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 |
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proofwiki-17200 | Monotone Function is of Bounded Variation | Let $a, b$ be real numbers with $a < b$.
Let $f : \closedint a b \to \R$ be a monotone function.
Then $f$ is of bounded variation. | We use the notation from the definition of bounded variation.
Let $P = \set {x_0, x_1, \ldots, x_n}$ be a finite subdivision of $\closedint a b$.
As $f$ is monotone, it is either increasing or decreasing.
First consider the case of $f$ increasing, then:
:$\map f {x_i} \ge \map f {x_{i - 1} }$
for all $i \in \N$ with $... | Let $a, b$ be [[Definition:Real Number|real numbers]] with $a < b$.
Let $f : \closedint a b \to \R$ be a [[Definition:Monotone (Order Theory)/Real Function|monotone function]].
Then $f$ is of [[Definition:Bounded Variation (Closed Bounded Interval)|bounded variation]]. | We use the notation from the [[Definition:Bounded Variation (Closed Bounded Interval)|definition of bounded variation]].
Let $P = \set {x_0, x_1, \ldots, x_n}$ be a [[Definition:Finite Subdivision|finite subdivision]] of $\closedint a b$.
As $f$ is [[Definition:Monotone (Order Theory)/Real Function|monotone]], it is ... | Monotone Function is of Bounded Variation | https://proofwiki.org/wiki/Monotone_Function_is_of_Bounded_Variation | https://proofwiki.org/wiki/Monotone_Function_is_of_Bounded_Variation | [
"Bounded Variation"
] | [
"Definition:Real Number",
"Definition:Monotone (Order Theory)/Real Function",
"Definition:Bounded Variation/Closed Bounded Interval"
] | [
"Definition:Bounded Variation/Closed Bounded Interval",
"Definition:Subdivision of Interval/Finite",
"Definition:Monotone (Order Theory)/Real Function",
"Definition:Increasing/Real Function",
"Definition:Decreasing/Real Function",
"Definition:Increasing/Real Function",
"Definition:Absolute Value",
"Te... |
proofwiki-17201 | Differentiable Function with Bounded Derivative is of Bounded Variation | Let $a, b$ be real numbers with $a < b$.
Let $f : \closedint a b \to \R$ be a continuous function.
Let $f$ be differentiable on $\openint a b$, with bounded derivative.
Then $f$ is of bounded variation. | For each finite subdivision $P$ of $\closedint a b$, write:
:$P = \set {x_0, x_1, \ldots, x_n}$
with:
:$a = x_0 < x_1 < x_2 < \cdots < x_{n - 1} < x_n = b$
Since the derivative of $f$ is bounded, there exists some $M \in \R$ such that:
:$\size {\map {f'} x} \le M$
for all $x \in \openint a b$.
By the Mean Value Theor... | Let $a, b$ be [[Definition:Real Number|real numbers]] with $a < b$.
Let $f : \closedint a b \to \R$ be a [[Definition:Continuous Real Function|continuous function]].
Let $f$ be [[Definition:Differentiable Real Function|differentiable]] on $\openint a b$, with [[Definition:Bounded Real-Valued Function|bounded]] [[Defi... | For each [[Definition:Finite Subdivision|finite subdivision]] $P$ of $\closedint a b$, write:
:$P = \set {x_0, x_1, \ldots, x_n}$
with:
:$a = x_0 < x_1 < x_2 < \cdots < x_{n - 1} < x_n = b$
Since the [[Definition:Derivative|derivative]] of $f$ is [[Definition:Bounded Real-Valued Function|bounded]], there exists so... | Differentiable Function with Bounded Derivative is of Bounded Variation | https://proofwiki.org/wiki/Differentiable_Function_with_Bounded_Derivative_is_of_Bounded_Variation | https://proofwiki.org/wiki/Differentiable_Function_with_Bounded_Derivative_is_of_Bounded_Variation | [
"Bounded Variation"
] | [
"Definition:Real Number",
"Definition:Continuous Real Function",
"Definition:Differentiable Mapping/Real Function",
"Definition:Bounded Mapping/Real-Valued",
"Definition:Derivative",
"Definition:Bounded Variation/Closed Bounded Interval"
] | [
"Definition:Subdivision of Interval/Finite",
"Definition:Derivative",
"Definition:Bounded Mapping/Real-Valued",
"Mean Value Theorem",
"Definition:Bounded Mapping/Real-Valued",
"Definition:Bounded Variation/Closed Bounded Interval",
"Telescoping Series/Example 2",
"Definition:Subdivision of Interval/Fi... |
proofwiki-17202 | Function of Bounded Variation is Bounded | Let $a, b$ be real numbers with $a < b$.
Let $f : \closedint a b \to \R$ be a function of bounded variation.
Then $f$ is bounded. | We use the notation from the definition of bounded variation.
Since $f$ is of bounded variation, there exists $M \ge 0$ such that:
:$\map {V_f} {P ; \closedint a b} \le M$
for all finite subdivisions $P$ of $\closedint a b$.
Let $x$ be a real number with:
:$a < x < b$
Then $\set {a, x, b}$ is a finite subdivision of $\... | Let $a, b$ be [[Definition:Real Number|real numbers]] with $a < b$.
Let $f : \closedint a b \to \R$ be a [[Definition:Real Function|function]] of [[Definition:Bounded Variation (Closed Bounded Interval)|bounded variation]].
Then $f$ is [[Definition:Bounded Real-Valued Function|bounded]]. | We use the notation from the [[Definition:Bounded Variation (Closed Bounded Interval)|definition of bounded variation]].
Since $f$ is of [[Definition:Bounded Variation (Closed Bounded Interval)|bounded variation]], there exists $M \ge 0$ such that:
:$\map {V_f} {P ; \closedint a b} \le M$
for all [[Definition:Finite... | Function of Bounded Variation is Bounded | https://proofwiki.org/wiki/Function_of_Bounded_Variation_is_Bounded | https://proofwiki.org/wiki/Function_of_Bounded_Variation_is_Bounded | [
"Bounded Variation"
] | [
"Definition:Real Number",
"Definition:Real Function",
"Definition:Bounded Variation/Closed Bounded Interval",
"Definition:Bounded Mapping/Real-Valued"
] | [
"Definition:Bounded Variation/Closed Bounded Interval",
"Definition:Bounded Variation/Closed Bounded Interval",
"Definition:Subdivision of Interval/Finite",
"Definition:Real Number",
"Definition:Subdivision of Interval/Finite",
"Reverse Triangle Inequality/Real and Complex Fields",
"Definition:Bounded M... |
proofwiki-17203 | Total Variation is Non-Negative | Let $a, b$ be real numbers with $a < b$.
Let $f : \closedint a b \to \R$ be a function of bounded variation.
Let $\map {V_f} {\closedint a b}$ be the total variation of $f$ on $\closedint a b$.
Then:
:$\map {V_f} {\closedint a b} \ge 0$
with equality {{iff}} $f$ is constant. | We use the notation from the definition of bounded variation.
Note that by the definition of absolute value, we have:
:$\size x \ge 0$
for all $x \in \R$.
Let $P$ be a finite subdivision of $\closedint a b$.
Write:
:$P = \set {x_0, x_1, \ldots, x_n}$
with:
:$a = x_0 < x_1 < x_2 < \cdots < x_{n - 1} < x_n = b$
Then:
... | Let $a, b$ be [[Definition:Real Number|real numbers]] with $a < b$.
Let $f : \closedint a b \to \R$ be a [[Definition:Real Function|function]] of [[Definition:Bounded Variation (Closed Bounded Interval)|bounded variation]].
Let $\map {V_f} {\closedint a b}$ be the [[Definition:Total Variation of Real Function on Clos... | We use the notation from the [[Definition:Bounded Variation (Closed Bounded Interval)|definition of bounded variation]].
Note that by the definition of [[Definition:Absolute Value|absolute value]], we have:
:$\size x \ge 0$
for all $x \in \R$.
Let $P$ be a [[Definition:Finite Subdivision|finite subdivision]] of $\... | Total Variation is Non-Negative | https://proofwiki.org/wiki/Total_Variation_is_Non-Negative | https://proofwiki.org/wiki/Total_Variation_is_Non-Negative | [
"Total Variation",
"Total Variation of Real Function",
"Total Variation of Real Function"
] | [
"Definition:Real Number",
"Definition:Real Function",
"Definition:Bounded Variation/Closed Bounded Interval",
"Definition:Total Variation/Real Function/Closed Bounded Interval",
"Definition:Constant Mapping"
] | [
"Definition:Bounded Variation/Closed Bounded Interval",
"Definition:Absolute Value",
"Definition:Subdivision of Interval/Finite",
"Definition:Supremum of Mapping/Real-Valued Function",
"Definition:Constant Mapping",
"Definition:Constant Mapping",
"Definition:Bounded Variation/Closed Bounded Interval"
] |
proofwiki-17204 | Norm Equivalence is Equivalence | Let $X$ be a vector space.
Let $\norm {\, \cdot \,}_a$ and $\norm {\, \cdot \,}_b$ be equivalent norms on $X$.
Denote this relation by $\sim$:
:$\norm {\, \cdot \,}_a \sim \norm {\, \cdot \,}_b$.
Then $\sim$ is an equivalence relation. | === Reflexivity ===
Let $\norm {\, \cdot \,}$ be a norm on $X$.
Then for all $x \in X$ we have that:
:$\norm x = 1 \cdot \norm {x}$
Therefore:
:$1 \cdot \norm x \le \norm x \le 1 \cdot \norm x$
Hence:
:$\norm {\, \cdot \,} \sim \norm {\, \cdot \,}$.
{{qed|lemma}} | Let $X$ be a [[Definition:Vector Space|vector space]].
Let $\norm {\, \cdot \,}_a$ and $\norm {\, \cdot \,}_b$ be [[Definition:Equivalence of Norms|equivalent norms]] on $X$.
Denote this [[Definition:Relation|relation]] by $\sim$:
:$\norm {\, \cdot \,}_a \sim \norm {\, \cdot \,}_b$.
Then $\sim$ is an [[Definition... | === Reflexivity ===
Let $\norm {\, \cdot \,}$ be a [[Definition:Norm|norm]] on $X$.
Then for all $x \in X$ we have that:
:$\norm x = 1 \cdot \norm {x}$
Therefore:
:$1 \cdot \norm x \le \norm x \le 1 \cdot \norm x$
Hence:
:$\norm {\, \cdot \,} \sim \norm {\, \cdot \,}$.
{{qed|lemma}} | Norm Equivalence is Equivalence | https://proofwiki.org/wiki/Norm_Equivalence_is_Equivalence | https://proofwiki.org/wiki/Norm_Equivalence_is_Equivalence | [
"Examples of Equivalence Relations",
"Norm Theory",
"Vector Spaces"
] | [
"Definition:Vector Space",
"Definition:Equivalence of Norms",
"Definition:Relation",
"Definition:Equivalence Relation"
] | [
"Definition:Norm"
] |
proofwiki-17205 | Continuous Non-Negative Real Function with Zero Integral is Zero Function | Let $a, b$ be real numbers with $a < b$.
Let $f : \closedint a b \to \R$ be a continuous function.
Let:
:$\map f x \ge 0$
for all $x \in \closedint a b$.
Let:
:$\ds \int_a^b \map f x \rd x = 0$
Then $\map f x = 0$ for all $x \in \closedint a b$. | From Definite Integral of Constant, if $\map f x = 0$ for all $x \in \closedint a b$, then:
:$\ds \int_a^b \map f x \rd x = 0$
We want to show that if:
:$\ds \int_a^b \map f x \rd x = 0$
then:
:$\map f x = 0$ for all $x \in \closedint a b$.
Since $\map f x \ge 0$, by Relative Sizes of Definite Integrals:
:$\ds \int_a^... | Let $a, b$ be [[Definition:Real Number|real numbers]] with $a < b$.
Let $f : \closedint a b \to \R$ be a [[Definition:Continuous Real Function|continuous function]].
Let:
:$\map f x \ge 0$
for all $x \in \closedint a b$.
Let:
:$\ds \int_a^b \map f x \rd x = 0$
Then $\map f x = 0$ for all $x \in \closedint a... | From [[Definite Integral of Constant]], if $\map f x = 0$ for all $x \in \closedint a b$, then:
:$\ds \int_a^b \map f x \rd x = 0$
We want to show that if:
:$\ds \int_a^b \map f x \rd x = 0$
then:
:$\map f x = 0$ for all $x \in \closedint a b$.
Since $\map f x \ge 0$, by [[Relative Sizes of Definite Integrals]]:... | Continuous Non-Negative Real Function with Zero Integral is Zero Function/Proof 1 | https://proofwiki.org/wiki/Continuous_Non-Negative_Real_Function_with_Zero_Integral_is_Zero_Function | https://proofwiki.org/wiki/Continuous_Non-Negative_Real_Function_with_Zero_Integral_is_Zero_Function/Proof_1 | [
"Definite Integrals",
"Continuous Non-Negative Real Function with Zero Integral is Zero Function"
] | [
"Definition:Real Number",
"Definition:Continuous Real Function"
] | [
"Integral of Constant/Definite",
"Relative Sizes of Definite Integrals",
"Definition:Continuous Real Function/One Side",
"Definition:Real Interval/Endpoints",
"Definition:Continuous Real Function/Point",
"Sum of Integrals on Adjacent Intervals for Continuous Functions",
"Relative Sizes of Definite Integ... |
proofwiki-17206 | Continuous Non-Negative Real Function with Zero Integral is Zero Function | Let $a, b$ be real numbers with $a < b$.
Let $f : \closedint a b \to \R$ be a continuous function.
Let:
:$\map f x \ge 0$
for all $x \in \closedint a b$.
Let:
:$\ds \int_a^b \map f x \rd x = 0$
Then $\map f x = 0$ for all $x \in \closedint a b$. | From Continuous Real Function is Darboux Integrable, $f$ is Darboux integrable on $\closedint a b$.
Let $F : \closedint a b \to \R$ be a real function defined by:
:$\ds \map F x = \int_a^x \map f x \rd x$
We are assured that this function is well-defined, since $f$ is integrable on $\closedint a b$.
From Fundamental T... | Let $a, b$ be [[Definition:Real Number|real numbers]] with $a < b$.
Let $f : \closedint a b \to \R$ be a [[Definition:Continuous Real Function|continuous function]].
Let:
:$\map f x \ge 0$
for all $x \in \closedint a b$.
Let:
:$\ds \int_a^b \map f x \rd x = 0$
Then $\map f x = 0$ for all $x \in \closedint a... | From [[Continuous Real Function is Darboux Integrable]], $f$ is [[Definition:Darboux Integrable Function|Darboux integrable]] on $\closedint a b$.
Let $F : \closedint a b \to \R$ be a [[Definition:Real Function|real function]] defined by:
:$\ds \map F x = \int_a^x \map f x \rd x$
We are assured that this function i... | Continuous Non-Negative Real Function with Zero Integral is Zero Function/Proof 2 | https://proofwiki.org/wiki/Continuous_Non-Negative_Real_Function_with_Zero_Integral_is_Zero_Function | https://proofwiki.org/wiki/Continuous_Non-Negative_Real_Function_with_Zero_Integral_is_Zero_Function/Proof_2 | [
"Definite Integrals",
"Continuous Non-Negative Real Function with Zero Integral is Zero Function"
] | [
"Definition:Real Number",
"Definition:Continuous Real Function"
] | [
"Continuous Real Function is Darboux Integrable",
"Definition:Darboux Integrable Function",
"Definition:Real Function",
"Definition:Well-Defined",
"Definition:Darboux Integrable Function",
"Fundamental Theorem of Calculus/First Part",
"Definition:Continuous Real Function",
"Definition:Differentiable M... |
proofwiki-17207 | Sequential Continuity is Equivalent to Continuity in the Reals/Sufficient Condition | Let $A \subseteq \R$ be a subset of the real numbers.
Let $c \in A$.
Let $f: A \to \R$ be a real function.
Then if $f$ is continuous at $c$:
:for each sequence $\sequence {x_n}$ in $A$ that converges to $c$, the sequence $\sequence {\map f {x_n} }$ converges to $\map f c$. | It suffices to show that if $f$ is discontinuous at $c$:
:there exists a real sequence $\sequence {x_n}$ in $A$ such that $\sequence {x_n}$ converges to $c$ but $\sequence {\map f {x_n} }$ does not converge to $\map f c$.
As $f$ is discontinuous, there exists some $\varepsilon > 0$ such that for all $\delta > 0$:
:ther... | Let $A \subseteq \R$ be a [[Definition:Subset|subset]] of the [[Definition:Real Number|real numbers]].
Let $c \in A$.
Let $f: A \to \R$ be a [[Definition:Real Function|real function]].
Then if $f$ is [[Definition:Continuous Real Function|continuous]] at $c$:
:for each [[Definition:Real Sequence|sequence]] $\sequenc... | It suffices to show that if $f$ is [[Definition:Discontinuous Real Function|discontinuous]] at $c$:
:there exists a [[Definition:Real Sequence|real sequence]] $\sequence {x_n}$ in $A$ such that $\sequence {x_n}$ [[Definition:Convergent Real Sequence|converges]] to $c$ but $\sequence {\map f {x_n} }$ does not [[Definit... | Sequential Continuity is Equivalent to Continuity in the Reals/Sufficient Condition | https://proofwiki.org/wiki/Sequential_Continuity_is_Equivalent_to_Continuity_in_the_Reals/Sufficient_Condition | https://proofwiki.org/wiki/Sequential_Continuity_is_Equivalent_to_Continuity_in_the_Reals/Sufficient_Condition | [
"Sequential Continuity is Equivalent to Continuity in the Reals"
] | [
"Definition:Subset",
"Definition:Real Number",
"Definition:Real Function",
"Definition:Continuous Real Function",
"Definition:Real Sequence",
"Definition:Convergent Sequence/Real Numbers",
"Definition:Real Sequence",
"Definition:Convergent Sequence/Real Numbers"
] | [
"Definition:Discontinuous Mapping/Real Function",
"Definition:Real Sequence",
"Definition:Convergent Sequence/Real Numbers",
"Definition:Convergent Sequence/Real Numbers",
"Definition:Discontinuous Mapping/Real Function",
"Definition:Real Sequence",
"Squeeze Theorem/Sequences/Real Numbers",
"Definitio... |
proofwiki-17208 | Sequential Continuity is Equivalent to Continuity in the Reals/Necessary Condition | Let $A \subseteq \R$ be a subset of the real numbers.
Let $c \in A$.
Let $f : A \to \R$ be a real function.
Then if $f$ is continuous at $c$:
:for each sequence $\sequence {x_n}$ in $A$ that converges to $c$, the sequence $\sequence {\map f {x_n} }$ converges to $\map f c$. | Let $c \in \R$.
Let $\sequence {x_n}$ be a sequence in $A$ that converges to $c$.
Let $\varepsilon \in \R_{> 0}$.
Since $f$ is continuous at $c$, there exists $\delta > 0$ such that:
:for all $x \in A$ with $\size {x - c} < \delta$, we have $\size {\map f x - \map f c} < \varepsilon$.
Additionally, since $\sequence ... | Let $A \subseteq \R$ be a [[Definition:Subset|subset]] of the [[Definition:Real Number|real numbers]].
Let $c \in A$.
Let $f : A \to \R$ be a [[Definition:Real Function|real function]].
Then if $f$ is [[Definition:Continuous Real Function|continuous]] at $c$:
:for each [[Definition:Real Sequence|sequence]] $\seque... | Let $c \in \R$.
Let $\sequence {x_n}$ be a [[Definition:Real Sequence|sequence]] in $A$ that [[Definition:Convergent Real Sequence|converges]] to $c$.
Let $\varepsilon \in \R_{> 0}$.
Since $f$ is [[Definition:Continuous Real Function|continuous]] at $c$, there exists $\delta > 0$ such that:
:for all $x \in A$ wi... | Sequential Continuity is Equivalent to Continuity in the Reals/Necessary Condition | https://proofwiki.org/wiki/Sequential_Continuity_is_Equivalent_to_Continuity_in_the_Reals/Necessary_Condition | https://proofwiki.org/wiki/Sequential_Continuity_is_Equivalent_to_Continuity_in_the_Reals/Necessary_Condition | [
"Sequential Continuity is Equivalent to Continuity in the Reals"
] | [
"Definition:Subset",
"Definition:Real Number",
"Definition:Real Function",
"Definition:Continuous Real Function",
"Definition:Real Sequence",
"Definition:Convergent Sequence/Real Numbers"
] | [
"Definition:Real Sequence",
"Definition:Convergent Sequence/Real Numbers",
"Definition:Continuous Real Function",
"Definition:Convergent Sequence/Real Numbers",
"Definition:Convergent Sequence/Real Numbers"
] |
proofwiki-17209 | Metric Closure and Topological Closure of Subset are Equivalent | Let $M = \struct{A, d}$ be a metric space.
Let $T = \struct{A, \tau}$ be the topological space with the topology induced by $d$.
Let $H \subseteq A$.
Then:
:the metric closure of $H$ in $M$ equals the topological closure of $H$ in $T$ | Let $H^i$ be the set of isolated points of $H$ in $M$.
From Isolated Point in Metric Space iff Isolated Point in Topological Space:
:$H^i$ equals the set of isolated points of $H$ in the topological space $T$.
Let $H'$ be the set of limit points of $H$ in $M$.
From Limit Point in Metric Space iff Limit Point in Topolog... | Let $M = \struct{A, d}$ be a [[Definition:Metric Space|metric space]].
Let $T = \struct{A, \tau}$ be the [[Definition:Topological Space|topological space]] with the [[Definition:Topology Induced by Metric|topology induced]] by $d$.
Let $H \subseteq A$.
Then:
:the [[Definition:Closure (Metric Space)|metric closure]]... | Let $H^i$ be the [[Definition:Set|set]] of [[Definition:Isolated Point of Subset of Metric Space|isolated points]] of $H$ in $M$.
From [[Isolated Point in Metric Space iff Isolated Point in Topological Space]]:
:$H^i$ equals the [[Definition:Set|set]] of [[Definition:Isolated Point (Topology)|isolated points]] of $H$ ... | Metric Closure and Topological Closure of Subset are Equivalent | https://proofwiki.org/wiki/Metric_Closure_and_Topological_Closure_of_Subset_are_Equivalent | https://proofwiki.org/wiki/Metric_Closure_and_Topological_Closure_of_Subset_are_Equivalent | [
"Set Closures"
] | [
"Definition:Metric Space",
"Definition:Topological Space",
"Definition:Topology Induced by Metric",
"Definition:Closure (Topology)/Metric Space",
"Definition:Closure (Topology)"
] | [
"Definition:Set",
"Definition:Isolated Point (Metric Space)/Subset",
"Isolated Point in Metric Space iff Isolated Point in Topological Space",
"Definition:Set",
"Definition:Isolated Point (Topology)",
"Definition:Topological Space",
"Definition:Set",
"Definition:Limit Point/Metric Space",
"Limit Poi... |
proofwiki-17210 | Set together with Omega-Accumulation Points is not necessarily Closed | Let $T = \struct {S, \tau}$ be a topological space.
Let $H \subseteq S$.
Let $\Omega$ denote the set of $\omega$-accumulation points of $H$.
Then it is not necessarily the case that $H \cup \Omega$ is a closed set of $T$. | ;Proof by Counterexample
Let $T = \struct {\R, \tau}$ denote the right order topology on $\R$.
Let $H \subseteq \R$ be a non-empty finite subset of $\R$.
Let $\Omega$ denote the set of $\omega$-accumulation points of $H$.
Since $H$ is finite, it has no $\omega$-accumulation points, that is, $\Omega = \O$.
Therefore $H ... | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $H \subseteq S$.
Let $\Omega$ denote the [[Definition:Set|set]] of [[Definition:Omega-Accumulation Point|$\omega$-accumulation points]] of $H$.
Then it is not necessarily the case that $H \cup \Omega$ is a [[Definition:Closed S... | ;[[Proof by Counterexample]]
Let $T = \struct {\R, \tau}$ denote the [[Definition:Right Order Topology on Real Numbers|right order topology on $\R$]].
Let $H \subseteq \R$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Finite Set|finite]] [[Definition:Subset|subset]] of $\R$.
Let $\Omega$ denote the [[Defi... | Set together with Omega-Accumulation Points is not necessarily Closed | https://proofwiki.org/wiki/Set_together_with_Omega-Accumulation_Points_is_not_necessarily_Closed | https://proofwiki.org/wiki/Set_together_with_Omega-Accumulation_Points_is_not_necessarily_Closed | [
"Set Closures",
"Omega-Accumulation Points"
] | [
"Definition:Topological Space",
"Definition:Set",
"Definition:Omega-Accumulation Point",
"Definition:Closed Set/Topology"
] | [
"Proof by Counterexample",
"Definition:Right Order Topology on Real Numbers",
"Definition:Non-Empty Set",
"Definition:Finite Set",
"Definition:Subset",
"Definition:Set",
"Definition:Omega-Accumulation Point",
"Definition:Finite Set",
"Definition:Omega-Accumulation Point",
"Definition:Finite Set",
... |
proofwiki-17211 | Set together with Condensation Points is not necessarily Closed | Let $T = \struct {S, \tau}$ be a topological space.
Let $H \subseteq S$.
Let $\CC$ denote the set of condensation points of $H$.
Then it is not necessarily the case that $H \cup \CC$ is a closed set of $T$. | ;Proof by Counterexample
Let $T = \struct {\R, \tau}$ denote the right order topology on $\R$.
Let $H \subseteq \R$ be a non-empty finite subset of $\R$.
Let $\CC$ denote the set of condensation points of $H$.
Since $H$ is finite, it has no condensation points, that is, $\CC = \O$.
Therefore $H \cup \CC = H$ is finite.... | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $H \subseteq S$.
Let $\CC$ denote the [[Definition:Set|set]] of [[Definition:Condensation Point|condensation points]] of $H$.
Then it is not necessarily the case that $H \cup \CC$ is a [[Definition:Closed Set (Topology)|closed ... | ;[[Proof by Counterexample]]
Let $T = \struct {\R, \tau}$ denote the [[Definition:Right Order Topology on Real Numbers|right order topology on $\R$]].
Let $H \subseteq \R$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Finite Set|finite]] [[Definition:Subset|subset]] of $\R$.
Let $\CC$ denote the [[Definit... | Set together with Condensation Points is not necessarily Closed | https://proofwiki.org/wiki/Set_together_with_Condensation_Points_is_not_necessarily_Closed | https://proofwiki.org/wiki/Set_together_with_Condensation_Points_is_not_necessarily_Closed | [
"Set Closures",
"Condensation Points"
] | [
"Definition:Topological Space",
"Definition:Set",
"Definition:Condensation Point",
"Definition:Closed Set/Topology"
] | [
"Proof by Counterexample",
"Definition:Right Order Topology on Real Numbers",
"Definition:Non-Empty Set",
"Definition:Finite Set",
"Definition:Subset",
"Definition:Set",
"Definition:Condensation Point",
"Definition:Finite Set",
"Definition:Condensation Point",
"Definition:Finite Set",
"Closed Se... |
proofwiki-17212 | Basis Test for Isolated Point | Let $T = \struct {S, \tau}$ be a topological space.
Let $\BB$ be a synthetic basis of $T$.
Let $H \subseteq S$.
Then $x \in H$ is an isolated point of $H$ {{iff}}:
:$\exists U \in \BB : U \cap H = \set x$ | === Necessary Condition ===
Let $x \in H$ be an isolated point of $H$.
By definition of an isolated point:
:$\exists U \in \tau: U \cap H = \set x$
By definition of a synthetic basis of $T$:
:$\exists V \in \BB: x \in V \subseteq U$
From Set Intersection Preserves Subsets:
:$V \cap H \subseteq U \cap H = \set x$
From... | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $\BB$ be a [[Definition:Synthetic Basis|synthetic basis]] of $T$.
Let $H \subseteq S$.
Then $x \in H$ is an [[Definition:Isolated Point (Topology)|isolated point]] of $H$ {{iff}}:
:$\exists U \in \BB : U \cap H = \set x$ | === Necessary Condition ===
Let $x \in H$ be an [[Definition:Isolated Point (Topology)|isolated point]] of $H$.
By definition of an [[Definition:Isolated Point (Topology)|isolated point]]:
:$\exists U \in \tau: U \cap H = \set x$
By definition of a [[Definition:Synthetic Basis|synthetic basis]] of $T$:
:$\exists V ... | Basis Test for Isolated Point | https://proofwiki.org/wiki/Basis_Test_for_Isolated_Point | https://proofwiki.org/wiki/Basis_Test_for_Isolated_Point | [
"Isolated Points"
] | [
"Definition:Topological Space",
"Definition:Basis (Topology)/Synthetic Basis",
"Definition:Isolated Point (Topology)"
] | [
"Definition:Isolated Point (Topology)",
"Definition:Isolated Point (Topology)",
"Definition:Basis (Topology)/Synthetic Basis",
"Set Intersection Preserves Subsets",
"Singleton of Element is Subset",
"Definition:Set Equality",
"Definition:Basis (Topology)/Synthetic Basis",
"Definition:Isolated Point (T... |
proofwiki-17213 | Basis Test for Limit Point | Let $T = \struct {S, \tau}$ be a topological space.
Let $\BB$ be a synthetic basis of $T$.
Let $H \subseteq S$.
Then $x \in S$ is a limit point of $H$ {{iff}}:
:$\forall U \in \BB : x \in U$ satisfies $H \cap U \setminus \set x \ne \O$ | === Necessary Condition ===
Let $x \in S$ be a limit point of $H$.
By definition of a limit point of $H$:
:$\forall U \in \tau : x \in U$ satisfies $H \cap U \setminus \set x \ne \O$
By definition of a synthetic basis of $T$:
:$\BB \subseteq \tau$
The result follows.
{{qed|lemma}} | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $\BB$ be a [[Definition:Synthetic Basis|synthetic basis]] of $T$.
Let $H \subseteq S$.
Then $x \in S$ is a [[Definition:Limit Point (Topology)|limit point]] of $H$ {{iff}}:
:$\forall U \in \BB : x \in U$ satisfies $H \cap U \se... | === Necessary Condition ===
Let $x \in S$ be a [[Definition:Limit Point (Topology)|limit point]] of $H$.
By definition of a [[Definition:Limit Point (Topology)|limit point]] of $H$:
:$\forall U \in \tau : x \in U$ satisfies $H \cap U \setminus \set x \ne \O$
By definition of a [[Definition:Synthetic Basis|synthetic ... | Basis Test for Limit Point | https://proofwiki.org/wiki/Basis_Test_for_Limit_Point | https://proofwiki.org/wiki/Basis_Test_for_Limit_Point | [
"Limit Points"
] | [
"Definition:Topological Space",
"Definition:Basis (Topology)/Synthetic Basis",
"Definition:Limit Point/Topology"
] | [
"Definition:Limit Point/Topology",
"Definition:Limit Point/Topology",
"Definition:Basis (Topology)/Synthetic Basis",
"Definition:Basis (Topology)/Synthetic Basis",
"Definition:Limit Point/Topology"
] |
proofwiki-17214 | Basis Test for Adherent Point | Let $T = \struct {S, \tau}$ be a topological space.
Let $\BB$ be a synthetic basis of $T$.
Let $H \subseteq S$.
Then $x \in S$ is an adherent point of $H$ {{iff}}:
:$\forall U \in \BB : x \in U$ satisfies $H \cap U \ne \O$ | === Necessary Condition ===
Let $x \in S$ be an adherent point of $H$.
By definition of an adherent point of $H$:
:$\forall U \in \tau : x \in U$ satisfies $H \cap U \ne \O$
By definition of a synthetic basis of $T$:
:$\BB \subseteq \tau$
The result follows.
{{qed|lemma}} | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $\BB$ be a [[Definition:Synthetic Basis|synthetic basis]] of $T$.
Let $H \subseteq S$.
Then $x \in S$ is an [[Definition:Adherent Point of Set|adherent point]] of $H$ {{iff}}:
:$\forall U \in \BB : x \in U$ satisfies $H \cap U ... | === Necessary Condition ===
Let $x \in S$ be an [[Definition:Adherent Point of Set|adherent point]] of $H$.
By definition of an [[Definition:Adherent Point of Set|adherent point]] of $H$:
:$\forall U \in \tau : x \in U$ satisfies $H \cap U \ne \O$
By definition of a [[Definition:Synthetic Basis|synthetic basis]] of ... | Basis Test for Adherent Point | https://proofwiki.org/wiki/Basis_Test_for_Adherent_Point | https://proofwiki.org/wiki/Basis_Test_for_Adherent_Point | [
"Adherent Points of Sets"
] | [
"Definition:Topological Space",
"Definition:Basis (Topology)/Synthetic Basis",
"Definition:Adherent Point of Set"
] | [
"Definition:Adherent Point of Set",
"Definition:Adherent Point of Set",
"Definition:Basis (Topology)/Synthetic Basis",
"Definition:Basis (Topology)/Synthetic Basis",
"Definition:Adherent Point of Set"
] |
proofwiki-17215 | Set of Liouville Numbers is Uncountable | The set of Liouville numbers is uncountable. | By {{Corollary|Liouville's Constant is Transcendental}}, all numbers of the form:
{{begin-eqn}}
{{eqn | l = \sum_{n \mathop \ge 1} \frac {a_n} {10^{n!} }
| r = \frac {a_1} {10^1} + \frac {a_2} {10^2} + \frac {a_3} {10^6} + \frac {a_4} {10^{24} } + \cdots
| c =
}}
{{end-eqn}}
where
:$a_1, a_2, a_3, \ldots \... | The set of [[Definition:Liouville Number|Liouville numbers]] is [[Definition:Uncountable Set|uncountable]]. | By {{Corollary|Liouville's Constant is Transcendental}}, all numbers of the form:
{{begin-eqn}}
{{eqn | l = \sum_{n \mathop \ge 1} \frac {a_n} {10^{n!} }
| r = \frac {a_1} {10^1} + \frac {a_2} {10^2} + \frac {a_3} {10^6} + \frac {a_4} {10^{24} } + \cdots
| c =
}}
{{end-eqn}}
where
:$a_1, a_2, a_3, \ldots \... | Set of Liouville Numbers is Uncountable | https://proofwiki.org/wiki/Set_of_Liouville_Numbers_is_Uncountable | https://proofwiki.org/wiki/Set_of_Liouville_Numbers_is_Uncountable | [
"Transcendental Numbers"
] | [
"Definition:Liouville Number",
"Definition:Uncountable/Set"
] | [
"Definition:Liouville Number",
"Definition:Sequence",
"Definition:Liouville Number",
"Set of Infinite Sequences is Uncountable",
"Definition:Uncountable/Set",
"Definition:Sequence",
"Definition:Liouville Number",
"Definition:Uncountable/Set",
"Definition:Subset",
"Definition:Uncountable/Set",
"S... |
proofwiki-17216 | Isolated Point in Metric Space iff Isolated Point in Topological Space | Let $M = \struct {A, d}$ be a metric space.
Let $T = \struct {A, \tau}$ be the topological space with the topology induced by $d$.
Let $H \subseteq A$.
Let $x \in H$
Then:
:$x$ is an isolated point of $H$ in $M$ {{iff}} $x$ is an isolated point of $H$ in $T$ | From Open Balls form Local Basis for Point of Metric Space, the set:
:$\BB_x = \set {\map {B_\epsilon} x : \epsilon \in \R_{>0} }$
is a local basis of $x$.
From Local Basis Test for Isolated Point:
:$x$ is an isolated point of $H$ in $T$ {{iff}} $\exists \epsilon \in \R_{>0}: \map {B_\epsilon} x \cap H = \set x$
By def... | Let $M = \struct {A, d}$ be a [[Definition:Metric Space|metric space]].
Let $T = \struct {A, \tau}$ be the [[Definition:Topological Space|topological space]] with the [[Definition:Topology Induced by Metric|topology induced]] by $d$.
Let $H \subseteq A$.
Let $x \in H$
Then:
:$x$ is an [[Definition:Isolated Point (M... | From [[Open Balls form Local Basis for Point of Metric Space]], the [[Definition:Set|set]]:
:$\BB_x = \set {\map {B_\epsilon} x : \epsilon \in \R_{>0} }$
is a [[Definition:Local Basis|local basis]] of $x$.
From [[Local Basis Test for Isolated Point]]:
:$x$ is an [[Definition:Isolated Point (Topology)|isolated point]]... | Isolated Point in Metric Space iff Isolated Point in Topological Space | https://proofwiki.org/wiki/Isolated_Point_in_Metric_Space_iff_Isolated_Point_in_Topological_Space | https://proofwiki.org/wiki/Isolated_Point_in_Metric_Space_iff_Isolated_Point_in_Topological_Space | [
"Isolated Points"
] | [
"Definition:Metric Space",
"Definition:Topological Space",
"Definition:Topology Induced by Metric",
"Definition:Isolated Point (Metric Space)",
"Definition:Isolated Point (Topology)"
] | [
"Open Balls form Local Basis for Point of Metric Space",
"Definition:Set",
"Definition:Local Basis",
"Local Basis Test for Isolated Point",
"Definition:Isolated Point (Topology)",
"Definition:Isolated Point (Metric Space)",
"Definition:Isolated Point (Topology)",
"Definition:Isolated Point (Metric Spa... |
proofwiki-17217 | Limit Point in Metric Space iff Limit Point in Topological Space | Let $M = \struct {A, d}$ be a metric space.
Let $T = \struct {A, \tau}$ be the topological space with the topology induced by $d$.
Let $H \subseteq A$.
Then:
:$x \in H$ is a limit point in $M$ {{iff}} $x$ is a limit point in $T$ | From Open Balls form Local Basis for Point of Metric Space, the set:
:$\BB_x = \set{\map {B_\epsilon} x : \epsilon \in \R_{>0}}$
is a local basis of $x$.
From Local Basis Test for Limit Point:
:$x$ is a limit point of $H$ in $T$ {{iff}} $\forall \epsilon \in \R_{>0}: H \cap \map {B_\epsilon} x \setminus \set x \ne \O$
... | Let $M = \struct {A, d}$ be a [[Definition:Metric Space|metric space]].
Let $T = \struct {A, \tau}$ be the [[Definition:Topological Space|topological space]] with the [[Definition:Topology Induced by Metric|topology induced]] by $d$.
Let $H \subseteq A$.
Then:
:$x \in H$ is a [[Definition:Limit Point (Metric Space)... | From [[Open Balls form Local Basis for Point of Metric Space]], the [[Definition:Set|set]]:
:$\BB_x = \set{\map {B_\epsilon} x : \epsilon \in \R_{>0}}$
is a [[Definition:Local Basis|local basis]] of $x$.
From [[Local Basis Test for Limit Point]]:
:$x$ is a [[Definition:Limit Point (Topology)|limit point]] of $H$ in $... | Limit Point in Metric Space iff Limit Point in Topological Space | https://proofwiki.org/wiki/Limit_Point_in_Metric_Space_iff_Limit_Point_in_Topological_Space | https://proofwiki.org/wiki/Limit_Point_in_Metric_Space_iff_Limit_Point_in_Topological_Space | [
"Limit Points"
] | [
"Definition:Metric Space",
"Definition:Topological Space",
"Definition:Topology Induced by Metric",
"Definition:Limit Point/Metric Space",
"Definition:Limit Point/Topology"
] | [
"Open Balls form Local Basis for Point of Metric Space",
"Definition:Set",
"Definition:Local Basis",
"Local Basis Test for Limit Point",
"Definition:Limit Point/Topology",
"Definition:Limit Point/Metric Space",
"Definition:Limit Point/Topology",
"Definition:Limit Point/Metric Space",
"Category:Limit... |
proofwiki-17218 | Boundary of Boundary is not necessarily Equal to Boundary | Let $T = \struct {S, \tau}$ be a topological space.
Let $H \subseteq S$.
Let $\partial H$ denote the boundary of $H$.
While it is true that:
:$\map \partial {\partial H} \subseteq \partial H$
it is not necessarily the case that:
:$\map \partial {\partial H} = \partial H$ | From Boundary of Boundary is Contained in Boundary, we have that:
:$\map \partial {\partial H} \subseteq \partial H$
It remains to be proved that the equality does not always hold.
Proof by Counterexample:
Let $T = \struct {S, \set {\O, S} }$ be an indiscrete topological space.
Let $H \subseteq S$ such that $H \ne \O$ ... | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $H \subseteq S$.
Let $\partial H$ denote the [[Definition:Boundary (Topology)|boundary]] of $H$.
While it is true that:
:$\map \partial {\partial H} \subseteq \partial H$
it is not necessarily the case that:
:$\map \partial {\... | From [[Boundary of Boundary is Contained in Boundary]], we have that:
:$\map \partial {\partial H} \subseteq \partial H$
It remains to be proved that the equality does not always hold.
[[Proof by Counterexample]]:
Let $T = \struct {S, \set {\O, S} }$ be an [[Definition:Indiscrete Space|indiscrete topological space... | Boundary of Boundary is not necessarily Equal to Boundary | https://proofwiki.org/wiki/Boundary_of_Boundary_is_not_necessarily_Equal_to_Boundary | https://proofwiki.org/wiki/Boundary_of_Boundary_is_not_necessarily_Equal_to_Boundary | [
"Set Boundaries"
] | [
"Definition:Topological Space",
"Definition:Boundary (Topology)"
] | [
"Boundary of Boundary is Contained in Boundary",
"Proof by Counterexample",
"Definition:Indiscrete Topology",
"Boundary of Subset of Indiscrete Space",
"Boundary of Boundary of Subset of Indiscrete Space"
] |
proofwiki-17219 | Sum of Functions of Bounded Variation is of Bounded Variation | Let $a, b$ be real numbers with $a < b$.
Let $f, g : \closedint a b \to \R$ be functions of bounded variation.
Let $\map {V_f} {\closedint a b}$ and $\map {V_g} {\closedint a b}$ be the total variations of $f$ and $g$ respectively, on $\closedint a b$.
Then $f + g$ is of bounded variation on $\closedint a b$ with:
:$\... | For each finite subdivision $P$ of $\closedint a b$, write:
:$P = \set {x_0, x_1, \ldots, x_n }$
with:
:$a = x_0 < x_1 < x_2 < \cdots < x_{n - 1} < x_n = b$
Then:
{{begin-eqn}}
{{eqn | l = \map {V_{f + g} } {P ; \closedint a b}
| r = \sum_{i \mathop = 1}^n \size {\map {\paren {f + g} } {x_i} - \map {\paren {f + g} } ... | Let $a, b$ be [[Definition:Real Number|real numbers]] with $a < b$.
Let $f, g : \closedint a b \to \R$ be [[Definition:Real Function|functions]] of [[Definition:Bounded Variation/Closed Bounded Interval|bounded variation]].
Let $\map {V_f} {\closedint a b}$ and $\map {V_g} {\closedint a b}$ be the [[Definition:Total ... | For each [[Definition:Finite Subdivision|finite subdivision]] $P$ of $\closedint a b$, write:
:$P = \set {x_0, x_1, \ldots, x_n }$
with:
:$a = x_0 < x_1 < x_2 < \cdots < x_{n - 1} < x_n = b$
Then:
{{begin-eqn}}
{{eqn | l = \map {V_{f + g} } {P ; \closedint a b}
| r = \sum_{i \mathop = 1}^n \size {\map {\paren {f... | Sum of Functions of Bounded Variation is of Bounded Variation | https://proofwiki.org/wiki/Sum_of_Functions_of_Bounded_Variation_is_of_Bounded_Variation | https://proofwiki.org/wiki/Sum_of_Functions_of_Bounded_Variation_is_of_Bounded_Variation | [
"Total Variation of Real Function",
"Bounded Variation"
] | [
"Definition:Real Number",
"Definition:Real Function",
"Definition:Bounded Variation/Closed Bounded Interval",
"Definition:Total Variation/Real Function/Closed Bounded Interval",
"Definition:Bounded Variation/Closed Bounded Interval",
"Definition:Total Variation/Real Function/Closed Bounded Interval"
] | [
"Definition:Subdivision of Interval/Finite",
"Definition:Bounded Variation/Closed Bounded Interval",
"Triangle Inequality",
"Definition:Bounded Variation/Closed Bounded Interval",
"Definition:Subdivision of Interval/Finite",
"Definition:Bounded Variation/Closed Bounded Interval"
] |
proofwiki-17220 | Open Sets in Vector Spaces with Equivalent Norms Coincide | Let $M_a = \struct {X, \norm {\, \cdot \, }_a}$ and $M_b = \struct {X, \norm {\, \cdot \,}_b}$ be normed vector spaces.
Let $U \subseteq X$ be an open set in $M_a$.
Suppose, $\norm {\, \cdot \, }_a$ and $\norm {\, \cdot \,}_b$ are equivalent norms, i.e. $\norm {\, \cdot \, }_a \sim \norm {\, \cdot \,}_b$.
Then $U$ is a... | By definition of equivalent norms:
:$\exists m,M \in \R_{> 0} : m \le M : \forall x \in X: m \norm x_b \le \norm x_a \le M \norm x_b$
Since $U$ is open in $M_a$:
:$\forall x \in U : \exists \epsilon_a \in \R_{> 0} : \map {B_{\epsilon_a}} x \subseteq U$
where $\map {B_{\epsilon_a}} x$ stands for an open ball, defined a... | Let $M_a = \struct {X, \norm {\, \cdot \, }_a}$ and $M_b = \struct {X, \norm {\, \cdot \,}_b}$ be [[Definition:Normed Vector Space|normed vector spaces]].
Let $U \subseteq X$ be an [[Definition:Open Set in Normed Vector Space|open set]] in $M_a$.
Suppose, $\norm {\, \cdot \, }_a$ and $\norm {\, \cdot \,}_b$ are [[Def... | By definition of [[Definition:Equivalence of Norms|equivalent norms]]:
:$\exists m,M \in \R_{> 0} : m \le M : \forall x \in X: m \norm x_b \le \norm x_a \le M \norm x_b$
Since $U$ is [[Definition:Open Set in Normed Vector Space|open]] in $M_a$:
:$\forall x \in U : \exists \epsilon_a \in \R_{> 0} : \map {B_{\epsilon... | Open Sets in Vector Spaces with Equivalent Norms Coincide | https://proofwiki.org/wiki/Open_Sets_in_Vector_Spaces_with_Equivalent_Norms_Coincide | https://proofwiki.org/wiki/Open_Sets_in_Vector_Spaces_with_Equivalent_Norms_Coincide | [
"Equivalence Relations",
"Norm Theory",
"Vector Spaces"
] | [
"Definition:Normed Vector Space",
"Definition:Open Set/Normed Vector Space",
"Definition:Equivalence of Norms",
"Definition:Open Set/Normed Vector Space"
] | [
"Definition:Equivalence of Norms",
"Definition:Open Set/Normed Vector Space",
"Definition:Open Ball/Normed Vector Space",
"Definition:Open Ball/Normed Vector Space"
] |
proofwiki-17221 | Local Basis Test for Isolated Point | Let $T = \struct {S, \tau}$ be a topological space.
Let $H \subseteq S$.
Let $x \in H$.
Let $\BB_x$ be a local basis of $x$.
Then $x$ is an isolated point of $H$ {{iff}}:
:$\exists U \in \BB_x : U \cap H = \set x$ | === Necessary Condition ===
Let $x \in H$ be an isolated point of $H$.
By definition of an isolated point:
:$\exists U \in \tau: U \cap H = \set x$
By definition of a local basis of $T$:
:$\exists V \in \BB_x : x \in V \subseteq U$
From Set Intersection Preserves Subsets:
:$V \cap H \subseteq U \cap H = \set x$
From ... | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $H \subseteq S$.
Let $x \in H$.
Let $\BB_x$ be a [[Definition:Local Basis|local basis]] of $x$.
Then $x$ is an [[Definition:Isolated Point (Topology)|isolated point]] of $H$ {{iff}}:
:$\exists U \in \BB_x : U \cap H = \set x$ | === Necessary Condition ===
Let $x \in H$ be an [[Definition:Isolated Point (Topology)|isolated point]] of $H$.
By definition of an [[Definition:Isolated Point (Topology)|isolated point]]:
:$\exists U \in \tau: U \cap H = \set x$
By definition of a [[Definition:Local Basis|local basis]] of $T$:
:$\exists V \in \BB_... | Local Basis Test for Isolated Point | https://proofwiki.org/wiki/Local_Basis_Test_for_Isolated_Point | https://proofwiki.org/wiki/Local_Basis_Test_for_Isolated_Point | [
"Isolated Points"
] | [
"Definition:Topological Space",
"Definition:Local Basis",
"Definition:Isolated Point (Topology)"
] | [
"Definition:Isolated Point (Topology)",
"Definition:Isolated Point (Topology)",
"Definition:Local Basis",
"Set Intersection Preserves Subsets",
"Singleton of Element is Subset",
"Definition:Set Equality",
"Definition:Local Basis",
"Definition:Isolated Point (Topology)"
] |
proofwiki-17222 | Local Basis Test for Limit Point | Let $T = \struct {S, \tau}$ be a topological space.
Let $H \subseteq S$.
Let $x \in S$.
Let $\BB_x$ be a local basis of $x$.
Then $x \in S$ is a limit point of $H$ {{iff}}:
:$\forall U \in \BB_x : H \cap U \setminus \set x \ne \O$ | === Necessary Condition ===
Let $x \in S$ be a limit point of $H$.
By definition of a limit point of $H$:
:$\forall U \in \tau : x \in U$ satisfies $H \cap U \setminus \set x \ne \O$
By definition of a local basis of $T$:
:$\BB_x \subseteq \tau$
The result follows.
{{qed|lemma}} | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $H \subseteq S$.
Let $x \in S$.
Let $\BB_x$ be a [[Definition:Local Basis|local basis]] of $x$.
Then $x \in S$ is a [[Definition:Limit Point (Topology)|limit point]] of $H$ {{iff}}:
:$\forall U \in \BB_x : H \cap U \setminus \... | === Necessary Condition ===
Let $x \in S$ be a [[Definition:Limit Point (Topology)|limit point]] of $H$.
By definition of a [[Definition:Limit Point (Topology)|limit point]] of $H$:
:$\forall U \in \tau : x \in U$ satisfies $H \cap U \setminus \set x \ne \O$
By definition of a [[Definition:Local Basis|local basis]] ... | Local Basis Test for Limit Point | https://proofwiki.org/wiki/Local_Basis_Test_for_Limit_Point | https://proofwiki.org/wiki/Local_Basis_Test_for_Limit_Point | [
"Limit Points"
] | [
"Definition:Topological Space",
"Definition:Local Basis",
"Definition:Limit Point/Topology"
] | [
"Definition:Limit Point/Topology",
"Definition:Limit Point/Topology",
"Definition:Local Basis",
"Definition:Local Basis",
"Definition:Limit Point/Topology"
] |
proofwiki-17223 | Local Basis Test for Adherent Point | Let $T = \struct {S, \tau}$ be a topological space.
Let $H \subseteq S$.
Let $x \in S$.
Let $\BB_x$ be a local basis of $x$.
Then $x \in S$ is an adherent point of $H$ {{iff}}:
:$\forall U \in \BB_x : H \cap U \ne \O$ | === Necessary Condition ===
Let $x \in S$ be an adherent point of $H$.
By definition of an adherent point:
:$\forall U \in \tau : x \in U$ satisfies $H \cap U \ne \O$
By definition of a local basis of $T$:
:$\BB_x \subseteq \tau$
The result follows.
{{qed|lemma}} | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $H \subseteq S$.
Let $x \in S$.
Let $\BB_x$ be a [[Definition:Local Basis|local basis]] of $x$.
Then $x \in S$ is an [[Definition:Adherent Point of Set|adherent point of $H$]] {{iff}}:
:$\forall U \in \BB_x : H \cap U \ne \O$ | === Necessary Condition ===
Let $x \in S$ be an [[Definition:Adherent Point of Set|adherent point of $H$]].
By definition of an [[Definition:Adherent Point of Set|adherent point]]:
:$\forall U \in \tau : x \in U$ satisfies $H \cap U \ne \O$
By definition of a [[Definition:Local Basis|local basis]] of $T$:
:$\BB_x \s... | Local Basis Test for Adherent Point | https://proofwiki.org/wiki/Local_Basis_Test_for_Adherent_Point | https://proofwiki.org/wiki/Local_Basis_Test_for_Adherent_Point | [
"Adherent Points of Sets",
"Local Bases"
] | [
"Definition:Topological Space",
"Definition:Local Basis",
"Definition:Adherent Point of Set"
] | [
"Definition:Adherent Point of Set",
"Definition:Adherent Point of Set",
"Definition:Local Basis",
"Definition:Local Basis",
"Definition:Adherent Point of Set"
] |
proofwiki-17224 | Closure of Subset of Metric Space is Closed | Let $M = \struct {A, d}$ be a metric space.
Let $H \subseteq A$ be a subset of $A$.
Let $H^-$ denote the closure of $H$.
Then $H^-$ is a closed set of $M$. | Let $\overline {\paren {H^-} }$ denote the complement of $H^-$.
Let $x \in \overline {\paren {H^-} }$.
By definition of the closure of $H$:
:$x$ is not a limit point of $H$.
So:
:$\exists \epsilon \in \R_{> 0}: \paren {\map {B_\epsilon} x \setminus \set x} \cap H = \O$
From Intersection with Set Difference is Set Diffe... | Let $M = \struct {A, d}$ be a [[Definition:Metric Space|metric space]].
Let $H \subseteq A$ be a [[Definition:Subset|subset]] of $A$.
Let $H^-$ denote the [[Definition:Closure (Metric Space)|closure]] of $H$.
Then $H^-$ is a [[Definition:Closed Set (Metric Space)|closed set]] of $M$. | Let $\overline {\paren {H^-} }$ denote the [[Definition:Relative Complement|complement]] of $H^-$.
Let $x \in \overline {\paren {H^-} }$.
By definition of the [[Definition:Closure (Metric Space)|closure]] of $H$:
:$x$ is not a [[Definition:Limit Point (Topology)|limit point]] of $H$.
So:
:$\exists \epsilon \in \R_{... | Closure of Subset of Metric Space is Closed | https://proofwiki.org/wiki/Closure_of_Subset_of_Metric_Space_is_Closed | https://proofwiki.org/wiki/Closure_of_Subset_of_Metric_Space_is_Closed | [
"Set Closures"
] | [
"Definition:Metric Space",
"Definition:Subset",
"Definition:Closure (Topology)/Metric Space",
"Definition:Closed Set/Metric Space"
] | [
"Definition:Relative Complement",
"Definition:Closure (Topology)/Metric Space",
"Definition:Limit Point/Topology",
"Intersection with Set Difference is Set Difference with Intersection",
"Set Difference with Superset is Empty Set",
"Definition:Closure (Topology)/Metric Space",
"Definition:Isolated Point... |
proofwiki-17225 | Convergent Sequences in Vector Spaces with Equivalent Norms Coincide | Let $M_a = \struct {X, \norm {\, \cdot \, }_a}$ and $M_b = \struct {X, \norm {\, \cdot \,}_b}$ be normed vector spaces.
Let $\sequence {x_n}_{n \mathop \in \N}$ be an convergent sequence in $M_a$.
Suppose, $\norm {\, \cdot \, }_a$ and $\norm {\, \cdot \,}_b$ are equivalent norms, i.e. $\norm {\, \cdot \, }_a \sim \norm... | Let $L \in X$.
Then:
:$\forall \epsilon_a \in \R_{> 0} : \exists N \in \N : \forall n \in \N : n > N \implies \norm {x_n - L}_a < \epsilon_a$
By equivalence of norms:
:$\exists M \in \R_{> 0} : \norm {x_n - L}_b \le M \norm {x_n - L}_a < M \epsilon_a$
Let $\epsilon_b := M \epsilon_a$
Then:
:$\forall \epsilon_b \in \R_{... | Let $M_a = \struct {X, \norm {\, \cdot \, }_a}$ and $M_b = \struct {X, \norm {\, \cdot \,}_b}$ be [[Definition:Normed Vector Space|normed vector spaces]].
Let $\sequence {x_n}_{n \mathop \in \N}$ be an [[Definition:Convergent Sequence in Normed Vector Space|convergent sequence]] in $M_a$.
Suppose, $\norm {\, \cdot \,... | Let $L \in X$.
Then:
:$\forall \epsilon_a \in \R_{> 0} : \exists N \in \N : \forall n \in \N : n > N \implies \norm {x_n - L}_a < \epsilon_a$
By [[Definition:Equivalence of Norms|equivalence of norms]]:
:$\exists M \in \R_{> 0} : \norm {x_n - L}_b \le M \norm {x_n - L}_a < M \epsilon_a$
Let $\epsilon_b := M \epsil... | Convergent Sequences in Vector Spaces with Equivalent Norms Coincide | https://proofwiki.org/wiki/Convergent_Sequences_in_Vector_Spaces_with_Equivalent_Norms_Coincide | https://proofwiki.org/wiki/Convergent_Sequences_in_Vector_Spaces_with_Equivalent_Norms_Coincide | [
"Convergent Sequences (Normed Vector Spaces)",
"Equivalence Relations",
"Norm Theory",
"Vector Spaces",
"Convergent Sequences (Normed Vector Spaces)"
] | [
"Definition:Normed Vector Space",
"Definition:Convergent Sequence/Normed Vector Space",
"Definition:Equivalence of Norms",
"Definition:Convergent Sequence/Normed Vector Space"
] | [
"Definition:Equivalence of Norms",
"Definition:Convergent Sequence/Normed Vector Space"
] |
proofwiki-17226 | Open Mapping is not necessarily Closed Mapping | Let $T_1 = \struct {S_1, \tau_1}$ and $T_2 = \struct {S_2, \tau_2}$ be topological spaces.
Let $f: T_1 \to T_2$ be a mapping which is not a bijection.
Let $f$ be an open mapping.
Then it is not necessarily the case that $f$ is also a closed mapping. | Note that if $f$ is a bijection, the result Bijection is Open iff Closed applies.
It is to be shown that if $f$ is not a bijection, this is not necessarily the case.
This is achieved by Proof by Counterexample:
Let $\struct {\R^2, d}$ be the real number plane with the usual (Euclidean) topology.
Let $\rho: \R^2 \to \R$... | Let $T_1 = \struct {S_1, \tau_1}$ and $T_2 = \struct {S_2, \tau_2}$ be [[Definition:Topological Space|topological spaces]].
Let $f: T_1 \to T_2$ be a [[Definition:Mapping|mapping]] which is not a [[Definition:Bijection|bijection]].
Let $f$ be an [[Definition:Open Mapping|open mapping]].
Then it is not necessarily ... | Note that if $f$ is a [[Definition:Bijection|bijection]], the result [[Bijection is Open iff Closed]] applies.
It is to be shown that if $f$ is not a [[Definition:Bijection|bijection]], this is not necessarily the case.
This is achieved by [[Proof by Counterexample]]:
Let $\struct {\R^2, d}$ be the [[Definition:Rea... | Open Mapping is not necessarily Closed Mapping | https://proofwiki.org/wiki/Open_Mapping_is_not_necessarily_Closed_Mapping | https://proofwiki.org/wiki/Open_Mapping_is_not_necessarily_Closed_Mapping | [
"Open Mappings",
"Closed Mappings"
] | [
"Definition:Topological Space",
"Definition:Mapping",
"Definition:Bijection",
"Definition:Open Mapping",
"Definition:Closed Mapping"
] | [
"Definition:Bijection",
"Bijection is Open iff Closed",
"Definition:Bijection",
"Proof by Counterexample",
"Definition:Euclidean Space/Euclidean Topology/Real Number Plane",
"Definition:Projection (Mapping Theory)/First Projection",
"Projection on Real Euclidean Plane is Open Mapping",
"Definition:Ope... |
proofwiki-17227 | Primitive of Power of x by Cosine of a x/Corollary | :$\ds \int x^m \cos a x \rd x = \sum_{k \mathop = 1}^{m + 1} \paren {m^{\underline {k - 1} } \frac {x^{m + 1 - k} } {a^k} \map {\sin} {x + \dfrac {\pi} 2 \paren {k - 1} } }$
where $m^{\underline {k - 1} }$ denotes the $k - 1$th falling factorial of $m$. | {{begin-eqn}}
{{eqn | n = 1
| l = \int x^m \cos a x \rd x
| r = \frac {x^m \sin a x} a + \frac {m x^{m - 1} \cos a x} {a^2} - \frac {m \paren {m - 1} } {a^2} \int x^{m - 2} \cos a x \rd x
| c = Primitive of Power of x by Cosine of a x
}}
{{eqn | n = 2
| ll= \leadsto
| l = \int x^{m - 2} \c... | :$\ds \int x^m \cos a x \rd x = \sum_{k \mathop = 1}^{m + 1} \paren {m^{\underline {k - 1} } \frac {x^{m + 1 - k} } {a^k} \map {\sin} {x + \dfrac {\pi} 2 \paren {k - 1} } }$
where $m^{\underline {k - 1} }$ denotes the $k - 1$th [[Definition:Falling Factorial|falling factorial]] of $m$. | {{begin-eqn}}
{{eqn | n = 1
| l = \int x^m \cos a x \rd x
| r = \frac {x^m \sin a x} a + \frac {m x^{m - 1} \cos a x} {a^2} - \frac {m \paren {m - 1} } {a^2} \int x^{m - 2} \cos a x \rd x
| c = [[Primitive of Power of x by Cosine of a x]]
}}
{{eqn | n = 2
| ll= \leadsto
| l = \int x^{m - 2... | Primitive of Power of x by Cosine of a x/Corollary | https://proofwiki.org/wiki/Primitive_of_Power_of_x_by_Cosine_of_a_x/Corollary | https://proofwiki.org/wiki/Primitive_of_Power_of_x_by_Cosine_of_a_x/Corollary | [
"Primitives involving Cosine Function"
] | [
"Definition:Falling Factorial"
] | [
"Primitive of Power of x by Cosine of a x",
"Real Multiplication Distributes over Addition",
"Category:Primitives involving Cosine Function"
] |
proofwiki-17228 | Homeomorphism may Exist between Non-Comparable Topologies | Let $S$ be a set.
Let $T_1 = \struct {S, \tau_1}$ and $T_2 = \struct {S, \tau_2}$ be topological spaces defined on the underlying set $S$.
Let $\tau_1$ and $\tau_2$ be non-comparable.
Then it may possibly be the case that $T_1$ and $T_2$ are homeomorphic. | A counterexample is demonstrated in Homeomorphic Non-Comparable Particular Point Topologies.
{{qed}} | Let $S$ be a [[Definition:Set|set]].
Let $T_1 = \struct {S, \tau_1}$ and $T_2 = \struct {S, \tau_2}$ be [[Definition:Topological Space|topological spaces]] defined on the [[Definition:Underlying Set of Topological Space|underlying set]] $S$.
Let $\tau_1$ and $\tau_2$ be non-[[Definition:Comparable Topologies|comparab... | A counterexample is demonstrated in [[Homeomorphic Non-Comparable Particular Point Topologies]].
{{qed}} | Homeomorphism may Exist between Non-Comparable Topologies | https://proofwiki.org/wiki/Homeomorphism_may_Exist_between_Non-Comparable_Topologies | https://proofwiki.org/wiki/Homeomorphism_may_Exist_between_Non-Comparable_Topologies | [
"Homeomorphisms (Topological Spaces)",
"Comparable Topologies"
] | [
"Definition:Set",
"Definition:Topological Space",
"Definition:Underlying Set/Topological Space",
"Definition:Comparable Topologies",
"Definition:Homeomorphism/Topological Spaces"
] | [
"Homeomorphic Non-Comparable Particular Point Topologies"
] |
proofwiki-17229 | Equivalence of Definitions of Limit Point of Filter Basis | {{TFAE|def = Limit Point of Filter Basis}}
Let $T = \struct {S, \tau}$ be a topological space.
Let $\FF$ be a filter on the underlying set $S$ of $T$.
Let $\BB$ be a filter basis of $\FF$. | Let $\FF$ be a filter on $S$.
Let $\BB$ be a filter basis of $\FF$. | {{TFAE|def = Limit Point of Filter Basis}}
Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $\FF$ be a [[Definition:Filter on Set|filter]] on the [[Definition:Underlying Set of Topological Space|underlying set]] $S$ of $T$.
Let $\BB$ be a [[Definition:Filter Basis|filter basis... | Let $\FF$ be a [[Definition:Filter on Set|filter]] on $S$.
Let $\BB$ be a [[Definition:Filter Basis|filter basis]] of $\FF$. | Equivalence of Definitions of Limit Point of Filter Basis | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Limit_Point_of_Filter_Basis | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Limit_Point_of_Filter_Basis | [
"Limit Points of Filter Bases"
] | [
"Definition:Topological Space",
"Definition:Filter on Set",
"Definition:Underlying Set/Topological Space",
"Definition:Filter Basis"
] | [
"Definition:Filter on Set",
"Definition:Filter Basis",
"Definition:Filter Basis",
"Definition:Filter Basis",
"Definition:Filter on Set"
] |
proofwiki-17230 | Richert's Theorem | Let $S = \set {s_1, s_2, \dots}$ be an infinite set of (strictly) positive integers, with the property:
:$s_n < s_{n + 1}$ for every $n \in \N$
Suppose there exists some integers $N, k$ such that every integer $n$ with $N < n \le N + s_{k + 1}$:
:$n$ can be expressed as a sum of distinct elements in $\set {s_1, s_2, \d... | We prove this using First Principle of Mathematical Induction.
Let $\map P n$ be the proposition:
:For every integer $m$ with $N < m \le N + s_{n + 1}$:
::$m$ can be expressed as a sum of distinct elements in $\set {s_1, s_2, \dots, s_n}$. | Let $S = \set {s_1, s_2, \dots}$ be an [[Definition:Infinite Set|infinite set]] of [[Definition:Strictly Positive Integer|(strictly) positive integers]], with the property:
:$s_n < s_{n + 1}$ for every $n \in \N$
Suppose there exists some [[Definition:Integer|integers]] $N, k$ such that every [[Definition:Integer|inte... | We prove this using [[First Principle of Mathematical Induction]].
Let $\map P n$ be the proposition:
:For every [[Definition:Integer|integer]] $m$ with $N < m \le N + s_{n + 1}$:
::$m$ can be expressed as a [[Definition:Integer Addition|sum]] of [[Definition:Distinct|distinct]] elements in $\set {s_1, s_2, \dots, s... | Richert's Theorem | https://proofwiki.org/wiki/Richert's_Theorem | https://proofwiki.org/wiki/Richert's_Theorem | [
"Number Theory"
] | [
"Definition:Infinite Set",
"Definition:Strictly Positive/Integer",
"Definition:Integer",
"Definition:Integer",
"Definition:Addition/Integers",
"Definition:Distinct",
"Definition:Addition/Integers",
"Definition:Distinct"
] | [
"Principle of Mathematical Induction",
"Definition:Integer",
"Definition:Addition/Integers",
"Definition:Distinct",
"Definition:Integer",
"Definition:Addition/Integers",
"Definition:Distinct",
"Definition:Integer",
"Definition:Addition/Integers",
"Definition:Distinct",
"Definition:Integer",
"D... |
proofwiki-17231 | Equivalence of Definitions of Filter Basis | {{TFAE|def = Filter Basis}}
Let $S$ be a set.
Let $\FF$ be a filter on $S$. | === $(1)$ implies $(2)$ ===
Let $\BB$ be a filter basis of $\FF$ by definition $1$.
{{Recall|Filter Basis|index = 1}}
{{:Definition:Filter Basis/Definition 1}}
Let $U \in \FF$.
Then by definition of $\FF$:
:$\exists V \in \BB: V \subseteq U$
Thus $\BB$ is a filter basis of $\FF$ by definition $2$.
{{qed|lemma}} | {{TFAE|def = Filter Basis}}
Let $S$ be a [[Definition:Set|set]].
Let $\FF$ be a [[Definition:Filter on Set|filter]] on $S$. | === $(1)$ implies $(2)$ ===
Let $\BB$ be a [[Definition:Filter Basis/Definition 1|filter basis of $\FF$ by definition $1$]].
{{Recall|Filter Basis|index = 1}}
{{:Definition:Filter Basis/Definition 1}}
Let $U \in \FF$.
Then by definition of $\FF$:
:$\exists V \in \BB: V \subseteq U$
Thus $\BB$ is a [[Definition:Fil... | Equivalence of Definitions of Filter Basis | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Filter_Basis | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Filter_Basis | [
"Filter Bases"
] | [
"Definition:Set",
"Definition:Filter on Set"
] | [
"Definition:Filter Basis/Definition 1",
"Definition:Filter Basis/Definition 2",
"Definition:Filter Basis/Definition 2",
"Definition:Filter Basis/Definition 1"
] |
proofwiki-17232 | Number as Sum of Distinct Primes | For $n \ne 1, 4, 6$, $n$ can be expressed as the sum of distinct primes. | Let $S = \set {s_n}_{n \mathop \in N}$ be the set of primes.
Then $S = \set {2, 3, 5, 7, 11, 13, \dots}$.
By Bertrand-Chebyshev Theorem:
:$s_{n + 1} \le 2 s_n$ for all $n \in \N$.
We observe that every integer $n$ where $6 < n \le 6 + s_6 = 19$ can be expressed as a sum of distinct elements in $\set {s_1, \dots, s_5} =... | For $n \ne 1, 4, 6$, $n$ can be expressed as the [[Definition:Integer Addition|sum]] of [[Definition:Distinct|distinct]] [[Definition:Prime Number|primes]]. | Let $S = \set {s_n}_{n \mathop \in N}$ be the set of [[Definition:Prime Number|primes]].
Then $S = \set {2, 3, 5, 7, 11, 13, \dots}$.
By [[Bertrand-Chebyshev Theorem]]:
:$s_{n + 1} \le 2 s_n$ for all $n \in \N$.
We observe that every [[Definition:Integer|integer]] $n$ where $6 < n \le 6 + s_6 = 19$ can be expressed... | Number as Sum of Distinct Primes | https://proofwiki.org/wiki/Number_as_Sum_of_Distinct_Primes | https://proofwiki.org/wiki/Number_as_Sum_of_Distinct_Primes | [
"Prime Numbers"
] | [
"Definition:Addition/Integers",
"Definition:Distinct",
"Definition:Prime Number"
] | [
"Definition:Prime Number",
"Bertrand-Chebyshev Theorem",
"Definition:Integer",
"Definition:Addition/Integers",
"Definition:Distinct",
"Richert's Theorem",
"Category:Prime Numbers"
] |
proofwiki-17233 | Multiple of Function of Bounded Variation is of Bounded Variation | Let $a, b, k$ be real numbers with $a < b$.
Let $f : \closedint a b \to \R$ be a functions of bounded variation.
Let the total variation of $f$ on $\closedint a b$ be $\map {V_f} {\closedint a b}$.
Then $k f$ is of bounded variation with:
:$\map {V_{k f} } {\closedint a b} = \size k \map {V_f} {\closedint a b}$
where ... | For each finite subdivision $P$ of $\closedint a b$, write:
:$P = \set {x_0, x_1, \ldots, x_n }$
with:
:$a = x_0 < x_1 < x_2 < \cdots < x_{n - 1} < x_n = b$
Then:
{{begin-eqn}}
{{eqn | l = \map {V_{k f} } {P ; \closedint a b}
| r = \sum_{i \mathop = 1}^n \size {k \map f {x_i} - k \map f {x_{i - 1} } }
| c = using th... | Let $a, b, k$ be [[Definition:Real Number|real numbers]] with $a < b$.
Let $f : \closedint a b \to \R$ be a [[Definition:Real Function|functions]] of [[Definition:Bounded Variation (Closed Bounded Interval)|bounded variation]].
Let the [[Definition:Total Variation of Real Function on Closed Bounded Interval|total var... | For each [[Definition:Finite Subdivision|finite subdivision]] $P$ of $\closedint a b$, write:
:$P = \set {x_0, x_1, \ldots, x_n }$
with:
:$a = x_0 < x_1 < x_2 < \cdots < x_{n - 1} < x_n = b$
Then:
{{begin-eqn}}
{{eqn | l = \map {V_{k f} } {P ; \closedint a b}
| r = \sum_{i \mathop = 1}^n \size {k \map f {x_i} - ... | Multiple of Function of Bounded Variation is of Bounded Variation | https://proofwiki.org/wiki/Multiple_of_Function_of_Bounded_Variation_is_of_Bounded_Variation | https://proofwiki.org/wiki/Multiple_of_Function_of_Bounded_Variation_is_of_Bounded_Variation | [
"Bounded Variation",
"Total Variation of Real Function"
] | [
"Definition:Real Number",
"Definition:Real Function",
"Definition:Bounded Variation/Closed Bounded Interval",
"Definition:Total Variation/Real Function/Closed Bounded Interval",
"Definition:Bounded Variation/Closed Bounded Interval",
"Definition:Total Variation/Real Function/Closed Bounded Interval"
] | [
"Definition:Subdivision of Interval/Finite",
"Definition:Bounded Variation/Closed Bounded Interval",
"Definition:Bounded Variation/Closed Bounded Interval",
"Definition:Subdivision of Interval/Finite",
"Definition:Bounded Variation/Closed Bounded Interval",
"Multiple of Supremum",
"Category:Bounded Vari... |
proofwiki-17234 | Difference of Functions of Bounded Variation is of Bounded Variation | Let $a, b$ be real numbers with $a < b$.
Let $f, g : \closedint a b \to \R$ be functions of bounded variation.
Let $\map {V_f} {\closedint a b}$ and $\map {V_g} {\closedint a b}$ be the total variations of $f$ and $g$ respectively, on $\closedint a b$.
Then $f - g$ is of bounded variation with:
:$\map {V_{f - g} } {\c... | By Multiple of Function of Bounded Variation is of Bounded Variation, we have that:
:$-g$ is of bounded variation.
So, by Sum of Functions of Bounded Variation is of Bounded Variation, we have that:
:$f + \paren {-g} = f - g$ is of bounded variation
with:
:$\map {V_{f - g} } {\closedint a b} \le \map {V_f} {\closedi... | Let $a, b$ be [[Definition:Real Number|real numbers]] with $a < b$.
Let $f, g : \closedint a b \to \R$ be [[Definition:Real Function|functions]] of [[Definition:Bounded Variation (Closed Bounded Interval)|bounded variation]].
Let $\map {V_f} {\closedint a b}$ and $\map {V_g} {\closedint a b}$ be the [[Definition:Tota... | By [[Multiple of Function of Bounded Variation is of Bounded Variation]], we have that:
:$-g$ is of [[Definition:Bounded Variation (Closed Bounded Interval)|bounded variation]].
So, by [[Sum of Functions of Bounded Variation is of Bounded Variation]], we have that:
:$f + \paren {-g} = f - g$ is of [[Definition:Bou... | Difference of Functions of Bounded Variation is of Bounded Variation | https://proofwiki.org/wiki/Difference_of_Functions_of_Bounded_Variation_is_of_Bounded_Variation | https://proofwiki.org/wiki/Difference_of_Functions_of_Bounded_Variation_is_of_Bounded_Variation | [
"Total Variation of Real Function",
"Bounded Variation"
] | [
"Definition:Real Number",
"Definition:Real Function",
"Definition:Bounded Variation/Closed Bounded Interval",
"Definition:Total Variation/Real Function/Closed Bounded Interval",
"Definition:Bounded Variation/Closed Bounded Interval",
"Definition:Total Variation/Real Function/Closed Bounded Interval"
] | [
"Multiple of Function of Bounded Variation is of Bounded Variation",
"Definition:Bounded Variation/Closed Bounded Interval",
"Sum of Functions of Bounded Variation is of Bounded Variation",
"Definition:Bounded Variation/Closed Bounded Interval",
"Definition:Total Variation/Real Function/Closed Bounded Inter... |
proofwiki-17235 | Product of Functions of Bounded Variation is of Bounded Variation | Let $a, b$ be real numbers with $a < b$.
Let $f, g : \closedint a b \to \R$ be functions of bounded variation.
Let $\map {V_f} {\closedint a b}$ and $\map {V_g} {\closedint a b}$ be the total variations of $f$ and $g$ respectively.
Then the pointwise product $f \cdot g$ is of bounded variation with:
:$\map {V_{f \cdot... | For each finite subdivision $P$ of $\closedint a b$, write:
:$P = \set {x_0, x_1, \ldots, x_n }$
with:
:$a = x_0 < x_1 < x_2 < \cdots < x_{n - 1} < x_n = b$
By Function of Bounded Variation is Bounded:
:$f$ and $g$ are bounded.
So, there exists $A, B \in \R$ such that:
:$\size {\map f x} \le B$
:$\size {\map g x} \... | Let $a, b$ be [[Definition:Real Number|real numbers]] with $a < b$.
Let $f, g : \closedint a b \to \R$ be [[Definition:Real Function|functions]] of [[Definition:Bounded Variation (Closed Bounded Interval)|bounded variation]].
Let $\map {V_f} {\closedint a b}$ and $\map {V_g} {\closedint a b}$ be the [[Definition:Tota... | For each [[Definition:Finite Subdivision|finite subdivision]] $P$ of $\closedint a b$, write:
:$P = \set {x_0, x_1, \ldots, x_n }$
with:
:$a = x_0 < x_1 < x_2 < \cdots < x_{n - 1} < x_n = b$
By [[Function of Bounded Variation is Bounded]]:
:$f$ and $g$ are [[Definition:Bounded Real-Valued Function|bounded]].
S... | Product of Functions of Bounded Variation is of Bounded Variation | https://proofwiki.org/wiki/Product_of_Functions_of_Bounded_Variation_is_of_Bounded_Variation | https://proofwiki.org/wiki/Product_of_Functions_of_Bounded_Variation_is_of_Bounded_Variation | [
"Total Variation of Real Function",
"Bounded Variation"
] | [
"Definition:Real Number",
"Definition:Real Function",
"Definition:Bounded Variation/Closed Bounded Interval",
"Definition:Total Variation/Real Function/Closed Bounded Interval",
"Definition:Pointwise Multiplication of Real-Valued Functions",
"Definition:Bounded Variation/Closed Bounded Interval",
"Defin... | [
"Definition:Subdivision of Interval/Finite",
"Function of Bounded Variation is Bounded",
"Definition:Bounded Mapping/Real-Valued",
"Definition:Bounded Variation/Closed Bounded Interval",
"Triangle Inequality",
"Definition:Bounded Variation/Closed Bounded Interval",
"Definition:Subdivision of Interval/Fi... |
proofwiki-17236 | Reciprocal of Function of Bounded Variation Bounded away from Zero is of Bounded Variation | Let $a, b$ be real numbers with $a < b$.
Let $f: \closedint a b \to \R$ be functions of bounded variation.
Let $f$ be bounded away from zero.
That is, there exists $M \in \R$ such that:
:$\size {\map f x} \ge M > 0$
for all $x \in \closedint a b$.
Let $\map {V_f} {\closedint a b}$ be the total variation of $f$ on $\... | For each finite subdivision $P$ of $\closedint a b$, write:
:$P = \set {x_0, x_1, \ldots, x_n }$
with:
:$a = x_0 < x_1 < x_2 < \cdots < x_{n - 1} < x_n = b$
Note that:
:$\dfrac 1 {\map f x} \le \dfrac 1 M$
for all $x \in \closedint a b$.
We then have:
{{begin-eqn}}
{{eqn | l = \map {V_{1 / f} } {P ; \closedint a b}
... | Let $a, b$ be [[Definition:Real Number|real numbers]] with $a < b$.
Let $f: \closedint a b \to \R$ be [[Definition:Real Function|functions]] of [[Definition:Bounded Variation (Closed Bounded Interval)|bounded variation]].
Let $f$ be [[Definition:Bounded Real-Valued Function|bounded]] away from zero.
That is, there ... | For each [[Definition:Finite Subdivision|finite subdivision]] $P$ of $\closedint a b$, write:
:$P = \set {x_0, x_1, \ldots, x_n }$
with:
:$a = x_0 < x_1 < x_2 < \cdots < x_{n - 1} < x_n = b$
Note that:
:$\dfrac 1 {\map f x} \le \dfrac 1 M$
for all $x \in \closedint a b$.
We then have:
{{begin-eqn}}
{{eqn | l... | Reciprocal of Function of Bounded Variation Bounded away from Zero is of Bounded Variation | https://proofwiki.org/wiki/Reciprocal_of_Function_of_Bounded_Variation_Bounded_away_from_Zero_is_of_Bounded_Variation | https://proofwiki.org/wiki/Reciprocal_of_Function_of_Bounded_Variation_Bounded_away_from_Zero_is_of_Bounded_Variation | [
"Total Variation of Real Function",
"Bounded Variation"
] | [
"Definition:Real Number",
"Definition:Real Function",
"Definition:Bounded Variation/Closed Bounded Interval",
"Definition:Bounded Mapping/Real-Valued",
"Definition:Total Variation/Real Function/Closed Bounded Interval",
"Definition:Bounded Variation/Closed Bounded Interval",
"Definition:Total Variation/... | [
"Definition:Subdivision of Interval/Finite",
"Definition:Bounded Variation/Closed Bounded Interval",
"Definition:Bounded Variation/Closed Bounded Interval",
"Definition:Subdivision of Interval/Finite",
"Definition:Bounded Variation/Closed Bounded Interval",
"Multiple of Supremum"
] |
proofwiki-17237 | Cauchy Sequences in Vector Spaces with Equivalent Norms Coincide | Let $M_a = \struct {X, \norm {\, \cdot \, }_a}$ and $M_b = \struct {X, \norm {\, \cdot \,}_b}$ be normed vector spaces.
Let $\sequence {x_n}_{n \mathop \in \N}$ be a Cauchy sequence in $M_a$.
Suppose, $\norm {\, \cdot \, }_a$ and $\norm {\, \cdot \,}_b$ are equivalent norms, that is:
:$\norm {\, \cdot \, }_a \sim \norm... | We have that $\sequence {x_n}_{n \mathop \in \N}$ is a Cauchy sequence in $M_a$.
Then:
:$\forall \epsilon_a \in \R_{> 0} : \exists N \in \N : \forall n, m \in \N : n, m > N \implies \norm {x_n - x_m}_a < \epsilon_a$
By equivalence of norms:
:$\exists M \in \R_{> 0} : \norm {x_n - x_m}_b \le M \norm {x_n - x_m}_a < M \e... | Let $M_a = \struct {X, \norm {\, \cdot \, }_a}$ and $M_b = \struct {X, \norm {\, \cdot \,}_b}$ be [[Definition:Normed Vector Space|normed vector spaces]].
Let $\sequence {x_n}_{n \mathop \in \N}$ be a [[Definition:Cauchy Sequence in Normed Vector Space|Cauchy sequence]] in $M_a$.
Suppose, $\norm {\, \cdot \, }_a$ and... | We have that $\sequence {x_n}_{n \mathop \in \N}$ is a [[Definition:Cauchy Sequence in Normed Vector Space|Cauchy sequence]] in $M_a$.
Then:
:$\forall \epsilon_a \in \R_{> 0} : \exists N \in \N : \forall n, m \in \N : n, m > N \implies \norm {x_n - x_m}_a < \epsilon_a$
By [[Definition:Equivalence of Norms|equivalenc... | Cauchy Sequences in Vector Spaces with Equivalent Norms Coincide | https://proofwiki.org/wiki/Cauchy_Sequences_in_Vector_Spaces_with_Equivalent_Norms_Coincide | https://proofwiki.org/wiki/Cauchy_Sequences_in_Vector_Spaces_with_Equivalent_Norms_Coincide | [
"Equivalence Relations",
"Norm Theory",
"Vector Spaces",
"Cauchy Sequences"
] | [
"Definition:Normed Vector Space",
"Definition:Cauchy Sequence/Normed Vector Space",
"Definition:Equivalence of Norms",
"Definition:Cauchy Sequence/Normed Vector Space"
] | [
"Definition:Cauchy Sequence/Normed Vector Space",
"Definition:Equivalence of Norms",
"Definition:Cauchy Sequence/Normed Vector Space"
] |
proofwiki-17238 | Differentiable Function of Bounded Variation may not have Bounded Derivative | Let $a, b$ be real numbers with $a < b$.
Let $f : \closedint a b \to \R$ be a continuous function of bounded variation.
Let $f$ be differentiable on $\openint a b$.
Then $f'$ is not necessarily bounded. | Proof by Counterexample:
Take $a = 0$, $b = 1$.
Let $f : \closedint 0 1 \to \R$ have:
:$\map f x = \sqrt x$
for all $x \in \closedint 0 1$.
Note that $f$ is increasing, so by Monotone Function is of Bounded Variation:
:$f$ is of bounded variation.
By Derivative of Power, $f$ is differentiable on $\openint 0 1$ with... | Let $a, b$ be [[Definition:Real Number|real numbers]] with $a < b$.
Let $f : \closedint a b \to \R$ be a [[Definition:Continuous Real Function|continuous function]] of [[Definition:Bounded Variation (Closed Bounded Interval)|bounded variation]].
Let $f$ be [[Definition:Differentiable Real Function|differentiable]] on... | [[Proof by Counterexample]]:
Take $a = 0$, $b = 1$.
Let $f : \closedint 0 1 \to \R$ have:
:$\map f x = \sqrt x$
for all $x \in \closedint 0 1$.
Note that $f$ is [[Definition:Increasing Real Function|increasing]], so by [[Monotone Function is of Bounded Variation]]:
:$f$ is of [[Definition:Bounded Variation (C... | Differentiable Function of Bounded Variation may not have Bounded Derivative | https://proofwiki.org/wiki/Differentiable_Function_of_Bounded_Variation_may_not_have_Bounded_Derivative | https://proofwiki.org/wiki/Differentiable_Function_of_Bounded_Variation_may_not_have_Bounded_Derivative | [
"Bounded Variation"
] | [
"Definition:Real Number",
"Definition:Continuous Real Function",
"Definition:Bounded Variation/Closed Bounded Interval",
"Definition:Differentiable Mapping/Real Function",
"Definition:Bounded Mapping/Real-Valued"
] | [
"Proof by Counterexample",
"Definition:Increasing/Real Function",
"Monotone Function is of Bounded Variation",
"Definition:Bounded Variation/Closed Bounded Interval",
"Power Rule for Derivatives",
"Definition:Differentiable Mapping/Real Function",
"Definition:Derivative/Real Function",
"Definition:Bou... |
proofwiki-17239 | Existence of Urysohn Function does not guarantee Normal Space | Let $T = \struct {S, \tau}$ be a regular space.
Let $T$ have the property that:
:For all closed sets $A, B \subseteq S$ of $T$ such that $A \cap B = \O$, there exists an Urysohn function for $A$ and $B$.
Then it is not necessarily the case that $T$ is a normal space. | Let $T$ have the specified property.
By definition of a normal space, for $T$ to be normal, it has to be both $T_4$ space and a $T_1$ space.
From Urysohn's Lemma Converse, $T$ is a $T_4$ space.
It remains to be shown that $T$ is not necessarily a $T_1$ space.
;Proof by Counterexample
Let $S$ be a set and let $\PP$ be a... | Let $T = \struct {S, \tau}$ be a [[Definition:Regular Space|regular space]].
Let $T$ have the property that:
:For all [[Definition:Closed Set (Topology)|closed sets]] $A, B \subseteq S$ of $T$ such that $A \cap B = \O$, there exists an [[Definition:Urysohn Function|Urysohn function]] for $A$ and $B$.
Then it is not... | Let $T$ have the specified property.
By definition of a [[Definition:Normal Space|normal space]], for $T$ to be [[Definition:Normal Space|normal]], it has to be both [[Definition:T4 Space|$T_4$ space]] and a [[Definition:T1 Space|$T_1$ space]].
From [[Urysohn's Lemma Converse]], $T$ is a [[Definition:T4 Space|$T_4$ s... | Existence of Urysohn Function does not guarantee Normal Space | https://proofwiki.org/wiki/Existence_of_Urysohn_Function_does_not_guarantee_Normal_Space | https://proofwiki.org/wiki/Existence_of_Urysohn_Function_does_not_guarantee_Normal_Space | [
"Normal Spaces",
"Urysohn Functions"
] | [
"Definition:Regular Space",
"Definition:Closed Set/Topology",
"Definition:Urysohn Function",
"Definition:Normal Space"
] | [
"Definition:Normal Space",
"Definition:Normal Space",
"Definition:T4 Space",
"Definition:T1 Space",
"Urysohn's Lemma Converse",
"Definition:T4 Space",
"Definition:T1 Space",
"Proof by Counterexample",
"Definition:Set",
"Definition:Set Partition",
"Definition:Trivial Partition/Partition of Single... |
proofwiki-17240 | T3.5 Space is not necessarily T2 | Let $T = \struct {S, \tau}$ be a be a $T_{3 \frac 1 2}$ space.
Then it is not necessarily the case that $T$ is a $T_2$ (Hausdorff) space. | ;Proof by Counterexample
Let $S$ be a set.
Pet $\PP$ be a partition on $S$ which is specifically not the (trivial) partition of singletons.
Let $T = \struct {S, \tau}$ be the partition space whose basis is $\PP$.
From Partition Space is $T_{3 \frac 1 2}$, we have that $T$ is a $T_{3 \frac 1 2}$ space.
From Partition Sp... | Let $T = \struct {S, \tau}$ be a be a [[Definition:T3.5 Space|$T_{3 \frac 1 2}$ space]].
Then it is not necessarily the case that $T$ is a [[Definition:T2 Space|$T_2$ (Hausdorff) space]]. | ;[[Proof by Counterexample]]
Let $S$ be a [[Definition:Set|set]].
Pet $\PP$ be a [[Definition:Partition (Set Theory)|partition]] on $S$ which is specifically not the [[Definition:Partition of Singletons|(trivial) partition of singletons]].
Let $T = \struct {S, \tau}$ be the [[Definition:Partition Space|partition spa... | T3.5 Space is not necessarily T2 | https://proofwiki.org/wiki/T3.5_Space_is_not_necessarily_T2 | https://proofwiki.org/wiki/T3.5_Space_is_not_necessarily_T2 | [
"T3.5 Spaces",
"Hausdorff Spaces"
] | [
"Definition:T3.5 Space",
"Definition:T2 Space"
] | [
"Proof by Counterexample",
"Definition:Set",
"Definition:Set Partition",
"Definition:Trivial Partition/Partition of Singletons",
"Definition:Partition Topology",
"Basis for Partition Topology",
"Partition Space is T3.5",
"Definition:T3.5 Space",
"Partition Space is not T2",
"Definition:T2 Space"
] |
proofwiki-17241 | T0 Space is not necessarily T1 | Let $T = \struct {S, \tau}$ be a be a $T_0$ space.
Then it is not necessarily the case that $T$ is a $T_1$ space. | ;Proof by Counterexample
Let $T = \struct {S, \tau_p}$ be a particular point space such that $S$ is not a singleton.
From Particular Point Space is $T_0$, we have that $T$ is a $T_0$ space.
From Non-Trivial Particular Point Space is not $T_1$, $T$ is not a $T_1$ space.
The result follows.
{{qed}} | Let $T = \struct {S, \tau}$ be a be a [[Definition:T0 Space|$T_0$ space]].
Then it is not necessarily the case that $T$ is a [[Definition:T1 Space|$T_1$ space]]. | ;[[Proof by Counterexample]]
Let $T = \struct {S, \tau_p}$ be a [[Definition:Particular Point Space|particular point space]] such that $S$ is not a [[Definition:Singleton|singleton]].
From [[Particular Point Space is T0 |Particular Point Space is $T_0$]], we have that $T$ is a [[Definition:T0 Space|$T_0$ space]].
F... | T0 Space is not necessarily T1/Proof 1 | https://proofwiki.org/wiki/T0_Space_is_not_necessarily_T1 | https://proofwiki.org/wiki/T0_Space_is_not_necessarily_T1/Proof_1 | [
"T0 Space is not necessarily T1",
"T0 Spaces",
"T1 Spaces"
] | [
"Definition:T0 Space",
"Definition:T1 Space"
] | [
"Proof by Counterexample",
"Definition:Particular Point Topology",
"Definition:Singleton",
"Particular Point Space is T0 ",
"Definition:T0 Space",
"Non-Trivial Particular Point Space is not T1",
"Definition:T1 Space"
] |
proofwiki-17242 | T0 Space is not necessarily T1 | Let $T = \struct {S, \tau}$ be a be a $T_0$ space.
Then it is not necessarily the case that $T$ is a $T_1$ space. | ;Proof by Counterexample
Let $T$ be the overlapping interval space.
From Overlapping Interval Space fulfils no Separation Axioms but $T_0$, we have that $T$ satisfies none of the Tychonoff separation axioms except for the $T_0$ axiom.
That is:
:$T$ is a $T_0$ space
but:
:$T$ is not a $T_1$ space.
The result follows.
{{... | Let $T = \struct {S, \tau}$ be a be a [[Definition:T0 Space|$T_0$ space]].
Then it is not necessarily the case that $T$ is a [[Definition:T1 Space|$T_1$ space]]. | ;[[Proof by Counterexample]]
Let $T$ be the [[Definition:Overlapping Interval Topology|overlapping interval space]].
From [[Overlapping Interval Space fulfils no Separation Axioms but T0|Overlapping Interval Space fulfils no Separation Axioms but $T_0$]], we have that $T$ satisfies none of the [[Definition:Tychonoff ... | T0 Space is not necessarily T1/Proof 2 | https://proofwiki.org/wiki/T0_Space_is_not_necessarily_T1 | https://proofwiki.org/wiki/T0_Space_is_not_necessarily_T1/Proof_2 | [
"T0 Space is not necessarily T1",
"T0 Spaces",
"T1 Spaces"
] | [
"Definition:T0 Space",
"Definition:T1 Space"
] | [
"Proof by Counterexample",
"Definition:Overlapping Interval Topology",
"Overlapping Interval Space fulfils no Separation Axioms but T0",
"Definition:Tychonoff Separation Axioms",
"Definition:T0 Space",
"Definition:T0 Space",
"Definition:T1 Space"
] |
proofwiki-17243 | Existence of Compact Space which Satisfies No Separation Axioms | There exists at least one example of a compact space for which none of the Tychonoff separation axioms are satisfied. | Let $T_S = \struct {S, \tau_S}$ be a finite complement topology on an infinite set $S$.
Let $D = \struct {A, \set {\O, A} }$ be the indiscrete space on an arbitrary doubleton $A = \set {a, b}$.
Let $T = T_S \times D$ be the double pointed topology on $T_S$.
From Double Pointed Finite Complement Topology is Compact, $T$... | There exists at least one example of a [[Definition:Compact Topological Space|compact space]] for which none of the [[Definition:Tychonoff Separation Axioms|Tychonoff separation axioms]] are satisfied. | Let $T_S = \struct {S, \tau_S}$ be a [[Definition:Finite Complement Topology|finite complement topology]] on an [[Definition:Infinite Set|infinite set]] $S$.
Let $D = \struct {A, \set {\O, A} }$ be the [[Definition:Indiscrete Space|indiscrete space]] on an [[Definition:Arbitrary|arbitrary]] [[Definition:Doubleton|doub... | Existence of Compact Space which Satisfies No Separation Axioms | https://proofwiki.org/wiki/Existence_of_Compact_Space_which_Satisfies_No_Separation_Axioms | https://proofwiki.org/wiki/Existence_of_Compact_Space_which_Satisfies_No_Separation_Axioms | [
"Separation Axioms",
"Compact Topological Spaces"
] | [
"Definition:Compact Topological Space",
"Definition:Tychonoff Separation Axioms"
] | [
"Definition:Finite Complement Topology",
"Definition:Infinite Set",
"Definition:Indiscrete Topology",
"Definition:Arbitrary",
"Definition:Doubleton",
"Definition:Double Pointed Topology",
"Double Pointed Finite Complement Topology is Compact",
"Definition:Compact Topological Space",
"Double Pointed ... |
proofwiki-17244 | Everywhere Dense iff Interior of Complement is Empty | Let $T = \struct {S, \tau}$ be a topological space.
Let $A \subset S$.
Then $A$ is everywhere dense {{iff}}:
:$\paren {\relcomp S A}^\circ = \O$
where $A^\circ$ is the interior of $A$. | By definition of everywhere dense, $A$ is everywhere dense {{iff}}:
:$A^- = S$
where $A^-$ is the closure of $A$.
That happens {{iff}}:
{{begin-eqn}}
{{eqn | l = \paren {\relcomp S A}^\circ
| r = \relcomp S {A^-}
| c = Complement of Interior equals Closure of Complement
}}
{{eqn | r = \relcomp S S
}}
{{eqn ... | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $A \subset S$.
Then $A$ is [[Definition:Everywhere Dense|everywhere dense]] {{iff}}:
:$\paren {\relcomp S A}^\circ = \O$
where $A^\circ$ is the [[Definition:Interior (Topology)|interior]] of $A$. | By definition of [[Definition:Everywhere Dense|everywhere dense]], $A$ is [[Definition:Everywhere Dense|everywhere dense]] {{iff}}:
:$A^- = S$
where $A^-$ is the [[Definition:Closure (Topology)|closure]] of $A$.
That happens {{iff}}:
{{begin-eqn}}
{{eqn | l = \paren {\relcomp S A}^\circ
| r = \relcomp S {A^-}
... | Everywhere Dense iff Interior of Complement is Empty | https://proofwiki.org/wiki/Everywhere_Dense_iff_Interior_of_Complement_is_Empty | https://proofwiki.org/wiki/Everywhere_Dense_iff_Interior_of_Complement_is_Empty | [
"Denseness",
"Set Interiors"
] | [
"Definition:Topological Space",
"Definition:Everywhere Dense",
"Definition:Interior (Topology)"
] | [
"Definition:Everywhere Dense",
"Definition:Everywhere Dense",
"Definition:Closure (Topology)",
"Complement of Interior equals Closure of Complement",
"Relative Complement with Self is Empty Set",
"Category:Denseness",
"Category:Set Interiors"
] |
proofwiki-17245 | Greatest Set is Unique | Let $S$ be a set.
Let $\powerset S$ be the power set of $S$.
Let $\TT \subseteq \powerset S$ be a subset of $\powerset S$.
Then the greatest set of $\TT$, if it exists, must be unique. | Let $A, B \in \TT$ both be greatest sets of $\TT$.
Since $A$ is the greatest set:
:$B \subseteq A$
Since $B$ is the greatest set:
:$A \subseteq B$
Hence, by definition of set equality:
:$A = B$
Therefore the greatest set of $\TT$ is unique.
{{qed}}
Category:Greatest Set
Category:Greatest Elements
Category:Set Theory
oi... | Let $S$ be a [[Definition:Set|set]].
Let $\powerset S$ be the [[Definition:Power Set|power set]] of $S$.
Let $\TT \subseteq \powerset S$ be a [[Definition:Subset|subset]] of $\powerset S$.
Then the [[Definition:Greatest Set by Set Inclusion|greatest set]] of $\TT$, if it exists, must be [[Definition:Unique|unique]]... | Let $A, B \in \TT$ both be [[Definition:Greatest Set by Set Inclusion|greatest sets]] of $\TT$.
Since $A$ is the [[Definition:Greatest Set by Set Inclusion|greatest set]]:
:$B \subseteq A$
Since $B$ is the [[Definition:Greatest Set by Set Inclusion|greatest set]]:
:$A \subseteq B$
Hence, by definition of [[Definiti... | Greatest Set is Unique | https://proofwiki.org/wiki/Greatest_Set_is_Unique | https://proofwiki.org/wiki/Greatest_Set_is_Unique | [
"Greatest Set",
"Greatest Elements",
"Set Theory"
] | [
"Definition:Set",
"Definition:Power Set",
"Definition:Subset",
"Definition:Greatest Set by Set Inclusion",
"Definition:Unique"
] | [
"Definition:Greatest Set by Set Inclusion",
"Definition:Greatest Set by Set Inclusion",
"Definition:Greatest Set by Set Inclusion",
"Definition:Set Equality",
"Definition:Greatest Set by Set Inclusion",
"Definition:Unique",
"Category:Greatest Set",
"Category:Greatest Elements",
"Category:Set Theory"... |
proofwiki-17246 | Greatest Set may not Exist | Let $S$ be a set.
Let $\powerset S$ be the power set of $S$.
Let $\TT \subseteq \powerset S$ be a subset of $\powerset S$.
The greatest set of $\TT$ may not exist. | Let $S = \set {0, 1}$ and $\TT = \set {\set 0, \set 1} \in \powerset S$.
Then since $\set 0 \nsubseteq \set 1$:
:$\set 1$ is not the greatest set of $\TT$.
Similarly, since $\set 1 \nsubseteq \set 0$:
:$\set 0$ is not the greatest set of $\TT$.
Therefore $\TT$ has no greatest set.
{{qed}}
Category:Greatest Set
Category... | Let $S$ be a [[Definition:Set|set]].
Let $\powerset S$ be the [[Definition:Power Set|power set]] of $S$.
Let $\TT \subseteq \powerset S$ be a [[Definition:Subset|subset]] of $\powerset S$.
The [[Definition:Greatest Set by Set Inclusion|greatest set]] of $\TT$ may not exist. | Let $S = \set {0, 1}$ and $\TT = \set {\set 0, \set 1} \in \powerset S$.
Then since $\set 0 \nsubseteq \set 1$:
:$\set 1$ is not the [[Definition:Greatest Set by Set Inclusion|greatest set]] of $\TT$.
Similarly, since $\set 1 \nsubseteq \set 0$:
:$\set 0$ is not the [[Definition:Greatest Set by Set Inclusion|greatest... | Greatest Set may not Exist | https://proofwiki.org/wiki/Greatest_Set_may_not_Exist | https://proofwiki.org/wiki/Greatest_Set_may_not_Exist | [
"Greatest Set",
"Greatest Elements",
"Set Theory"
] | [
"Definition:Set",
"Definition:Power Set",
"Definition:Subset",
"Definition:Greatest Set by Set Inclusion"
] | [
"Definition:Greatest Set by Set Inclusion",
"Definition:Greatest Set by Set Inclusion",
"Definition:Greatest Set by Set Inclusion",
"Category:Greatest Set",
"Category:Greatest Elements",
"Category:Set Theory"
] |
proofwiki-17247 | Smallest Set is Unique | Let $S$ be a set.
Let $\powerset S$ be the power set of $S$.
Let $\TT \subseteq \powerset S$ be a subset of $\powerset S$.
Then the smallest set of $\TT$, if it exists, must be unique. | Let $A, B \in \TT$ both be smallest sets of $\TT$.
Since $A$ is the smallest set:
:$A \subseteq B$
Since $B$ is the smallest set:
:$B \subseteq A$
Hence, by definition of set equality:
:$A = B$
Therefore the smallest set of $\TT$ is unique.
{{qed}}
Category:Set Theory
Category:Smallest Elements
d0d9b0rrqd44jwrzu9n6a5t4... | Let $S$ be a [[Definition:Set|set]].
Let $\powerset S$ be the [[Definition:Power Set|power set]] of $S$.
Let $\TT \subseteq \powerset S$ be a [[Definition:Subset|subset]] of $\powerset S$.
Then the [[Definition:Smallest Set by Set Inclusion|smallest set]] of $\TT$, if it exists, must be unique. | Let $A, B \in \TT$ both be [[Definition:Smallest Set by Set Inclusion|smallest sets]] of $\TT$.
Since $A$ is the [[Definition:Smallest Set by Set Inclusion|smallest set]]:
:$A \subseteq B$
Since $B$ is the [[Definition:Smallest Set by Set Inclusion|smallest set]]:
:$B \subseteq A$
Hence, by definition of [[Definiti... | Smallest Set is Unique | https://proofwiki.org/wiki/Smallest_Set_is_Unique | https://proofwiki.org/wiki/Smallest_Set_is_Unique | [
"Set Theory",
"Smallest Elements"
] | [
"Definition:Set",
"Definition:Power Set",
"Definition:Subset",
"Definition:Smallest Set by Set Inclusion"
] | [
"Definition:Smallest Set by Set Inclusion",
"Definition:Smallest Set by Set Inclusion",
"Definition:Smallest Set by Set Inclusion",
"Definition:Set Equality",
"Definition:Smallest Set by Set Inclusion",
"Definition:Unique",
"Category:Set Theory",
"Category:Smallest Elements"
] |
proofwiki-17248 | Smallest Set may not Exist | Let $S$ be a set.
Let $\powerset S$ be the power set of $S$.
Let $\TT \subseteq \powerset S$ be a subset of $\powerset S$.
The smallest set of $\TT$ may not exist. | Let $S = \set {0, 1}$ and $\TT = \set {\set 0, \set 1} \in \powerset S$.
Then since $\set 0 \nsubseteq \set 1$:
:$\set 0$ is not the smallest set of $\TT$.
Similarly, since $\set 1 \nsubseteq \set 0$:
:$\set 1$ is not the smallest set of $\TT$.
Therefore $\TT$ has no smallest set.
{{qed}}
Category:Set Theory
Category:S... | Let $S$ be a [[Definition:Set|set]].
Let $\powerset S$ be the [[Definition:Power Set|power set]] of $S$.
Let $\TT \subseteq \powerset S$ be a [[Definition:Subset|subset]] of $\powerset S$.
The [[Definition:Smallest Set by Set Inclusion|smallest set]] of $\TT$ may not exist. | Let $S = \set {0, 1}$ and $\TT = \set {\set 0, \set 1} \in \powerset S$.
Then since $\set 0 \nsubseteq \set 1$:
:$\set 0$ is not the [[Definition:Smallest Set by Set Inclusion|smallest set]] of $\TT$.
Similarly, since $\set 1 \nsubseteq \set 0$:
:$\set 1$ is not the [[Definition:Smallest Set by Set Inclusion|smallest... | Smallest Set may not Exist | https://proofwiki.org/wiki/Smallest_Set_may_not_Exist | https://proofwiki.org/wiki/Smallest_Set_may_not_Exist | [
"Set Theory",
"Smallest Elements"
] | [
"Definition:Set",
"Definition:Power Set",
"Definition:Subset",
"Definition:Smallest Set by Set Inclusion"
] | [
"Definition:Smallest Set by Set Inclusion",
"Definition:Smallest Set by Set Inclusion",
"Definition:Smallest Set by Set Inclusion",
"Category:Set Theory",
"Category:Smallest Elements"
] |
proofwiki-17249 | Mapping is Surjection iff Direct Image Mapping is Surjection | Let $f: S \to T$ be a mapping.
Let $f^\to: \powerset S \to \powerset T$ be the direct image mapping of $f$.
Then:
:$f^\to$ is a surjection.
{{iff}}
:$f: S \to T$ is also a surjection. | === Necessary Condition ===
Follows from Direct Image Mapping of Surjection is Surjection.
{{qed|lemma}} | Let $f: S \to T$ be a [[Definition:Mapping|mapping]].
Let $f^\to: \powerset S \to \powerset T$ be the [[Definition:Direct Image Mapping of Mapping|direct image mapping]] of $f$.
Then:
:$f^\to$ is a [[Definition:Surjection|surjection]].
{{iff}}
:$f: S \to T$ is also a [[Definition:Surjection|surjection]]. | === Necessary Condition ===
Follows from [[Direct Image Mapping of Surjection is Surjection]].
{{qed|lemma}} | Mapping is Surjection iff Direct Image Mapping is Surjection | https://proofwiki.org/wiki/Mapping_is_Surjection_iff_Direct_Image_Mapping_is_Surjection | https://proofwiki.org/wiki/Mapping_is_Surjection_iff_Direct_Image_Mapping_is_Surjection | [
"Surjections",
"Direct Image Mappings"
] | [
"Definition:Mapping",
"Definition:Direct Image Mapping/Mapping",
"Definition:Surjection",
"Definition:Surjection"
] | [
"Direct Image Mapping of Surjection is Surjection"
] |
proofwiki-17250 | Cauchy Sequence in Metric Space is not necessarily Convergent | Let $M = \struct {A, d}$ be a metric space.
Let $\sequence {x_n}$ be a Cauchy sequence in $M$.
Then it is not necessarily the case that $M$ is a convergent sequence in $M$. | Let $A \subseteq \R$ be the set of all points on $\R$ defined as:
:$A := \set {\dfrac 1 n : n \in \Z_{>0} }$
Let $M = \struct {A, \tau_d}$ be the integer reciprocal space under the usual (Euclidean) topology.
Let $\sequence {x_n}$ be a sequence in $A$ that converges to the limit $l \in A$.
From Integer Reciprocal Space... | Let $M = \struct {A, d}$ be a [[Definition:Metric Space|metric space]].
Let $\sequence {x_n}$ be a [[Definition:Cauchy Sequence in Metric Space|Cauchy sequence]] in $M$.
Then it is not necessarily the case that $M$ is a [[Definition:Convergent Sequence in Metric Space|convergent sequence]] in $M$. | Let $A \subseteq \R$ be the [[Definition:Set|set]] of all points on $\R$ defined as:
:$A := \set {\dfrac 1 n : n \in \Z_{>0} }$
Let $M = \struct {A, \tau_d}$ be the [[Definition:Integer Reciprocal Space|integer reciprocal space]] under the [[Definition:Euclidean Topology on Real Number Line|usual (Euclidean) topology]... | Cauchy Sequence in Metric Space is not necessarily Convergent | https://proofwiki.org/wiki/Cauchy_Sequence_in_Metric_Space_is_not_necessarily_Convergent | https://proofwiki.org/wiki/Cauchy_Sequence_in_Metric_Space_is_not_necessarily_Convergent | [
"Metric Spaces",
"Cauchy Sequences"
] | [
"Definition:Metric Space",
"Definition:Cauchy Sequence/Metric Space",
"Definition:Convergent Sequence/Metric Space"
] | [
"Definition:Set",
"Definition:Integer Reciprocal Space",
"Definition:Euclidean Space/Euclidean Topology/Real Number Line",
"Definition:Sequence",
"Definition:Convergent Sequence/Metric Space",
"Definition:Limit of Sequence/Metric Space",
"Integer Reciprocal Space contains Cauchy Sequence with no Limit",... |
proofwiki-17251 | Nested Sequences in Complete Metric Space not Tending to Zero may be Disjoint | Let $M = \struct {A, d}$ be a complete metric space.
Let $\family {S_k}_{k \mathop \in \N}$ be a nested sequence of closed balls in $M$.
Let the radii of $\family {S_k}_{k \mathop \in \N}$ be convergent in $M$, but not to zero.
Then it is not necessarily the case that their intersection $\ds \bigcap S_k$ is non-empty. | Let $M = \struct {A, d}$ be Sierpiński's metric space.
{{Recall|Sierpiński's Metric Space|Sierpiński's metric space}}
{{:Definition:Sierpiński's Metric Space}}
From Sierpiński's Metric Space is Complete, $M$ is a complete metric space.
Let $S_k = \set {y \in A: \map d {y, x_k} \le 1 + \dfrac 1 {2 n} }$.
From Nested Seq... | Let $M = \struct {A, d}$ be a [[Definition:Complete Metric Space|complete metric space]].
Let $\family {S_k}_{k \mathop \in \N}$ be a [[Definition:Nested Sequence|nested sequence]] of [[Definition:Closed Ball|closed balls]] in $M$.
Let the [[Definition:Radius of Closed Ball|radii]] of $\family {S_k}_{k \mathop \in \N... | Let $M = \struct {A, d}$ be [[Definition:Sierpiński's Metric Space|Sierpiński's metric space]].
{{Recall|Sierpiński's Metric Space|Sierpiński's metric space}}
{{:Definition:Sierpiński's Metric Space}}
From [[Sierpiński's Metric Space is Complete]], $M$ is a [[Definition:Complete Metric Space|complete metric space]].
... | Nested Sequences in Complete Metric Space not Tending to Zero may be Disjoint | https://proofwiki.org/wiki/Nested_Sequences_in_Complete_Metric_Space_not_Tending_to_Zero_may_be_Disjoint | https://proofwiki.org/wiki/Nested_Sequences_in_Complete_Metric_Space_not_Tending_to_Zero_may_be_Disjoint | [
"Complete Metric Spaces",
"Nested Sequences"
] | [
"Definition:Complete Metric Space",
"Definition:Nested Sequence",
"Definition:Closed Ball",
"Definition:Closed Ball/Metric Space/Radius",
"Definition:Convergent Sequence/Metric Space",
"Definition:Zero (Number)",
"Definition:Set Intersection/Family of Sets",
"Definition:Non-Empty Set"
] | [
"Definition:Sierpiński's Metric Space",
"Sierpiński's Metric Space is Complete",
"Definition:Complete Metric Space",
"Nested Sequence of Closed Balls in Sierpiński's Metric Space with Empty Intersection"
] |
proofwiki-17252 | Union of Interiors is Subset of Interior of Union | Let $T$ be a topological space.
Let $\H$ be a set of subsets of $T$.
That is, let $\H \subseteq \powerset T$ where $\powerset T$ is the power set of $T$.
Then the union of the interiors of the elements of $\H$ is a subset of the interior of the union of $\H$:
:$\ds \bigcup_{H \mathop \in \H} H^\circ \subseteq \paren {\... | Let $\mathbb U$ be the set of all open subsets of $\bigcup \H$.
Then by definition of interior:
:$\ds \paren {\bigcup_{H \mathop \in \H} H}^\circ = \paren {\bigcup \H}^\circ = \bigcup \mathbb U$
As $\mathbb U$ contains all open subsets of $\bigcup \H$, and $H^\circ$ is open for any $H \in \H$:
:$\ds \set {H^\circ : H \... | Let $T$ be a [[Definition:Topological Space|topological space]].
Let $\H$ be a [[Definition:Set|set]] of [[Definition:Subset|subsets]] of $T$.
That is, let $\H \subseteq \powerset T$ where $\powerset T$ is the [[Definition:Power Set|power set]] of $T$.
Then the [[Definition:Set Union|union]] of the [[Definition:In... | Let $\mathbb U$ be the [[Definition:Set|set]] of all [[Definition:Open Set (Topology)|open]] [[Definition:Subset|subsets]] of $\bigcup \H$.
Then by definition of [[Definition:Interior (Topology)/Definition 1|interior]]:
:$\ds \paren {\bigcup_{H \mathop \in \H} H}^\circ = \paren {\bigcup \H}^\circ = \bigcup \mathbb U$... | Union of Interiors is Subset of Interior of Union/Proof 2 | https://proofwiki.org/wiki/Union_of_Interiors_is_Subset_of_Interior_of_Union | https://proofwiki.org/wiki/Union_of_Interiors_is_Subset_of_Interior_of_Union/Proof_2 | [
"Union of Interiors is Subset of Interior of Union",
"Set Interiors",
"Set Union"
] | [
"Definition:Topological Space",
"Definition:Set",
"Definition:Subset",
"Definition:Power Set",
"Definition:Set Union",
"Definition:Interior (Topology)",
"Definition:Element",
"Definition:Subset",
"Definition:Interior (Topology)",
"Definition:Set Union"
] | [
"Definition:Set",
"Definition:Open Set/Topology",
"Definition:Subset",
"Definition:Interior (Topology)/Definition 1",
"Definition:Open Set/Topology",
"Definition:Subset",
"Definition:Open Set/Topology"
] |
proofwiki-17253 | Closed Ball is Closed/Normed Vector Space | Let $M = \struct {X, \norm {\, \cdot \,}}$ be a normed vector space.
Let $x \in X$.
Let $\epsilon \in \R_{> 0}$.
Let $\map {B_\epsilon^-} x$ be the closed $\epsilon$-ball of $x$ in $M$.
Then $\map {B_\epsilon^-} x$ is a closed set of $M$. | We show that the complement $X \setminus \map {B_\epsilon^-} x$ is open in $M$.
Let $y \in X \setminus \map {B_\epsilon^-} x$.
Then by definition of closed ball:
:$\norm {x - y} > \epsilon$
Put:
:$\delta := \norm {x - y} - \epsilon > 0$
Then:
:$\norm {x - y} - \delta = \epsilon$
Let $z \in \map {B_\delta} y$.
Then:
{{b... | Let $M = \struct {X, \norm {\, \cdot \,}}$ be a [[Definition:Normed Vector Space|normed vector space]].
Let $x \in X$.
Let $\epsilon \in \R_{> 0}$.
Let $\map {B_\epsilon^-} x$ be the [[Definition:Closed Ball in Normed Vector Space|closed $\epsilon$-ball]] of $x$ in $M$.
Then $\map {B_\epsilon^-} x$ is a [[Definiti... | We show that the [[Definition:Set Complement|complement]] $X \setminus \map {B_\epsilon^-} x$ is [[Definition:Open Set in Normed Vector Space|open]] in $M$.
Let $y \in X \setminus \map {B_\epsilon^-} x$.
Then by definition of [[Definition:Closed Ball in Normed Vector Space|closed ball]]:
:$\norm {x - y} > \epsilon$
... | Closed Ball is Closed/Normed Vector Space | https://proofwiki.org/wiki/Closed_Ball_is_Closed/Normed_Vector_Space | https://proofwiki.org/wiki/Closed_Ball_is_Closed/Normed_Vector_Space | [
"Normed Vector Spaces",
"Closed Balls",
"Closed Ball is Closed",
"Closed Sets (Normed Vector Spaces)",
"Closed Ball is Closed",
"Closed Sets (Normed Vector Spaces)"
] | [
"Definition:Normed Vector Space",
"Definition:Closed Ball/Normed Vector Space",
"Definition:Closed Set/Normed Vector Space"
] | [
"Definition:Set Complement",
"Definition:Open Set/Normed Vector Space",
"Definition:Closed Ball/Normed Vector Space",
"Reverse Triangle Inequality",
"Definition:Open Set/Normed Vector Space",
"Definition:Closed Set/Normed Vector Space",
"Definition:Closed Set/Normed Vector Space"
] |
proofwiki-17254 | Unit Sphere is Closed/Normed Vector Space | Let $M = \struct {X, \norm {\, \cdot \,} }$ be a normed vector space.
Let $\Bbb S := \set {x \in X : \norm x = 1}$ be a unit sphere in $M$.
Then $\Bbb S$ is closed in $M$. | Let $\map {B_1} 0 = \set {x \in X : \norm x < 1}$ be an open ball.
Let $\map { {B_1}^-} 0 = \set {x \in X : \norm x \le 1}$ be a closed ball.
Then:
:$\ds X = \Bbb S \bigcup \relcomp X {\Bbb S}$
where
:$\ds \relcomp X {\Bbb S} = \map {B_1} 0 \bigcup \paren {X \setminus \map { {B_1}^-} 0}$
is the relative complement of $... | Let $M = \struct {X, \norm {\, \cdot \,} }$ be a [[Definition:Normed Vector Space|normed vector space]].
Let $\Bbb S := \set {x \in X : \norm x = 1}$ be a [[Definition:Unit Sphere (Normed Vector Space)|unit sphere]] in $M$.
Then $\Bbb S$ is [[Definition:Closed Set in Normed Vector Space|closed]] in $M$. | Let $\map {B_1} 0 = \set {x \in X : \norm x < 1}$ be an [[Definition:Open Ball in Normed Vector Space|open ball]].
Let $\map { {B_1}^-} 0 = \set {x \in X : \norm x \le 1}$ be a [[Definition:Closed Ball in Normed Vector Space|closed ball]].
Then:
:$\ds X = \Bbb S \bigcup \relcomp X {\Bbb S}$
where
:$\ds \relcomp X ... | Unit Sphere is Closed/Normed Vector Space | https://proofwiki.org/wiki/Unit_Sphere_is_Closed/Normed_Vector_Space | https://proofwiki.org/wiki/Unit_Sphere_is_Closed/Normed_Vector_Space | [
"Closed Sets",
"Normed Vector Spaces",
"Closed Sets (Normed Vector Spaces)",
"Closed Sets (Normed Vector Spaces)"
] | [
"Definition:Normed Vector Space",
"Definition:Unit Sphere/Normed Vector Space",
"Definition:Closed Set/Normed Vector Space"
] | [
"Definition:Open Ball/Normed Vector Space",
"Definition:Closed Ball/Normed Vector Space",
"Definition:Relative Complement",
"Closed Ball is Closed/Normed Vector Space",
"Definition:Closed Set/Normed Vector Space",
"Definition:Open Set/Normed Vector Space",
"Open Ball is Open Set/Normed Vector Space",
... |
proofwiki-17255 | Continued Fraction Expansion of Euler's Number/Proof 1/Lemma | :For $n \in \Z , n \ge 0$:
{{begin-eqn}}
{{eqn | l = A_n
| r = q_{3 n} e - p_{3 n}
}}
{{eqn | l = B_n
| r = p_{3 n + 1} - q_{3 n + 1} e
}}
{{eqn | l = C_n
| r = p_{3 n + 2} - q_{3 n + 2} e
}}
{{end-eqn}} | To prove the assertion, we begin by demonstrating the relationships hold for the initial conditions at $n = 0$:
{{begin-eqn}}
{{eqn | l = A_0
| r = \int_0^1 e^x \rd x
}}
{{eqn | r = \bigintlimits {e^x} {x \mathop = 0} {x \mathop = 1}
| c = Primitive of Exponential Function
}}
{{eqn | r = e - 1
}}
{{eqn | r ... | :For $n \in \Z , n \ge 0$:
{{begin-eqn}}
{{eqn | l = A_n
| r = q_{3 n} e - p_{3 n}
}}
{{eqn | l = B_n
| r = p_{3 n + 1} - q_{3 n + 1} e
}}
{{eqn | l = C_n
| r = p_{3 n + 2} - q_{3 n + 2} e
}}
{{end-eqn}} | To prove the assertion, we begin by demonstrating the relationships hold for the initial conditions at $n = 0$:
{{begin-eqn}}
{{eqn | l = A_0
| r = \int_0^1 e^x \rd x
}}
{{eqn | r = \bigintlimits {e^x} {x \mathop = 0} {x \mathop = 1}
| c = [[Primitive of Exponential Function]]
}}
{{eqn | r = e - 1
}}
{{eqn... | Continued Fraction Expansion of Euler's Number/Proof 1/Lemma | https://proofwiki.org/wiki/Continued_Fraction_Expansion_of_Euler's_Number/Proof_1/Lemma | https://proofwiki.org/wiki/Continued_Fraction_Expansion_of_Euler's_Number/Proof_1/Lemma | [
"Continued Fraction Expansion of Euler's Number"
] | [] | [
"Primitive of Exponential Function",
"Primitive of x by Exponential of a x",
"Definition:Recursive Sequence",
"Definition:Derivative",
"Definition:Integration/Integrand",
"Definition:Integration/Integrand",
"Definition:Integration/Integrand",
"Definition:Integration/Integrand",
"Definition:Recursive... |
proofwiki-17256 | Limit Points of Open Set of Particular Point Space | Let $T = \struct {S, \tau_p}$ be a particular point space.
Let $U \subseteq S$ such that $p \in U$.
Let $x \in S$ such that $x \ne p$.
Then $x$ is a limit point of $U$. | Every open set of $T = \struct {S, \tau_p}$ except $\O$ contains the point $p$ by definition.
So every open set $U \in \tau_p$ such that $x \in U$ contains $p$.
So by definition of the limit point of a set, $x$ is a limit point of $U$.
{{qed}} | Let $T = \struct {S, \tau_p}$ be a [[Definition:Particular Point Topology|particular point space]].
Let $U \subseteq S$ such that $p \in U$.
Let $x \in S$ such that $x \ne p$.
Then $x$ is a [[Definition:Limit Point of Set|limit point]] of $U$. | Every [[Definition:Open Set (Topology)|open set]] of $T = \struct {S, \tau_p}$ except $\O$ contains the point $p$ by [[Definition:Particular Point Topology|definition]].
So every [[Definition:Open Set (Topology)|open set]] $U \in \tau_p$ such that $x \in U$ contains $p$.
So by definition of the [[Definition:Limit Poi... | Limit Points of Open Set of Particular Point Space/Proof 1 | https://proofwiki.org/wiki/Limit_Points_of_Open_Set_of_Particular_Point_Space | https://proofwiki.org/wiki/Limit_Points_of_Open_Set_of_Particular_Point_Space/Proof_1 | [
"Limit Points of Open Set of Particular Point Space",
"Limit Points in Particular Point Space",
"Examples of Limit Points"
] | [
"Definition:Particular Point Topology",
"Definition:Limit Point/Topology/Set"
] | [
"Definition:Open Set/Topology",
"Definition:Particular Point Topology",
"Definition:Open Set/Topology",
"Definition:Limit Point/Topology/Set",
"Definition:Limit Point/Topology/Set"
] |
proofwiki-17257 | Limit Points of Open Set of Particular Point Space | Let $T = \struct {S, \tau_p}$ be a particular point space.
Let $U \subseteq S$ such that $p \in U$.
Let $x \in S$ such that $x \ne p$.
Then $x$ is a limit point of $U$. | Follows directly from:
:Particular Point Topology is Closed Extension Topology of Discrete Topology
:Limit Points in Subset of Closed Extension Space
{{qed}} | Let $T = \struct {S, \tau_p}$ be a [[Definition:Particular Point Topology|particular point space]].
Let $U \subseteq S$ such that $p \in U$.
Let $x \in S$ such that $x \ne p$.
Then $x$ is a [[Definition:Limit Point of Set|limit point]] of $U$. | Follows directly from:
:[[Particular Point Topology is Closed Extension Topology of Discrete Topology]]
:[[Limit Points in Subset of Closed Extension Space]]
{{qed}} | Limit Points of Open Set of Particular Point Space/Proof 2 | https://proofwiki.org/wiki/Limit_Points_of_Open_Set_of_Particular_Point_Space | https://proofwiki.org/wiki/Limit_Points_of_Open_Set_of_Particular_Point_Space/Proof_2 | [
"Limit Points of Open Set of Particular Point Space",
"Limit Points in Particular Point Space",
"Examples of Limit Points"
] | [
"Definition:Particular Point Topology",
"Definition:Limit Point/Topology/Set"
] | [
"Particular Point Topology is Closed Extension Topology of Discrete Topology",
"Limit Points in Closed Extension Space/Subset"
] |
proofwiki-17258 | Limit Points in Closed Extension Space/Subset | Let $U \subseteq S^*_p$ such that $p \in U$.
Let $x \in S$.
Then $x$ is a limit point of $U$. | Every open set of $T^*_p = \struct {S^*_p, \tau^*_p}$ except $\O$ contains the point $p$ by definition.
So every open set $U \in \tau^*_p$ such that $x \in U$ contains $p$.
So by definition of the limit point of a set, $x$ is a limit point of $U$.
{{qed}} | Let $U \subseteq S^*_p$ such that $p \in U$.
Let $x \in S$.
Then $x$ is a [[Definition:Limit Point of Set|limit point]] of $U$. | Every [[Definition:Open Set (Topology)|open set]] of $T^*_p = \struct {S^*_p, \tau^*_p}$ except $\O$ contains the point $p$ by [[Definition:Closed Extension Topology|definition]].
So every [[Definition:Open Set (Topology)|open set]] $U \in \tau^*_p$ such that $x \in U$ contains $p$.
So by definition of the [[Definiti... | Limit Points in Closed Extension Space/Subset | https://proofwiki.org/wiki/Limit_Points_in_Closed_Extension_Space/Subset | https://proofwiki.org/wiki/Limit_Points_in_Closed_Extension_Space/Subset | [
"Limit Points in Closed Extension Space"
] | [
"Definition:Limit Point/Topology/Set"
] | [
"Definition:Open Set/Topology",
"Definition:Closed Extension Topology",
"Definition:Open Set/Topology",
"Definition:Limit Point/Topology/Set",
"Definition:Limit Point/Topology/Set"
] |
proofwiki-17259 | Convergent Sequence in Particular Point Space | Let $T = \struct {S, \tau_p}$ be a particular point space.
Let $\sequence {a_i}$ be a convergent sequence in $T$.
Except for a finite number of indices, the terms of $\sequence {a_i}$ for which $a_i \ne p$ are all equal. | {{Recall|Convergent Sequence (Topology)|convergent sequence}}
{{:Definition:Convergent Sequence/Topology/Definition 2}}
Let $\sequence {a_i}$ be a convergent sequence in $T$ whose limit is $\alpha$.
Then by definition every open set in $T$ containing $\alpha$ contains all but a finite number of terms of $\sequence {a_i... | Let $T = \struct {S, \tau_p}$ be a [[Definition:Particular Point Topology|particular point space]].
Let $\sequence {a_i}$ be a [[Definition:Convergent Sequence (Topology)|convergent sequence]] in $T$.
Except for a [[Definition:Finite Set|finite number]] of [[Definition:Index of Term of Sequence|indices]], the [[Defi... | {{Recall|Convergent Sequence (Topology)|convergent sequence}}
{{:Definition:Convergent Sequence/Topology/Definition 2}}
Let $\sequence {a_i}$ be a [[Definition:Convergent Sequence (Topology)|convergent sequence]] in $T$ whose [[Definition:Limit of Sequence (Topology)|limit]] is $\alpha$.
Then by definition every [[De... | Convergent Sequence in Particular Point Space | https://proofwiki.org/wiki/Convergent_Sequence_in_Particular_Point_Space | https://proofwiki.org/wiki/Convergent_Sequence_in_Particular_Point_Space | [
"Particular Point Topologies",
"Examples of Convergent Sequences (Topology)"
] | [
"Definition:Particular Point Topology",
"Definition:Convergent Sequence/Topology",
"Definition:Finite Set",
"Definition:Term of Sequence/Index",
"Definition:Term of Sequence"
] | [
"Definition:Convergent Sequence/Topology",
"Definition:Limit of Sequence/Topological Space",
"Definition:Open Set/Topology",
"Definition:Finite Set",
"Definition:Term of Sequence",
"Definition:Open Set/Topology",
"Definition:Finite Set",
"Definition:Term of Sequence",
"Definition:Finite Set",
"Def... |
proofwiki-17260 | Accumulation Points for Sequence in Particular Point Space | Let $T = \struct {S, \tau_p}$ be a particular point space.
Let $\sequence {a_i}$ be an infinite sequence in $T$.
Let $\beta$ be an accumulation point of $\sequence {a_i}$.
Then $\beta$ is such that an infinite number of terms of $\sequence {a_i}$ are equal either to $\beta$ or to $p$. | {{Recall|Accumulation Point of Sequence|accumulation point}}
{{:Definition:Accumulation Point of Sequence}}
Let $\beta$ be an accumulation point of $\sequence {a_i}$.
Then by definition:
:all open sets of $T$ which contain $\beta$ also contain an infinite number of terms of $\sequence {a_i}$.
This condition applies to ... | Let $T = \struct {S, \tau_p}$ be a [[Definition:Particular Point Topology|particular point space]].
Let $\sequence {a_i}$ be an [[Definition:Infinite Sequence|infinite sequence]] in $T$.
Let $\beta$ be an [[Definition:Accumulation Point of Sequence|accumulation point]] of $\sequence {a_i}$.
Then $\beta$ is such tha... | {{Recall|Accumulation Point of Sequence|accumulation point}}
{{:Definition:Accumulation Point of Sequence}}
Let $\beta$ be an [[Definition:Accumulation Point of Sequence|accumulation point]] of $\sequence {a_i}$.
Then by definition:
:all [[Definition:Open Set (Topology)|open sets]] of $T$ which contain $\beta$ also c... | Accumulation Points for Sequence in Particular Point Space | https://proofwiki.org/wiki/Accumulation_Points_for_Sequence_in_Particular_Point_Space | https://proofwiki.org/wiki/Accumulation_Points_for_Sequence_in_Particular_Point_Space | [
"Particular Point Topologies",
"Examples of Accumulation Points"
] | [
"Definition:Particular Point Topology",
"Definition:Sequence/Infinite Sequence",
"Definition:Accumulation Point/Sequence",
"Definition:Infinite Set",
"Definition:Term of Sequence"
] | [
"Definition:Accumulation Point/Sequence",
"Definition:Open Set/Topology",
"Definition:Infinite Set",
"Definition:Term of Sequence",
"Definition:Open Set/Topology",
"Definition:Infinite Set",
"Definition:Term of Sequence"
] |
proofwiki-17261 | Absolute Value of Absolutely Continuous Function is Absolutely Continuous | Let $I \subseteq \R$ be a real interval.
Let $f : I \to \R$ be an absolutely continuous function.
Then $\size f$ is absolutely continuous. | Let $\epsilon$ be a positive real number.
Since $f$ is absolutely continuous, there exists real $\delta > 0$ such that for all collections of disjoint closed real intervals $\closedint {a_1} {b_1}, \dotsc, \closedint {a_n} {b_n} \subseteq I$ with:
:$\ds \sum_{i \mathop = 1}^n \paren {b_i - a_i} < \delta$
we have:
:$\... | Let $I \subseteq \R$ be a [[Definition:Real Interval|real interval]].
Let $f : I \to \R$ be an [[Definition:Absolutely Continuous Real Function|absolutely continuous]] [[Definition:Real Function|function]].
Then $\size f$ is [[Definition:Absolutely Continuous Real Function|absolutely continuous]]. | Let $\epsilon$ be a [[Definition:Positive Real Number|positive real number]].
Since $f$ is [[Definition:Absolutely Continuous Real Function|absolutely continuous]], there exists [[Definition:Real Number|real]] $\delta > 0$ such that for all collections of [[Definition:Disjoint Sets|disjoint]] [[Definition:Closed Real ... | Absolute Value of Absolutely Continuous Function is Absolutely Continuous | https://proofwiki.org/wiki/Absolute_Value_of_Absolutely_Continuous_Function_is_Absolutely_Continuous | https://proofwiki.org/wiki/Absolute_Value_of_Absolutely_Continuous_Function_is_Absolutely_Continuous | [
"Absolute Value Function",
"Absolutely Continuous Real Functions"
] | [
"Definition:Real Interval",
"Definition:Absolute Continuity/Real Function",
"Definition:Real Function",
"Definition:Absolute Continuity/Real Function"
] | [
"Definition:Positive/Real Number",
"Definition:Absolute Continuity/Real Function",
"Definition:Real Number",
"Definition:Disjoint Sets",
"Definition:Real Interval/Closed",
"Reverse Triangle Inequality/Real and Complex Fields",
"Definition:Absolute Continuity/Real Function",
"Category:Absolute Value Fu... |
proofwiki-17262 | Constant Real Function is Absolutely Continuous | Let $I \subseteq \R$ be a real interval.
Let $f : I \to \R$ be an constant real function.
Then $f$ is absolutely continuous. | Let $\delta, \epsilon$ be positive real numbers.
Let $\closedint {a_1} {b_1}, \dotsc, \closedint {a_n} {b_n} \subseteq I$ be a collection of disjoint closed real intervals with:
:$\ds \sum_{i \mathop = 1}^n \paren {b_i - a_i} < \delta$
Since $f$ is constant, for all $i \in \set {1, 2, \ldots, n}$ we have:
:$\size {\ma... | Let $I \subseteq \R$ be a [[Definition:Real Interval|real interval]].
Let $f : I \to \R$ be an [[Definition:Constant Mapping|constant]] [[Definition:Real Function|real function]].
Then $f$ is [[Definition:Absolutely Continuous Real Function|absolutely continuous]]. | Let $\delta, \epsilon$ be [[Definition:Positive Real Number|positive real numbers]].
Let $\closedint {a_1} {b_1}, \dotsc, \closedint {a_n} {b_n} \subseteq I$ be a collection of [[Definition:Disjoint Sets|disjoint]] [[Definition:Closed Real Interval|closed real intervals]] with:
:$\ds \sum_{i \mathop = 1}^n \paren {b... | Constant Real Function is Absolutely Continuous | https://proofwiki.org/wiki/Constant_Real_Function_is_Absolutely_Continuous | https://proofwiki.org/wiki/Constant_Real_Function_is_Absolutely_Continuous | [
"Constant Mappings",
"Absolutely Continuous Real Functions"
] | [
"Definition:Real Interval",
"Definition:Constant Mapping",
"Definition:Real Function",
"Definition:Absolute Continuity/Real Function"
] | [
"Definition:Positive/Real Number",
"Definition:Disjoint Sets",
"Definition:Real Interval/Closed",
"Definition:Constant Mapping",
"Definition:Absolute Continuity/Real Function",
"Category:Constant Mappings",
"Category:Absolutely Continuous Real Functions"
] |
proofwiki-17263 | Sum of Absolutely Continuous Functions is Absolutely Continuous | Let $I \subseteq \R$ be a real interval.
Let $f, g : I \to \R$ be absolutely continuous real functions.
Then $f + g$ is absolutely continuous. | Let $\epsilon$ be a positive real number.
Since $f$ is absolutely continuous, there exists real $\delta_1 > 0$ such that for all sets of disjoint closed real intervals $\closedint {a_1} {b_1}, \dotsc, \closedint {a_n} {b_n} \subseteq I$ with:
:$\ds \sum_{i \mathop = 1}^n \paren {b_i - a_i} < \delta_1$
we have:
:$\ds ... | Let $I \subseteq \R$ be a [[Definition:Real Interval|real interval]].
Let $f, g : I \to \R$ be [[Definition:Absolutely Continuous Real Function|absolutely continuous]] [[Definition:Real Function|real functions]].
Then $f + g$ is [[Definition:Absolutely Continuous Real Function|absolutely continuous]]. | Let $\epsilon$ be a [[Definition:Positive Real Number|positive real number]].
Since $f$ is [[Definition:Absolutely Continuous Real Function|absolutely continuous]], there exists [[Definition:Real Number|real]] $\delta_1 > 0$ such that for all [[Definition:Set|sets]] of [[Definition:Disjoint Sets|disjoint]] [[Definitio... | Sum of Absolutely Continuous Functions is Absolutely Continuous | https://proofwiki.org/wiki/Sum_of_Absolutely_Continuous_Functions_is_Absolutely_Continuous | https://proofwiki.org/wiki/Sum_of_Absolutely_Continuous_Functions_is_Absolutely_Continuous | [
"Absolutely Continuous Real Functions"
] | [
"Definition:Real Interval",
"Definition:Absolute Continuity/Real Function",
"Definition:Real Function",
"Definition:Absolute Continuity/Real Function"
] | [
"Definition:Positive/Real Number",
"Definition:Absolute Continuity/Real Function",
"Definition:Real Number",
"Definition:Set",
"Definition:Disjoint Sets",
"Definition:Real Interval/Closed",
"Definition:Absolute Continuity/Real Function",
"Definition:Real Number",
"Definition:Set",
"Definition:Disj... |
proofwiki-17264 | Multiple of Absolutely Continuous Function is Absolutely Continuous | Let $k$ be a real number.
Let $I \subseteq \R$ be a real interval.
Let $f : I \to \R$ be an absolutely continuous real function.
Then $k f$ is absolutely continuous. | Note that if $k = 0$, then $k f$ is constant.
Hence, by Constant Real Function is Absolutely Continuous:
:$k f$ is absolutely continuous if $k = 0$.
Take now $k \ne 0$.
Let $\epsilon$ be a positive real number.
Since $f$ is absolutely continuous, there exists real $\delta > 0$ such that for every set of disjoint clo... | Let $k$ be a [[Definition:Real Number|real number]].
Let $I \subseteq \R$ be a [[Definition:Real Interval|real interval]].
Let $f : I \to \R$ be an [[Definition:Absolutely Continuous Real Function|absolutely continuous real function]].
Then $k f$ is [[Definition:Absolutely Continuous Real Function|absolutely contin... | Note that if $k = 0$, then $k f$ is [[Definition:Constant Mapping|constant]].
Hence, by [[Constant Real Function is Absolutely Continuous]]:
:$k f$ is [[Definition:Absolutely Continuous Real Function|absolutely continuous]] if $k = 0$.
Take now $k \ne 0$.
Let $\epsilon$ be a [[Definition:Positive Real Number|posi... | Multiple of Absolutely Continuous Function is Absolutely Continuous | https://proofwiki.org/wiki/Multiple_of_Absolutely_Continuous_Function_is_Absolutely_Continuous | https://proofwiki.org/wiki/Multiple_of_Absolutely_Continuous_Function_is_Absolutely_Continuous | [
"Absolutely Continuous Real Functions"
] | [
"Definition:Real Number",
"Definition:Real Interval",
"Definition:Absolute Continuity/Real Function",
"Definition:Absolute Continuity/Real Function"
] | [
"Definition:Constant Mapping",
"Constant Real Function is Absolutely Continuous",
"Definition:Absolute Continuity/Real Function",
"Definition:Positive/Real Number",
"Definition:Absolute Continuity/Real Function",
"Definition:Real Number",
"Definition:Set",
"Definition:Disjoint Sets",
"Definition:Rea... |
proofwiki-17265 | Norms on Finite-Dimensional Real Vector Space are Equivalent | Norms on finite-dimensional real vector space are equivalent. | We will prove that all norms are equivalent to $\norm {\, \cdot \,}_2$.
By definition, two norms are equivalent on $\R^d$ {{iff}}:
:$ \exists m, M \in \R_{> 0} : \forall \mathbf x \in \R^d : m \norm {\mathbf x}_a \le \norm {\mathbf x}_b \le M \norm {\mathbf x}_a$ | [[Definition:Norm on Vector Space|Norms]] on [[Definition:Finite Dimensional Vector Space|finite-dimensional]] [[Definition:Real Vector Space|real vector space]] are [[Definition:Equivalence of Norms|equivalent]]. | We will prove that all [[Definition:Norm on Vector Space|norms]] are [[Definition:Equivalence of Norms|equivalent]] to $\norm {\, \cdot \,}_2$.
By definition, two [[Definition:Norm on Vector Space|norms]] are [[Definition:Equivalence of Norms|equivalent]] on $\R^d$ {{iff}}:
:$ \exists m, M \in \R_{> 0} : \forall \mat... | Norms on Finite-Dimensional Real Vector Space are Equivalent | https://proofwiki.org/wiki/Norms_on_Finite-Dimensional_Real_Vector_Space_are_Equivalent | https://proofwiki.org/wiki/Norms_on_Finite-Dimensional_Real_Vector_Space_are_Equivalent | [
"Normed Vector Spaces",
"Equivalence Relations",
"Norm Theory",
"Vector Spaces",
"Finite Dimensional Vector Spaces"
] | [
"Definition:Norm/Vector Space",
"Definition:Dimension of Vector Space/Finite",
"Definition:Real Vector Space",
"Definition:Equivalence of Norms"
] | [
"Definition:Norm/Vector Space",
"Definition:Equivalence of Norms",
"Definition:Norm/Vector Space",
"Definition:Equivalence of Norms",
"Definition:Norm/Vector Space",
"Definition:Norm/Vector Space",
"Definition:Equivalence of Norms",
"Definition:Norm/Vector Space",
"Definition:Equivalence of Norms"
] |
proofwiki-17266 | Linear Combination of Absolutely Continuous Function is Absolutely Continuous | Let $\alpha, \beta$ be real numbers.
Let $I \subseteq \R$ be a real interval.
Let $f, g : I \to \R$ be absolutely continuous real functions.
Then $\alpha f + \beta g$ is absolutely continuous. | From Multiple of Absolutely Continuous Function is Absolutely Continuous, we have:
:$\alpha f$ and $\beta g$ are absolutely continuous.
We therefore have, by Sum of Absolutely Continuous Functions is Absolutely Continuous:
:$\alpha f + \beta g$ is absolutely continuous.
{{qed}}
Category:Absolutely Continuous Real Func... | Let $\alpha, \beta$ be [[Definition:Real Number|real numbers]].
Let $I \subseteq \R$ be a [[Definition:Real Interval|real interval]].
Let $f, g : I \to \R$ be [[Definition:Absolutely Continuous Real Function|absolutely continuous real functions]].
Then $\alpha f + \beta g$ is [[Definition:Absolutely Continuous Real... | From [[Multiple of Absolutely Continuous Function is Absolutely Continuous]], we have:
:$\alpha f$ and $\beta g$ are [[Definition:Absolutely Continuous Real Function|absolutely continuous]].
We therefore have, by [[Sum of Absolutely Continuous Functions is Absolutely Continuous]]:
:$\alpha f + \beta g$ is [[Definit... | Linear Combination of Absolutely Continuous Function is Absolutely Continuous | https://proofwiki.org/wiki/Linear_Combination_of_Absolutely_Continuous_Function_is_Absolutely_Continuous | https://proofwiki.org/wiki/Linear_Combination_of_Absolutely_Continuous_Function_is_Absolutely_Continuous | [
"Absolutely Continuous Real Functions"
] | [
"Definition:Real Number",
"Definition:Real Interval",
"Definition:Absolute Continuity/Real Function",
"Definition:Absolute Continuity/Real Function"
] | [
"Multiple of Absolutely Continuous Function is Absolutely Continuous",
"Definition:Absolute Continuity/Real Function",
"Sum of Absolutely Continuous Functions is Absolutely Continuous",
"Definition:Absolute Continuity/Real Function",
"Category:Absolutely Continuous Real Functions"
] |
proofwiki-17267 | Integral of Distribution Function | Let $\struct {X, \Sigma, \mu}$ be a measure space and $f$ be a $\mu$-measurable function. Let $p > 0, r \geq 0$.
For $\lambda > 0$, let $E_\lambda = \set {x \in X: \size {\map f x} > \lambda}$, so that $\map m \lambda = \map \mu {E_\lambda}$ is the distribution function of $f$.
Then:
:$\ds \int_0^\infty p \lambda^{p - ... | {{begin-eqn}}
{{eqn | l = \int_0^\infty p \lambda^{p - 1} \int_{E_\lambda} \size f^r \rd \mu \rd \lambda
| r = \int_0^\infty \int_{E_\lambda} p \lambda^{p - 1} \size f^r \rd \mu \rd \lambda
}}
{{eqn | r = \int_X \int_0^{\size {\map f x} } p \lambda^{p - 1} \size f^r \rd \lambda \rd \mu
| c = by Tonelli's Th... | Let $\struct {X, \Sigma, \mu}$ be a [[Definition:Measure Space|measure space]] and $f$ be a $\mu$-measurable function. Let $p > 0, r \geq 0$.
For $\lambda > 0$, let $E_\lambda = \set {x \in X: \size {\map f x} > \lambda}$, so that $\map m \lambda = \map \mu {E_\lambda}$ is the distribution function of $f$.
Then:
:$\... | {{begin-eqn}}
{{eqn | l = \int_0^\infty p \lambda^{p - 1} \int_{E_\lambda} \size f^r \rd \mu \rd \lambda
| r = \int_0^\infty \int_{E_\lambda} p \lambda^{p - 1} \size f^r \rd \mu \rd \lambda
}}
{{eqn | r = \int_X \int_0^{\size {\map f x} } p \lambda^{p - 1} \size f^r \rd \lambda \rd \mu
| c = by [[Tonelli's ... | Integral of Distribution Function | https://proofwiki.org/wiki/Integral_of_Distribution_Function | https://proofwiki.org/wiki/Integral_of_Distribution_Function | [
"Measure Theory"
] | [
"Definition:Measure Space"
] | [
"Tonelli's Theorem",
"Integral of Power",
"Category:Measure Theory"
] |
proofwiki-17268 | Limit Points in Open Extension Space/Subset | Let $U \subseteq S^*_p$.
Then $p$ is a limit point of $U$. | Every open set of $T^*_p = \struct {S^*_p, \tau^*_{\bar p} }$ except $S^*_p$ does not contain the point $p$ by definition.
So every open set $U \in \tau^*_{\bar p}$ such that $p \in U$ (there is only the one such open set) contains $x$.
So by definition of the limit point of a set, $p$ is a limit point of $U$.
{{qed}}
... | Let $U \subseteq S^*_p$.
Then $p$ is a [[Definition:Limit Point of Set|limit point]] of $U$. | Every [[Definition:Open Set (Topology)|open set]] of $T^*_p = \struct {S^*_p, \tau^*_{\bar p} }$ except $S^*_p$ does not contain the point $p$ by [[Definition:Open Extension Topology|definition]].
So every [[Definition:Open Set (Topology)|open set]] $U \in \tau^*_{\bar p}$ such that $p \in U$ (there is only the one su... | Limit Points in Open Extension Space/Subset | https://proofwiki.org/wiki/Limit_Points_in_Open_Extension_Space/Subset | https://proofwiki.org/wiki/Limit_Points_in_Open_Extension_Space/Subset | [
"Limit Points in Open Extension Space"
] | [
"Definition:Limit Point/Topology/Set"
] | [
"Definition:Open Set/Topology",
"Definition:Open Extension Topology",
"Definition:Open Set/Topology",
"Definition:Open Set/Topology",
"Definition:Limit Point/Topology/Set",
"Definition:Limit Point/Topology/Set",
"Category:Limit Points in Open Extension Space"
] |
proofwiki-17269 | Differentiable Function with Bounded Derivative is Absolutely Continuous | Let $a, b$ be real numbers with $a < b$.
Let $f: \closedint a b \to \R$ be a continuous function.
Let $f$ be differentiable on $\openint a b$, with bounded derivative.
Then $f$ is absolutely continuous. | Since the derivative of $f$ is bounded, there exists some $M \in \R_{> 0}$ such that:
:$\size {\map {f'} x} \le M$
for all $x \in \openint a b$.
Let $\closedint {a_1} {b_1}, \dotsc, \closedint {a_n} {b_n} \subseteq \closedint a b$ be a collection of disjoint closed real intervals.
Note that for each $i \in \set {1, 2,... | Let $a, b$ be [[Definition:Real Number|real numbers]] with $a < b$.
Let $f: \closedint a b \to \R$ be a [[Definition:Continuous Real Function|continuous function]].
Let $f$ be [[Definition:Differentiable Real Function|differentiable]] on $\openint a b$, with [[Definition:Bounded Real-Valued Function|bounded]] [[Defin... | Since the [[Definition:Derivative|derivative]] of $f$ is bounded, there exists some $M \in \R_{> 0}$ such that:
:$\size {\map {f'} x} \le M$
for all $x \in \openint a b$.
Let $\closedint {a_1} {b_1}, \dotsc, \closedint {a_n} {b_n} \subseteq \closedint a b$ be a collection of [[Definition:Disjoint Sets|disjoint]] [[... | Differentiable Function with Bounded Derivative is Absolutely Continuous | https://proofwiki.org/wiki/Differentiable_Function_with_Bounded_Derivative_is_Absolutely_Continuous | https://proofwiki.org/wiki/Differentiable_Function_with_Bounded_Derivative_is_Absolutely_Continuous | [
"Differentiable Real Functions",
"Absolutely Continuous Real Functions"
] | [
"Definition:Real Number",
"Definition:Continuous Real Function",
"Definition:Differentiable Mapping/Real Function",
"Definition:Bounded Mapping/Real-Valued",
"Definition:Derivative",
"Definition:Absolute Continuity/Real Function"
] | [
"Definition:Derivative",
"Definition:Disjoint Sets",
"Definition:Real Interval/Closed",
"Definition:Continuous Real Function",
"Definition:Differentiable Mapping/Real Function",
"Mean Value Theorem",
"Definition:Positive/Real Number",
"Definition:Disjoint Sets",
"Definition:Real Interval/Closed",
... |
proofwiki-17270 | Lipschitz Continuous Real Function is Absolutely Continuous | Let $I \subseteq \R$ be a real interval.
Let $f$ be Lipschitz continuous on $I$.
Then $f$ is absolutely continuous on $I$. | By the definition of Lipschitz continuity, there exists $K \in \R$ such that:
:$\size {\map f x - \map f y} \le K \size {x - y}$
for all $x, y \in I$.
If $K = 0$, then:
:$\size {\map f x - \map f y} = 0$
for all $x, y \in I$.
In this case, $f$ is constant.
Hence, by Constant Real Function is Absolutely Continuous:
:... | Let $I \subseteq \R$ be a [[Definition:Real Interval|real interval]].
Let $f$ be [[Definition:Lipschitz Continuous Real Function|Lipschitz continuous]] on $I$.
Then $f$ is [[Definition:Absolutely Continuous Real Function|absolutely continuous]] on $I$. | By the [[Definition:Lipschitz Continuous Real Function|definition of Lipschitz continuity]], there exists $K \in \R$ such that:
:$\size {\map f x - \map f y} \le K \size {x - y}$
for all $x, y \in I$.
If $K = 0$, then:
:$\size {\map f x - \map f y} = 0$
for all $x, y \in I$.
In this case, $f$ is [[Definition:C... | Lipschitz Continuous Real Function is Absolutely Continuous | https://proofwiki.org/wiki/Lipschitz_Continuous_Real_Function_is_Absolutely_Continuous | https://proofwiki.org/wiki/Lipschitz_Continuous_Real_Function_is_Absolutely_Continuous | [
"Lipschitz Continuity",
"Absolutely Continuous Real Functions"
] | [
"Definition:Real Interval",
"Definition:Lipschitz Continuity/Real Function",
"Definition:Absolute Continuity/Real Function"
] | [
"Definition:Lipschitz Continuity/Real Function",
"Definition:Constant Mapping",
"Constant Real Function is Absolutely Continuous",
"Definition:Absolute Continuity/Real Function",
"Definition:Disjoint Sets",
"Definition:Real Interval/Closed",
"Definition:Positive/Real Number",
"Definition:Disjoint Sets... |
proofwiki-17271 | Either-Or Topology is Compact | Let $T = \struct {S, \tau}$ be the either-or space.
Then $T$ is a compact space. | Any open cover $\CC$ of $T$ must contain an open set of $T$ which contains $0$.
So $\openint {-1} 1$ will always be covered by one set in $\CC$, leaving just $-1$ and $1$ possibly needing to be included in at most two other sets.
So $\CC$ has a subcover containing at most three sets.
Hence $T$ is a compact space by def... | Let $T = \struct {S, \tau}$ be the [[Definition:Either-Or Topology|either-or space]].
Then $T$ is a [[Definition:Compact Topological Space|compact space]]. | Any [[Definition:Open Cover|open cover]] $\CC$ of $T$ must contain an [[Definition:Open Set (Topology)|open set]] of $T$ which contains $0$.
So $\openint {-1} 1$ will always be [[Definition:Cover of Set|covered]] by one [[Definition:Set|set]] in $\CC$, leaving just $-1$ and $1$ possibly needing to be included in at mo... | Either-Or Topology is Compact | https://proofwiki.org/wiki/Either-Or_Topology_is_Compact | https://proofwiki.org/wiki/Either-Or_Topology_is_Compact | [
"Either-Or Topology",
"Examples of Compact Topological Spaces"
] | [
"Definition:Either-Or Topology",
"Definition:Compact Topological Space"
] | [
"Definition:Open Cover",
"Definition:Open Set/Topology",
"Definition:Cover of Set",
"Definition:Set",
"Definition:Set",
"Definition:Subcover",
"Definition:Set",
"Definition:Compact Topological Space"
] |
proofwiki-17272 | Compact Space is Lindelöf | Every '''compact space''' is also a '''Lindelöf space'''. | We have:
:Compact Space is $\sigma$-Compact Space
:$\sigma$-Compact Space is Lindelöf Space
Hence the result.
{{qed}}
Category:Compact Topological Spaces
Category:Lindelöf Spaces
57oc7idrcsgesuxsrsxafj25g7doa01 | Every '''[[Definition:Compact Topological Space|compact space]]''' is also a '''[[Definition:Lindelöf Space|Lindelöf space]]'''. | We have:
:[[Compact Space is Sigma-Compact Space|Compact Space is $\sigma$-Compact Space]]
:[[Sigma-Compact Space is Lindelöf Space|$\sigma$-Compact Space is Lindelöf Space]]
Hence the result.
{{qed}}
[[Category:Compact Topological Spaces]]
[[Category:Lindelöf Spaces]]
57oc7idrcsgesuxsrsxafj25g7doa01 | Compact Space is Lindelöf | https://proofwiki.org/wiki/Compact_Space_is_Lindelöf | https://proofwiki.org/wiki/Compact_Space_is_Lindelöf | [
"Compact Topological Spaces",
"Lindelöf Spaces"
] | [
"Definition:Compact Topological Space",
"Definition:Lindelöf Space"
] | [
"Compact Space is Sigma-Compact",
"Sigma-Compact Space is Lindelöf",
"Category:Compact Topological Spaces",
"Category:Lindelöf Spaces"
] |
proofwiki-17273 | Product of Absolutely Continuous Functions is Absolutely Continuous | Let $a, b$ be real numbers with $a < b$.
Let $f, g : \closedint a b \to \R$ be absolutely continuous real functions.
Then $f \times g$ is absolutely continuous. | From Absolutely Continuous Real Function is Continuous:
:$f$ and $g$ are continuous.
From Closed Real Interval is Compact in Metric Space:
:$\closedint a b$ is compact.
Therefore, by Continuous Function on Compact Subspace of Euclidean Space is Bounded:
:$f$ and $g$ are bounded.
That is, there exists $M_f, M_g \in \R_{... | Let $a, b$ be [[Definition:Real Number|real numbers]] with $a < b$.
Let $f, g : \closedint a b \to \R$ be [[Definition:Absolutely Continuous Real Function|absolutely continuous real functions]].
Then $f \times g$ is [[Definition:Absolutely Continuous Real Function|absolutely continuous]]. | From [[Absolutely Continuous Real Function is Continuous]]:
:$f$ and $g$ are [[Definition:Continuous Real Function|continuous]].
From [[Closed Real Interval is Compact in Metric Space]]:
:$\closedint a b$ is [[Definition:Compact Subset of Real Numbers|compact]].
Therefore, by [[Continuous Function on Compact Subspa... | Product of Absolutely Continuous Functions is Absolutely Continuous | https://proofwiki.org/wiki/Product_of_Absolutely_Continuous_Functions_is_Absolutely_Continuous | https://proofwiki.org/wiki/Product_of_Absolutely_Continuous_Functions_is_Absolutely_Continuous | [
"Absolutely Continuous Real Functions"
] | [
"Definition:Real Number",
"Definition:Absolute Continuity/Real Function",
"Definition:Absolute Continuity/Real Function"
] | [
"Absolutely Continuous Real Function is Continuous",
"Definition:Continuous Real Function",
"Closed Real Interval is Compact Space/Metric Space",
"Definition:Compact Space/Real Analysis",
"Continuous Function on Compact Subspace of Euclidean Space is Bounded",
"Definition:Bounded Mapping",
"Definition:S... |
proofwiki-17274 | Accumulation Point of Sequence is not necessarily Limit Point | Let $T = \struct {S, \tau}$ be a topological space.
Let $\sequence {a_n}$ be a sequence in $T$.
Let $q \in S$ be an accumulation point of $\sequence {a_n}$.
Then it is not necessarily the case that $q$ is also a limit of $\sequence {a_n}$. | Proof by Counterexample:
Let $\struct {\R, \tau_d}$ be the real number line with the usual (Euclidean) topology.
Let $\sequence {a_n}$ be the sequence defined as:
{{begin-eqn}}
{{eqn | l = \sequence {a_n}
| r = \begin {cases} 1 & : \text {$n$ odd} \\ n / 2 & : \text {$n$ even} \end {cases}
| c =
}}
{{eqn |... | Let $T = \struct {S, \tau}$ be a [[Definition:Topological Space|topological space]].
Let $\sequence {a_n}$ be a [[Definition:Sequence|sequence]] in $T$.
Let $q \in S$ be an [[Definition:Accumulation Point of Sequence|accumulation point]] of $\sequence {a_n}$.
Then it is not necessarily the case that $q$ is also a [... | [[Proof by Counterexample]]:
Let $\struct {\R, \tau_d}$ be the [[Definition:Real Number Line with Euclidean Topology|real number line with the usual (Euclidean) topology]].
Let $\sequence {a_n}$ be the [[Definition:Sequence|sequence]] defined as:
{{begin-eqn}}
{{eqn | l = \sequence {a_n}
| r = \begin {cases} 1... | Accumulation Point of Sequence is not necessarily Limit Point | https://proofwiki.org/wiki/Accumulation_Point_of_Sequence_is_not_necessarily_Limit_Point | https://proofwiki.org/wiki/Accumulation_Point_of_Sequence_is_not_necessarily_Limit_Point | [
"Accumulation Points",
"Limit Points"
] | [
"Definition:Topological Space",
"Definition:Sequence",
"Definition:Accumulation Point/Sequence",
"Definition:Limit of Sequence/Topological Space"
] | [
"Proof by Counterexample",
"Definition:Euclidean Space/Euclidean Topology/Real Number Line",
"Definition:Sequence",
"Definition:Unique",
"Definition:Accumulation Point/Sequence",
"Definition:Limit of Sequence/Topological Space",
"Definition:Limit of Sequence/Topological Space"
] |
proofwiki-17275 | Limit Point of Underlying Set of Sequence of Reciprocals and Reciprocals + 1 | Let $\sequence {a_n}$ denote the sequence defined as:
{{begin-eqn}}
{{eqn | l = a_n
| r = \begin {cases} \dfrac 2 {n + 1} & : \text {$n$ odd} \\ \\ 1 + \dfrac 2 n & : \text {$n$ even} \end {cases}
| c =
}}
{{eqn | r = \sequence {\dfrac 1 1, 1 + \dfrac 1 1, \dfrac 1 2, 1 + \dfrac 1 2, \dfrac 1 3, 1 + \dfrac... | Let $\epsilon \in \R_{>0}$ be a (strictly) positive real number.
Then the open interval $\openint {-\epsilon} \epsilon$ contains $0$ and all elements $a_m$ of $S$ such that $0 < \dfrac 1 m < \epsilon$.
Hence $\dfrac 1 m \in \openint {-\epsilon} \epsilon$.
Hence the result by definition of limit point of $S$.
{{qed}} | Let $\sequence {a_n}$ denote the [[Definition:Sequence|sequence]] defined as:
{{begin-eqn}}
{{eqn | l = a_n
| r = \begin {cases} \dfrac 2 {n + 1} & : \text {$n$ odd} \\ \\ 1 + \dfrac 2 n & : \text {$n$ even} \end {cases}
| c =
}}
{{eqn | r = \sequence {\dfrac 1 1, 1 + \dfrac 1 1, \dfrac 1 2, 1 + \dfrac 1 ... | Let $\epsilon \in \R_{>0}$ be a [[Definition:Strictly Positive Real Number|(strictly) positive real number]].
Then the [[Definition:Open Real Interval|open interval]] $\openint {-\epsilon} \epsilon$ contains $0$ and all elements $a_m$ of $S$ such that $0 < \dfrac 1 m < \epsilon$.
Hence $\dfrac 1 m \in \openint {-\eps... | Limit Point of Underlying Set of Sequence of Reciprocals and Reciprocals + 1 | https://proofwiki.org/wiki/Limit_Point_of_Underlying_Set_of_Sequence_of_Reciprocals_and_Reciprocals_+_1 | https://proofwiki.org/wiki/Limit_Point_of_Underlying_Set_of_Sequence_of_Reciprocals_and_Reciprocals_+_1 | [
"Sequences",
"Examples of Limit Points"
] | [
"Definition:Sequence",
"Definition:Real Number/Real Number Line",
"Definition:Euclidean Space/Euclidean Topology/Real Number Line",
"Definition:Set",
"Definition:Term of Sequence",
"Definition:Subset",
"Definition:Limit Point/Topology/Set"
] | [
"Definition:Strictly Positive/Real Number",
"Definition:Real Interval/Open",
"Definition:Limit Point/Topology/Set"
] |
proofwiki-17276 | Omega-Accumulation Point of Underlying Set of Sequence of Reciprocals and Reciprocals + 1 | Let $\sequence {a_n}$ denote the sequence defined as:
{{begin-eqn}}
{{eqn | l = a_n
| r = \begin {cases} \dfrac 2 {n + 1} & : \text {$n$ odd} \\ \\ 1 + \dfrac 2 n & : \text {$n$ even} \end {cases}
| c =
}}
{{eqn | r = \sequence {\dfrac 1 1, 1 + \dfrac 1 1, \dfrac 1 2, 1 + \dfrac 1 2, \dfrac 1 3, 1 + \dfrac... | Let $\epsilon \in \R_{>0}$ be a (strictly) positive real number.
Then the open interval $\openint {-\epsilon} \epsilon$ contains $0$ and all elements $a_m$ of $S$ such that $0 < \dfrac 1 m < \epsilon$.
We have that:
:$\forall n \in \N: n \ge m$
have the property that $0 < \dfrac 1 n < \epsilon$.
Hence there are a count... | Let $\sequence {a_n}$ denote the [[Definition:Sequence|sequence]] defined as:
{{begin-eqn}}
{{eqn | l = a_n
| r = \begin {cases} \dfrac 2 {n + 1} & : \text {$n$ odd} \\ \\ 1 + \dfrac 2 n & : \text {$n$ even} \end {cases}
| c =
}}
{{eqn | r = \sequence {\dfrac 1 1, 1 + \dfrac 1 1, \dfrac 1 2, 1 + \dfrac 1 ... | Let $\epsilon \in \R_{>0}$ be a [[Definition:Strictly Positive Real Number|(strictly) positive real number]].
Then the [[Definition:Open Real Interval|open interval]] $\openint {-\epsilon} \epsilon$ contains $0$ and all elements $a_m$ of $S$ such that $0 < \dfrac 1 m < \epsilon$.
We have that:
:$\forall n \in \N: n \... | Omega-Accumulation Point of Underlying Set of Sequence of Reciprocals and Reciprocals + 1 | https://proofwiki.org/wiki/Omega-Accumulation_Point_of_Underlying_Set_of_Sequence_of_Reciprocals_and_Reciprocals_+_1 | https://proofwiki.org/wiki/Omega-Accumulation_Point_of_Underlying_Set_of_Sequence_of_Reciprocals_and_Reciprocals_+_1 | [
"Sequences",
"Examples of Omega-Accumulation Points"
] | [
"Definition:Sequence",
"Definition:Real Number/Real Number Line",
"Definition:Euclidean Space/Euclidean Topology/Real Number Line",
"Definition:Set",
"Definition:Term of Sequence",
"Definition:Subset",
"Definition:Omega-Accumulation Point"
] | [
"Definition:Strictly Positive/Real Number",
"Definition:Real Interval/Open",
"Definition:Countably Infinite/Set",
"Definition:Term of Sequence",
"Definition:Omega-Accumulation Point"
] |
proofwiki-17277 | Accumulation Point of Sequence of Reciprocals and Reciprocals + 1 | Let $\struct {\R, \tau}$ denote the real number line under the usual (Euclidean) topology.
Let $\sequence {a_n}$ denote the sequence in $\struct {\R, \tau}$ defined as:
{{begin-eqn}}
{{eqn | l = a_n
| r = \begin {cases} \dfrac 2 {n + 1} & : \text {$n$ odd} \\ 1 + \dfrac 2 n & : \text {$n$ even} \end {cases}
... | Let $\epsilon \in \R_{>0}$ be a (strictly) positive real number.
Then the open interval $\openint {-\epsilon} \epsilon$ contains $0$ and all elements $a_m$ of $S$ such that $0 < \dfrac 1 m < \epsilon$.
We have that:
:$\forall n \in \N: n \ge m$
have the property that $0 < \dfrac 1 n < \epsilon$.
Hence there are a count... | Let $\struct {\R, \tau}$ denote the [[Definition:Real Number Line|real number line]] under the [[Definition:Euclidean Topology on Real Number Line|usual (Euclidean) topology]].
Let $\sequence {a_n}$ denote the [[Definition:Sequence|sequence]] in $\struct {\R, \tau}$ defined as:
{{begin-eqn}}
{{eqn | l = a_n
| r... | Let $\epsilon \in \R_{>0}$ be a [[Definition:Strictly Positive Real Number|(strictly) positive real number]].
Then the [[Definition:Open Real Interval|open interval]] $\openint {-\epsilon} \epsilon$ contains $0$ and all elements $a_m$ of $S$ such that $0 < \dfrac 1 m < \epsilon$.
We have that:
:$\forall n \in \N: n \... | Accumulation Point of Sequence of Reciprocals and Reciprocals + 1 | https://proofwiki.org/wiki/Accumulation_Point_of_Sequence_of_Reciprocals_and_Reciprocals_+_1 | https://proofwiki.org/wiki/Accumulation_Point_of_Sequence_of_Reciprocals_and_Reciprocals_+_1 | [
"Sequences",
"Examples of Accumulation Points"
] | [
"Definition:Real Number/Real Number Line",
"Definition:Euclidean Space/Euclidean Topology/Real Number Line",
"Definition:Sequence",
"Definition:Accumulation Point/Sequence"
] | [
"Definition:Strictly Positive/Real Number",
"Definition:Real Interval/Open",
"Definition:Countably Infinite/Set",
"Definition:Term of Sequence",
"Definition:Accumulation Point/Sequence"
] |
proofwiki-17278 | Zero is not a Limit Point of Sequence of Reciprocals and Reciprocals + 1 | Let $\struct {\R, \tau}$ denote the real number line under the usual (Euclidean) topology.
Let $\sequence {a_n}$ denote the sequence in $\struct {\R, \tau}$ defined as:
{{begin-eqn}}
{{eqn | l = a_n
| r = \begin {cases} \dfrac 2 {n + 1} & : \text {$n$ odd} \\ \\ 1 + \dfrac 2 n & : \text {$n$ even} \end {cases}
... | The open interval $\openint {-1} 1$ contains $0$, and also contains all terms of $\sequence {a_n}$ with odd indices greater than $1$.
However, all terms of $\sequence {a_n}$ with even indices are outside $\openint {-1} 1$.
Hence $0$ cannot be a limit of $\sequence {a_n}$.
{{qed}} | Let $\struct {\R, \tau}$ denote the [[Definition:Real Number Line|real number line]] under the [[Definition:Euclidean Topology on Real Number Line|usual (Euclidean) topology]].
Let $\sequence {a_n}$ denote the [[Definition:Sequence|sequence]] in $\struct {\R, \tau}$ defined as:
{{begin-eqn}}
{{eqn | l = a_n
| r... | The [[Definition:Open Real Interval|open interval]] $\openint {-1} 1$ contains $0$, and also contains all [[Definition:Term of Sequence|terms]] of $\sequence {a_n}$ with [[Definition:Odd Integer|odd]] [[Definition:Index of Term of Sequence|indices]] greater than $1$.
However, all [[Definition:Term of Sequence|terms]] ... | Zero is not a Limit Point of Sequence of Reciprocals and Reciprocals + 1 | https://proofwiki.org/wiki/Zero_is_not_a_Limit_Point_of_Sequence_of_Reciprocals_and_Reciprocals_+_1 | https://proofwiki.org/wiki/Zero_is_not_a_Limit_Point_of_Sequence_of_Reciprocals_and_Reciprocals_+_1 | [
"Sequences",
"Reciprocals",
"Examples of Limit Points"
] | [
"Definition:Real Number/Real Number Line",
"Definition:Euclidean Space/Euclidean Topology/Real Number Line",
"Definition:Sequence",
"Definition:Limit of Sequence/Topological Space"
] | [
"Definition:Real Interval/Open",
"Definition:Term of Sequence",
"Definition:Odd Integer",
"Definition:Term of Sequence/Index",
"Definition:Term of Sequence",
"Definition:Even Integer",
"Definition:Term of Sequence/Index",
"Definition:Limit of Sequence/Topological Space"
] |
proofwiki-17279 | Lagrange's Theorem (Number Theory) | Let $\struct {\Z_p, +_p, \times_p}$ be the ring of integers modulo $p$ for some prime $p$.
Let $f$ be a polynomial in one variable of degree $n$ over $\Z_p$.
Then $f$ has at most $n$ roots in $\Z_p$. | Proof by induction on $n$: | Let $\struct {\Z_p, +_p, \times_p}$ be the [[Definition:Ring of Integers Modulo m|ring of integers modulo $p$]] for some [[Definition:Prime Number|prime]] $p$.
Let $f$ be a [[Definition:Polynomial over Field|polynomial in one variable]] of [[Definition:Degree of Polynomial|degree]] $n$ over $\Z_p$.
Then $f$ has at mo... | Proof by [[Principle of Mathematical Induction|induction]] on $n$: | Lagrange's Theorem (Number Theory) | https://proofwiki.org/wiki/Lagrange's_Theorem_(Number_Theory) | https://proofwiki.org/wiki/Lagrange's_Theorem_(Number_Theory) | [
"Lagrange's Theorem (Number Theory)",
"Number Theory",
"Polynomial Theory",
"Proofs by Induction"
] | [
"Definition:Ring of Integers Modulo m",
"Definition:Prime Number",
"Definition:Polynomial over Ring",
"Definition:Degree of Polynomial",
"Definition:Root of Polynomial"
] | [
"Principle of Mathematical Induction",
"Principle of Mathematical Induction"
] |
proofwiki-17280 | 1 plus Power of 2 is not Perfect Power except 9 | The only solution to:
:$1 + 2^n = a^b$
is:
:$\tuple {n, a, b} = \tuple {3, 3, 2}$
for positive integers $n, a, b$ with $b > 1$. | It suffices to prove the result for prime values of $b$.
For $n = 0$, it is clear that $1 + 2^0 = 2$ is not a perfect power.
For $n > 0$, $1 + 2^n$ is odd.
Hence for the equation to hold $a$ must be odd as well.
Writing $a = 2 m + 1$ we have:
{{begin-eqn}}
{{eqn | l = 1 + 2^n
| r = \paren {2 m + 1}^b
}}
{{eqn | r... | The only solution to:
:$1 + 2^n = a^b$
is:
:$\tuple {n, a, b} = \tuple {3, 3, 2}$
for [[Definition:Positive Integer|positive integers]] $n, a, b$ with $b > 1$. | It suffices to prove the result for [[Definition:Prime Number|prime]] values of $b$.
For $n = 0$, it is clear that $1 + 2^0 = 2$ is not a [[Definition:Perfect Power|perfect power]].
For $n > 0$, $1 + 2^n$ is [[Definition:Odd Integer|odd]].
Hence for the equation to hold $a$ must be [[Definition:Odd Integer|odd]] as... | 1 plus Power of 2 is not Perfect Power except 9 | https://proofwiki.org/wiki/1_plus_Power_of_2_is_not_Perfect_Power_except_9 | https://proofwiki.org/wiki/1_plus_Power_of_2_is_not_Perfect_Power_except_9 | [
"Number Theory"
] | [
"Definition:Positive/Integer"
] | [
"Definition:Prime Number",
"Definition:Perfect Power",
"Definition:Odd Integer",
"Definition:Odd Integer",
"Binomial Theorem",
"Binomial Coefficient with Zero",
"Definition:Divisor (Algebra)/Integer",
"Definition:Power (Algebra)/Integer",
"Definition:Power (Algebra)/Integer",
"Definition:Positive/... |
proofwiki-17281 | Fermat Number is not Perfect Power | There exist no Fermat numbers which are perfect powers. | Each Fermat number is in the form of $1 + 2^n$ for some $n \in \Z$.
This $n$ must also be a power of $2$.
From 1 plus Power of 2 is not Perfect Power except 9 we have:
:$1 + 2^n = a^b$
has only one solution $\tuple {n, a, b} = \tuple {3, 3, 2}$.
But $3$ is not a power of $2$.
Hence no Fermat numbers are perfect powers.... | There exist no [[Definition:Fermat Number|Fermat numbers]] which are [[Definition:Perfect Power|perfect powers]]. | Each [[Definition:Fermat Number|Fermat number]] is in the form of $1 + 2^n$ for some $n \in \Z$.
This $n$ must also be a [[Definition:Integer Power|power]] of $2$.
From [[1 plus Power of 2 is not Perfect Power except 9]] we have:
:$1 + 2^n = a^b$
has only one solution $\tuple {n, a, b} = \tuple {3, 3, 2}$.
But $3$... | Fermat Number is not Perfect Power | https://proofwiki.org/wiki/Fermat_Number_is_not_Perfect_Power | https://proofwiki.org/wiki/Fermat_Number_is_not_Perfect_Power | [
"Fermat Numbers"
] | [
"Definition:Fermat Number",
"Definition:Perfect Power"
] | [
"Definition:Fermat Number",
"Definition:Power (Algebra)/Integer",
"1 plus Power of 2 is not Perfect Power except 9",
"Definition:Power (Algebra)/Integer",
"Definition:Fermat Number",
"Definition:Perfect Power",
"Category:Fermat Numbers"
] |
proofwiki-17282 | Uncountable Closed Ordinal Space is Countably Compact | Let $\Omega$ denote the first uncountable ordinal.
Let $\closedint 0 \Omega$ denote the closed ordinal space on $\Omega$.
Then $\closedint 0 \Omega$ is a countably compact space. | We have:
:Closed Ordinal Space is Compact
:Compact Space is Countably Compact
{{qed}} | Let $\Omega$ denote the first [[Definition:Uncountable Ordinal|uncountable ordinal]].
Let $\closedint 0 \Omega$ denote the [[Definition:Uncountable Closed Ordinal Space|closed ordinal space]] on $\Omega$.
Then $\closedint 0 \Omega$ is a [[Definition:Countably Compact Space|countably compact space]]. | We have:
:[[Closed Ordinal Space is Compact]]
:[[Compact Space is Countably Compact]]
{{qed}} | Uncountable Closed Ordinal Space is Countably Compact | https://proofwiki.org/wiki/Uncountable_Closed_Ordinal_Space_is_Countably_Compact | https://proofwiki.org/wiki/Uncountable_Closed_Ordinal_Space_is_Countably_Compact | [
"Uncountable Closed Ordinal Spaces",
"Examples of Countably Compact Spaces"
] | [
"Definition:Uncountable Ordinal",
"Definition:Ordinal Space/Closed/Uncountable",
"Definition:Countably Compact Space"
] | [
"Closed Ordinal Space is Compact",
"Compact Space is Countably Compact"
] |
proofwiki-17283 | Sum of Euler Numbers by Binomial Coefficients Vanishes | $\forall n \in \Z_{>0}: \ds \sum_{k \mathop = 0}^n \binom {2 n} {2 k} E_{2 k} = 0$
where $E_k$ denotes the $k$th Euler number. | {{begin-eqn}}
{{eqn | l = \sum_{n \mathop = 0}^\infty \frac {E_n x^n} {n!}
| r = \frac {2 e^x} {e^{2 x} + 1}
| c = {{Defof|Euler Numbers}}
}}
{{eqn | r = \paren {\frac {2 e^x} {e^{2 x} + 1 } } \paren {\frac {e^{-x} } {e^{-x} } }
| c = multiplying top and bottom by $e^{-x}$
}}
{{eqn | r = \paren {\frac... | $\forall n \in \Z_{>0}: \ds \sum_{k \mathop = 0}^n \binom {2 n} {2 k} E_{2 k} = 0$
where $E_k$ denotes the $k$th [[Definition:Euler Numbers|Euler number]]. | {{begin-eqn}}
{{eqn | l = \sum_{n \mathop = 0}^\infty \frac {E_n x^n} {n!}
| r = \frac {2 e^x} {e^{2 x} + 1}
| c = {{Defof|Euler Numbers}}
}}
{{eqn | r = \paren {\frac {2 e^x} {e^{2 x} + 1 } } \paren {\frac {e^{-x} } {e^{-x} } }
| c = multiplying [[Definition:Numerator|top]] and [[Definition:Denominat... | Sum of Euler Numbers by Binomial Coefficients Vanishes | https://proofwiki.org/wiki/Sum_of_Euler_Numbers_by_Binomial_Coefficients_Vanishes | https://proofwiki.org/wiki/Sum_of_Euler_Numbers_by_Binomial_Coefficients_Vanishes | [
"Sum of Euler Numbers by Binomial Coefficients Vanishes",
"Euler Numbers",
"Binomial Coefficients"
] | [
"Definition:Euler Numbers"
] | [
"Definition:Fraction/Numerator",
"Definition:Fraction/Denominator",
"Power Series Expansion for Hyperbolic Cosine Function",
"Product of Absolutely Convergent Series",
"Definition:Subtraction",
"Definition:Even Integer",
"Definition:Power (Algebra)",
"Definition:Coefficient of Polynomial",
"Definiti... |
proofwiki-17284 | Uncountable Open Ordinal Space is Countably Compact | Let $\Omega$ denote the first uncountable ordinal.
Let $\hointr 0 \Omega$ denote the open ordinal space on $\Omega$.
Then $\hointr 0 \Omega$ is a countably compact space. | Let $\closedint 0 \Omega$ denote the closed ordinal space on $\Omega$.
From Uncountable Closed Ordinal Space is Countably Compact, $\closedint 0 \Omega$ is a countably compact space.
So every sequence in $\hointr 0 \Omega$ has an accumulation point in $\closedint 0 \Omega$.
{{LinkWanted|sequence in $\hointr 0 \Omega$ h... | Let $\Omega$ denote the first [[Definition:Uncountable Ordinal|uncountable ordinal]].
Let $\hointr 0 \Omega$ denote the [[Definition:Uncountable Open Ordinal Space|open ordinal space]] on $\Omega$.
Then $\hointr 0 \Omega$ is a [[Definition:Countably Compact Space|countably compact space]]. | Let $\closedint 0 \Omega$ denote the [[Definition:Uncountable Closed Ordinal Space|closed ordinal space]] on $\Omega$.
From [[Uncountable Closed Ordinal Space is Countably Compact]], $\closedint 0 \Omega$ is a [[Definition:Countably Compact Space|countably compact space]].
So every [[Definition:Sequence|sequence]] in... | Uncountable Open Ordinal Space is Countably Compact | https://proofwiki.org/wiki/Uncountable_Open_Ordinal_Space_is_Countably_Compact | https://proofwiki.org/wiki/Uncountable_Open_Ordinal_Space_is_Countably_Compact | [
"Uncountable Open Ordinal Spaces",
"Examples of Countably Compact Spaces"
] | [
"Definition:Uncountable Ordinal",
"Definition:Ordinal Space/Open/Uncountable",
"Definition:Countably Compact Space"
] | [
"Definition:Ordinal Space/Closed/Uncountable",
"Uncountable Closed Ordinal Space is Countably Compact",
"Definition:Countably Compact Space",
"Definition:Sequence",
"Definition:Accumulation Point/Sequence",
"Definition:Sequence",
"Definition:Accumulation Point/Sequence",
"Definition:Accumulation Point... |
proofwiki-17285 | Uncountable Open Ordinal Space is not Metacompact | Let $\Omega$ denote the first uncountable ordinal.
Let $\hointr 0 \Omega$ denote the open ordinal space on $\Omega$.
Then $\hointr 0 \Omega$ is not a metacompact space. | {{AimForCont}} $\hointr 0 \Omega$ is a metacompact space.
From Uncountable Open Ordinal Space is Countably Compact we have that $\hointr 0 \Omega$ is a countably compact space.
From Metacompact Countably Compact Space is Compact it follows that $\hointr 0 \Omega$ is a compact space.
But from Open Ordinal Space is not C... | Let $\Omega$ denote the first [[Definition:Uncountable Ordinal|uncountable ordinal]].
Let $\hointr 0 \Omega$ denote the [[Definition:Uncountable Open Ordinal Space|open ordinal space]] on $\Omega$.
Then $\hointr 0 \Omega$ is not a [[Definition:Metacompact Space|metacompact space]]. | {{AimForCont}} $\hointr 0 \Omega$ is a [[Definition:Metacompact Space|metacompact space]].
From [[Uncountable Open Ordinal Space is Countably Compact]] we have that $\hointr 0 \Omega$ is a [[Definition:Countably Compact Space|countably compact space]].
From [[Metacompact Countably Compact Space is Compact]] it follow... | Uncountable Open Ordinal Space is not Metacompact | https://proofwiki.org/wiki/Uncountable_Open_Ordinal_Space_is_not_Metacompact | https://proofwiki.org/wiki/Uncountable_Open_Ordinal_Space_is_not_Metacompact | [
"Uncountable Open Ordinal Spaces",
"Examples of Metacompact Spaces"
] | [
"Definition:Uncountable Ordinal",
"Definition:Ordinal Space/Open/Uncountable",
"Definition:Metacompact Space"
] | [
"Definition:Metacompact Space",
"Uncountable Open Ordinal Space is Countably Compact",
"Definition:Countably Compact Space",
"Metacompact Countably Compact Space is Compact",
"Definition:Compact Topological Space",
"Open Ordinal Space is not Compact Space",
"Definition:Contradiction",
"Definition:Comp... |
proofwiki-17286 | Uncountable Open Ordinal Space is not Paracompact | Let $\Omega$ denote the first uncountable ordinal.
Let $\hointr 0 \Omega$ denote the open ordinal space on $\Omega$.
Then $\hointr 0 \Omega$ is not a paracompact space. | {{AimForCont}} $\hointr 0 \Omega$ is a paracompact space.
From Paracompact Space is Metacompact, it follows that $\hointr 0 \Omega$ is a metacompact space.
But from Uncountable Open Ordinal Space is not Metacompact this contradicts the fact that $\hointr 0 \Omega$ is not a metacompact space.
Hence the result by Proof b... | Let $\Omega$ denote the first [[Definition:Uncountable Ordinal|uncountable ordinal]].
Let $\hointr 0 \Omega$ denote the [[Definition:Uncountable Open Ordinal Space|open ordinal space]] on $\Omega$.
Then $\hointr 0 \Omega$ is not a [[Definition:Paracompact Space|paracompact space]]. | {{AimForCont}} $\hointr 0 \Omega$ is a [[Definition:Paracompact Space|paracompact space]].
From [[Paracompact Space is Metacompact]], it follows that $\hointr 0 \Omega$ is a [[Definition:Metacompact Space|metacompact space]].
But from [[Uncountable Open Ordinal Space is not Metacompact]] this [[Definition:Contradicti... | Uncountable Open Ordinal Space is not Paracompact | https://proofwiki.org/wiki/Uncountable_Open_Ordinal_Space_is_not_Paracompact | https://proofwiki.org/wiki/Uncountable_Open_Ordinal_Space_is_not_Paracompact | [
"Uncountable Open Ordinal Spaces",
"Examples of Paracompact Spaces"
] | [
"Definition:Uncountable Ordinal",
"Definition:Ordinal Space/Open/Uncountable",
"Definition:Paracompact Space"
] | [
"Definition:Paracompact Space",
"Paracompact Space is Metacompact",
"Definition:Metacompact Space",
"Uncountable Open Ordinal Space is not Metacompact",
"Definition:Contradiction",
"Definition:Metacompact Space",
"Proof by Contradiction"
] |
proofwiki-17287 | Uncountable Open Ordinal Space is not Lindelöf | Let $\Omega$ denote the first uncountable ordinal.
Let $\hointr 0 \Omega$ denote the open ordinal space on $\Omega$.
Then $\hointr 0 \Omega$ is not a Lindelöf space. | {{AimForCont}} $\hointr 0 \Omega$ is a Lindelöf space.
From Ordinal Space is Completely Normal, $\hointr 0 \Omega$ is a completely normal.
From Sequence of Implications of Separation Axioms, $\hointr 0 \Omega$ is a $T_3$ space.
From $T_3$ Lindelöf Space is Paracompact, it follows that $\hointr 0 \Omega$ is a paracompac... | Let $\Omega$ denote the first [[Definition:Uncountable Ordinal|uncountable ordinal]].
Let $\hointr 0 \Omega$ denote the [[Definition:Uncountable Open Ordinal Space|open ordinal space]] on $\Omega$.
Then $\hointr 0 \Omega$ is not a [[Definition:Lindelöf Space|Lindelöf space]]. | {{AimForCont}} $\hointr 0 \Omega$ is a [[Definition:Lindelöf Space|Lindelöf space]].
From [[Ordinal Space is Completely Normal]], $\hointr 0 \Omega$ is a [[Definition:Completely Normal Space|completely normal]].
From [[Sequence of Implications of Separation Axioms]], $\hointr 0 \Omega$ is a [[Definition:T3 Space|$T_3... | Uncountable Open Ordinal Space is not Lindelöf | https://proofwiki.org/wiki/Uncountable_Open_Ordinal_Space_is_not_Lindelöf | https://proofwiki.org/wiki/Uncountable_Open_Ordinal_Space_is_not_Lindelöf | [
"Uncountable Open Ordinal Spaces",
"Examples of Lindelöf Spaces"
] | [
"Definition:Uncountable Ordinal",
"Definition:Ordinal Space/Open/Uncountable",
"Definition:Lindelöf Space"
] | [
"Definition:Lindelöf Space",
"Ordinal Space is Completely Normal",
"Definition:Completely Normal Space",
"Sequence of Implications of Separation Axioms",
"Definition:T3 Space",
"T3 Lindelöf Space is Paracompact",
"Definition:Paracompact Space",
"Definition:Contradiction",
"Uncountable Open Ordinal S... |
proofwiki-17288 | Uncountable Open Ordinal Space is not Sigma-Compact | Let $\Omega$ denote the first uncountable ordinal.
Let $\hointr 0 \Omega$ denote the open ordinal space on $\Omega$.
Then $\hointr 0 \Omega$ is not a $\sigma$-compact space. | {{AimForCont}} $\hointr 0 \Omega$ is a $\sigma$-compact space.
From Sigma-Compact Space is Lindelöf Space, $\hointr 0 \Omega$ is a Lindelöf space.
But this contradicts the fact that from Uncountable Open Ordinal Space is not Lindelöf, $\hointr 0 \Omega$ is not a Lindelöf space.
Hence the result by Proof by Contradictio... | Let $\Omega$ denote the first [[Definition:Uncountable Ordinal|uncountable ordinal]].
Let $\hointr 0 \Omega$ denote the [[Definition:Uncountable Open Ordinal Space|open ordinal space]] on $\Omega$.
Then $\hointr 0 \Omega$ is not a [[Definition:Sigma-Compact Space|$\sigma$-compact space]]. | {{AimForCont}} $\hointr 0 \Omega$ is a [[Definition:Sigma-Compact Space|$\sigma$-compact space]].
From [[Sigma-Compact Space is Lindelöf Space]], $\hointr 0 \Omega$ is a [[Definition:Lindelöf Space|Lindelöf space]].
But this [[Definition:Contradiction|contradicts]] the fact that from [[Uncountable Open Ordinal Space ... | Uncountable Open Ordinal Space is not Sigma-Compact | https://proofwiki.org/wiki/Uncountable_Open_Ordinal_Space_is_not_Sigma-Compact | https://proofwiki.org/wiki/Uncountable_Open_Ordinal_Space_is_not_Sigma-Compact | [
"Uncountable Open Ordinal Spaces",
"Examples of Sigma-Compact Spaces"
] | [
"Definition:Uncountable Ordinal",
"Definition:Ordinal Space/Open/Uncountable",
"Definition:Sigma-Compact Space"
] | [
"Definition:Sigma-Compact Space",
"Sigma-Compact Space is Lindelöf",
"Definition:Lindelöf Space",
"Definition:Contradiction",
"Uncountable Open Ordinal Space is not Lindelöf",
"Definition:Lindelöf Space",
"Proof by Contradiction"
] |
proofwiki-17289 | Uncountable Closed Ordinal Space is Lindelöf | Let $\Omega$ denote the first uncountable ordinal.
Let $\closedint 0 \Omega$ denote the closed ordinal space on $\Omega$.
Then $\closedint 0 \Omega$ is a Lindelöf space. | We have:
:Closed Ordinal Space is Compact
:Compact Space is Lindelöf
{{qed}} | Let $\Omega$ denote the first [[Definition:Uncountable Ordinal|uncountable ordinal]].
Let $\closedint 0 \Omega$ denote the [[Definition:Uncountable Closed Ordinal Space|closed ordinal space]] on $\Omega$.
Then $\closedint 0 \Omega$ is a [[Definition:Lindelöf Space|Lindelöf space]]. | We have:
:[[Closed Ordinal Space is Compact]]
:[[Compact Space is Lindelöf]]
{{qed}} | Uncountable Closed Ordinal Space is Lindelöf | https://proofwiki.org/wiki/Uncountable_Closed_Ordinal_Space_is_Lindelöf | https://proofwiki.org/wiki/Uncountable_Closed_Ordinal_Space_is_Lindelöf | [
"Uncountable Closed Ordinal Spaces",
"Examples of Lindelöf Spaces"
] | [
"Definition:Uncountable Ordinal",
"Definition:Ordinal Space/Closed/Uncountable",
"Definition:Lindelöf Space"
] | [
"Closed Ordinal Space is Compact",
"Compact Space is Lindelöf"
] |
proofwiki-17290 | Uncountable Closed Ordinal Space is Sigma-Compact | Let $\Omega$ denote the first uncountable ordinal.
Let $\closedint 0 \Omega$ denote the closed ordinal space on $\Omega$.
Then $\closedint 0 \Omega$ is a $\sigma$-compact space. | We have:
:Closed Ordinal Space is Compact
:Compact Space is $\sigma$-Compact Space
{{qed}} | Let $\Omega$ denote the first [[Definition:Uncountable Ordinal|uncountable ordinal]].
Let $\closedint 0 \Omega$ denote the [[Definition:Uncountable Closed Ordinal Space|closed ordinal space]] on $\Omega$.
Then $\closedint 0 \Omega$ is a [[Definition:Sigma-Compact Space|$\sigma$-compact space]]. | We have:
:[[Closed Ordinal Space is Compact]]
:[[Compact Space is Sigma-Compact Space|Compact Space is $\sigma$-Compact Space]]
{{qed}} | Uncountable Closed Ordinal Space is Sigma-Compact | https://proofwiki.org/wiki/Uncountable_Closed_Ordinal_Space_is_Sigma-Compact | https://proofwiki.org/wiki/Uncountable_Closed_Ordinal_Space_is_Sigma-Compact | [
"Uncountable Closed Ordinal Spaces",
"Examples of Sigma-Compact Spaces"
] | [
"Definition:Uncountable Ordinal",
"Definition:Ordinal Space/Closed/Uncountable",
"Definition:Sigma-Compact Space"
] | [
"Closed Ordinal Space is Compact",
"Compact Space is Sigma-Compact"
] |
proofwiki-17291 | Uncountable Open Ordinal Space is Sequentially Compact Space | Let $\Omega$ denote the first uncountable ordinal.
Let $\hointr 0 \Omega$ denote the open ordinal space on $\Omega$.
Then $\hointr 0 \Omega$ is a sequentially compact space. | We have that:
:Uncountable Open Ordinal Space is First-Countable
:Uncountable Open Ordinal Space is Countably Compact
The result follows from First-Countable Space is Sequentially Compact iff Countably Compact.
{{qed}} | Let $\Omega$ denote the first [[Definition:Uncountable Ordinal|uncountable ordinal]].
Let $\hointr 0 \Omega$ denote the [[Definition:Uncountable Open Ordinal Space|open ordinal space]] on $\Omega$.
Then $\hointr 0 \Omega$ is a [[Definition:Sequentially Compact Space|sequentially compact space]]. | We have that:
:[[Uncountable Open Ordinal Space is First-Countable]]
:[[Uncountable Open Ordinal Space is Countably Compact]]
The result follows from [[First-Countable Space is Sequentially Compact iff Countably Compact]].
{{qed}} | Uncountable Open Ordinal Space is Sequentially Compact Space | https://proofwiki.org/wiki/Uncountable_Open_Ordinal_Space_is_Sequentially_Compact_Space | https://proofwiki.org/wiki/Uncountable_Open_Ordinal_Space_is_Sequentially_Compact_Space | [
"Uncountable Open Ordinal Spaces",
"Examples of Sequentially Compact Spaces"
] | [
"Definition:Uncountable Ordinal",
"Definition:Ordinal Space/Open/Uncountable",
"Definition:Sequentially Compact Space"
] | [
"Uncountable Open Ordinal Space is First-Countable",
"Uncountable Open Ordinal Space is Countably Compact",
"First-Countable Space is Sequentially Compact iff Countably Compact"
] |
proofwiki-17292 | Recurrence Relation for Euler Numbers | Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Then:
{{begin-eqn}}
{{eqn | l = E_{2 n}
| r = -\sum_{k \mathop = 0}^{n - 1} \dbinom {2 n} {2 k} E_{2 k}
| c =
}}
{{eqn | r = -\paren {\binom {2 n} 0 E_0 + \binom {2 n} 2 E_2 + \binom {2 n} 4 E_4 + \cdots + \binom {2 n} {2 n - 2} E_{2 n - 2} }
| c ... | {{begin-eqn}}
{{eqn | q = \forall n \in \Z_{>0}
| l = \sum_{k \mathop = 0}^n \binom {2 n} {2 k} E_{2 k}
| r = 0
| c = Sum of Euler Numbers by Binomial Coefficients Vanishes
}}
{{eqn | ll= \leadsto
| l = \sum_{k \mathop = 0}^{n - 1} \dbinom {2 n} {2 k} E_{2 k} + \dbinom {2 n} {2 n} E_{2 n}
... | Let $n \in \Z_{>0}$ be a [[Definition:Strictly Positive Integer|(strictly) positive integer]].
Then:
{{begin-eqn}}
{{eqn | l = E_{2 n}
| r = -\sum_{k \mathop = 0}^{n - 1} \dbinom {2 n} {2 k} E_{2 k}
| c =
}}
{{eqn | r = -\paren {\binom {2 n} 0 E_0 + \binom {2 n} 2 E_2 + \binom {2 n} 4 E_4 + \cdots + \bino... | {{begin-eqn}}
{{eqn | q = \forall n \in \Z_{>0}
| l = \sum_{k \mathop = 0}^n \binom {2 n} {2 k} E_{2 k}
| r = 0
| c = [[Sum of Euler Numbers by Binomial Coefficients Vanishes]]
}}
{{eqn | ll= \leadsto
| l = \sum_{k \mathop = 0}^{n - 1} \dbinom {2 n} {2 k} E_{2 k} + \dbinom {2 n} {2 n} E_{2 n}
... | Recurrence Relation for Euler Numbers | https://proofwiki.org/wiki/Recurrence_Relation_for_Euler_Numbers | https://proofwiki.org/wiki/Recurrence_Relation_for_Euler_Numbers | [
"Recurrence Relation for Euler Numbers",
"Euler Numbers",
"Recurrence Relations"
] | [
"Definition:Strictly Positive/Integer",
"Definition:Euler Numbers"
] | [
"Sum of Euler Numbers by Binomial Coefficients Vanishes",
"Binomial Coefficient with Self"
] |
proofwiki-17293 | Integer to Power of Multiple of Order/Corollary | Then $\map \phi n$ is a multiple of $c$, where $\map \phi n$ is the Euler phi function of $n$. | From Euler's Theorem (Number Theory):
:$a^{\map \phi n} \equiv 1 \pmod n$
Applying Integer to Power of Multiple of Order we see that $\map \phi n$ is a multiple of $c$.
{{qed}} | Then $\map \phi n$ is a [[Definition:Multiple of Integer|multiple]] of $c$, where $\map \phi n$ is the [[Definition:Euler Phi Function|Euler phi function]] of $n$. | From [[Euler's Theorem (Number Theory)]]:
:$a^{\map \phi n} \equiv 1 \pmod n$
Applying [[Integer to Power of Multiple of Order]] we see that $\map \phi n$ is a [[Definition:Multiple of Integer|multiple]] of $c$.
{{qed}} | Integer to Power of Multiple of Order/Corollary | https://proofwiki.org/wiki/Integer_to_Power_of_Multiple_of_Order/Corollary | https://proofwiki.org/wiki/Integer_to_Power_of_Multiple_of_Order/Corollary | [
"Integer to Power of Multiple of Order",
"Number Theory"
] | [
"Definition:Multiple/Integer",
"Definition:Euler Phi Function"
] | [
"Euler's Theorem (Number Theory)",
"Integer to Power of Multiple of Order",
"Definition:Multiple/Integer"
] |
proofwiki-17294 | Divisor of Fermat Number/Euler's Result | Then $m$ is in the form:
:$k \, 2^{n + 1} + 1$
where $k \in \Z_{>0}$ is an integer. | It is sufficient to prove the result for prime divisors.
The general argument for all divisors follows from the argument:
:$\paren {a \, 2^c + 1} \paren {b \, 2^c + 1} = a b \, 2^{2 c} + \paren {a + b} \, 2^c + 1 = \paren {a b \, 2^c + a + b} \, 2^c + 1$
So the product of two factors of the form preserves that form.
Le... | Then $m$ is in the form:
:$k \, 2^{n + 1} + 1$
where $k \in \Z_{>0}$ is an [[Definition:Integer|integer]]. | It is sufficient to prove the result for [[Definition:Prime Divisor|prime divisors]].
The general argument for all [[Definition:Divisor of Integer|divisors]] follows from the argument:
:$\paren {a \, 2^c + 1} \paren {b \, 2^c + 1} = a b \, 2^{2 c} + \paren {a + b} \, 2^c + 1 = \paren {a b \, 2^c + a + b} \, 2^c + 1$
... | Divisor of Fermat Number/Euler's Result | https://proofwiki.org/wiki/Divisor_of_Fermat_Number/Euler's_Result | https://proofwiki.org/wiki/Divisor_of_Fermat_Number/Euler's_Result | [
"Divisor of Fermat Number"
] | [
"Definition:Integer"
] | [
"Definition:Prime Factor",
"Definition:Divisor (Algebra)/Integer",
"Definition:Divisor (Algebra)/Integer",
"Definition:Prime Factor",
"Congruence of Powers",
"Integer to Power of Multiple of Order",
"Definition:Multiplicative Order of Integer",
"Euler Phi Function of Prime"
] |
proofwiki-17295 | Divisor of Fermat Number/Refinement by Lucas | Let $n \ge 2$.
Then $m$ is in the form:
:$k \, 2^{n + 2} + 1$ | It is sufficient to prove the result for prime divisors.
The general argument for all divisors follows from the argument:
:$\paren {a \, 2^c + 1} \paren {b \, 2^c + 1} = a b \, 2^{2 c} + \paren {a + b} \, 2^c + 1 = \paren {a b \, 2^c + a + b} \, 2^c + 1$
So the product of two factors of the form preserves that form.
Le... | Let $n \ge 2$.
Then $m$ is in the form:
:$k \, 2^{n + 2} + 1$ | It is sufficient to prove the result for [[Definition:Prime Divisor|prime divisors]].
The general argument for all [[Definition:Divisor of Integer|divisors]] follows from the argument:
:$\paren {a \, 2^c + 1} \paren {b \, 2^c + 1} = a b \, 2^{2 c} + \paren {a + b} \, 2^c + 1 = \paren {a b \, 2^c + a + b} \, 2^c + 1$
... | Divisor of Fermat Number/Refinement by Lucas | https://proofwiki.org/wiki/Divisor_of_Fermat_Number/Refinement_by_Lucas | https://proofwiki.org/wiki/Divisor_of_Fermat_Number/Refinement_by_Lucas | [
"Divisor of Fermat Number"
] | [] | [
"Definition:Prime Factor",
"Definition:Divisor (Algebra)/Integer",
"Definition:Divisor (Algebra)/Integer",
"Definition:Prime Factor",
"Divisor of Fermat Number/Euler's Result",
"Definition:Divisor (Algebra)/Integer",
"Second Supplement to Law of Quadratic Reciprocity",
"Definition:Quadratic Residue",
... |
proofwiki-17296 | Set Closure is Smallest Closed Set/Normed Vector Space | Let $\struct {X, \norm {\, \cdot \,} }$ be a normed vector space.
Let $S$ be a subset of $X$:
:$S \subseteq X$
Let $S^-$ be the closure of $S$.
Then $S^-$ is the smallest closed set which contains $S$. | Let $F$ be a closed set in $X$.
Suppose $S \subseteq F$. | Let $\struct {X, \norm {\, \cdot \,} }$ be a [[Definition:Normed Vector Space|normed vector space]].
Let $S$ be a [[Definition:Subset|subset]] of $X$:
:$S \subseteq X$
Let $S^-$ be the [[Definition:Closure in Normed Vector Space|closure]] of $S$.
Then $S^-$ is the [[Definition:Smallest Set by Set Inclusion|smalles... | Let $F$ be a [[Definition:Closed Set in Normed Vector Space|closed set]] in $X$.
Suppose $S \subseteq F$. | Set Closure is Smallest Closed Set/Normed Vector Space | https://proofwiki.org/wiki/Set_Closure_is_Smallest_Closed_Set/Normed_Vector_Space | https://proofwiki.org/wiki/Set_Closure_is_Smallest_Closed_Set/Normed_Vector_Space | [
"Closed Sets",
"Normed Vector Spaces",
"Set Closure is Smallest Closed Set"
] | [
"Definition:Normed Vector Space",
"Definition:Subset",
"Definition:Closure/Normed Vector Space",
"Definition:Smallest Set by Set Inclusion",
"Definition:Closed Set/Normed Vector Space"
] | [
"Definition:Closed Set/Normed Vector Space",
"Definition:Closed Set/Normed Vector Space",
"Definition:Closed Set/Normed Vector Space",
"Definition:Closed Set/Normed Vector Space"
] |
proofwiki-17297 | Bounded Real Function may not be of Bounded Variation | Let $a, b$ be real numbers with $a < b$.
Let $f : \closedint a b \to \R$ be a bounded function.
Then $f$ is not necessarily of bounded variation. | Let $a = 0$, $b = 1$.
Define $f : \closedint 0 1 \to \R$ by:
:$\map f x = \begin{cases} 1 & x \in \Q \\ 0 & x \not \in \Q \end{cases}$
For each finite subdivision $P$ of $\closedint 0 1$, write:
:$P = \set {x_0, x_1, \ldots, x_n }$
with:
:$0 = x_0 < x_1 < x_2 < \cdots < x_{n - 1} < x_n = 1$
For each such subdivision ... | Let $a, b$ be [[Definition:Real Number|real numbers]] with $a < b$.
Let $f : \closedint a b \to \R$ be a [[Definition:Bounded Mapping|bounded]] [[Definition:Real Function|function]].
Then $f$ is not necessarily of [[Definition:Bounded Variation (Closed Bounded Interval)|bounded variation]]. | Let $a = 0$, $b = 1$.
Define $f : \closedint 0 1 \to \R$ by:
:$\map f x = \begin{cases} 1 & x \in \Q \\ 0 & x \not \in \Q \end{cases}$
For each [[Definition:Finite Subdivision|finite subdivision]] $P$ of $\closedint 0 1$, write:
:$P = \set {x_0, x_1, \ldots, x_n }$
with:
:$0 = x_0 < x_1 < x_2 < \cdots < x_{n - ... | Bounded Real Function may not be of Bounded Variation | https://proofwiki.org/wiki/Bounded_Real_Function_may_not_be_of_Bounded_Variation | https://proofwiki.org/wiki/Bounded_Real_Function_may_not_be_of_Bounded_Variation | [
"Bounded Variation"
] | [
"Definition:Real Number",
"Definition:Bounded Mapping",
"Definition:Real Function",
"Definition:Bounded Variation/Closed Bounded Interval"
] | [
"Definition:Subdivision of Interval/Finite",
"Definition:Subdivision of Interval/Finite",
"Definition:Sequence",
"Definition:Subdivision of Interval/Finite",
"Definition:Bounded Variation/Closed Bounded Interval",
"Definition:Subdivision of Interval/Finite",
"Definition:Real Sequence",
"Between two Ra... |
proofwiki-17298 | Element is Loop iff Member of Closure of Empty Set | Let $M = \struct{S, \mathscr I}$ be a matroid.
Let $x \in S$.
Then:
:$x$ is a loop {{iff}} $x \in \map \sigma \O$
where $\map \sigma \O$ denotes the closure of the empty set. | From Element is Loop iff Rank is Zero:
:$x$ is a loop {{iff}} $\map \rho {\set x} = 0$
where $\rho$ is the rank function of $M$.
Now:
{{begin-eqn}}
{{eqn | r = x \in \map \sigma \O
| o =
}}
{{eqn | ll= \leadstoandfrom
| r = x \sim \O
| o =
| c = {{Defof|Closure Operator (Matroid)|Closure Operato... | Let $M = \struct{S, \mathscr I}$ be a [[Definition:Matroid|matroid]].
Let $x \in S$.
Then:
:$x$ is a [[Definition:Loop (Matroid)|loop]] {{iff}} $x \in \map \sigma \O$
where $\map \sigma \O$ denotes the [[Definition:Closure Operator (Matroid)|closure]] of the [[Definition:Empty Set|empty set]]. | From [[Element is Loop iff Rank is Zero]]:
:$x$ is a [[Definition:Loop (Matroid)|loop]] {{iff}} $\map \rho {\set x} = 0$
where $\rho$ is the [[Definition:Rank Function (Matroid)|rank function]] of $M$.
Now:
{{begin-eqn}}
{{eqn | r = x \in \map \sigma \O
| o =
}}
{{eqn | ll= \leadstoandfrom
| r = x \sim \O... | Element is Loop iff Member of Closure of Empty Set | https://proofwiki.org/wiki/Element_is_Loop_iff_Member_of_Closure_of_Empty_Set | https://proofwiki.org/wiki/Element_is_Loop_iff_Member_of_Closure_of_Empty_Set | [
"Matroid Loops",
"Matroid Closure"
] | [
"Definition:Matroid",
"Definition:Loop (Matroid)",
"Definition:Closure Operator (Matroid)",
"Definition:Empty Set"
] | [
"Singleton is Dependent implies Rank is Zero/Corollary",
"Definition:Loop (Matroid)",
"Definition:Rank Function (Matroid)",
"Rank of Empty Set is Zero"
] |
proofwiki-17299 | Singleton is Dependent implies Rank is Zero/Corollary | :$x$ is a loop {{iff}} $\map \rho {\set x} = 0$ | By definition of a loop:
:$x$ is a loop {{iff}} $\set x \notin \mathscr I$
From Singleton is Dependent implies Rank is Zero:
:if $\set x \notin \mathscr I$ then $\map \rho {\set x} = 0$
From Singleton is Independent implies Rank is One:
:if $\set x \in \mathscr I$ then $\map \rho {\set x} = 1$
It follows that:
:$\set x... | :$x$ is a [[Definition:Loop (Matroid)|loop]] {{iff}} $\map \rho {\set x} = 0$ | By definition of a [[Definition:Loop (Matroid)|loop]]:
:$x$ is a [[Definition:Loop (Matroid)|loop]] {{iff}} $\set x \notin \mathscr I$
From [[Singleton is Dependent implies Rank is Zero]]:
:if $\set x \notin \mathscr I$ then $\map \rho {\set x} = 0$
From [[Singleton is Independent implies Rank is One]]:
:if $\set x \... | Singleton is Dependent implies Rank is Zero/Corollary | https://proofwiki.org/wiki/Singleton_is_Dependent_implies_Rank_is_Zero/Corollary | https://proofwiki.org/wiki/Singleton_is_Dependent_implies_Rank_is_Zero/Corollary | [
"Matroid Dependent Subsets",
"Matroid Rank Functions"
] | [
"Definition:Loop (Matroid)"
] | [
"Definition:Loop (Matroid)",
"Definition:Loop (Matroid)",
"Singleton is Dependent implies Rank is Zero",
"Singleton is Independent implies Rank is One"
] |
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