematical Induction"
] |
proofwiki-4824 | Natural Number Addition Commutativity with Successor | Let $\N$ be the natural numbers.
Then:
:$\forall m, n \in \N: m^+ + n = \paren {m + n}^+$ | Proof by induction:
From definition of addition:
{{begin-eqn}}
{{eqn | q = \forall m, n \in \N
| l = m + 0
| r = m
| c =
}}
{{eqn | l = m + n^+
| r = \paren {m + n}^+
| c =
}}
{{end-eqn}}
For all $n \in \N$, let $\map P n$ be the proposition:
:$\forall m \in \N: m^+ + n = \paren {m + n}^... | Let $\N$ be the [[Definition:Natural Numbers|natural numbers]].
Then:
:$\forall m, n \in \N: m^+ + n = \paren {m + n}^+$ | Proof by [[Principle of Mathematical Induction|induction]]:
From definition of [[Definition:Addition in Minimally Inductive Set|addition]]:
{{begin-eqn}}
{{eqn | q = \forall m, n \in \N
| l = m + 0
| r = m
| c =
}}
{{eqn | l = m + n^+
| r = \paren {m + n}^+
| c =
}}
{{end-eqn}}
For ... | Natural Number Addition Commutativity with Successor/Proof 1 | https://proofwiki.org/wiki/Natural_Number_Addition_Commutativity_with_Successor | https://proofwiki.org/wiki/Natural_Number_Addition_Commutativity_with_Successor/Proof_1 | [
"Natural Number Addition",
"Natural Number Addition Commutativity with Successor"
] | [
"Definition:Natural Numbers"
] | [
"Principle of Mathematical Induction",
"Definition:Addition in Minimally Inductive Set",
"Definition:Proposition",
"Definition:Addition in Minimally Inductive Set",
"Definition:Basis for the Induction",
"Definition:Induction Hypothesis",
"Principle of Mathematical Induction",
"Natural Number Addition ... |
proofwiki-4825 | Natural Number Addition Commutativity with Successor | Let $\N$ be the natural numbers.
Then:
:$\forall m, n \in \N: m^+ + n = \paren {m + n}^+$ | Using the following axioms:
{{:Axiom:Axiomatization of 1-Based Natural Numbers}}
Proof by induction:
From Axiomatization of $1$-Based Natural Numbers, we have by definition that:
{{begin-eqn}}
{{eqn | q = \forall m, n \in \N
| l = m + 0
| r = m
}}
{{eqn | l = \paren {m + n}^+
| r = m + n^+
}}
{{end-eq... | Let $\N$ be the [[Definition:Natural Numbers|natural numbers]].
Then:
:$\forall m, n \in \N: m^+ + n = \paren {m + n}^+$ | Using the [[Axiom:Axiomatization of 1-Based Natural Numbers|following axioms]]:
{{:Axiom:Axiomatization of 1-Based Natural Numbers}}
Proof by [[Principle of Mathematical Induction|induction]]:
From [[Axiom:Axiomatization of 1-Based Natural Numbers|Axiomatization of $1$-Based Natural Numbers]], we have by definition ... | Natural Number Addition Commutativity with Successor/Proof 2 | https://proofwiki.org/wiki/Natural_Number_Addition_Commutativity_with_Successor | https://proofwiki.org/wiki/Natural_Number_Addition_Commutativity_with_Successor/Proof_2 | [
"Natural Number Addition",
"Natural Number Addition Commutativity with Successor"
] | [
"Definition:Natural Numbers"
] | [
"Axiom:Axiomatization of 1-Based Natural Numbers",
"Principle of Mathematical Induction",
"Axiom:Axiomatization of 1-Based Natural Numbers",
"Definition:Proposition",
"Definition:Basis for the Induction",
"Definition:Induction Hypothesis",
"Definition:Induction Step",
"Natural Number Addition is Assoc... |
proofwiki-4826 | Natural Numbers are Comparable | Let $\N$ be the natural numbers.
Let $m, n \in \N$.
Then $m$ and $n$ are comparable by the ordering relation $\le$.
That is, either:
:$(1): \quad m \le n$
or:
:$(2): \quad n \le m$
or possibly both. | Let $\N$ be defined as the minimally inductive set $\omega$.
By definition of the ordering on minimally inductive set:
:$m \le n \iff \begin {cases} m = n & \text {or} \\ m \in n & \end {cases}$
Thus it is sufficient to prove that exactly one of the following is the case:
:$(1): \quad m \in n$
:$(2): \quad m = n$
:$(3)... | Let $\N$ be the [[Definition:Natural Numbers|natural numbers]].
Let $m, n \in \N$.
Then $m$ and $n$ are [[Definition:Comparable Elements|comparable]] by the [[Definition:Ordering on Natural Numbers|ordering relation]] $\le$.
That is, either:
:$(1): \quad m \le n$
or:
:$(2): \quad n \le m$
or possibly both. | Let $\N$ be defined as the [[Definition:Minimally Inductive Set|minimally inductive set]] $\omega$.
By definition of the [[Definition:Ordering on Minimally Inductive Set|ordering on minimally inductive set]]:
:$m \le n \iff \begin {cases} m = n & \text {or} \\ m \in n & \end {cases}$
Thus it is sufficient to prove t... | Natural Numbers are Comparable/Proof | https://proofwiki.org/wiki/Natural_Numbers_are_Comparable | https://proofwiki.org/wiki/Natural_Numbers_are_Comparable/Proof | [
"Ordering on Natural Numbers",
"Natural Numbers are Comparable"
] | [
"Definition:Natural Numbers",
"Definition:Comparable Elements",
"Definition:Ordering on Natural Numbers"
] | [
"Definition:Minimally Inductive Set",
"Definition:Ordering on Natural Numbers/Minimally Inductive Set",
"Principle of Finite Induction",
"Definition:Set",
"Definition:Comparable Elements",
"Definition:Set",
"Definition:Successor Mapping/Successor Set",
"Definition:Successor Mapping/Successor Set",
"... |
proofwiki-4827 | Natural Numbers are Comparable | Let $\N$ be the natural numbers.
Let $m, n \in \N$.
Then $m$ and $n$ are comparable by the ordering relation $\le$.
That is, either:
:$(1): \quad m \le n$
or:
:$(2): \quad n \le m$
or possibly both. | Let $\N$ be defined as the von Neumann construction $\omega$.
By definition of the ordering on von Neumann construction:
:$m \le n \iff m \subseteq n$
From Von Neumann Construction of Natural Numbers is Minimally Inductive, $\omega$ is minimally inductive class under the successor mapping.
Then from Minimally Inductive... | Let $\N$ be the [[Definition:Natural Numbers|natural numbers]].
Let $m, n \in \N$.
Then $m$ and $n$ are [[Definition:Comparable Elements|comparable]] by the [[Definition:Ordering on Natural Numbers|ordering relation]] $\le$.
That is, either:
:$(1): \quad m \le n$
or:
:$(2): \quad n \le m$
or possibly both. | Let $\N$ be defined as the [[Definition:Von Neumann Construction of Natural Numbers|von Neumann construction]] $\omega$.
By definition of the [[Definition:Ordering on Von Neumann Construction of Natural Numbers|ordering on von Neumann construction]]:
:$m \le n \iff m \subseteq n$
From [[Von Neumann Construction of N... | Natural Numbers are Comparable/Strong Result/Proof 1 | https://proofwiki.org/wiki/Natural_Numbers_are_Comparable | https://proofwiki.org/wiki/Natural_Numbers_are_Comparable/Strong_Result/Proof_1 | [
"Ordering on Natural Numbers",
"Natural Numbers are Comparable"
] | [
"Definition:Natural Numbers",
"Definition:Comparable Elements",
"Definition:Ordering on Natural Numbers"
] | [
"Definition:Natural Numbers/Von Neumann Construction",
"Definition:Ordering on Natural Numbers/Von Neumann Construction",
"Von Neumann Construction of Natural Numbers is Minimally Inductive",
"Definition:Minimally Inductive Class under General Mapping",
"Definition:Natural Numbers/Von Neumann Construction/S... |
proofwiki-4828 | Natural Numbers are Comparable | Let $\N$ be the natural numbers.
Let $m, n \in \N$.
Then $m$ and $n$ are comparable by the ordering relation $\le$.
That is, either:
:$(1): \quad m \le n$
or:
:$(2): \quad n \le m$
or possibly both. | {{ProofWanted|Proof using Minimally Inductive Class under Slowly Progressing Mapping is Nest by exploiting Successor Mapping is Slowly Progressing.}} | Let $\N$ be the [[Definition:Natural Numbers|natural numbers]].
Let $m, n \in \N$.
Then $m$ and $n$ are [[Definition:Comparable Elements|comparable]] by the [[Definition:Ordering on Natural Numbers|ordering relation]] $\le$.
That is, either:
:$(1): \quad m \le n$
or:
:$(2): \quad n \le m$
or possibly both. | {{ProofWanted|Proof using [[Minimally Inductive Class under Slowly Progressing Mapping is Nest]] by exploiting [[Successor Mapping is Slowly Progressing]].}} | Natural Numbers are Comparable/Strong Result/Proof 2 | https://proofwiki.org/wiki/Natural_Numbers_are_Comparable | https://proofwiki.org/wiki/Natural_Numbers_are_Comparable/Strong_Result/Proof_2 | [
"Ordering on Natural Numbers",
"Natural Numbers are Comparable"
] | [
"Definition:Natural Numbers",
"Definition:Comparable Elements",
"Definition:Ordering on Natural Numbers"
] | [
"Minimally Inductive Class under Slowly Progressing Mapping is Nest",
"Successor Mapping is Slowly Progressing"
] |
proofwiki-4829 | Zero is Zero Element for Natural Number Multiplication | Let $\N$ be the natural numbers.
Then $0$ is a zero element for multiplication:
:$\forall n \in \N: 0 \times n = 0 = n \times 0$ | Proof by induction.
For all $n \in \N$, let $\map P n$ be the proposition:
:$0 \times n = 0 = n \times 0$ | Let $\N$ be the [[Definition:Natural Numbers|natural numbers]].
Then $0$ is a [[Definition:Zero Element|zero element]] for [[Definition:Natural Number Multiplication|multiplication]]:
:$\forall n \in \N: 0 \times n = 0 = n \times 0$ | Proof by [[Principle of Mathematical Induction|induction]].
For all $n \in \N$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:$0 \times n = 0 = n \times 0$ | Zero is Zero Element for Natural Number Multiplication | https://proofwiki.org/wiki/Zero_is_Zero_Element_for_Natural_Number_Multiplication | https://proofwiki.org/wiki/Zero_is_Zero_Element_for_Natural_Number_Multiplication | [
"Natural Numbers"
] | [
"Definition:Natural Numbers",
"Definition:Zero Element",
"Definition:Multiplication/Natural Numbers"
] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Principle of Mathematical Induction"
] |
proofwiki-4830 | Hilbert Space Isomorphism is Equivalence Relation | Hilbert space isomorphism is an equivalence relation. | Checking the three defining properties of an equivalence relation in turn: | [[Definition:Isomorphism (Hilbert Spaces)|Hilbert space isomorphism]] is an [[Definition:Equivalence Relation|equivalence relation]]. | Checking the three defining properties of an [[Definition:Equivalence Relation|equivalence relation]] in turn: | Hilbert Space Isomorphism is Equivalence Relation | https://proofwiki.org/wiki/Hilbert_Space_Isomorphism_is_Equivalence_Relation | https://proofwiki.org/wiki/Hilbert_Space_Isomorphism_is_Equivalence_Relation | [
"Hilbert Spaces",
"Examples of Equivalence Relations"
] | [
"Definition:Isomorphism (Hilbert Spaces)",
"Definition:Equivalence Relation"
] | [
"Definition:Equivalence Relation",
"Definition:Equivalence Relation"
] |
proofwiki-4831 | Element of Finite Ordinal iff Subset | Let $m, n$ be distinct finite ordinals.
Then:
:$m \in n \iff m \subseteq n$ | Let $m \in n$.
Since $n$ is an ordinal, it is transitive.
Therefore, it follows directly that $m \subseteq n$.
Now let $m \subseteq n$.
We have by hypothesis that $m \ne n$.
From Natural Numbers are Comparable it follows that either $m \in n$ or $n \in m$.
Suppose $n \in m$.
Then $\exists x \in m: m \subseteq x$ which ... | Let $m, n$ be [[Definition:Distinct Objects|distinct]] [[Definition:Finite Ordinal|finite ordinals]].
Then:
:$m \in n \iff m \subseteq n$ | Let $m \in n$.
Since $n$ is an [[Definition:Ordinal|ordinal]], it is [[Definition:Transitive Set|transitive]].
Therefore, it follows directly that $m \subseteq n$.
Now let $m \subseteq n$.
We have by hypothesis that $m \ne n$.
From [[Natural Numbers are Comparable]] it follows that either $m \in n$ or $n \in m$.
... | Element of Finite Ordinal iff Subset | https://proofwiki.org/wiki/Element_of_Finite_Ordinal_iff_Subset | https://proofwiki.org/wiki/Element_of_Finite_Ordinal_iff_Subset | [
"Finite Ordinals",
"Subsets"
] | [
"Definition:Distinct/Plural",
"Definition:Finite Ordinal"
] | [
"Definition:Ordinal",
"Definition:Transitive Class",
"Natural Numbers are Comparable",
"Finite Ordinal is not Subset of one of its Elements"
] |
proofwiki-4832 | Ordering on Natural Numbers is Compatible with Addition | Let $m, n, k \in \N$ where $\N$ is the set of natural numbers.
Then:
:$m < n \iff m + k < n + k$ | Proof by induction:
For all $k \in \N$, let $\map P k$ be the proposition:
:$m < n \iff m + k < n + k$
$\map P 0$ is true, as this just says $m + 0 = m < n = n + 0$.
This is our basis for the induction. | Let $m, n, k \in \N$ where $\N$ is the [[Definition:Natural Numbers|set of natural numbers]].
Then:
:$m < n \iff m + k < n + k$ | Proof by [[Principle of Mathematical Induction|induction]]:
For all $k \in \N$, let $\map P k$ be the [[Definition:Proposition|proposition]]:
:$m < n \iff m + k < n + k$
$\map P 0$ is true, as this just says $m + 0 = m < n = n + 0$.
This is our [[Definition:Basis for the Induction|basis for the induction]]. | Ordering on Natural Numbers is Compatible with Addition | https://proofwiki.org/wiki/Ordering_on_Natural_Numbers_is_Compatible_with_Addition | https://proofwiki.org/wiki/Ordering_on_Natural_Numbers_is_Compatible_with_Addition | [
"Natural Number Addition",
"Examples of Orderings"
] | [
"Definition:Natural Numbers"
] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Definition:Basis for the Induction",
"Principle of Mathematical Induction"
] |
proofwiki-4833 | Hilbert Space Isomorphism is Bijection | Let $H, K$ be Hilbert spaces.
Denote by $\innerprod \cdot \cdot_H$ and $\innerprod \cdot \cdot_K$ their respective inner products.
Let $U: H \to K$ be an isomorphism.
Then $U$ is a bijection. | As $U$ is an isomorphism, it is necessarily surjective.
Suppose now that $g, h \in H$ are such that $\map U g = \map U h$.
Then as $U$ is a linear map, it follows that $\map U {g - h} = \bszero_K$, the zero vector of $K$.
From property $(3)$ of an isomorphism, conclude that:
:$0 = \innerprod {\bszero_K} {\bszero_K}_K =... | Let $H, K$ be [[Definition:Hilbert Space|Hilbert spaces]].
Denote by $\innerprod \cdot \cdot_H$ and $\innerprod \cdot \cdot_K$ their respective [[Definition:Inner Product|inner products]].
Let $U: H \to K$ be an [[Definition:Isomorphism (Hilbert Spaces)|isomorphism]].
Then $U$ is a [[Definition:Bijection|bijection... | As $U$ is an [[Definition:Isomorphism (Hilbert Spaces)|isomorphism]], it is necessarily [[Definition:Surjection|surjective]].
Suppose now that $g, h \in H$ are such that $\map U g = \map U h$.
Then as $U$ is a [[Definition:Linear Mapping|linear map]], it follows that $\map U {g - h} = \bszero_K$, the [[Definition:Ze... | Hilbert Space Isomorphism is Bijection | https://proofwiki.org/wiki/Hilbert_Space_Isomorphism_is_Bijection | https://proofwiki.org/wiki/Hilbert_Space_Isomorphism_is_Bijection | [
"Hilbert Spaces"
] | [
"Definition:Hilbert Space",
"Definition:Inner Product",
"Definition:Isomorphism (Hilbert Spaces)",
"Definition:Bijection"
] | [
"Definition:Isomorphism (Hilbert Spaces)",
"Definition:Surjection",
"Definition:Linear Transformation",
"Definition:Zero Vector",
"Definition:Isomorphism (Hilbert Spaces)",
"Definition:Inner Product",
"Definition:Injection",
"Definition:Injection",
"Definition:Surjection",
"Definition:Bijection",
... |
proofwiki-4834 | Surjection that Preserves Inner Product is Linear | Let $H, K$ be Hilbert spaces, and denote by ${\innerprod \cdot \cdot}_H$ and ${\innerprod \cdot \cdot}_K$ their respective inner products.
Let $U: H \to K$ be a surjection such that:
:$\forall g, h \in H: {\innerprod g h}_H = {\innerprod {Ug} {Uh} }_K$
Then $U$ is a linear map, and hence an isomorphism. | Let $x, y \in H$.
Let $\alpha \in \GF$.
By surjectivity of $U$, choose $z \in H$ such that $Uz = \map U {\alpha x + y} - \paren { \alpha Ux + Uy }$.
Then:
{{begin-eqn}}
{{eqn | l = {\innerprod {Uz} {Uz} }_K
| r = {\innerprod {\map U {\alpha x + y} - \paren{\alpha Ux + Uy } } {Uz} }_K
}}
{{eqn | r = {\innerprod {\... | Let $H, K$ be [[Definition:Hilbert Space|Hilbert spaces]], and denote by ${\innerprod \cdot \cdot}_H$ and ${\innerprod \cdot \cdot}_K$ their respective [[Definition:Inner Product|inner products]].
Let $U: H \to K$ be a [[Definition:Surjection|surjection]] such that:
:$\forall g, h \in H: {\innerprod g h}_H = {\innerp... | Let $x, y \in H$.
Let $\alpha \in \GF$.
By [[Definition:Surjection|surjectivity]] of $U$, choose $z \in H$ such that $Uz = \map U {\alpha x + y} - \paren { \alpha Ux + Uy }$.
Then:
{{begin-eqn}}
{{eqn | l = {\innerprod {Uz} {Uz} }_K
| r = {\innerprod {\map U {\alpha x + y} - \paren{\alpha Ux + Uy } } {Uz} }_K
... | Surjection that Preserves Inner Product is Linear | https://proofwiki.org/wiki/Surjection_that_Preserves_Inner_Product_is_Linear | https://proofwiki.org/wiki/Surjection_that_Preserves_Inner_Product_is_Linear | [
"Hilbert Spaces",
"Linear Transformations on Hilbert Spaces"
] | [
"Definition:Hilbert Space",
"Definition:Inner Product",
"Definition:Surjection",
"Definition:Linear Transformation",
"Definition:Isomorphism (Hilbert Spaces)"
] | [
"Definition:Surjection",
"Definition:Linear Transformation",
"Definition:Inner Product",
"Definition:Linear Transformation",
"Definition:Positiveness",
"Definition:Linear Transformation"
] |
proofwiki-4835 | Linear Operator on General Logarithm | Let $\phi: \R^\R \to \R^\R, y \mapsto \map \phi y$ be a linear operator on the space of functions from $\R\to\R$.
Let $y$ be a real function such that:
:$\forall x \in \R: \map y x > \map \bszero x = 0$.
Let $\log_a y$ be the logarithm of $y$ to base $a$.
Then:
:$\map \phi {\log_a y} = \dfrac 1 {\ln a} \paren {\map \ph... | {{begin-eqn}}
{{eqn | l = \map \phi {\log_a y}
| r = \map \phi {\frac {\ln y}{\ln a} }
| c = Change of Base of Logarithm
}}
{{eqn | r = \frac 1 {\ln a} \paren {\map \phi {\ln y} }
| c = {{Defof|Linear Operator}}
}}
{{end-eqn}}
{{qed}}
Category:Linear Operators
Category:Logarithms
jrexc2dlq51lr59wq4z0h... | Let $\phi: \R^\R \to \R^\R, y \mapsto \map \phi y$ be a [[Definition:Linear Operator|linear operator]] on the space of functions from $\R\to\R$.
Let $y$ be a [[Definition:Real Function|real function]] such that:
:$\forall x \in \R: \map y x > \map \bszero x = 0$.
Let $\log_a y$ be the [[Definition:General Logarithm|... | {{begin-eqn}}
{{eqn | l = \map \phi {\log_a y}
| r = \map \phi {\frac {\ln y}{\ln a} }
| c = [[Change of Base of Logarithm]]
}}
{{eqn | r = \frac 1 {\ln a} \paren {\map \phi {\ln y} }
| c = {{Defof|Linear Operator}}
}}
{{end-eqn}}
{{qed}}
[[Category:Linear Operators]]
[[Category:Logarithms]]
jrexc2dl... | Linear Operator on General Logarithm | https://proofwiki.org/wiki/Linear_Operator_on_General_Logarithm | https://proofwiki.org/wiki/Linear_Operator_on_General_Logarithm | [
"Linear Operators",
"Logarithms"
] | [
"Definition:Linear Operator",
"Definition:Real Function",
"Definition:General Logarithm",
"Definition:Natural Logarithm"
] | [
"Change of Base of Logarithm",
"Category:Linear Operators",
"Category:Logarithms"
] |
proofwiki-4836 | Ordering on Natural Numbers is Compatible with Multiplication | Let $m, n, k \in \N$ where $\N$ is the set of natural numbers.
Let $k \ne 0$.
Then:
:$m < n \iff m \times k < n \times k$ | Proof by induction:
First we note that if $k = 0$ then $m \times k = 0 = n \times k$ whatever $m$ and $n$ are, and the proposition clearly does not hold.
So, for all $k \in \N \setminus \set 0$, let $\map P k$ be the proposition:
:$m < n \iff m \times k < n \times k$
From Identity Element of Natural Number Multiplicati... | Let $m, n, k \in \N$ where $\N$ is the [[Definition:Natural Numbers|set of natural numbers]].
Let $k \ne 0$.
Then:
:$m < n \iff m \times k < n \times k$ | Proof by [[Principle of Mathematical Induction|induction]]:
First we note that if $k = 0$ then $m \times k = 0 = n \times k$ whatever $m$ and $n$ are, and the proposition clearly does not hold.
So, for all $k \in \N \setminus \set 0$, let $\map P k$ be the [[Definition:Proposition|proposition]]:
:$m < n \iff m \time... | Ordering on Natural Numbers is Compatible with Multiplication | https://proofwiki.org/wiki/Ordering_on_Natural_Numbers_is_Compatible_with_Multiplication | https://proofwiki.org/wiki/Ordering_on_Natural_Numbers_is_Compatible_with_Multiplication | [
"Natural Number Multiplication",
"Examples of Orderings"
] | [
"Definition:Natural Numbers"
] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Identity Element of Natural Number Multiplication is One",
"Definition:Basis for the Induction",
"Principle of Mathematical Induction"
] |
proofwiki-4837 | Natural Number Ordering is Transitive | Let $m, n, k \in \N$ where $\N$ is the set of natural numbers.
Let $<$ be the relation defined on $\N$ such that:
:$m < n \iff m \in n$
where $\N$ is defined as the minimally inductive set $\omega$.
Then:
:$k < m, m < n \implies k < n$
That is: $<$ is a transitive relation. | Let $k < m, m < n$.
By definition it follows that $k \in m, m \in n$.
We have from Element of Finite Ordinal iff Subset that:
:$k \in m \iff k \subseteq m$
:$m \in n \iff m \subseteq n$
It follows from Subset Relation is Transitive that $k \subseteq n$.
Hence the result.
{{qed}}
Category:Natural Numbers
dkmzf5dhzfezvno... | Let $m, n, k \in \N$ where $\N$ is the [[Definition:Natural Numbers|set of natural numbers]].
Let $<$ be the [[Natural Numbers are Comparable|relation defined on $\N$]] such that:
:$m < n \iff m \in n$
where $\N$ is defined as the [[Definition:Minimally Inductive Set|minimally inductive set]] $\omega$.
Then:
:$k < m... | Let $k < m, m < n$.
By definition it follows that $k \in m, m \in n$.
We have from [[Element of Finite Ordinal iff Subset]] that:
:$k \in m \iff k \subseteq m$
:$m \in n \iff m \subseteq n$
It follows from [[Subset Relation is Transitive]] that $k \subseteq n$.
Hence the result.
{{qed}}
[[Category:Natural Numbers]... | Natural Number Ordering is Transitive | https://proofwiki.org/wiki/Natural_Number_Ordering_is_Transitive | https://proofwiki.org/wiki/Natural_Number_Ordering_is_Transitive | [
"Natural Numbers"
] | [
"Definition:Natural Numbers",
"Natural Numbers are Comparable",
"Definition:Minimally Inductive Set",
"Definition:Transitive Relation"
] | [
"Element of Finite Ordinal iff Subset",
"Subset Relation is Transitive",
"Category:Natural Numbers"
] |
proofwiki-4838 | Proper Subset of Finite Ordinal is Equivalent to Smaller Ordinal | Let $n$ be a finite ordinal.
Let $x \subsetneq n$.
Then for some finite ordinal $m < n$:
:$m \sim x$
where $m \sim x$ denotes that $m$ is (set) equivalent to $x$.
That is, every proper subset of a finite ordinal $n$ is equivalent to some finite ordinal smaller than $n$. | Proof by induction:
For all finite ordinals $n$, let $\map P n$ be the proposition:
:$x \subsetneq n \implies \exists m \in \N: m < n: m \sim x$
$\map P 0$ is vacuously true, as there are no proper subsets of $0 = \O$.
This is our basis for the induction. | Let $n$ be a [[Definition:Finite Ordinal|finite ordinal]].
Let $x \subsetneq n$.
Then for some [[Definition:Finite Ordinal|finite ordinal]] $m < n$:
:$m \sim x$
where $m \sim x$ denotes that $m$ is [[Definition:Set Equivalence|(set) equivalent]] to $x$.
That is, every [[Definition:Proper Subset|proper subset]] of ... | Proof by [[Principle of Mathematical Induction|induction]]:
For all [[Definition:Finite Ordinal|finite ordinals]] $n$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:$x \subsetneq n \implies \exists m \in \N: m < n: m \sim x$
$\map P 0$ is [[Definition:Vacuous Truth|vacuously true]], as there are no ... | Proper Subset of Finite Ordinal is Equivalent to Smaller Ordinal | https://proofwiki.org/wiki/Proper_Subset_of_Finite_Ordinal_is_Equivalent_to_Smaller_Ordinal | https://proofwiki.org/wiki/Proper_Subset_of_Finite_Ordinal_is_Equivalent_to_Smaller_Ordinal | [
"Finite Ordinals",
"Proofs by Induction"
] | [
"Definition:Finite Ordinal",
"Definition:Finite Ordinal",
"Definition:Set Equivalence",
"Definition:Proper Subset",
"Definition:Finite Ordinal",
"Definition:Set Equivalence"
] | [
"Principle of Mathematical Induction",
"Definition:Finite Ordinal",
"Definition:Proposition",
"Definition:Vacuous Truth",
"Definition:Proper Subset",
"Definition:Basis for the Induction",
"Definition:Finite Ordinal",
"Principle of Mathematical Induction"
] |
proofwiki-4839 | Hilbert Space Direct Sum is Hilbert Space | Let $\sequence {H_i}_{i \mathop \in I}$ be a $I$-indexed family of Hilbert spaces over $\Bbb F \in \set {\R, \C}$.
Let $H = \ds \bigoplus_{i \mathop \in I} H_i$ be their Hilbert space direct sum.
Then $H$ is a Hilbert space. | {{Proofread}}
{{wtd|I think some of the partial results are up on their own pages now, and thus can be replaced by a reference (or this should become the case).
Furthermore, this is *so* long, transclusions and/or foldable templates are a must}} | Let $\sequence {H_i}_{i \mathop \in I}$ be a [[Definition:Indexing Set|$I$-indexed family]] of [[Definition:Hilbert Space|Hilbert spaces]] over $\Bbb F \in \set {\R, \C}$.
Let $H = \ds \bigoplus_{i \mathop \in I} H_i$ be their [[Definition:Hilbert Space Direct Sum|Hilbert space direct sum]].
Then $H$ is a [[Definiti... | {{Proofread}}
{{wtd|I think some of the partial results are up on their own pages now, and thus can be replaced by a reference (or this should become the case).
Furthermore, this is *so* long, transclusions and/or foldable templates are a must}} | Hilbert Space Direct Sum is Hilbert Space | https://proofwiki.org/wiki/Hilbert_Space_Direct_Sum_is_Hilbert_Space | https://proofwiki.org/wiki/Hilbert_Space_Direct_Sum_is_Hilbert_Space | [
"Hilbert Spaces"
] | [
"Definition:Indexing Set",
"Definition:Hilbert Space",
"Definition:Hilbert Space Direct Sum",
"Definition:Hilbert Space"
] | [] |
proofwiki-4840 | Morera's Theorem | Let $D$ be a simply connected domain in $\C$.
Let $f: D \to \C$ be a continuous function.
Let $f$ be such that:
:$\ds \int_\gamma \map f z \rd z = 0$
for every simple closed contour $\gamma$ in $D$
Then $f$ is analytic on $D$. | For a fixed $z_0 \in D$ and $z \in D$ we consider the function:
:$\ds \map F z = \int_\gamma \map f w \rd w$
where $\gamma$ is any (simple) contour starting at $z_0$ and ending at $z$.
By Primitive of Function on Connected Domain, $F$ is a primitive of $f$.
Since $F$ is analytic and $F' = f$, we conclude that $f$ is an... | Let $D$ be a [[Definition:Simply Connected Domain|simply connected domain]] in $\C$.
Let $f: D \to \C$ be a [[Definition:Continuous Complex Function|continuous function]].
Let $f$ be such that:
:$\ds \int_\gamma \map f z \rd z = 0$
for every [[Definition:Simple Contour (Complex Plane)|simple]] [[Definition:Closed ... | For a fixed $z_0 \in D$ and $z \in D$ we consider the function:
:$\ds \map F z = \int_\gamma \map f w \rd w$
where $\gamma$ is any [[Definition:Simple Contour (Complex Plane)|(simple) contour]] starting at $z_0$ and ending at $z$.
By [[Primitive of Function on Connected Domain]], $F$ is a [[Definition:Complex Primiti... | Morera's Theorem | https://proofwiki.org/wiki/Morera's_Theorem | https://proofwiki.org/wiki/Morera's_Theorem | [
"Calculus",
"Complex Analysis"
] | [
"Definition:Connected Domain (Complex Analysis)/Simply Connected Domain",
"Definition:Continuous Complex Function",
"Definition:Contour/Simple/Complex Plane",
"Definition:Contour/Closed/Complex Plane",
"Definition:Analytic Function/Complex Plane"
] | [
"Definition:Contour/Simple/Complex Plane",
"Primitive of Function on Connected Domain",
"Definition:Primitive (Calculus)/Complex",
"Definition:Analytic Function/Complex Plane",
"Definition:Analytic Function/Complex Plane"
] |
proofwiki-4841 | Relation is Antisymmetric iff Intersection with Inverse is Coreflexive | Let $\RR$ be a relation on $S$.
Then:
:$\RR$ is antisymmetric
{{iff}}:
:$\RR \cap \RR^{-1}$ is coreflexive
where $\RR^{-1}$ is the inverse of $\RR$.
That is, {{iff}}:
:$\RR \cap \RR^{-1} \subseteq \Delta_S$ | === Necessary Condition ===
Let $\RR$ be an antisymmetric relation.
Let $\tuple {a, b} \in \RR \cap \RR^{-1}$.
That means:
:$\tuple {a, b} \in \RR$
and
:$\tuple {a, b} \in \RR^{-1}$
which means, by definition of inverse relation:
:$\tuple {b, a} \in \RR$
But as $\RR$ is antisymmetric, that means $a = b$.
Thus:
:$\tuple... | Let $\RR$ be a [[Definition:Relation|relation]] on $S$.
Then:
:$\RR$ is [[Definition:Antisymmetric Relation|antisymmetric]]
{{iff}}:
:$\RR \cap \RR^{-1}$ is [[Definition:Coreflexive Relation|coreflexive]]
where $\RR^{-1}$ is the [[Definition:Inverse Relation|inverse]] of $\RR$.
That is, {{iff}}:
:$\RR \cap \RR^{-1} ... | === Necessary Condition ===
Let $\RR$ be an [[Definition:Antisymmetric Relation|antisymmetric relation]].
Let $\tuple {a, b} \in \RR \cap \RR^{-1}$.
That means:
:$\tuple {a, b} \in \RR$
and
:$\tuple {a, b} \in \RR^{-1}$
which means, by definition of [[Definition:Inverse Relation|inverse relation]]:
:$\tuple {b, a} \... | Relation is Antisymmetric iff Intersection with Inverse is Coreflexive | https://proofwiki.org/wiki/Relation_is_Antisymmetric_iff_Intersection_with_Inverse_is_Coreflexive | https://proofwiki.org/wiki/Relation_is_Antisymmetric_iff_Intersection_with_Inverse_is_Coreflexive | [
"Antisymmetric Relations",
"Coreflexive Relations"
] | [
"Definition:Relation",
"Definition:Antisymmetric Relation",
"Definition:Coreflexive Relation",
"Definition:Inverse Relation"
] | [
"Definition:Antisymmetric Relation",
"Definition:Inverse Relation",
"Definition:Antisymmetric Relation",
"Definition:Diagonal Relation",
"Definition:Subset",
"Definition:Coreflexive Relation/Definition 2",
"Definition:Coreflexive Relation",
"Definition:Coreflexive Relation/Definition 2",
"Definition... |
proofwiki-4842 | Min Operation on Toset forms Semigroup | Let $\struct {S, \preceq}$ be a totally ordered set.
Let $\map \min {x, y}$ denote the min operation on $x, y \in S$.
Then $\struct {S, \min}$ is a semigroup. | By the definition of the min operation, either:
:$\map \min {x, y}= x$
or
:$\map \min {x, y}= y$
So $\min$ is closed on $S$.
From Min Operation is Associative:
:$\forall x, y, z \in S: \map \min {x, \map \min {y, z} } = \map \min {\map \min {x, y}, z}$
Hence the result, by definition of semigroup.
{{qed}} | Let $\struct {S, \preceq}$ be a [[Definition:Totally Ordered Set|totally ordered set]].
Let $\map \min {x, y}$ denote the [[Definition:Min Operation|min operation]] on $x, y \in S$.
Then $\struct {S, \min}$ is a [[Definition:Semigroup|semigroup]]. | By the definition of the [[Definition:Min Operation|min operation]], either:
:$\map \min {x, y}= x$
or
:$\map \min {x, y}= y$
So $\min$ is [[Definition:Closed Algebraic Structure|closed on $S$]].
From [[Min Operation is Associative]]:
:$\forall x, y, z \in S: \map \min {x, \map \min {y, z} } = \map \min {\map \min ... | Min Operation on Toset forms Semigroup | https://proofwiki.org/wiki/Min_Operation_on_Toset_forms_Semigroup | https://proofwiki.org/wiki/Min_Operation_on_Toset_forms_Semigroup | [
"Min Operation"
] | [
"Definition:Totally Ordered Set",
"Definition:Min Operation",
"Definition:Semigroup"
] | [
"Definition:Min Operation",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Min Operation is Associative",
"Definition:Semigroup"
] |
proofwiki-4843 | Max Semigroup is Commutative | Let $\struct {S, \preceq}$ be a totally ordered set.
Then the semigroup $\struct {S, \max}$ is commutative. | Let $x, y \in S$.
From Max Operation is Commutative:
:$\map \max {x, y} = \map \max {y, x}$
Hence the result, by definition of commutative semigroup.
{{qed}} | Let $\struct {S, \preceq}$ be a [[Definition:Totally Ordered Set|totally ordered set]].
Then the [[Max Operation on Toset forms Semigroup|semigroup]] $\struct {S, \max}$ is [[Definition:Commutative Semigroup|commutative]]. | Let $x, y \in S$.
From [[Max Operation is Commutative]]:
:$\map \max {x, y} = \map \max {y, x}$
Hence the result, by definition of [[Definition:Commutative Semigroup|commutative semigroup]].
{{qed}} | Max Semigroup is Commutative | https://proofwiki.org/wiki/Max_Semigroup_is_Commutative | https://proofwiki.org/wiki/Max_Semigroup_is_Commutative | [
"Max Operation",
"Examples of Commutative Semigroups"
] | [
"Definition:Totally Ordered Set",
"Max Operation on Toset forms Semigroup",
"Definition:Commutative Semigroup"
] | [
"Max Operation is Commutative",
"Definition:Commutative Semigroup"
] |
proofwiki-4844 | Relation is Connected iff Union with Inverse and Diagonal is Trivial Relation | Let $\RR$ be a relation on $S$.
Then $\RR$ is a connected relation {{iff}}:
:$\RR \cup \RR^{-1} \cup \Delta_S = S \times S$
where $\RR^{-1}$ is the inverse of $\RR$ and $\Delta_S$ is the diagonal relation. | === Necessary Condition ===
Let $\RR$ be a connected relation.
By definition of relation:
:$\RR \subseteq S \times S$
:$\RR^{-1} \subseteq S \times S$
:$\Delta_S \subseteq S \times S$
So from Union is Smallest Superset (and indeed, trivially):
:$\RR \cup \RR^{-1} \cup \Delta_S \subseteq S \times S$
Let $\tuple {a, b} \... | Let $\RR$ be a [[Definition:Relation|relation]] on $S$.
Then $\RR$ is a [[Definition:Connected Relation|connected relation]] {{iff}}:
:$\RR \cup \RR^{-1} \cup \Delta_S = S \times S$
where $\RR^{-1}$ is the [[Definition:Inverse Relation|inverse]] of $\RR$ and $\Delta_S$ is the [[Definition:Diagonal Relation|diagonal r... | === Necessary Condition ===
Let $\RR$ be a [[Definition:Connected Relation|connected relation]].
By definition of [[Definition:Relation|relation]]:
:$\RR \subseteq S \times S$
:$\RR^{-1} \subseteq S \times S$
:$\Delta_S \subseteq S \times S$
So from [[Union is Smallest Superset]] (and indeed, trivially):
:$\RR \cup ... | Relation is Connected iff Union with Inverse and Diagonal is Trivial Relation | https://proofwiki.org/wiki/Relation_is_Connected_iff_Union_with_Inverse_and_Diagonal_is_Trivial_Relation | https://proofwiki.org/wiki/Relation_is_Connected_iff_Union_with_Inverse_and_Diagonal_is_Trivial_Relation | [
"Connected Relations",
"Inverse Relations",
"Trivial Relation"
] | [
"Definition:Relation",
"Definition:Connected Relation",
"Definition:Inverse Relation",
"Definition:Diagonal Relation"
] | [
"Definition:Connected Relation",
"Definition:Relation",
"Union is Smallest Superset",
"Definition:Diagonal Relation",
"Set is Subset of Union",
"Definition:Connected Relation",
"Definition:Inverse Relation",
"Set is Subset of Union",
"Definition:Subset",
"Definition:Set Equality/Definition 2",
"... |
proofwiki-4845 | Triangle Inequality for Generalized Sums | Let $V$ be a Banach space.
Let $\norm {\,\cdot\,}$ denote the norm on $V$.
Let $\family {v_i}_{i \mathop \in I}$ be an indexed subset of $V$.
Let the generalized sum $\ds \sum \set {v_i: i \in I}$ converge absolutely.
Then:
:$(1): \quad \ds \norm {\sum \set {v_i: i \in I} } \le \sum \set {\norm {v_i}: i \in I}$ | First of all, note that Absolutely Convergent Generalized Sum Converges assures us that the {{LHS}} in $(1)$ is defined.
{{AimForCont}} there exists an $\epsilon > 0$ such that:
:$\ds \norm {\sum \set {v_i: i \in I} } > \sum \set {\norm {v_i}: i \in I} + \epsilon$
This supposition is seen to be equivalent to:
:$\ds \no... | Let $V$ be a [[Definition:Banach Space|Banach space]].
Let $\norm {\,\cdot\,}$ denote the [[Definition:Norm on Vector Space|norm]] on $V$.
Let $\family {v_i}_{i \mathop \in I}$ be an [[Definition:Indexed Set|indexed]] [[Definition:Subset|subset]] of $V$.
Let the [[Definition:Generalized Sum|generalized sum]] $\ds \s... | First of all, note that [[Absolutely Convergent Generalized Sum Converges]] assures us that the {{LHS}} in $(1)$ is defined.
{{AimForCont}} there exists an $\epsilon > 0$ such that:
:$\ds \norm {\sum \set {v_i: i \in I} } > \sum \set {\norm {v_i}: i \in I} + \epsilon$
This supposition is seen to be equivalent to:
:$... | Triangle Inequality for Generalized Sums | https://proofwiki.org/wiki/Triangle_Inequality_for_Generalized_Sums | https://proofwiki.org/wiki/Triangle_Inequality_for_Generalized_Sums | [
"Generalized Sums",
"Banach Spaces",
"Triangle Inequality"
] | [
"Definition:Banach Space",
"Definition:Norm/Vector Space",
"Definition:Indexing Set/Indexed Set",
"Definition:Subset",
"Definition:Generalized Sum",
"Definition:Absolutely Convergent Generalized Sum"
] | [
"Absolutely Convergent Generalized Sum Converges",
"Definition:Generalized Sum",
"Definition:Finite Set",
"Definition:Subset",
"Generalized Sum is Monotone",
"Category:Generalized Sums",
"Category:Banach Spaces",
"Category:Triangle Inequality"
] |
proofwiki-4846 | Absolutely Convergent Generalized Sum Converges | Let $V$ be a Banach space.
Let $\norm {\, \cdot \,}$ denote the norm on $V$.
Let $d$ denote the corresponding induced metric.
Let $\family {v_i}_{i \mathop \in I}$ be an indexed subset of $V$ such that the generalized sum $\ds \sum \set {v_i: i \in I }$ converges absolutely.
Then the generalized sum $\ds \sum \set {v_i... | The proof proceeds in two stages:
:$(1): \quad$ Finding a candidate $v \in V$ where the sum might converge to
:$(2): \quad$ Showing that the candidate is indeed sought limit.
That $\ds \sum \set {v_i: i \mathop \in I}$ converges absolutely means that $\ds \sum \set {\norm {v_i}: i \mathop \in I}$ converges.
Now, for al... | Let $V$ be a [[Definition:Banach Space|Banach space]].
Let $\norm {\, \cdot \,}$ denote the [[Definition:Norm on Vector Space|norm]] on $V$.
Let $d$ denote the corresponding [[Definition:Metric Induced by Norm|induced metric]].
Let $\family {v_i}_{i \mathop \in I}$ be an [[Definition:Indexed Set|indexed]] [[Definiti... | The proof proceeds in two stages:
:$(1): \quad$ Finding a candidate $v \in V$ where the sum might converge to
:$(2): \quad$ Showing that the candidate is indeed sought limit.
That $\ds \sum \set {v_i: i \mathop \in I}$ [[Definition:Generalized Sum/Absolute Net Convergence|converges absolutely]] means that $\ds \sum ... | Absolutely Convergent Generalized Sum Converges | https://proofwiki.org/wiki/Absolutely_Convergent_Generalized_Sum_Converges | https://proofwiki.org/wiki/Absolutely_Convergent_Generalized_Sum_Converges | [
"Absolute Convergence",
"Generalized Sums",
"Banach Spaces"
] | [
"Definition:Banach Space",
"Definition:Norm/Vector Space",
"Definition:Metric Induced by Norm",
"Definition:Indexing Set/Indexed Set",
"Definition:Subset",
"Definition:Generalized Sum",
"Definition:Generalized Sum/Absolute Net Convergence",
"Definition:Generalized Sum",
"Definition:Convergent Net"
] | [
"Definition:Generalized Sum/Absolute Net Convergence",
"Definition:Generalized Sum",
"Definition:Finite Set",
"Definition:Finite Set",
"Definition:Sequence",
"Definition:Cauchy Sequence",
"Definition:Norm/Vector Space",
"Union with Relative Complement",
"Generalized Sum is Monotone",
"Definition:C... |
proofwiki-4847 | Generalized Sum is Monotone | Let $I$ be an indexing set.
Let $\family {a_i}_{i \mathop \in I}$ be an $I$-indexed family of positive real numbers.
That is, let $a_i \in \R_{\ge 0}$ for all $i \in I$.
Then, for every finite subset $F$ of $I$:
:$\ds \sum_{i \mathop \in F} a_i \le \sum_{i \mathop \in I} a_i$
provided the generalized sum on the right c... | Let $\struct {\FF, \subseteq}$ be the set of finite subsets of $I$, ordered by inclusion.
From Finite Subsets form Directed Set, $\struct {\FF, \subseteq}$ is directed.
For brevity write:
:$\ds s_X = \sum_{i \mathop \in X} a_i$
for $X \in \FF$.
{{AimForCont}} that:
:$\ds \sum_{i \mathop \in F} a_i > \sum_{i \mathop \... | Let $I$ be an [[Definition:Indexing Set|indexing set]].
Let $\family {a_i}_{i \mathop \in I}$ be an [[Definition:Indexed Family|$I$-indexed family]] of [[Definition:Positive Real Number|positive real numbers]].
That is, let $a_i \in \R_{\ge 0}$ for all $i \in I$.
Then, for every [[Definition:Finite Set|finite]] [[D... | Let $\struct {\FF, \subseteq}$ be the [[Definition:Set|set]] of [[Definition:Finite Subset|finite subsets]] of $I$, ordered by [[Definition:Set Inclusion|inclusion]].
From [[Finite Subsets form Directed Set]], $\struct {\FF, \subseteq}$ is [[Definition:Directed Set|directed]].
For brevity write:
:$\ds s_X = \sum_{i \... | Generalized Sum is Monotone | https://proofwiki.org/wiki/Generalized_Sum_is_Monotone | https://proofwiki.org/wiki/Generalized_Sum_is_Monotone | [
"Generalized Sums"
] | [
"Definition:Indexing Set",
"Definition:Indexing Set/Family",
"Definition:Positive/Real Number",
"Definition:Finite Set",
"Definition:Subset",
"Definition:Generalized Sum",
"Definition:Generalized Sum/Net Convergence"
] | [
"Definition:Set",
"Definition:Finite Subset",
"Definition:Subset",
"Finite Subsets form Directed Set",
"Definition:Directed Preordering",
"Definition:Finite Set",
"Definition:Subset",
"Definition:Convergent Net",
"Definition:Finite Set",
"Definition:Finite Set",
"Category:Generalized Sums"
] |
proofwiki-4848 | Convergent Generalized Sum of Positive Reals has Countably Many Non-Zero Terms | Let $\family {a_i}_{i \mathop \in I}$ be an $I$-indexed family of positive real numbers.
That is, let $a_i \in \R_{\ge 0}$ for all $i \in I$.
Suppose that $\ds \sum \set {a_i: i \in I}$ converges.
Then the set $I_{>0} := \set {i \in I: a_i > 0}$ is countable. | For $\lambda \in \R$, let $I_{>\lambda} := \set {i \in I: a_i > \lambda}$.
Then as $\ds \sum \set {a_i: i \mathop \in I}$ converges, necessarily all of the sets $I_{> \frac 1 n}$ are finite.
However, we have that $\ds I_{>0} = \bigcup_{n \mathop = 1}^\infty I_{> \frac 1 n}$.
From Countable Union of Countable Sets is Co... | Let $\family {a_i}_{i \mathop \in I}$ be an [[Definition:Indexed Set|$I$-indexed family]] of [[Definition:Positive Real Number|positive real numbers]].
That is, let $a_i \in \R_{\ge 0}$ for all $i \in I$.
Suppose that $\ds \sum \set {a_i: i \in I}$ converges.
Then the set $I_{>0} := \set {i \in I: a_i > 0}$ is [[De... | For $\lambda \in \R$, let $I_{>\lambda} := \set {i \in I: a_i > \lambda}$.
Then as $\ds \sum \set {a_i: i \mathop \in I}$ converges, necessarily all of the sets $I_{> \frac 1 n}$ are [[Definition:Finite Set|finite]].
However, we have that $\ds I_{>0} = \bigcup_{n \mathop = 1}^\infty I_{> \frac 1 n}$.
From [[Countab... | Convergent Generalized Sum of Positive Reals has Countably Many Non-Zero Terms | https://proofwiki.org/wiki/Convergent_Generalized_Sum_of_Positive_Reals_has_Countably_Many_Non-Zero_Terms | https://proofwiki.org/wiki/Convergent_Generalized_Sum_of_Positive_Reals_has_Countably_Many_Non-Zero_Terms | [
"Generalized Sums",
"Real Analysis"
] | [
"Definition:Indexing Set/Indexed Set",
"Definition:Positive/Real Number",
"Definition:Countable Set"
] | [
"Definition:Finite Set",
"Countable Union of Countable Sets is Countable",
"Definition:Countable Set",
"Category:Generalized Sums",
"Category:Real Analysis"
] |
proofwiki-4849 | Mappings Partially Ordered by Extension | Let $S$ and $T$ be sets.
Let $F$ be the set of all mappings from $S$ to $T$.
Let $\RR \subseteq F \times F$ be the relation defined as:
:$\tuple {f, g} \in \RR \iff \Dom f \subseteq \Dom g \land \forall x \in \Dom f: \map f x = \map g x$
That is, $f \mathrel \RR g$ {{iff}} $g$ is an extension of $f$.
Then $\RR$ is an o... | Let $x \in \Dom f$ such that $\map f x = y$.
Then by definition $x \in \Dom g$ and $\map g x = y$.
Thus by definition of subset, $f \subseteq g$.
We have that Subset Relation is Ordering.
Hence the result.
{{qed}} | Let $S$ and $T$ be [[Definition:Set|sets]].
Let $F$ be the set of all [[Definition:Mapping|mappings]] from $S$ to $T$.
Let $\RR \subseteq F \times F$ be the [[Definition:Relation|relation]] defined as:
:$\tuple {f, g} \in \RR \iff \Dom f \subseteq \Dom g \land \forall x \in \Dom f: \map f x = \map g x$
That is, $f ... | Let $x \in \Dom f$ such that $\map f x = y$.
Then by definition $x \in \Dom g$ and $\map g x = y$.
Thus by definition of [[Definition:Subset|subset]], $f \subseteq g$.
We have that [[Subset Relation is Ordering]].
Hence the result.
{{qed}} | Mappings Partially Ordered by Extension | https://proofwiki.org/wiki/Mappings_Partially_Ordered_by_Extension | https://proofwiki.org/wiki/Mappings_Partially_Ordered_by_Extension | [
"Mapping Theory",
"Order Theory"
] | [
"Definition:Set",
"Definition:Mapping",
"Definition:Relation",
"Definition:Extension of Mapping",
"Definition:Ordering"
] | [
"Definition:Subset",
"Subset Relation is Ordering"
] |
proofwiki-4850 | Finite Subsets form Directed Set | Let $I$ be a set.
Denote with $\FF$ the set of finite subsets of $I$.
Let $\subseteq$ be the subset relation on $\FF$.
Then $\struct {\FF, \subseteq}$ is a directed set. | A fortiori from Finite Subsets form Directed Ordering.
{{qed}}
Category:Preorder Theory
Category:Subset Relation
3fty8n41x4782vos1tgeqauwf8cdlw6 | Let $I$ be a [[Definition:Set|set]].
Denote with $\FF$ the set of [[Definition:Finite Subset|finite subsets]] of $I$.
Let $\subseteq$ be the [[Subset Relation is Ordering|subset relation]] on $\FF$.
Then $\struct {\FF, \subseteq}$ is a [[Definition:Directed Set|directed set]]. | [[Definition:A Fortiori|A fortiori]] from [[Finite Subsets form Directed Ordering]].
{{qed}}
[[Category:Preorder Theory]]
[[Category:Subset Relation]]
3fty8n41x4782vos1tgeqauwf8cdlw6 | Finite Subsets form Directed Set | https://proofwiki.org/wiki/Finite_Subsets_form_Directed_Set | https://proofwiki.org/wiki/Finite_Subsets_form_Directed_Set | [
"Preorder Theory",
"Subset Relation"
] | [
"Definition:Set",
"Definition:Finite Subset",
"Subset Relation is Ordering",
"Definition:Directed Preordering"
] | [
"Definition:A Fortiori",
"Finite Subsets form Directed Ordering",
"Category:Preorder Theory",
"Category:Subset Relation"
] |
proofwiki-4851 | Strictly Precedes is Strict Ordering | Let $\struct {S, \preceq}$ be an ordered set.
Let $\prec$ be the relation on $S$ defined as:
:$a \prec b \iff \paren {a \ne b} \land \paren {a \preceq b}$
That is, $a \prec b$ {{iff}} $a$ strictly precedes $b$.
Then:
:$a \preceq b \iff \paren {a = b} \lor \paren {a \prec b}$
and $\prec$ is a strict ordering on $S$. | We are given that $\struct {S, \preceq}$ is an ordered set. | Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
Let $\prec$ be the [[Definition:Relation|relation]] on $S$ defined as:
:$a \prec b \iff \paren {a \ne b} \land \paren {a \preceq b}$
That is, $a \prec b$ {{iff}} $a$ [[Definition:Strictly Precede|strictly precedes]] $b$.
Then:
:$a \preceq b \i... | We are given that $\struct {S, \preceq}$ is an [[Definition:Ordered Set|ordered set]]. | Strictly Precedes is Strict Ordering | https://proofwiki.org/wiki/Strictly_Precedes_is_Strict_Ordering | https://proofwiki.org/wiki/Strictly_Precedes_is_Strict_Ordering | [
"Strict Orderings"
] | [
"Definition:Ordered Set",
"Definition:Relation",
"Definition:Strictly Precede",
"Definition:Strict Ordering"
] | [
"Definition:Ordered Set"
] |
proofwiki-4852 | Smallest Element is Unique | Let $\struct {S, \preceq}$ be an ordered set.
If $S$ has a smallest element, then it can have only one.
That is, if $a$ and $b$ are both smallest elements of $S$, then $a = b$. | Let $a$ and $b$ both be smallest elements of $S$.
Then by definition:
:$\forall y \in S: a \preceq y$
:$\forall y \in S: b \preceq y$
Thus it follows that:
:$a \preceq b$
:$b \preceq a$
But as $\preceq$ is an ordering, it is antisymmetric.
Hence by definition of antisymmetric, $a = b$.
{{qed}} | Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
If $S$ has a [[Definition:Smallest Element|smallest element]], then it can have only one.
That is, if $a$ and $b$ are both [[Definition:Smallest Element|smallest elements]] of $S$, then $a = b$. | Let $a$ and $b$ both be [[Definition:Smallest Element|smallest elements]] of $S$.
Then by definition:
:$\forall y \in S: a \preceq y$
:$\forall y \in S: b \preceq y$
Thus it follows that:
:$a \preceq b$
:$b \preceq a$
But as $\preceq$ is an [[Definition:Ordering|ordering]], it is [[Definition:Antisymmetric Relation|... | Smallest Element is Unique | https://proofwiki.org/wiki/Smallest_Element_is_Unique | https://proofwiki.org/wiki/Smallest_Element_is_Unique | [
"Smallest Elements",
"Smallest Element is Unique"
] | [
"Definition:Ordered Set",
"Definition:Smallest Element",
"Definition:Smallest Element"
] | [
"Definition:Smallest Element",
"Definition:Ordering",
"Definition:Antisymmetric Relation",
"Definition:Antisymmetric Relation"
] |
proofwiki-4853 | Greatest Element is Unique | Let $\struct {S, \preceq}$ be a ordered set.
If $S$ has a greatest element, then it can have only one.
That is, if $a$ and $b$ are both greatest elements of $S$, then $a = b$. | Let $a$ and $b$ both be greatest elements of $S$.
Then by definition:
:$\forall y \in S: y \preceq a$
:$\forall y \in S: y \preceq b$
Thus it follows that:
:$b \preceq a$
:$a \preceq b$
But as $\preceq$ is an ordering, it is antisymmetric.
Hence, by definition of antisymmetric relation, $a = b$.
{{qed}} | Let $\struct {S, \preceq}$ be a [[Definition:Ordered Set|ordered set]].
If $S$ has a [[Definition:Greatest Element|greatest element]], then it can have only one.
That is, if $a$ and $b$ are both [[Definition:Greatest Element|greatest elements]] of $S$, then $a = b$. | Let $a$ and $b$ both be [[Definition:Greatest Element|greatest elements]] of $S$.
Then by definition:
:$\forall y \in S: y \preceq a$
:$\forall y \in S: y \preceq b$
Thus it follows that:
:$b \preceq a$
:$a \preceq b$
But as $\preceq$ is an [[Definition:Ordering|ordering]], it is [[Definition:Antisymmetric Relation|... | Greatest Element is Unique | https://proofwiki.org/wiki/Greatest_Element_is_Unique | https://proofwiki.org/wiki/Greatest_Element_is_Unique | [
"Greatest Elements",
"Greatest Element is Unique"
] | [
"Definition:Ordered Set",
"Definition:Greatest Element",
"Definition:Greatest Element"
] | [
"Definition:Greatest Element",
"Definition:Ordering",
"Definition:Antisymmetric Relation",
"Definition:Antisymmetric Relation"
] |
proofwiki-4854 | Smallest Element is Lower Bound | Let $\struct {S, \preceq}$ be an ordered set.
Let $T \subseteq S$.
Let $T$ have a smallest element $m \in T$.
Then $m$ is a lower bound of $T$.
It follows by definition that $T$ is bounded below. | Let $m \in T$ be a smallest element of $T$.
By definition:
:$\forall y \in T: m \preceq y$
But as $T \subseteq S$, it follows that $m \in S$.
Hence:
:$\exists m \in S: \forall y \in T: m \preceq y$
Thus $T$ is bounded below by the lower bound $m$.
{{qed}} | Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
Let $T \subseteq S$.
Let $T$ have a [[Definition:Smallest Element|smallest element]] $m \in T$.
Then $m$ is a [[Definition:Lower Bound of Set|lower bound]] of $T$.
It follows by definition that $T$ is [[Definition:Bounded Below Set|bounded b... | Let $m \in T$ be a [[Definition:Smallest Element|smallest element]] of $T$.
By definition:
:$\forall y \in T: m \preceq y$
But as $T \subseteq S$, it follows that $m \in S$.
Hence:
:$\exists m \in S: \forall y \in T: m \preceq y$
Thus $T$ is [[Definition:Bounded Below Set|bounded below]] by the [[Definition:Lower B... | Smallest Element is Lower Bound | https://proofwiki.org/wiki/Smallest_Element_is_Lower_Bound | https://proofwiki.org/wiki/Smallest_Element_is_Lower_Bound | [
"Smallest Elements",
"Boundedness"
] | [
"Definition:Ordered Set",
"Definition:Smallest Element",
"Definition:Lower Bound of Set",
"Definition:Bounded Below Set"
] | [
"Definition:Smallest Element",
"Definition:Bounded Below Set",
"Definition:Lower Bound of Set"
] |
proofwiki-4855 | Greatest Element is Upper Bound | Let $\struct {S, \preceq}$ be an ordered set.
Let $T \subseteq S$.
Let $T$ have a greatest element $M \in T$.
Then $M$ is an upper bound of $T$.
It follows by definition that $T$ is bounded above. | Let $M \in T$ be a greatest element of $T$.
By definition:
:$\forall y \in T: y \preceq M$
But as $T \subseteq S$, it follows that $M \in S$.
Hence:
:$\exists M \in S: \forall y \in T: y \preceq M$
Thus $T$ is bounded above by the upper bound $M$.
{{qed}} | Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
Let $T \subseteq S$.
Let $T$ have a [[Definition:Greatest Element|greatest element]] $M \in T$.
Then $M$ is an [[Definition:Upper Bound of Set|upper bound]] of $T$.
It follows by definition that $T$ is [[Definition:Bounded Above Set|bounded ... | Let $M \in T$ be a [[Definition:Greatest Element|greatest element]] of $T$.
By definition:
:$\forall y \in T: y \preceq M$
But as $T \subseteq S$, it follows that $M \in S$.
Hence:
:$\exists M \in S: \forall y \in T: y \preceq M$
Thus $T$ is [[Definition:Bounded Above Set|bounded above]] by the [[Definition:Upper B... | Greatest Element is Upper Bound | https://proofwiki.org/wiki/Greatest_Element_is_Upper_Bound | https://proofwiki.org/wiki/Greatest_Element_is_Upper_Bound | [
"Boundedness",
"Greatest Elements"
] | [
"Definition:Ordered Set",
"Definition:Greatest Element",
"Definition:Upper Bound of Set",
"Definition:Bounded Above Set"
] | [
"Definition:Greatest Element",
"Definition:Bounded Above Set",
"Definition:Upper Bound of Set"
] |
proofwiki-4856 | Intersection of Subset with Lower Bounds | Let $\struct {S, \preceq}$ be an ordered set.
Let $T \subseteq S$.
Let $T_*$ be the set of all lower bounds of $T$ in $S$.
Then $T_* \cap T \ne \O$ {{iff}}:
:$T$ has a smallest element $m$
and
:$T_* \cap T$ is a singleton such that $T_* \cap T = \set m$ | Suppose $T_* \cap T = \O$, where $\O$ denotes the empty set.
That means $T$ contains none of its lower bounds, if indeed it has any.
From Smallest Element is Lower Bound, if $T$ had a smallest element, it would be a lower bound contained in $T$.
It follows that $T$ can have no smallest element.
Otherwise $T_* \cap T \n... | Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
Let $T \subseteq S$.
Let $T_*$ be the [[Definition:Set|set]] of all [[Definition:Lower Bound of Set|lower bounds]] of $T$ in $S$.
Then $T_* \cap T \ne \O$ {{iff}}:
:$T$ has a [[Definition:Smallest Element|smallest element]] $m$
and
:$T_* \cap ... | Suppose $T_* \cap T = \O$, where $\O$ denotes the [[Definition:Empty Set|empty set]].
That means $T$ contains none of its [[Definition:Lower Bound of Set|lower bounds]], if indeed it has any.
From [[Smallest Element is Lower Bound]], if $T$ had a [[Definition:Smallest Element|smallest element]], it would be a [[Defin... | Intersection of Subset with Lower Bounds | https://proofwiki.org/wiki/Intersection_of_Subset_with_Lower_Bounds | https://proofwiki.org/wiki/Intersection_of_Subset_with_Lower_Bounds | [
"Order Theory"
] | [
"Definition:Ordered Set",
"Definition:Set",
"Definition:Lower Bound of Set",
"Definition:Smallest Element",
"Definition:Singleton"
] | [
"Definition:Empty Set",
"Definition:Lower Bound of Set",
"Smallest Element is Lower Bound",
"Definition:Smallest Element",
"Definition:Lower Bound of Set",
"Definition:Smallest Element",
"Definition:Lower Bound of Set",
"Intersection is Subset",
"Definition:Smallest Element",
"Smallest Element is ... |
proofwiki-4857 | Intersection of Subset with Upper Bounds | Let $\struct {S, \preceq}$ be an ordered set.
Let $T \subseteq S$.
Let $T^*$ be the set of all upper bounds of $T$ in $S$.
Then $T^* \cap T \ne \O$ {{iff}}:
:$T$ has a greatest element $M$
and
:$T^* \cap T$ is a singleton such that $T^* \cap T = \set M$ | Suppose $T^* \cap T = \O$, where $\O$ denotes the empty set.
That means $T$ contains none of its upper bounds, if indeed it has any.
From Greatest Element is Upper Bound, if $T$ had a greatest element, it would be an upper bound contained in $T$.
It follows that $T$ can have no greatest element.
Otherwise $T^* \cap T \... | Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
Let $T \subseteq S$.
Let $T^*$ be the [[Definition:Set|set]] of all [[Definition:Upper Bound of Set|upper bounds]] of $T$ in $S$.
Then $T^* \cap T \ne \O$ {{iff}}:
:$T$ has a [[Definition:Greatest Element|greatest element]] $M$
and
:$T^* \cap ... | Suppose $T^* \cap T = \O$, where $\O$ denotes the [[Definition:Empty Set|empty set]].
That means $T$ contains none of its [[Definition:Upper Bound of Set|upper bounds]], if indeed it has any.
From [[Greatest Element is Upper Bound]], if $T$ had a [[Definition:Greatest Element|greatest element]], it would be an [[Defi... | Intersection of Subset with Upper Bounds | https://proofwiki.org/wiki/Intersection_of_Subset_with_Upper_Bounds | https://proofwiki.org/wiki/Intersection_of_Subset_with_Upper_Bounds | [
"Order Theory"
] | [
"Definition:Ordered Set",
"Definition:Set",
"Definition:Upper Bound of Set",
"Definition:Greatest Element",
"Definition:Singleton"
] | [
"Definition:Empty Set",
"Definition:Upper Bound of Set",
"Greatest Element is Upper Bound",
"Definition:Greatest Element",
"Definition:Upper Bound of Set",
"Definition:Greatest Element",
"Definition:Upper Bound of Set",
"Intersection is Subset",
"Definition:Greatest Element",
"Greatest Element is ... |
proofwiki-4858 | Nesbitt's Inequality | Let $a$, $b$ and $c$ be (strictly) positive real numbers.
Then:
:$\dfrac a {b + c} + \dfrac b {a + c} + \dfrac c {a + b} \ge \dfrac 3 2$ | {{begin-eqn}}
{{eqn | l = \frac a {b + c} + \frac b {a + c} + \frac c {a + b}
| o = \ge
| r = \dfrac 3 2
}}
{{eqn | ll= \leadstoandfrom
| l = \frac {a + b + c} {b + c} + \frac {a + b + c} {a + c} + \frac {a + b + c} {a + b}
| o = \ge
| r = \frac 9 2
| c = by adding $3$
}}
{{eqn | ll=... | Let $a$, $b$ and $c$ be [[Definition:Strictly Positive Real Number|(strictly) positive real numbers]].
Then:
:$\dfrac a {b + c} + \dfrac b {a + c} + \dfrac c {a + b} \ge \dfrac 3 2$ | {{begin-eqn}}
{{eqn | l = \frac a {b + c} + \frac b {a + c} + \frac c {a + b}
| o = \ge
| r = \dfrac 3 2
}}
{{eqn | ll= \leadstoandfrom
| l = \frac {a + b + c} {b + c} + \frac {a + b + c} {a + c} + \frac {a + b + c} {a + b}
| o = \ge
| r = \frac 9 2
| c = by adding $3$
}}
{{eqn | ll=... | Nesbitt's Inequality/Proof 1 | https://proofwiki.org/wiki/Nesbitt's_Inequality | https://proofwiki.org/wiki/Nesbitt's_Inequality/Proof_1 | [
"Nesbitt's Inequality",
"Algebra",
"Inequalities"
] | [
"Definition:Strictly Positive/Real Number"
] | [
"Definition:Arithmetic Mean",
"Definition:Harmonic Mean",
"AM-HM Inequality",
"Definition:Inequality",
"Nesbitt's Inequality"
] |
proofwiki-4859 | Nesbitt's Inequality | Let $a$, $b$ and $c$ be (strictly) positive real numbers.
Then:
:$\dfrac a {b + c} + \dfrac b {a + c} + \dfrac c {a + b} \ge \dfrac 3 2$ | Let:
{{begin-eqn}}
{{eqn | l = S
| r = \frac a {b + c} + \frac b {c + a} + \frac c {a + b}
}}
{{eqn | l = M
| r = \frac b {b + c} + \frac c {c + a} + \frac a {a + b}
}}
{{eqn | l = N
| r = \frac c {b + c} + \frac a {c + a} + \frac b {a + b}
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | n = 1
| l = ... | Let $a$, $b$ and $c$ be [[Definition:Strictly Positive Real Number|(strictly) positive real numbers]].
Then:
:$\dfrac a {b + c} + \dfrac b {a + c} + \dfrac c {a + b} \ge \dfrac 3 2$ | Let:
{{begin-eqn}}
{{eqn | l = S
| r = \frac a {b + c} + \frac b {c + a} + \frac c {a + b}
}}
{{eqn | l = M
| r = \frac b {b + c} + \frac c {c + a} + \frac a {a + b}
}}
{{eqn | l = N
| r = \frac c {b + c} + \frac a {c + a} + \frac b {a + b}
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | n = 1
| l ... | Nesbitt's Inequality/Proof 2 | https://proofwiki.org/wiki/Nesbitt's_Inequality | https://proofwiki.org/wiki/Nesbitt's_Inequality/Proof_2 | [
"Nesbitt's Inequality",
"Algebra",
"Inequalities"
] | [
"Definition:Strictly Positive/Real Number"
] | [
"Cauchy's Mean Theorem",
"Cauchy's Mean Theorem"
] |
proofwiki-4860 | Greatest Element is Maximal | Let $\struct {S, \preceq}$ be an ordered set which has a greatest element.
Let $M$ be the greatest element of $\struct {S, \preceq}$.
Then $M$ is a maximal element. | By definition of greatest element:
:$\forall y \in S: y \preceq M$
Suppose $M \preceq y$.
As $\preceq$ is an ordering, $\preceq$ is by definition antisymmetric.
Thus it follows by definition of antisymmetry that $M = y$.
That is:
:$M \preceq y \implies M = y$
which is precisely the definition of a maximal element.
{{qe... | Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]] which has a [[Definition:Greatest Element|greatest element]].
Let $M$ be the [[Definition:Greatest Element|greatest element]] of $\struct {S, \preceq}$.
Then $M$ is a [[Definition:Maximal Element|maximal element]]. | By definition of [[Definition:Greatest Element|greatest element]]:
:$\forall y \in S: y \preceq M$
Suppose $M \preceq y$.
As $\preceq$ is an [[Definition:Ordering|ordering]], $\preceq$ is by definition [[Definition:Antisymmetric Relation|antisymmetric]].
Thus it follows by definition of [[Definition:Antisymmetric Re... | Greatest Element is Maximal | https://proofwiki.org/wiki/Greatest_Element_is_Maximal | https://proofwiki.org/wiki/Greatest_Element_is_Maximal | [
"Greatest Elements",
"Maximal Elements"
] | [
"Definition:Ordered Set",
"Definition:Greatest Element",
"Definition:Greatest Element",
"Definition:Maximal/Element"
] | [
"Definition:Greatest Element",
"Definition:Ordering",
"Definition:Antisymmetric Relation",
"Definition:Antisymmetric Relation",
"Definition:Maximal/Element"
] |
proofwiki-4861 | Smallest Element is Minimal | Let $\struct {S, \preceq}$ be an ordered set which has a smallest element.
Let $m$ be the smallest element of $\struct {S, \preceq}$.
Then $m$ is a minimal element. | By definition of smallest element:
:$\forall y \in S: m \preceq y$
Suppose $y \preceq m$.
As $\preceq$ is an ordering, $\preceq$ is by definition antisymmetric.
Thus it follows by definition of antisymmetry that $m = y$.
That is:
:$y \preceq m \implies m = y$
which is precisely the definition of a minimal element.
{{q... | Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]] which has a [[Definition:Smallest Element|smallest element]].
Let $m$ be the [[Definition:Smallest Element|smallest element]] of $\struct {S, \preceq}$.
Then $m$ is a [[Definition:Minimal Element|minimal element]]. | By definition of [[Definition:Smallest Element|smallest element]]:
:$\forall y \in S: m \preceq y$
Suppose $y \preceq m$.
As $\preceq$ is an [[Definition:Ordering|ordering]], $\preceq$ is by definition [[Definition:Antisymmetric Relation|antisymmetric]].
Thus it follows by definition of [[Definition:Antisymmetric R... | Smallest Element is Minimal | https://proofwiki.org/wiki/Smallest_Element_is_Minimal | https://proofwiki.org/wiki/Smallest_Element_is_Minimal | [
"Smallest Elements",
"Minimal Elements"
] | [
"Definition:Ordered Set",
"Definition:Smallest Element",
"Definition:Smallest Element",
"Definition:Minimal/Element"
] | [
"Definition:Smallest Element",
"Definition:Ordering",
"Definition:Antisymmetric Relation",
"Definition:Antisymmetric Relation",
"Definition:Minimal/Element"
] |
proofwiki-4862 | Minimal Element in Toset is Unique and Smallest | Let $\struct {S, \preceq}$ be a totally ordered set.
Let $m$ be a minimal element of $\struct {S, \preceq}$.
Then:
:$(1): \quad m$ is the smallest element of $\struct {S, \preceq}$
:$(2): \quad m$ is the only minimal element of $\struct {S, \preceq}$. | By definition of minimal element:
:$\forall y \in S: y \preceq m \implies m = y$
As $\struct {S, \preceq}$ is a totally ordered set, by definition $\preceq$ is connected.
That is:
:$\forall x, y \in S: y \preceq x \lor x \preceq y$
It follows that:
:$\forall y \in S: m = y \lor m \preceq y$
But as $m = y \implies m \pr... | Let $\struct {S, \preceq}$ be a [[Definition:Totally Ordered Set|totally ordered set]].
Let $m$ be a [[Definition:Minimal Element|minimal element]] of $\struct {S, \preceq}$.
Then:
:$(1): \quad m$ is the [[Definition:Smallest Element|smallest element]] of $\struct {S, \preceq}$
:$(2): \quad m$ is the only [[Definit... | By definition of [[Definition:Minimal Element|minimal element]]:
:$\forall y \in S: y \preceq m \implies m = y$
As $\struct {S, \preceq}$ is a [[Definition:Totally Ordered Set|totally ordered set]], by definition $\preceq$ is [[Definition:Connected Relation|connected]].
That is:
:$\forall x, y \in S: y \preceq x \lor... | Minimal Element in Toset is Unique and Smallest | https://proofwiki.org/wiki/Minimal_Element_in_Toset_is_Unique_and_Smallest | https://proofwiki.org/wiki/Minimal_Element_in_Toset_is_Unique_and_Smallest | [
"Total Orderings",
"Minimal Elements",
"Smallest Elements"
] | [
"Definition:Totally Ordered Set",
"Definition:Minimal/Element",
"Definition:Smallest Element",
"Definition:Minimal/Element"
] | [
"Definition:Minimal/Element",
"Definition:Totally Ordered Set",
"Definition:Connected Relation",
"Definition:Smallest Element",
"Definition:Minimal/Element",
"Definition:Smallest Element",
"Smallest Element is Unique"
] |
proofwiki-4863 | Maximal Element in Toset is Unique and Greatest | Let $\struct {S, \preceq}$ be a totally ordered set.
Let $M$ be a maximal element of $\struct {S, \preceq}$.
Then:
:$(1): \quad M$ is the greatest element of $\struct {S, \preceq}$.
:$(2): \quad M$ is the only maximal element of $\struct {S, \preceq}$. | By definition of maximal element:
:$\forall y \in S: M \preceq y \implies M = y$
As $\struct {S, \preceq}$ is a totally ordered set, by definition $\preceq$ is a connected.
That is:
:$\forall x, y \in S: y \preceq x \lor x \preceq y$
It follows that:
:$\forall y \in S: M = y \lor y \preceq M$
But as $M = y \implies y \... | Let $\struct {S, \preceq}$ be a [[Definition:Totally Ordered Set|totally ordered set]].
Let $M$ be a [[Definition:Maximal Element|maximal element]] of $\struct {S, \preceq}$.
Then:
:$(1): \quad M$ is the [[Definition:Greatest Element|greatest element]] of $\struct {S, \preceq}$.
:$(2): \quad M$ is the only [[Definit... | By definition of [[Definition:Maximal Element|maximal element]]:
:$\forall y \in S: M \preceq y \implies M = y$
As $\struct {S, \preceq}$ is a [[Definition:Totally Ordered Set|totally ordered set]], by definition $\preceq$ is a [[Definition:Connected Relation|connected]].
That is:
:$\forall x, y \in S: y \preceq x \l... | Maximal Element in Toset is Unique and Greatest | https://proofwiki.org/wiki/Maximal_Element_in_Toset_is_Unique_and_Greatest | https://proofwiki.org/wiki/Maximal_Element_in_Toset_is_Unique_and_Greatest | [
"Total Orderings",
"Maximal Elements",
"Greatest Elements"
] | [
"Definition:Totally Ordered Set",
"Definition:Maximal/Element",
"Definition:Greatest Element",
"Definition:Maximal/Element"
] | [
"Definition:Maximal/Element",
"Definition:Totally Ordered Set",
"Definition:Connected Relation",
"Definition:Greatest Element",
"Definition:Maximal/Element",
"Definition:Greatest Element",
"Greatest Element is Unique"
] |
proofwiki-4864 | Whitney Embedding Theorem | Every smooth $m$-dimensional manifold admits a smooth embedding into Euclidean space $\R^{2 m}$. | {{ProofWanted}}
{{Namedfor|Hassler Whitney|cat = Whitney, H.}} | Every [[Definition:Smooth Manifold|smooth $m$-dimensional manifold]] admits a [[Definition:Smooth Embedding|smooth embedding]] into [[Definition:Euclidean Space|Euclidean space $\R^{2 m}$]]. | {{ProofWanted}}
{{Namedfor|Hassler Whitney|cat = Whitney, H.}} | Whitney Embedding Theorem | https://proofwiki.org/wiki/Whitney_Embedding_Theorem | https://proofwiki.org/wiki/Whitney_Embedding_Theorem | [
"Smooth Manifolds",
"Embeddings (Topology)"
] | [
"Definition:Topological Manifold/Smooth Manifold",
"Definition:Smooth Embedding",
"Definition:Euclidean Space"
] | [] |
proofwiki-4865 | Max Semigroup is Idempotent | Let $\struct {S, \preceq}$ be a totally ordered set.
Then the semigroup $\struct {S, \max}$ is an idempotent semigroup. | The fact that $\struct {S, \max}$ is a semigroup is demonstrated in Max Operation on Toset forms Semigroup.
Then the max operation is idempotent:
:$\forall x \in S: \max \set {x, x} = x$
The result follows by the definition of idempotent semigroup.
{{qed}} | Let $\struct {S, \preceq}$ be a [[Definition:Totally Ordered Set|totally ordered set]].
Then the [[Definition:Semigroup|semigroup]] $\struct {S, \max}$ is an [[Definition:Idempotent Semigroup|idempotent semigroup]]. | The fact that $\struct {S, \max}$ is a [[Definition:Semigroup|semigroup]] is demonstrated in [[Max Operation on Toset forms Semigroup]].
Then the [[Max and Min are Idempotent|max operation is idempotent]]:
:$\forall x \in S: \max \set {x, x} = x$
The result follows by the definition of [[Definition:Idempotent Semigro... | Max Semigroup is Idempotent | https://proofwiki.org/wiki/Max_Semigroup_is_Idempotent | https://proofwiki.org/wiki/Max_Semigroup_is_Idempotent | [
"Max Operation",
"Examples of Idempotent Semigroups"
] | [
"Definition:Totally Ordered Set",
"Definition:Semigroup",
"Definition:Idempotent Semigroup"
] | [
"Definition:Semigroup",
"Max Operation on Toset forms Semigroup",
"Max and Min are Idempotent",
"Definition:Idempotent Semigroup"
] |
proofwiki-4866 | Continuity of Linear Transformations | Let $H, K$ be Hilbert spaces, and let $A: H \to K$ be a linear transformation.
{{TFAE}}
{{begin-itemize}}
{{item|(1):|$A$ is continuous}}
{{item|(2):|$A$ is continuous at $\mathbf 0_H$}}
{{item|(3):|$A$ is continuous at some point}}
{{item|(4):|$\exists c > 0: \forall h \in H: \norm {\map A h}_K \le c \norm h_H$}}
{{en... | It is clear that $(1) \implies (2) \implies (3)$.
For $(4) \implies (2)$:
For any $\epsilon > 0$, there exists $\delta = \dfrac \epsilon c$, such that when $\norm {\mathbf 0_H - h}_H < \delta$:
:$\norm {\map A h - \map A {\mathbf 0_H} }_K \le c \norm h_H < c\delta = \epsilon$
Now we prove $(3) \implies (1)$:
Let $A$ be... | Let $H, K$ be [[Definition:Hilbert Space|Hilbert spaces]], and let $A: H \to K$ be a [[Definition:Linear Transformation|linear transformation]].
{{TFAE}}
{{begin-itemize}}
{{item|(1):|$A$ is [[Definition:Everywhere Continuous Mapping (Topology)|continuous]]}}
{{item|(2):|$A$ is [[Definition:Continuous at Point of Top... | It is clear that $(1) \implies (2) \implies (3)$.
For $(4) \implies (2)$:
For any $\epsilon > 0$, there exists $\delta = \dfrac \epsilon c$, such that when $\norm {\mathbf 0_H - h}_H < \delta$:
:$\norm {\map A h - \map A {\mathbf 0_H} }_K \le c \norm h_H < c\delta = \epsilon$
Now we prove $(3) \implies (1)$:
Let... | Continuity of Linear Transformations | https://proofwiki.org/wiki/Continuity_of_Linear_Transformations | https://proofwiki.org/wiki/Continuity_of_Linear_Transformations | [
"Continuous Linear Transformations",
"Linear Transformations on Hilbert Spaces"
] | [
"Definition:Hilbert Space",
"Definition:Linear Transformation",
"Definition:Continuous Mapping (Topology)/Everywhere",
"Definition:Continuous Mapping (Topology)/Point",
"Definition:Continuous Mapping (Topology)/Point"
] | [
"Definition:Continuous Mapping (Topology)/Point",
"Definition:Continuous Mapping (Topology)/Everywhere",
"Definition:Continuous Mapping (Topology)/Point",
"Definition:Open Ball",
"Definition:Strictly Positive/Real Number",
"Definition:Open Ball/Radius",
"Definition:Open Ball/Center",
"Definition:Image... |
proofwiki-4867 | Equivalence of Definitions of Norm of Linear Transformation | Let $H, K$ be Hilbert spaces.
Let $A: H \to K$ be a bounded linear transformation.
{{TFAE|def = Norm on Bounded Linear Transformation}} | Let:
{{begin-eqn}}
{{eqn | l = \lambda_1
| r = \sup \set {\norm {A h}_K: \norm h_H \le 1}
}}
{{eqn | l = \lambda_2
| r = \sup \set {\dfrac {\norm {A h}_K} {\norm h_H}: h \in H, h \ne 0_H}
}}
{{eqn | l = \lambda_3
| r = \sup \set {\norm {A h}_K: \norm h_H = 1}
}}
{{eqn | l = \lambda_4
| r = \inf ... | Let $H, K$ be [[Definition:Hilbert Space|Hilbert spaces]].
Let $A: H \to K$ be a [[Definition:Bounded Linear Transformation|bounded linear transformation]].
{{TFAE|def = Norm on Bounded Linear Transformation}} | Let:
{{begin-eqn}}
{{eqn | l = \lambda_1
| r = \sup \set {\norm {A h}_K: \norm h_H \le 1}
}}
{{eqn | l = \lambda_2
| r = \sup \set {\dfrac {\norm {A h}_K} {\norm h_H}: h \in H, h \ne 0_H}
}}
{{eqn | l = \lambda_3
| r = \sup \set {\norm {A h}_K: \norm h_H = 1}
}}
{{eqn | l = \lambda_4
| r = \inf... | Equivalence of Definitions of Norm of Linear Transformation | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Norm_of_Linear_Transformation | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Norm_of_Linear_Transformation | [
"Linear Transformations on Hilbert Spaces",
"Equivalence of Definitions of Norm of Linear Transformation"
] | [
"Definition:Hilbert Space",
"Definition:Bounded Linear Transformation"
] | [
"Norm on Bounded Linear Transformation is Finite"
] |
proofwiki-4868 | Max Semigroup on Toset forms Semilattice | Let $\struct {S, \preceq}$ be a totally ordered set.
Then the max semigroup $\struct {S, \max}$ is a semilattice. | The Max Semigroup is Commutative and idempotent.
Hence the result, by definition of a semilattice.
{{qed}} | Let $\struct {S, \preceq}$ be a [[Definition:Totally Ordered Set|totally ordered set]].
Then the [[Max Operation on Toset forms Semigroup|max semigroup]] $\struct {S, \max}$ is a [[Definition:Semilattice|semilattice]]. | The [[Max Semigroup is Commutative]] and [[Max Semigroup is Idempotent|idempotent]].
Hence the result, by definition of a [[Definition:Semilattice|semilattice]].
{{qed}} | Max Semigroup on Toset forms Semilattice | https://proofwiki.org/wiki/Max_Semigroup_on_Toset_forms_Semilattice | https://proofwiki.org/wiki/Max_Semigroup_on_Toset_forms_Semilattice | [
"Max Operation",
"Semilattices"
] | [
"Definition:Totally Ordered Set",
"Max Operation on Toset forms Semigroup",
"Definition:Semilattice"
] | [
"Max Semigroup is Commutative",
"Max Semigroup is Idempotent",
"Definition:Semilattice"
] |
proofwiki-4869 | Max Operation Yields Supremum of Parameters | Let $\struct {S, \preceq}$ be a totally ordered set.
Let $x, y \in S$.
Then:
:$\max \set {x, y} = \sup \set {x, y}$
where:
:$\max$ denotes the max operation
:$\sup$ denotes the supremum. | As $\struct {S, \preceq}$ be a totally ordered set, all elements of $S$ are $\preceq$-comparable.
Therefore there are two cases to consider: | Let $\struct {S, \preceq}$ be a [[Definition:Totally Ordered Set|totally ordered set]].
Let $x, y \in S$.
Then:
:$\max \set {x, y} = \sup \set {x, y}$
where:
:$\max$ denotes the [[Definition:Max Operation|max operation]]
:$\sup$ denotes the [[Definition:Supremum of Set|supremum]]. | As $\struct {S, \preceq}$ be a [[Definition:Totally Ordered Set|totally ordered set]], all [[Definition:Element|elements]] of $S$ are [[Definition:Comparable Elements|$\preceq$-comparable]].
Therefore there are two cases to consider: | Max Operation Yields Supremum of Parameters | https://proofwiki.org/wiki/Max_Operation_Yields_Supremum_of_Parameters | https://proofwiki.org/wiki/Max_Operation_Yields_Supremum_of_Parameters | [
"Max Operation"
] | [
"Definition:Totally Ordered Set",
"Definition:Max Operation",
"Definition:Supremum of Set"
] | [
"Definition:Totally Ordered Set",
"Definition:Element",
"Definition:Comparable Elements"
] |
proofwiki-4870 | Mappings to Vector Space form Vector Space | Let $X$ be a non-empty set.
Let $V$ be a vector space over a field (or division ring) $K$.
Let $V^X$ denote the set of all mappings from $X$ to $V$.
Let $+$ denote pointwise addition on $V^X$.
Let $\circ$ denote pointwise ($K$)-scalar multiplication on $V^X$.
Then $\struct {V^X, +, \circ}_K$ is a vector space over $K$. | {{ProofWanted}}
Category:Vector Spaces
jibm3x64dds1jlq3x2ncpdml0gfa8az | Let $X$ be a [[Definition:Non-Empty Set|non-empty set]].
Let $V$ be a [[Definition:Vector Space|vector space]] over a [[Definition:Field (Abstract Algebra)|field]] (or [[Definition:Division Ring|division ring]]) $K$.
Let $V^X$ denote the [[Definition:Set of All Mappings|set of all mappings]] from $X$ to $V$.
Let $+$... | {{ProofWanted}}
[[Category:Vector Spaces]]
jibm3x64dds1jlq3x2ncpdml0gfa8az | Mappings to Vector Space form Vector Space | https://proofwiki.org/wiki/Mappings_to_Vector_Space_form_Vector_Space | https://proofwiki.org/wiki/Mappings_to_Vector_Space_form_Vector_Space | [
"Vector Spaces"
] | [
"Definition:Non-Empty Set",
"Definition:Vector Space",
"Definition:Field (Abstract Algebra)",
"Definition:Division Ring",
"Definition:Set of All Mappings",
"Definition:Pointwise Addition of Mappings",
"Definition:Pointwise Scalar Multiplication of Mappings",
"Definition:Vector Space"
] | [
"Category:Vector Spaces"
] |
proofwiki-4871 | Space of Bounded Linear Transformations is Banach Space | Let $\GF \in \set {\R, \C}$.
Let $\struct {X, \norm {\, \cdot \,}_X}$ be a normed vector spaces over $\GF$.
Let $\struct {Y, \norm {\, \cdot \,}_Y}$ be a Banach space over $\GF$.
Let $\map B {X, Y}$ be the space of bounded linear transformations from $X$ to $Y$ endowed with pointwise addition and ($\Bbb F$)-scalar m... | Let $\sequence {A_n}_{n \mathop \in \N}$ be a Cauchy sequence in $\map B {X, Y}$.
Then for each $\epsilon > 0$ there exists $N \in \N$ such that:
:$\norm {A_n - A_m}_{\map B {X, Y} } < \epsilon$
for $n, m \ge N$.
Then for each $x \in X$ with $\norm x_X \le 1$ we have:
:$\norm {A_n x - A_m x}_Y < \epsilon$
for eac... | Let $\GF \in \set {\R, \C}$.
Let $\struct {X, \norm {\, \cdot \,}_X}$ be a [[Definition:Normed Vector Space|normed vector spaces]] over $\GF$.
Let $\struct {Y, \norm {\, \cdot \,}_Y}$ be a [[Definition:Banach Space|Banach space]] over $\GF$.
Let $\map B {X, Y}$ be the [[Definition:Space of Bounded Linear Transfor... | Let $\sequence {A_n}_{n \mathop \in \N}$ be a [[Definition:Cauchy Sequence|Cauchy sequence]] in $\map B {X, Y}$.
Then for each $\epsilon > 0$ there exists $N \in \N$ such that:
:$\norm {A_n - A_m}_{\map B {X, Y} } < \epsilon$
for $n, m \ge N$.
Then for each $x \in X$ with $\norm x_X \le 1$ we have:
:$\norm {A... | Space of Bounded Linear Transformations is Banach Space | https://proofwiki.org/wiki/Space_of_Bounded_Linear_Transformations_is_Banach_Space | https://proofwiki.org/wiki/Space_of_Bounded_Linear_Transformations_is_Banach_Space | [
"Banach Spaces",
"Space of Bounded Linear Transformations"
] | [
"Definition:Normed Vector Space",
"Definition:Banach Space",
"Definition:Space of Bounded Linear Transformations",
"Definition:Pointwise Addition of Mappings",
"Definition:Pointwise Scalar Multiplication of Mappings",
"Definition:Norm/Bounded Linear Transformation",
"Definition:Banach Space"
] | [
"Definition:Cauchy Sequence",
"Definition:Cauchy Sequence",
"Definition:Cauchy Sequence",
"Definition:Banach Space",
"Definition:Bounded Linear Transformation",
"Sum Rule for Sequence in Normed Vector Space",
"Multiple Rule for Sequence in Normed Vector Space",
"Fundamental Property of Norm on Bounded... |
proofwiki-4872 | Equivalence of Formulations of Pasch's Axiom | The two forms of Pasch's Axiom in Tarski's Geometry are consistent.
That is, the expressions:
:$(1): \quad \forall a, b, c, p, q: \exists x: \mathsf B a p c \land \mathsf B b q c \implies \mathsf B p x b \land \mathsf B q x a$
and:
:$(2): \quad \forall a, b, c, p, q: \exists x: \mathsf B a p c \land \mathsf B q c b \im... | {{proof wanted|It is important that we identify which other axioms this equivalence is based on for foundational purposes (and possible generalisations)}} | The two forms of [[Axiom:Pasch's Axiom (Tarski's Axioms)|Pasch's Axiom]] in [[Definition:Tarski's Geometry|Tarski's Geometry]] are consistent.
That is, the expressions:
:$(1): \quad \forall a, b, c, p, q: \exists x: \mathsf B a p c \land \mathsf B b q c \implies \mathsf B p x b \land \mathsf B q x a$
and:
:$(2): \q... | {{proof wanted|It is important that we identify which other axioms this equivalence is based on for foundational purposes (and possible generalisations)}} | Equivalence of Formulations of Pasch's Axiom | https://proofwiki.org/wiki/Equivalence_of_Formulations_of_Pasch's_Axiom | https://proofwiki.org/wiki/Equivalence_of_Formulations_of_Pasch's_Axiom | [
"Tarski's Geometry"
] | [
"Axiom:Pasch's Axiom (Tarski's Axioms)",
"Definition:Tarski's Geometry",
"Definition:Logical Equivalence"
] | [] |
proofwiki-4873 | Square-Summable Indexed Sets Closed Under Addition | Let $\family {a_i}_{i \mathop \in I}, \family {b_i}_{i \mathop \in I}$ be $I$-indexed families of real numbers.
Let:
{{begin-eqn}}
{{eqn | l = \sum \set { {a_i}^2: i \in I}
| o = <
| r = \infty
}}
{{eqn | l = \sum \set { {b_i}^2: i \in I}
| o = <
| r = \infty
}}
{{end-eqn}}
where $\ds \sum$ deno... | First we note that:
:$\forall i \in I: {a_i}^2 > 0$ and ${b_i}^2 > 0$
This will be used throughout implicitly.
There are two cases to consider: $a_i b_i < 0$ and $a_i b_i > 0$. | Let $\family {a_i}_{i \mathop \in I}, \family {b_i}_{i \mathop \in I}$ be [[Definition:Indexed Set|$I$-indexed families]] of [[Definition:Real Number|real numbers]].
Let:
{{begin-eqn}}
{{eqn | l = \sum \set { {a_i}^2: i \in I}
| o = <
| r = \infty
}}
{{eqn | l = \sum \set { {b_i}^2: i \in I}
| o = ... | First we note that:
:$\forall i \in I: {a_i}^2 > 0$ and ${b_i}^2 > 0$
This will be used throughout implicitly.
There are two cases to consider: $a_i b_i < 0$ and $a_i b_i > 0$. | Square-Summable Indexed Sets Closed Under Addition | https://proofwiki.org/wiki/Square-Summable_Indexed_Sets_Closed_Under_Addition | https://proofwiki.org/wiki/Square-Summable_Indexed_Sets_Closed_Under_Addition | [
"Generalized Sums"
] | [
"Definition:Indexing Set/Indexed Set",
"Definition:Real Number",
"Definition:Generalized Sum"
] | [] |
proofwiki-4874 | Generalized Sum Preserves Inequality | Let $\family {a_i}_{i \mathop \in I}, \family {b_i}_{i \mathop \in I}$ be $I$-indexed families of positive real numbers.
That is, let $a_i, b_i \in \R_{\ge 0}$ for all $i \in I$.
Suppose that for all $i \in I$, $a_i \le b_i$.
Furthermore, suppose that $\ds \sum \set {b_i: i \in I}$ converges.
Then:
:$\ds \sum \set {a_i... | First, it is proven that $\ds \sum \set {a_i: i \in I}$ converges.
Then, the inequality $\ds \sum \set {a_i: i \in I} \le \sum \set {b_i: i \in I}$ is well-defined, and hence can be proven. | Let $\family {a_i}_{i \mathop \in I}, \family {b_i}_{i \mathop \in I}$ be [[Definition:Indexed Set|$I$-indexed families]] of [[Definition:Positive Real Number|positive real numbers]].
That is, let $a_i, b_i \in \R_{\ge 0}$ for all $i \in I$.
Suppose that for all $i \in I$, $a_i \le b_i$.
Furthermore, suppose that $\... | First, it is proven that $\ds \sum \set {a_i: i \in I}$ [[Definition:Generalized Sum|converges]].
Then, the inequality $\ds \sum \set {a_i: i \in I} \le \sum \set {b_i: i \in I}$ is well-defined, and hence can be proven. | Generalized Sum Preserves Inequality | https://proofwiki.org/wiki/Generalized_Sum_Preserves_Inequality | https://proofwiki.org/wiki/Generalized_Sum_Preserves_Inequality | [
"Generalized Sums"
] | [
"Definition:Indexing Set/Indexed Set",
"Definition:Positive/Real Number",
"Definition:Generalized Sum",
"Definition:Generalized Sum"
] | [
"Definition:Generalized Sum",
"Definition:Generalized Sum",
"Definition:Generalized Sum",
"Definition:Generalized Sum"
] |
proofwiki-4875 | Generalized Sum is Linear | Let $\family {z_i}_{i \mathop \in I}$ and $\family {w_i}_{i \mathop \in I}$ be $I$-indexed families of complex numbers.
That is, let $z_i, w_i \in \C$ for all $i \in I$.
Let $\ds \sum \set {z_i: i \in I}$ and $\sum \set {w_i: i \mathop \in I}$ converge to $z, w \in \C$, respectively.
Then:
:$(1): \quad \ds \sum \set {z... | === Proof of $(1)$ ===
Let $\epsilon > 0$.
To verify the convergence, it is necessary to find a finite $F \subseteq I$ such that:
:$\ds \map d {\sum_{i \mathop\in G} z_i + w_i, z + w} < \epsilon$ for all finite $G$ with $F \subseteq G \subseteq I$
Now let $F_z, F_w \subseteq I$ be finite subsets of $I$ such that:
:$\ds... | Let $\family {z_i}_{i \mathop \in I}$ and $\family {w_i}_{i \mathop \in I}$ be [[Definition:Indexed Set|$I$-indexed families]] of [[Definition:Complex Number|complex numbers]].
That is, let $z_i, w_i \in \C$ for all $i \in I$.
Let $\ds \sum \set {z_i: i \in I}$ and $\sum \set {w_i: i \mathop \in I}$ [[Definition:Gen... | === Proof of $(1)$ ===
Let $\epsilon > 0$.
To verify the [[Definition:Generalized Sum|convergence]], it is necessary to find a [[Definition:Finite Subset|finite]] $F \subseteq I$ such that:
:$\ds \map d {\sum_{i \mathop\in G} z_i + w_i, z + w} < \epsilon$ for all [[Definition:Finite Subset|finite]] $G$ with $F \subs... | Generalized Sum is Linear | https://proofwiki.org/wiki/Generalized_Sum_is_Linear | https://proofwiki.org/wiki/Generalized_Sum_is_Linear | [
"Generalized Sums",
"Complex Numbers"
] | [
"Definition:Indexing Set/Indexed Set",
"Definition:Complex Number",
"Definition:Generalized Sum",
"Definition:Generalized Sum",
"Definition:Generalized Sum"
] | [
"Definition:Generalized Sum",
"Definition:Finite Subset",
"Definition:Finite Subset",
"Definition:Finite Subset",
"Definition:Finite Subset",
"Definition:Finite Subset",
"Definition:Finite Subset",
"Definition:Norm/Division Ring",
"Definition:Generalized Sum",
"Definition:Generalized Sum",
"Defi... |
proofwiki-4876 | Min Operation Yields Infimum of Parameters | Let $\struct {S, \preceq}$ be a totally ordered set.
Let $x, y \in S$.
Then:
:$\map \min {x, y} = \map \inf {\set {x, y} }$
where:
:$\min$ denotes the min operation
:$\inf$ denotes the infimum. | As $\struct {S, \preceq}$ be a totally ordered set, all elements of $S$ are $\preceq$-comparable.
Therefore there are two cases to consider: | Let $\struct {S, \preceq}$ be a [[Definition:Totally Ordered Set|totally ordered set]].
Let $x, y \in S$.
Then:
:$\map \min {x, y} = \map \inf {\set {x, y} }$
where:
:$\min$ denotes the [[Definition:Min Operation|min operation]]
:$\inf$ denotes the [[Definition:Infimum of Set|infimum]]. | As $\struct {S, \preceq}$ be a [[Definition:Totally Ordered Set|totally ordered set]], all elements of $S$ are [[Definition:Comparable Elements|$\preceq$-comparable]].
Therefore there are two cases to consider: | Min Operation Yields Infimum of Parameters | https://proofwiki.org/wiki/Min_Operation_Yields_Infimum_of_Parameters | https://proofwiki.org/wiki/Min_Operation_Yields_Infimum_of_Parameters | [
"Min Operation"
] | [
"Definition:Totally Ordered Set",
"Definition:Min Operation",
"Definition:Infimum of Set"
] | [
"Definition:Totally Ordered Set",
"Definition:Comparable Elements"
] |
proofwiki-4877 | Convergence of Generalized Sum of Complex Numbers | Let $\family {z_j}_{j \mathop \in I}$ be an $I$-indexed family of complex numbers.
That is, let $z_j \in \C$ for all $j \in I$.
Let $\map \Re {z_j}$ and $\map \Im {z_j}$ denote the families of real and imaginary parts of the family $z_j$.
{{TFAE}}
{{begin-itemize}}
{{item|(1):|$\ds \sum_{j \mathop \in I} z_j$ converges... | === $(2)$ implies $(1)$ ===
By Generalized Sum is Linear, the stated convergences lead to:
{{begin-eqn}}
{{eqn | l = z
| r = \map \Re z + i \map \Im z
| c = {{Defof|Real Part}} and {{Defof|Imaginary Part}}
}}
{{eqn | r = \sum_{j \mathop \in I} \map \Re {z_j} + i \sum_{j \mathop \in I} \map \Im {z_j}
|... | Let $\family {z_j}_{j \mathop \in I}$ be an [[Definition:Indexed Family|$I$-indexed family]] of [[Definition:Complex Number|complex numbers]].
That is, let $z_j \in \C$ for all $j \in I$.
Let $\map \Re {z_j}$ and $\map \Im {z_j}$ denote the [[Definition:Indexed Family|families]] of [[Definition:Real Part|real]] and ... | === $(2)$ implies $(1)$ ===
By [[Generalized Sum is Linear]], the stated convergences lead to:
{{begin-eqn}}
{{eqn | l = z
| r = \map \Re z + i \map \Im z
| c = {{Defof|Real Part}} and {{Defof|Imaginary Part}}
}}
{{eqn | r = \sum_{j \mathop \in I} \map \Re {z_j} + i \sum_{j \mathop \in I} \map \Im {z_j}
... | Convergence of Generalized Sum of Complex Numbers | https://proofwiki.org/wiki/Convergence_of_Generalized_Sum_of_Complex_Numbers | https://proofwiki.org/wiki/Convergence_of_Generalized_Sum_of_Complex_Numbers | [
"Generalized Sums",
"Complex Numbers"
] | [
"Definition:Indexing Set/Family",
"Definition:Complex Number",
"Definition:Indexing Set/Family",
"Definition:Complex Number/Real Part",
"Definition:Complex Number/Imaginary Part",
"Definition:Indexing Set/Family",
"Definition:Generalized Sum",
"Definition:Generalized Sum"
] | [
"Generalized Sum is Linear",
"Generalized Sum is Linear"
] |
proofwiki-4878 | Equidistance is Independent of Betweenness | Let $\GG$ be a formal systematic treatment of geometry containing only:
:The language and axioms of first-order logic, and the disciplines preceding it
:The undefined terms of Tarski's Geometry (excluding equidistance)
:Some or all of Tarski's Axioms of Geometry.
In $\GG$, equidistance $\equiv$ is necessarily an undef... | Our assertion is that $\equiv$ cannot be defined in terms of $\mathsf B$.
{{AimForCont}} that it can.
Call this assumption $\paren A$.
If $\paren A$ holds, it must hold in all systems.
Let one such system be $\tuple {\R^2, \mathsf B_1, \equiv_1}$ where:
:$\R^2$ is the cartesian product of the set of real numbers with i... | Let $\GG$ be a [[Definition:Formal System|formal systematic]] treatment of [[Definition:Geometry|geometry]] containing only:
:The [[Definition:Formal Language|language]] and [[Definition:Axiom|axioms]] of [[Definition:Predicate Logic|first-order logic]], and the [[Definition:Preceding Discipline|disciplines preceding ... | Our assertion is that $\equiv$ cannot be defined in terms of $\mathsf B$.
{{AimForCont}} that it can.
Call this assumption $\paren A$.
If $\paren A$ holds, it must hold in all systems.
Let one such system be $\tuple {\R^2, \mathsf B_1, \equiv_1}$ where:
:$\R^2$ is the [[Definition:Cartesian Product|cartesian produ... | Equidistance is Independent of Betweenness | https://proofwiki.org/wiki/Equidistance_is_Independent_of_Betweenness | https://proofwiki.org/wiki/Equidistance_is_Independent_of_Betweenness | [
"Geometry"
] | [
"Definition:Formal System",
"Definition:Geometry",
"Definition:Formal Language",
"Definition:Axiom",
"Definition:Predicate Logic",
"Definition:Preceding Discipline",
"Definition:Undefined Terms of Tarski's Geometry",
"Definition:Equidistance",
"Axiom:Tarski's Axioms",
"Definition:Equidistance",
... | [
"Definition:Cartesian Product",
"Definition:Real Number",
"Definition:Ternary",
"Definition:Relation",
"Definition:Between (Geometry)",
"Definition:Quaternary",
"Definition:Relation",
"Definition:Equidistance",
"Definition:Preceding Discipline",
"Definition:Isomorphism (Abstract Algebra)",
"Defi... |
proofwiki-4879 | Bounded Linear Transformation Induces Bounded Sesquilinear Form | Let $\Bbb F$ be a subfield of $\C$.
Let $\struct {V, \innerprod \cdot \cdot_V}$ and $\struct {U, \innerprod \cdot \cdot_U}$ be inner product spaces over $\Bbb F$.
Let $A : V \to U$ and $B : U \to V$ be bounded linear transformations.
Let $u, v: V \times U \to \C$ be defined by:
:$\map u {h, k} := \innerprod {Ah} k_U$... | We first show that $u$ and $v$ are sesquilinear, and then that they are bounded.
Let $\alpha \in \mathbb F$ and $h_1, h_2 \in V$ and $k \in U$.
We have:
{{begin-eqn}}
{{eqn | l = \map u {\alpha h_1 + h_2, k}
| r = \innerprod {\map A {\alpha h_1 + h_2} } k_U
}}
{{eqn | r = \innerprod {\alpha A h_1 + A h_2} k_U
... | Let $\Bbb F$ be a [[Definition:Subfield|subfield]] of $\C$.
Let $\struct {V, \innerprod \cdot \cdot_V}$ and $\struct {U, \innerprod \cdot \cdot_U}$ be [[Definition:Inner Product Space|inner product spaces]] over $\Bbb F$.
Let $A : V \to U$ and $B : U \to V$ be [[Definition:Bounded Linear Transformation|bounded line... | We first show that $u$ and $v$ are [[Definition:Sesquilinear Form|sesquilinear]], and then that they are [[Definition:Bounded Sesquilinear Form|bounded]].
Let $\alpha \in \mathbb F$ and $h_1, h_2 \in V$ and $k \in U$.
We have:
{{begin-eqn}}
{{eqn | l = \map u {\alpha h_1 + h_2, k}
| r = \innerprod {\map A {... | Bounded Linear Transformation Induces Bounded Sesquilinear Form | https://proofwiki.org/wiki/Bounded_Linear_Transformation_Induces_Bounded_Sesquilinear_Form | https://proofwiki.org/wiki/Bounded_Linear_Transformation_Induces_Bounded_Sesquilinear_Form | [
"Linear Transformations on Hilbert Spaces"
] | [
"Definition:Subfield",
"Definition:Inner Product Space",
"Definition:Bounded Linear Transformation",
"Definition:Bounded Sesquilinear Form"
] | [
"Definition:Sesquilinear Form",
"Definition:Bounded Sesquilinear Form",
"Inner Product is Sesquilinear",
"Inner Product is Sesquilinear",
"Inner Product is Sesquilinear",
"Inner Product is Sesquilinear",
"Definition:Sesquilinear Form",
"Definition:Bounded Sesquilinear Form",
"Definition:Inner Produc... |
proofwiki-4880 | Existence and Uniqueness of Adjoint | Let $\mathbb F \in \set {\R, \C}$.
{{explain|How sure are we that this does not hold for ALL subfields of $\C$, not just these ones? <br>At least, the Hilbert assumptions forces $\mathbb F$ to be complete.}}
Let $\struct {\HH, \innerprod \cdot \cdot_\HH}$ and $\struct {\KK, \innerprod \cdot \cdot_\KK}$ be Hilbert space... | We first show that such a unique mapping $B$ exists, without first insisting on a bounded linear transformation. | Let $\mathbb F \in \set {\R, \C}$.
{{explain|How sure are we that this does not hold for ALL subfields of $\C$, not just these ones? <br>At least, the Hilbert assumptions forces $\mathbb F$ to be complete.}}
Let $\struct {\HH, \innerprod \cdot \cdot_\HH}$ and $\struct {\KK, \innerprod \cdot \cdot_\KK}$ be [[Definitio... | We first show that such a unique [[Definition:Mapping|mapping]] $B$ exists, without first insisting on a [[Definition:Bounded Linear Transformation|bounded linear transformation]]. | Existence and Uniqueness of Adjoint | https://proofwiki.org/wiki/Existence_and_Uniqueness_of_Adjoint | https://proofwiki.org/wiki/Existence_and_Uniqueness_of_Adjoint | [
"Adjoints",
"Existence and Uniqueness of Adjoint"
] | [
"Definition:Hilbert Space",
"Definition:Bounded Linear Transformation",
"Definition:Bounded Linear Transformation",
"Definition:Bounded Linear Transformation",
"Definition:Hilbert Space",
"Definition:Adjoint Linear Transformation"
] | [
"Definition:Mapping",
"Definition:Bounded Linear Transformation",
"Definition:Bounded Linear Transformation",
"Definition:Bounded Linear Transformation"
] |
proofwiki-4881 | Linear Transformation is Isomorphism iff Inverse Equals Adjoint | Let $\struct {\HH, \innerprod \cdot \cdot_\HH}$ and $\struct {\KK, \innerprod \cdot \cdot_\KK}$ be Hilbert spaces.
Let $U : \HH \to \KK$ be a bounded linear transformation.
{{TFAE}}
{{begin-itemize}}
{{item|(1):|$U$ is an isomorphism}}
{{item|(2):|$U$ is invertible and $U^{-1} {{=}} U^*$, where $U^*$ denotes the adjoin... | === $(1)$ implies $(2)$ ===
Suppose that:
:$U$ is an isomorphism.
That is:
:$\innerprod {U g} {U h}_\KK = \innerprod g h_\HH$
for each $g, h \in \HH$.
Let $g, h \in \HH$.
From the definition of the adjoint, we have:
:$\innerprod g h_\HH = \innerprod g {U^* U h}_\KK$
So, from Inner Product is Sesquilinear, we have:
:$... | Let $\struct {\HH, \innerprod \cdot \cdot_\HH}$ and $\struct {\KK, \innerprod \cdot \cdot_\KK}$ be [[Definition:Hilbert Space|Hilbert spaces]].
Let $U : \HH \to \KK$ be a [[Definition:Bounded Linear Transformation|bounded linear transformation]].
{{TFAE}}
{{begin-itemize}}
{{item|(1):|$U$ is an [[Definition:Isomorph... | === $(1)$ implies $(2)$ ===
Suppose that:
:$U$ is an [[Definition:Isomorphism (Hilbert Spaces)|isomorphism]].
That is:
:$\innerprod {U g} {U h}_\KK = \innerprod g h_\HH$
for each $g, h \in \HH$.
Let $g, h \in \HH$.
From the definition of the [[Definition:Adjoint|adjoint]], we have:
:$\innerprod g h_\HH = \inn... | Linear Transformation is Isomorphism iff Inverse Equals Adjoint | https://proofwiki.org/wiki/Linear_Transformation_is_Isomorphism_iff_Inverse_Equals_Adjoint | https://proofwiki.org/wiki/Linear_Transformation_is_Isomorphism_iff_Inverse_Equals_Adjoint | [
"Linear Transformations on Hilbert Spaces",
"Adjoints"
] | [
"Definition:Hilbert Space",
"Definition:Bounded Linear Transformation",
"Definition:Isomorphism (Hilbert Spaces)",
"Definition:Inverse Mapping",
"Definition:Adjoint Linear Transformation"
] | [
"Definition:Isomorphism (Hilbert Spaces)",
"Definition:Adjoint",
"Inner Product is Sesquilinear",
"Definition:Positiveness",
"Definition:Inner Product",
"Definition:Identity Mapping",
"Hilbert Space Isomorphism is Bijection",
"Definition:Bijection",
"Definition:Inverse Linear Transformation",
"Def... |
proofwiki-4882 | Limit of Sequence is Limit of Real Function | Let $\sequence {a_n}$ be a real sequence.
Let $f: x \mapsto \map f x$ be a real function.
Suppose the limit:
:$\ds \lim_{x \mathop \to +\infty} \map f x$
exists.
If for every $n$ in the domain of $\sequence {a_n}$:
:$\map f n = a_n$
then:
:$\ds \lim_{n \mathop \to +\infty} \ a_n = \ds \lim_{x \mathop \to +\infty} \map ... | This is an instance of Limit of Function by Convergent Sequences, as the reals form a metric space.
{{qed}} | Let $\sequence {a_n}$ be a [[Definition:Real Sequence|real sequence]].
Let $f: x \mapsto \map f x$ be a [[Definition:Real Function|real function]].
Suppose the [[Definition:Limit at Infinity|limit]]:
:$\ds \lim_{x \mathop \to +\infty} \map f x$
exists.
If for every $n$ in the [[Definition:Domain of Mapping|domain]... | This is an instance of [[Limit of Function by Convergent Sequences]], as the [[Real Number Line is Complete Metric Space|reals form a metric space]].
{{qed}} | Limit of Sequence is Limit of Real Function | https://proofwiki.org/wiki/Limit_of_Sequence_is_Limit_of_Real_Function | https://proofwiki.org/wiki/Limit_of_Sequence_is_Limit_of_Real_Function | [
"Convergence",
"Sequences"
] | [
"Definition:Real Sequence",
"Definition:Real Function",
"Definition:Limit of Real Function/Limit at Infinity/Positive",
"Definition:Domain (Set Theory)/Mapping"
] | [
"Limit of Function by Convergent Sequences",
"Real Number Line is Complete Metric Space"
] |
proofwiki-4883 | Equality of Cartesian Products | Let $A, B, C, D$ be nonempty sets.
Then:
:$A \times B = C \times D \iff A = C \land B = D$
where $\times$ denotes cartesian product. | If $A = C$ and $B = D$, it is immediate that $A \times B = C \times D$.
Now suppose that $A \times B = C \times D$.
By definition of Cartesian product:
{{begin-eqn}}
{{eqn | l = x \in A, y \in B
| o = \leadstoandfrom
| r = \tuple {x, y} \in A \times B
}}
{{eqn | o = \leadstoandfrom
| r = \tuple {x, y}... | Let $A, B, C, D$ be [[Definition:Non-Empty Set|nonempty]] [[Definition:Set|sets]].
Then:
:$A \times B = C \times D \iff A = C \land B = D$
where $\times$ denotes [[Definition:Cartesian Product|cartesian product]]. | If $A = C$ and $B = D$, it is immediate that $A \times B = C \times D$.
Now suppose that $A \times B = C \times D$.
By definition of [[Definition:Cartesian Product|Cartesian product]]:
{{begin-eqn}}
{{eqn | l = x \in A, y \in B
| o = \leadstoandfrom
| r = \tuple {x, y} \in A \times B
}}
{{eqn | o = \lea... | Equality of Cartesian Products | https://proofwiki.org/wiki/Equality_of_Cartesian_Products | https://proofwiki.org/wiki/Equality_of_Cartesian_Products | [
"Cartesian Product",
"Equality"
] | [
"Definition:Non-Empty Set",
"Definition:Set",
"Definition:Cartesian Product"
] | [
"Definition:Cartesian Product"
] |
proofwiki-4884 | Adjoint of Composition of Linear Transformations is Composition of Adjoints | :$\paren {A B}^* = B^* A^*$ | Let ${\innerprod \cdot \cdot}_\HH$, ${\innerprod \cdot \cdot}_\KK$ and ${\innerprod \cdot \cdot}_\LL$ denote inner products over $\HH$, $\KK$ and $\LL$ respectively.
Let $h \in \HH$ and $l \in \LL$.
Then:
{{begin-eqn}}
{{eqn | l = \innerprod {\map {\paren {A B} } h} l_\LL
| r = \innerprod {\map B h} {\map {A^*} l... | :$\paren {A B}^* = B^* A^*$ | Let ${\innerprod \cdot \cdot}_\HH$, ${\innerprod \cdot \cdot}_\KK$ and ${\innerprod \cdot \cdot}_\LL$ denote [[Definition:Inner Product|inner products]] over $\HH$, $\KK$ and $\LL$ respectively.
Let $h \in \HH$ and $l \in \LL$.
Then:
{{begin-eqn}}
{{eqn | l = \innerprod {\map {\paren {A B} } h} l_\LL
| r = \i... | Adjoint of Composition of Linear Transformations is Composition of Adjoints | https://proofwiki.org/wiki/Adjoint_of_Composition_of_Linear_Transformations_is_Composition_of_Adjoints | https://proofwiki.org/wiki/Adjoint_of_Composition_of_Linear_Transformations_is_Composition_of_Adjoints | [
"Adjoints",
"Composite Mappings"
] | [] | [
"Definition:Inner Product",
"Definition:Adjoint Linear Transformation",
"Existence and Uniqueness of Adjoint"
] |
proofwiki-4885 | Adjoint is Involutive | :$A^{**} = A$ | Let $x \in \HH$ and $y \in \KK$ be vectors.
By the definition of the adjoint, we have:
:$\innerprod {A x} y_\KK = \innerprod x {A^* y}_\HH$
Taking complex conjugates, we have:
:$\overline {\innerprod {A x} y_\KK} = \overline {\innerprod x {A^* y}_\HH}$
Using the conjugate symmetry of the inner product, we have:
:$\... | :$A^{**} = A$ | Let $x \in \HH$ and $y \in \KK$ be [[Definition:Vector|vectors]].
By the definition of the [[Definition:Adjoint Linear Transformation|adjoint]], we have:
:$\innerprod {A x} y_\KK = \innerprod x {A^* y}_\HH$
Taking [[Definition:Complex Conjugate|complex conjugates]], we have:
:$\overline {\innerprod {A x} y_\KK} =... | Adjoint is Involutive | https://proofwiki.org/wiki/Adjoint_is_Involutive | https://proofwiki.org/wiki/Adjoint_is_Involutive | [
"Adjoints",
"Adjoint Linear Transformations",
"Involutions"
] | [] | [
"Definition:Vector",
"Definition:Adjoint Linear Transformation",
"Definition:Complex Conjugate",
"Definition:Conjugate Symmetric Mapping",
"Definition:Inner Product",
"Definition:Adjoint Linear Transformation",
"Existence and Uniqueness of Adjoint"
] |
proofwiki-4886 | Adjoining Commutes with Inverting | Let $\HH$ and $\KK$ be Hilbert spaces.
Let $\map \BB {\HH, \KK}$ be the set of bounded linear transformations from $\HH$ to $\KK$.
Let $A \in \map \BB {\HH, \KK}$ be a bounded linear transformation on $\HH$.
Let $A^{-1} \in \map \BB {\KK, \HH}$ be an inverse for $A$.
Let $A^*$ denote the adjoint of $A$.
Then $A^*$, is ... | By definition of inverse, one has $A A^{-1} = I_\KK$, where $I_\KK$ is the identity operator on $\KK$.
From Adjoint of Composition of Linear Transformations is Composition of Adjoints and Adjoint of Identity Transformation:
:$I_\KK = {I_\KK}^* = \paren {A A^{-1} }^* = \paren {A^{-1} }^*A^*$
Similarly:
:$I_\HH = {I_\HH}... | Let $\HH$ and $\KK$ be [[Definition:Hilbert Space|Hilbert spaces]].
Let $\map \BB {\HH, \KK}$ be the [[Definition:Set|set]] of [[Definition:Bounded Linear Transformation|bounded linear transformations]] from $\HH$ to $\KK$.
Let $A \in \map \BB {\HH, \KK}$ be a [[Definition:Bounded Linear Transformation|bounded linear... | By definition of [[Definition:Inverse (Bounded Linear Transformation)|inverse]], one has $A A^{-1} = I_\KK$, where $I_\KK$ is the [[Definition:Identity Operator|identity operator]] on $\KK$.
From [[Adjoint of Composition of Linear Transformations is Composition of Adjoints]] and [[Adjoint of Identity Transformation]]:... | Adjoining Commutes with Inverting | https://proofwiki.org/wiki/Adjoining_Commutes_with_Inverting | https://proofwiki.org/wiki/Adjoining_Commutes_with_Inverting | [
"Adjoints"
] | [
"Definition:Hilbert Space",
"Definition:Set",
"Definition:Bounded Linear Transformation",
"Definition:Bounded Linear Transformation",
"Definition:Invertible Bounded Linear Transformation",
"Definition:Adjoint Linear Transformation",
"Definition:Inverse Linear Transformation"
] | [
"Definition:Invertible Bounded Linear Transformation",
"Definition:Identity Mapping",
"Adjoint of Composition of Linear Transformations is Composition of Adjoints",
"Adjoint of Identity Transformation",
"Definition:Invertible Bounded Linear Transformation",
"Definition:Inverse Linear Transformation"
] |
proofwiki-4887 | Identity of Points | Two points share the same position {{iff}} they are the same points. | Let $a$ be a point with position $P_1$.
Let $b$ be a point with position $P_2$.
By hypothesis, $P_1 = P_2$.
By Leibniz's Law, two objects are the same object {{iff}} they share every property in common.
By the definition of point, the ''only'' property possessed by a point is position.
We have:
:$P_1 = P_2 \dashv \vdas... | Two [[Definition:Point|points]] share the same [[Definition:Position|position]] {{iff}} they are the [[Definition:Equals|same]] [[Definition:Point|points]]. | Let $a$ be a [[Definition:Point|point]] with [[Definition:Position|position]] $P_1$.
Let $b$ be a [[Definition:Point|point]] with [[Definition:Position|position]] $P_2$.
[[Definition:By Hypothesis|By hypothesis]], $P_1 = P_2$.
By [[Axiom:Leibniz's Law|Leibniz's Law]], two [[Definition:Object|objects]] are the same ... | Identity of Points | https://proofwiki.org/wiki/Identity_of_Points | https://proofwiki.org/wiki/Identity_of_Points | [
"Euclidean Geometry"
] | [
"Definition:Point",
"Definition:Position",
"Definition:Equals",
"Definition:Point"
] | [
"Definition:Point",
"Definition:Position",
"Definition:Point",
"Definition:Position",
"Definition:By Hypothesis",
"Axiom:Leibniz's Law",
"Definition:Object",
"Definition:Property",
"Definition:Point",
"Definition:Point",
"Definition:Position"
] |
proofwiki-4888 | Successor of Omega | :$\omega + 1 = \set {0, 1, 2, \ldots; \omega}$
where $\omega$ is the minimally inductive set and $\omega + 1$ is the successor of $\omega$.
Note the use of the semicolon; this is the notation for multipart infinite sets. | {{begin-eqn}}
{{eqn | l = \omega + 1
| r = \omega \cup \set {\omega}
| c = {{Defof|Successor Set}}
}}
{{eqn | r = \set {0, 1, 2, \ldots} \cup \set \omega
| c = {{Defof|Von Neumann Construction of Natural Numbers}}
}}
{{eqn | r = \set {0, 1, 2, \ldots; \omega}
| c = {{Defof|Set Union}}
}}
{{end-e... | :$\omega + 1 = \set {0, 1, 2, \ldots; \omega}$
where $\omega$ is the [[Definition:Minimally Inductive Set|minimally inductive set]] and $\omega + 1$ is the [[Definition:Successor Set|successor of $\omega$]].
Note the use of the semicolon; this is the notation for [[Definition:Multipart Infinite Set|multipart infinit... | {{begin-eqn}}
{{eqn | l = \omega + 1
| r = \omega \cup \set {\omega}
| c = {{Defof|Successor Set}}
}}
{{eqn | r = \set {0, 1, 2, \ldots} \cup \set \omega
| c = {{Defof|Von Neumann Construction of Natural Numbers}}
}}
{{eqn | r = \set {0, 1, 2, \ldots; \omega}
| c = {{Defof|Set Union}}
}}
{{end-e... | Successor of Omega | https://proofwiki.org/wiki/Successor_of_Omega | https://proofwiki.org/wiki/Successor_of_Omega | [
"Transfinite Arithmetic"
] | [
"Definition:Minimally Inductive Set",
"Definition:Successor Mapping/Successor Set",
"Definition:Set/Implicit Set Definition/Multipart Infinite Set"
] | [] |
proofwiki-4889 | Equivalence of Formulations of Axiom of Choice | The following formulations of the Axiom of Choice are equivalent:
=== Formulation 1 ===
{{:Axiom:Axiom of Choice/Formulation 1}}
=== Formulation 2 ===
{{:Axiom:Axiom of Choice/Formulation 2}}
=== Formulation 3 ===
{{:Axiom:Axiom of Choice/Formulation 3}}
=== Formulation 4 ===
{{:Axiom:Axiom of Choice/Formulation 4}} | === Formulation 1 implies Formulation 2 ===
{{:Equivalence of Formulations of Axiom of Choice/Formulation 1 implies Formulation 2}}{{qed|lemma}} | The following formulations of the [[Axiom:Axiom of Choice|Axiom of Choice]] are [[Definition:Logical Equivalence|equivalent]]:
=== [[Axiom:Axiom of Choice/Formulation 1|Formulation 1]] ===
{{:Axiom:Axiom of Choice/Formulation 1}}
=== [[Axiom:Axiom of Choice/Formulation 2|Formulation 2]] ===
{{:Axiom:Axiom of Choice/Fo... | === [[Equivalence of Formulations of Axiom of Choice/Formulation 1 implies Formulation 2|Formulation 1 implies Formulation 2]] ===
{{:Equivalence of Formulations of Axiom of Choice/Formulation 1 implies Formulation 2}}{{qed|lemma}} | Equivalence of Formulations of Axiom of Choice | https://proofwiki.org/wiki/Equivalence_of_Formulations_of_Axiom_of_Choice | https://proofwiki.org/wiki/Equivalence_of_Formulations_of_Axiom_of_Choice | [
"Axiom of Choice",
"Equivalence of Formulations of Axiom of Choice"
] | [
"Axiom:Axiom of Choice",
"Definition:Logical Equivalence",
"Axiom:Axiom of Choice/Formulation 1",
"Axiom:Axiom of Choice/Formulation 2",
"Axiom:Axiom of Choice/Formulation 3",
"Axiom:Axiom of Choice/Formulation 4"
] | [
"Equivalence of Formulations of Axiom of Choice/Formulation 1 implies Formulation 2"
] |
proofwiki-4890 | Norm of Adjoint | Let $H, K$ be Hilbert spaces.
Let $A \in \map B {H, K}$ be a bounded linear transformation.
Let $A^* \in \map B {K, H}$ be the adjoint of $A$.
Then $A$ and $A^*$ satisfy:
:$\norm A_{\map B {H, K} }^2 = \norm {A^*}_{\map B {K, H} }^2 = \norm {A^* A}_{\map B {H, H} }$
where:
:$\norm \cdot_{\map B {H, K} }$ denotes the op... | Let $h \in H$ such that $\norm h_H \le 1$.
Then:
{{begin-eqn}}
{{eqn | l = \norm {A h}_K^2
| r = \innerprod {A h} {A h}_K
| c = {{Defof|Inner Product Norm}}
}}
{{eqn | r = \innerprod {A^* A h} h_H
| c = {{Defof|Adjoint Linear Transformation}}
}}
{{eqn | o = \le
| r = \norm {A^*A h}_H \norm h_H
... | Let $H, K$ be [[Definition:Hilbert Space|Hilbert spaces]].
Let $A \in \map B {H, K}$ be a [[Definition:Bounded Linear Transformation|bounded linear transformation]].
Let $A^* \in \map B {K, H}$ be the [[Definition:Adjoint Linear Transformation|adjoint]] of $A$.
Then $A$ and $A^*$ satisfy:
:$\norm A_{\map B {H, K} }... | Let $h \in H$ such that $\norm h_H \le 1$.
Then:
{{begin-eqn}}
{{eqn | l = \norm {A h}_K^2
| r = \innerprod {A h} {A h}_K
| c = {{Defof|Inner Product Norm}}
}}
{{eqn | r = \innerprod {A^* A h} h_H
| c = {{Defof|Adjoint Linear Transformation}}
}}
{{eqn | o = \le
| r = \norm {A^*A h}_H \norm h_H... | Norm of Adjoint | https://proofwiki.org/wiki/Norm_of_Adjoint | https://proofwiki.org/wiki/Norm_of_Adjoint | [
"Adjoints"
] | [
"Definition:Hilbert Space",
"Definition:Bounded Linear Transformation",
"Definition:Adjoint Linear Transformation",
"Definition:Norm/Bounded Linear Transformation",
"Definition:Norm/Bounded Linear Transformation",
"Definition:Norm/Bounded Linear Transformation"
] | [
"Cauchy-Bunyakovsky-Schwarz Inequality",
"Fundamental Property of Norm on Bounded Linear Transformation",
"Norm on Bounded Linear Transformation is Submultiplicative",
"Definition:Norm/Bounded Linear Transformation",
"Adjoint is Involutive"
] |
proofwiki-4891 | Operator is Hermitian iff Numerical Range is Real | Let $\struct {\HH, \innerprod \cdot \cdot_\HH}$ be a Hilbert space over $\C$.
Let $A : \HH \to \HH$ be a bounded linear operator.
Let $\map W A$ be the numerical range of $A$.
Then $A$ is Hermitian {{iff}}:
:$\map W A \subseteq \R$
That is:
:$\forall h \in \HH: \innerprod {A h} h_\HH \mathop \in \R$ | === Necessary Condition ===
Suppose that $A$ is Hermitian.
Then:
:$A^* = A$
where $A^*$ denotes the adjoint of $A$.
Let $x \in \HH$.
Then, by the definition of the adjoint, we have:
:$\innerprod {A x} x_\HH = \innerprod x {A x}_\HH$
From the conjugate symmetry of the inner product, we have:
:$\innerprod x {A x}_\H... | Let $\struct {\HH, \innerprod \cdot \cdot_\HH}$ be a [[Definition:Hilbert Space|Hilbert space]] over $\C$.
Let $A : \HH \to \HH$ be a [[Definition:Bounded Linear Operator|bounded linear operator]].
Let $\map W A$ be the [[Definition:Numerical Range|numerical range]] of $A$.
Then $A$ is [[Definition:Hermitian Opera... | === Necessary Condition ===
Suppose that $A$ is [[Definition:Hermitian Operator|Hermitian]].
Then:
:$A^* = A$
where $A^*$ denotes the [[Definition:Adjoint Linear Transformation|adjoint]] of $A$.
Let $x \in \HH$.
Then, by the definition of the [[Definition:Adjoint Linear Transformation|adjoint]], we have:
:$\... | Operator is Hermitian iff Numerical Range is Real | https://proofwiki.org/wiki/Operator_is_Hermitian_iff_Numerical_Range_is_Real | https://proofwiki.org/wiki/Operator_is_Hermitian_iff_Numerical_Range_is_Real | [
"Hermitian Operators"
] | [
"Definition:Hilbert Space",
"Definition:Bounded Linear Operator",
"Definition:Numerical Range",
"Definition:Hermitian Operator"
] | [
"Definition:Hermitian Operator",
"Definition:Adjoint Linear Transformation",
"Definition:Adjoint Linear Transformation",
"Definition:Conjugate Symmetric Mapping",
"Definition:Inner Product",
"Complex Number equals Conjugate iff Wholly Real",
"Definition:Real Number",
"Definition:Real Number",
"Defin... |
proofwiki-4892 | Norm of Hermitian Operator | Let $\mathbb F \in \set {\R, \C}$.
Let $\struct {\HH, \innerprod \cdot \cdot_\HH}$ be a Hilbert space over $\mathbb F$.
Let $A : \HH \to \HH$ be a bounded Hermitian operator.
Let $\norm \cdot_\HH$ be the inner product norm on $\HH$.
Then the norm of $A$ satisfies:
:$\norm A = \sup \set {\size {\innerprod {A h} h_\HH}... | Let:
:$M = \sup \set {\size {\innerprod {A h} h_\HH}: h \in \HH, \norm h_\HH = 1}$
To show that $M = \norm A$ we first show that:
:$M \le \norm A$
We will then show that:
:$\norm A \le M$
Let $x \in \HH$ be such that:
:$\norm x_\HH = 1$.
Then we have:
{{begin-eqn}}
{{eqn | l = \size {\innerprod {A x} x_\HH}
... | Let $\mathbb F \in \set {\R, \C}$.
Let $\struct {\HH, \innerprod \cdot \cdot_\HH}$ be a [[Definition:Hilbert Space|Hilbert space]] over $\mathbb F$.
Let $A : \HH \to \HH$ be a [[Definition:Bounded Linear Operator|bounded]] [[Definition:Hermitian Operator|Hermitian operator]].
Let $\norm \cdot_\HH$ be the [[Definiti... | Let:
:$M = \sup \set {\size {\innerprod {A h} h_\HH}: h \in \HH, \norm h_\HH = 1}$
To show that $M = \norm A$ we first show that:
:$M \le \norm A$
We will then show that:
:$\norm A \le M$
Let $x \in \HH$ be such that:
:$\norm x_\HH = 1$.
Then we have:
{{begin-eqn}}
{{eqn | l = \size {\innerprod {A x} x_\... | Norm of Hermitian Operator | https://proofwiki.org/wiki/Norm_of_Hermitian_Operator | https://proofwiki.org/wiki/Norm_of_Hermitian_Operator | [
"Hermitian Operators",
"Norm of Hermitian Operator"
] | [
"Definition:Hilbert Space",
"Definition:Bounded Linear Operator",
"Definition:Hermitian Operator",
"Definition:Inner Product Norm",
"Definition:Norm/Bounded Linear Transformation"
] | [
"Cauchy-Bunyakovsky-Schwarz Inequality/Inner Product Spaces",
"Fundamental Property of Norm on Bounded Linear Transformation",
"Definition:Supremum of Set/Real Numbers",
"Definition:Linear Operator",
"Inner Product is Sesquilinear",
"Definition:Conjugate Symmetric Mapping",
"Definition:Inner Product",
... |
proofwiki-4893 | Operator Zero iff Inner Product Zero | Let $\HH$ be a Hilbert space over $\C$.
Let $A: \HH \to \HH$ be a bounded linear operator.
Then:
:$\forall h \in \HH: \innerprod {A h} h_\HH = 0$
{{Iff}}
:$A$ is the zero operator. | === Necessary Case ===
We have that $A$ is the zero operator.
Then:
:$\innerprod {A h} h = \innerprod 0 h = 0$
{{qed|lemma}} | Let $\HH$ be a [[Definition:Hilbert Space|Hilbert space]] over $\C$.
Let $A: \HH \to \HH$ be a [[Definition:Bounded Linear Operator|bounded linear operator]].
Then:
:$\forall h \in \HH: \innerprod {A h} h_\HH = 0$
{{Iff}}
:$A$ is the [[Definition:Zero Operator|zero operator]]. | === Necessary Case ===
We have that $A$ is the [[Definition:Zero Operator|zero operator]].
Then:
:$\innerprod {A h} h = \innerprod 0 h = 0$
{{qed|lemma}} | Operator Zero iff Inner Product Zero | https://proofwiki.org/wiki/Operator_Zero_iff_Inner_Product_Zero | https://proofwiki.org/wiki/Operator_Zero_iff_Inner_Product_Zero | [
"Linear Transformations on Hilbert Spaces"
] | [
"Definition:Hilbert Space",
"Definition:Bounded Linear Operator",
"Definition:Zero Operator"
] | [
"Definition:Zero Operator"
] |
proofwiki-4894 | Linear Operator is Sum of Real and Imaginary Parts | Let $\HH$ be a Hilbert space over $\C$.
Let $A \in \map B \HH$ be a bounded linear operator.
Let $B$ and $C$ be the real and imaginary parts of $A$, respectively.
Then $A = B + i C$. | {{begin-eqn}}
{{eqn|l = B + i C
|r = \frac 1 2 \paren {A + A^*} + i \frac 1 {2i} \paren {A - A^*}
|c = Definitions of $B, C$
}}
{{eqn|r = \frac 1 2 A + \frac 1 2 A^* + \frac 1 2 A - \frac 1 2 A^*
}}
{{eqn|r = A
}}
{{end-eqn}}
{{qed}} | Let $\HH$ be a [[Definition:Hilbert Space|Hilbert space]] over $\C$.
Let $A \in \map B \HH$ be a [[Definition:Bounded Linear Operator|bounded linear operator]].
Let $B$ and $C$ be the [[Definition:Real Part (Linear Operator)|real]] and [[Definition:Imaginary Part (Linear Operator)|imaginary parts]] of $A$, respectiv... | {{begin-eqn}}
{{eqn|l = B + i C
|r = \frac 1 2 \paren {A + A^*} + i \frac 1 {2i} \paren {A - A^*}
|c = Definitions of $B, C$
}}
{{eqn|r = \frac 1 2 A + \frac 1 2 A^* + \frac 1 2 A - \frac 1 2 A^*
}}
{{eqn|r = A
}}
{{end-eqn}}
{{qed}} | Linear Operator is Sum of Real and Imaginary Parts | https://proofwiki.org/wiki/Linear_Operator_is_Sum_of_Real_and_Imaginary_Parts | https://proofwiki.org/wiki/Linear_Operator_is_Sum_of_Real_and_Imaginary_Parts | [
"Linear Transformations on Hilbert Spaces"
] | [
"Definition:Hilbert Space",
"Definition:Bounded Linear Operator",
"Definition:Real Part (Linear Operator)",
"Definition:Imaginary Part (Linear Operator)"
] | [] |
proofwiki-4895 | Characterization of Normal Operators | Let $\GF \in \set {\R, \C}$.
Let $\HH$ be a Hilbert space over $\GF$.
Let $A$ be a bounded linear operator on $\HH$.
{{TFAE}}
{{begin-itemize}}
{{item|(1):|$A A^* {{=}} A^* A$, that is, $A$ is normal}}
{{item|(2):|$\forall h \in \HH: \norm {A h}_\HH {{=}} \norm {A^*h}_\HH$}}
{{end-itemize}}
where:
:$A^*$ denotes the ad... | === $(3)$ equivalent to $(1)$ ===
Suppose $\GF = \C$.
We have:
{{begin-eqn}}
{{eqn | l = \map \Re A \map \Im A
| r = \paren {\frac 1 2 \paren {A + A^\ast} } \paren {\frac 1 {2 i} \paren {A - A^\ast} }
}}
{{eqn | r = \frac 1 {4 i} \paren {A + A^\ast} \paren {A - A^\ast}
}}
{{eqn | r = \frac 1 {4 i} \paren {A^2 + A^... | Let $\GF \in \set {\R, \C}$.
Let $\HH$ be a [[Definition:Hilbert Space|Hilbert space]] over $\GF$.
Let $A$ be a [[Definition:Bounded Linear Operator|bounded linear operator]] on $\HH$.
{{TFAE}}
{{begin-itemize}}
{{item|(1):|$A A^* {{=}} A^* A$, that is, $A$ is [[Definition:Normal Operator|normal]]}}
{{item|(2):|$\f... | === $(3)$ equivalent to $(1)$ ===
Suppose $\GF = \C$.
We have:
{{begin-eqn}}
{{eqn | l = \map \Re A \map \Im A
| r = \paren {\frac 1 2 \paren {A + A^\ast} } \paren {\frac 1 {2 i} \paren {A - A^\ast} }
}}
{{eqn | r = \frac 1 {4 i} \paren {A + A^\ast} \paren {A - A^\ast}
}}
{{eqn | r = \frac 1 {4 i} \paren {A^2 +... | Characterization of Normal Operators | https://proofwiki.org/wiki/Characterization_of_Normal_Operators | https://proofwiki.org/wiki/Characterization_of_Normal_Operators | [
"Normal Operators",
"Linear Transformations on Hilbert Spaces"
] | [
"Definition:Hilbert Space",
"Definition:Bounded Linear Operator",
"Definition:Normal Operator",
"Definition:Adjoint",
"Definition:Inner Product Norm",
"Definition:Real Part (Linear Operator)",
"Definition:Imaginary Part (Linear Operator)",
"Definition:Commutative/Elements"
] | [] |
proofwiki-4896 | Bounded Linear Transformation is Isometry iff Adjoint is Left-Inverse | Let $\HH, \KK$ be Hilbert spaces.
Let $A \in \map B{\HH, \KK}$ be a bounded linear transformation.
Then $A$ is an isometry {{iff}}:
:$A^*A = I_\HH$
where $A^*$ denotes the adjoint of $A$, and $I_\HH$ the identity operator on $\HH$. | Let $g, h \in \HH$. Then by the definition of adjoint:
:${\innerprod {A g} {A h} }_\KK = {\innerprod {A^* A g} h}_\HH$
From the uniqueness of the adjoint, it follows that:
:${\innerprod {A g} {A h} }_\KK = {\innerprod g h}_\HH$
holds {{iff}} $A^*A = I_\HH$.
Hence the result by definition of isometry.
{{qed}} | Let $\HH, \KK$ be [[Definition:Hilbert Space|Hilbert spaces]].
Let $A \in \map B{\HH, \KK}$ be a [[Definition:Bounded Linear Transformation|bounded linear transformation]].
Then $A$ is an [[Definition:Isometry (Hilbert Spaces)|isometry]] {{iff}}:
:$A^*A = I_\HH$
where $A^*$ denotes the [[Definition:Adjoint Linear ... | Let $g, h \in \HH$. Then by the definition of [[Definition:Adjoint Linear Transformation|adjoint]]:
:${\innerprod {A g} {A h} }_\KK = {\innerprod {A^* A g} h}_\HH$
From the [[Definition:Adjoint Linear Transformation|uniqueness of the adjoint]], it follows that:
:${\innerprod {A g} {A h} }_\KK = {\innerprod g h}_\HH... | Bounded Linear Transformation is Isometry iff Adjoint is Left-Inverse | https://proofwiki.org/wiki/Bounded_Linear_Transformation_is_Isometry_iff_Adjoint_is_Left-Inverse | https://proofwiki.org/wiki/Bounded_Linear_Transformation_is_Isometry_iff_Adjoint_is_Left-Inverse | [
"Adjoints",
"Linear Transformations on Hilbert Spaces"
] | [
"Definition:Hilbert Space",
"Definition:Bounded Linear Transformation",
"Definition:Isometry (Hilbert Spaces)",
"Definition:Adjoint Linear Transformation",
"Definition:Identity Mapping"
] | [
"Definition:Adjoint Linear Transformation",
"Definition:Adjoint Linear Transformation",
"Definition:Isometry (Hilbert Spaces)"
] |
proofwiki-4897 | Characterization of Unitary Operators | Let $\HH$ be a Hilbert space.
Let $A$ be a bounded linear operator on $\HH$.
{{TFAE}}
{{begin-itemize}}
{{item|(1):|$A$ is a unitary operator}}
{{item|(2):|$A^* A {{=}} A A^* {{=}} I$, where $A^*$ denotes the adjoint of $A$, and $I$ denotes the identity operator}}
{{item|(3):|$A$ is a normal isometry}}
{{end-itemize}} | === $(3)$ implies $(2)$ ===
Suppose that $A$ is a normal isometry.
Then:
:$\innerprod {A x} {A x} = \innerprod x x$ for each $x \in \HH$.
From Adjoint is Involutive, we have:
:$\innerprod {A x} {\paren {A^\ast}^\ast x} = \innerprod x x$
So, from the definition of the adjoint, we have:
:$\innerprod {A^\ast A x} x = \in... | Let $\HH$ be a [[Definition:Hilbert Space|Hilbert space]].
Let $A$ be a [[Definition:Bounded Linear Operator|bounded linear operator]] on $\HH$.
{{TFAE}}
{{begin-itemize}}
{{item|(1):|$A$ is a [[Definition:Unitary Operator|unitary operator]]}}
{{item|(2):|$A^* A {{=}} A A^* {{=}} I$, where $A^*$ denotes the [[Defini... | === $(3)$ implies $(2)$ ===
Suppose that $A$ is a [[Definition:Normal Operator|normal]] [[Definition:Isometry (Hilbert Spaces)|isometry]].
Then:
:$\innerprod {A x} {A x} = \innerprod x x$ for each $x \in \HH$.
From [[Adjoint is Involutive]], we have:
:$\innerprod {A x} {\paren {A^\ast}^\ast x} = \innerprod x x$
So... | Characterization of Unitary Operators | https://proofwiki.org/wiki/Characterization_of_Unitary_Operators | https://proofwiki.org/wiki/Characterization_of_Unitary_Operators | [
"Linear Transformations on Hilbert Spaces"
] | [
"Definition:Hilbert Space",
"Definition:Bounded Linear Operator",
"Definition:Unitary Operator",
"Definition:Adjoint Linear Transformation",
"Definition:Identity Mapping",
"Definition:Normal Operator",
"Definition:Isometry (Hilbert Spaces)"
] | [
"Definition:Normal Operator",
"Definition:Isometry (Hilbert Spaces)",
"Adjoint is Involutive",
"Definition:Adjoint Linear Transformation",
"Operator with Zero Numerical Range is Zero Operator/Corollary",
"Definition:Normal Operator",
"Definition:Adjoint Linear Transformation",
"Adjoint is Involutive",... |
proofwiki-4898 | Kernel of Linear Transformation is Orthocomplement of Image of Adjoint | Let $\struct {\HH, \innerprod \cdot \cdot_\HH}$ and $\struct {\KK, \innerprod \cdot \cdot_\KK}$ be Hilbert spaces.
Let $\map \BB {\HH, \KK}$ denote the set of bounded linear transformations from $\HH$ to $\KK$.
Let $A \in \map \BB {\HH, \KK}$ be a bounded linear transformation.
Then $\ker A = \paren {\Img {A^*} }^\perp... | Let $x \in \HH$ be arbitrary.
First fix $x \in \map \ker A$ and let $y \in \Img {A^\ast}$ be arbitrary.
We then have:
:$y = A^\ast u$
for some $u \in \KK$.
We have:
:$\innerprod x y_\HH = \innerprod x {A^\ast u}_\HH$
By the definition of the adjoint, we have:
:$\innerprod x y_\HH = \innerprod {A x} u_\KK$
But $x \i... | Let $\struct {\HH, \innerprod \cdot \cdot_\HH}$ and $\struct {\KK, \innerprod \cdot \cdot_\KK}$ be [[Definition:Hilbert Space|Hilbert spaces]].
Let $\map \BB {\HH, \KK}$ denote the [[Definition:Set|set]] of [[Definition:Bounded Linear Transformation|bounded linear transformations]] from $\HH$ to $\KK$.
Let $A \in \ma... | Let $x \in \HH$ be arbitrary.
First fix $x \in \map \ker A$ and let $y \in \Img {A^\ast}$ be arbitrary.
We then have:
:$y = A^\ast u$
for some $u \in \KK$.
We have:
:$\innerprod x y_\HH = \innerprod x {A^\ast u}_\HH$
By the definition of the [[Definition:Adjoint Linear Transformation|adjoint]], we have:
:$\... | Kernel of Linear Transformation is Orthocomplement of Image of Adjoint | https://proofwiki.org/wiki/Kernel_of_Linear_Transformation_is_Orthocomplement_of_Image_of_Adjoint | https://proofwiki.org/wiki/Kernel_of_Linear_Transformation_is_Orthocomplement_of_Image_of_Adjoint | [
"Adjoints",
"Linear Transformations on Hilbert Spaces"
] | [
"Definition:Hilbert Space",
"Definition:Set",
"Definition:Bounded Linear Transformation",
"Definition:Bounded Linear Transformation",
"Definition:Adjoint Linear Transformation",
"Definition:Kernel of Linear Transformation",
"Definition:Image (Set Theory)/Mapping/Mapping",
"Definition:Orthogonal (Linea... | [
"Definition:Adjoint Linear Transformation",
"Definition:Subset",
"Definition:Adjoint Linear Transformation",
"Definition:Inner Product",
"Definition:Subset",
"Definition:Set Equality/Definition 1"
] |
proofwiki-4899 | Relation Contains Mapping is Equivalent to AoC | Let $S$ and $T$ be sets.
Let $\RR \subseteq S \times T$ be a relation on $S \times T$.
Then:
:there exists a mapping $f \subseteq \RR$ whose domain is the same as the preimage of $\RR$
{{iff}}
:the axiom of choice holds. | {{ProofWanted}}
{{AoC}} | Let $S$ and $T$ be [[Definition:Set|sets]].
Let $\RR \subseteq S \times T$ be a [[Definition:Relation|relation]] on $S \times T$.
Then:
:there exists a [[Definition:Mapping|mapping]] $f \subseteq \RR$ whose [[Definition:Domain of Mapping|domain]] is the same as the [[Definition:Preimage of Relation|preimage]] of $\R... | {{ProofWanted}}
{{AoC}} | Relation Contains Mapping is Equivalent to AoC | https://proofwiki.org/wiki/Relation_Contains_Mapping_is_Equivalent_to_AoC | https://proofwiki.org/wiki/Relation_Contains_Mapping_is_Equivalent_to_AoC | [] | [
"Definition:Set",
"Definition:Relation",
"Definition:Mapping",
"Definition:Domain (Set Theory)/Mapping",
"Definition:Preimage/Relation/Relation",
"Axiom:Axiom of Choice"
] | [] |
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