nyuuzyou/wikis huggingface dataset:
Equivalence of Definitions of Equivalent Division Ring Norms/Cauchy Sequence Equivalent implies Open Unit Ball Equivalent

== Theorem ==
Let $R$ be a division ring.
Let $\norm {\, \cdot \,}_1: R \to \R_{\ge 0}$ and $\norm {\, \cdot \,}_2: R \to \R_{\ge 0}$ be norms on $R$.
Let $\norm {\, \cdot \,}_1$ and $\norm {\, \cdot \,}_2$ satisfy:
:for all sequences $\sequence {x_n}$ in $R$: $\sequence {x_n}$ is a Cauchy sequence in $\norm {\, \cdot \,}_1$ $\sequence {x_n}$ is a Cauchy sequence in $\norm {\, \cdot \,}_2$
Then $\forall x \in R$:
:$\norm x_1 < 1 \iff \norm x_2 < 1$
== Proof ==
The contrapositive is proved.
Let there exist $x \in R$ such that $\norm x_1 < 1$ and $\norm x_2 \ge 1$.
Let $\sequence {x_n}$ be the sequence defined by: $\forall n: x_n = x^n$.
By Sequence of Powers of Number less than One in Normed Division Ring then $\sequence {x_n}$ is a null sequence in $\norm {\, \cdot \,}_1$.
By convergent sequence in normed division ring is a Cauchy sequence then $\sequence {x_n}$ is a Cauchy sequence in $\norm {\, \cdot \,}_1$.
Let $0_R$ be the zero of $R$ and $1_R$ be the unit of $R$.
By Norm of Unity of Division Ring and the assumption $\norm x_1 < 1$:
:$x \ne 1_R$
Hence:
:$x - 1_R \ne 0_R$
By norm axiom $(\text N 1)$: Positive Definiteness:
:$\norm {x - 1_R}_2 > 0$
Let $\epsilon = \dfrac {\norm {x - 1_R}_2} 2$.
Then $\norm {x - 1_R}_2 > \epsilon$.
We have that $\norm x_2 \ge 1$.
Hence for all $n \in \N$:
For all $n \in \N$:
 
So $\sequence {x_n}$ is not a Cauchy sequence in $\norm {\, \cdot \,}_2$.
The theorem now follows by the Rule of Transposition. 
== Sources ==
* : $\S 1.2$ Normed Fields, Proposition $1.10$ and Exercise $13$
Category:Equivalence of Definitions of Equivalent Division Ring Norms
Equivalence of Definitions of Equivalent Division Ring Norms/Convergently Equivalent implies Null Sequence Equivalent

== Theorem ==
Let $R$ be a division ring.
Let $\norm {\, \cdot \,}_1: R \to \R_{\ge 0}$ and $\norm {\, \cdot \,}_2: R \to \R_{\ge 0}$ be norms on $R$.
Let $\norm {\, \cdot \,}_1$ and $\norm {\, \cdot \,}_2$ satisfy:
:for all sequences $\sequence {x_n}$ in $R:\sequence {x_n}$ converges to $l$ in $\norm{\, \cdot \,}_1 \iff \sequence {x_n}$ is a converges to $l$ in $\norm {\, \cdot \,}_2$
Then for all sequences $\sequence {x_n}$ in $R$:
:$\sequence {x_n}$ is a null sequence in $\norm {\, \cdot \,}_1 \iff \sequence {x_n}$ is a null sequence in $\norm {\, \cdot \,}_2$ 
== Proof ==
Let $0_R$ be the zero of $R$, then:
:$\sequence {x_n}$ converges to $0_R$ in $\norm {\, \cdot \,}_1 \iff \sequence {x_n}$ converges to $0_R$ in $\norm {\, \cdot \,}_2$
Hence:
:$\sequence {x_n}$ is a null sequence in $\norm{\,\cdot\,}_1 \iff \sequence {x_n}$ is a null sequence in $\norm{\,\cdot\,}_2$ 
== Sources ==
* : $\S 3.1$ Absolute Values on $\Q$, Lemma 3.1.2 and Problem 66
Category:Equivalence of Definitions of Equivalent Division Ring Norms
Equivalence of Definitions of Equivalent Division Ring Norms/Norm is Power of Other Norm implies Cauchy Sequence Equivalent

== Theorem ==
Let $R$ be a division ring.
Let $\norm {\, \cdot \,}_1: R \to \R_{\ge 0}$ and $\norm {\, \cdot \,}_2: R \to \R_{\ge 0}$ be norms on $R$.
Let $\norm {\, \cdot \,}_1$ and $\norm {\, \cdot \,}_2$ satisfy:
:$\exists \alpha \in \R_{> 0}: \forall x \in R: \norm x_1 = \norm x_2^\alpha$
Then for all sequences $\sequence {x_n}$ in $R$:
:$\sequence {x_n}$ is a Cauchy sequence in $\norm {\, \cdot \,}_1$ $\sequence {x_n}$ is a Cauchy sequence in $\norm {\, \cdot \,}_2$
== Proof ==
Let $\sequence {x_n}$ be a Cauchy sequence in $\norm {\, \cdot \,}_1$.
Let $\epsilon > 0$ be given.
Since $\sequence {x_n}$ is a Cauchy sequence then:
:$\exists N \in \N: \forall n,m \ge N: \norm {x_n - x_m}_1 < \epsilon^\alpha$
Then:
:$\exists N \in \N: \forall n,m \ge N: \norm {x_n - x_m}_2^\alpha < \epsilon^\alpha$
Hence:
:$\exists N \in \N: \forall n,m \ge N: \norm {x_n - x_m}_2 < \epsilon$
So $\sequence {x_n}$ is a Cauchy sequence in $\norm {\, \cdot \,}_2$
It follows that for all sequences $\sequence {x_n}$ in $R$:
:$\sequence {x_n}$ is a Cauchy sequence in $\norm {\, \cdot \,}_1 \implies \sequence {x_n}$ is a Cauchy sequence in $\norm {\, \cdot \,}_2$
Let $\sequence {x_n}$ be a Cauchy sequence in $\norm {\, \cdot \,}_2$.
Let $\epsilon > 0$ be given.
Since $\sequence {x_n}$ is a Cauchy sequence then:
:$\exists N \in \N: \forall n,m \ge N: \norm {x_n - x_m}_2 \lt \epsilon^{1/\alpha}$
Then:
:$\exists N \in \N: \forall n, m \ge N: \norm {x_n - x_m}_2^\alpha < \epsilon$
Hence:
:$\exists N \in \N: \forall n, m \ge N: \norm {x_n - x_m}_1 < \epsilon$
So $\sequence {x_n}$ is a Cauchy sequence in $\norm {\, \cdot \,}_1$
It follows that for all sequences $\sequence {x_n}$ in $R$:
:$\sequence {x_n}$ is a Cauchy sequence in $\norm {\, \cdot \,}_2 \implies \sequence {x_n}$ is a Cauchy sequence in $\norm {\, \cdot \,}_1$
The result follows.
== Sources ==
* : $\S 1.2$ Normed Fields, Proposition $1.10$ and Exercise $13$
Category:Equivalence of Definitions of Equivalent Division Ring Norms
Equivalence of Definitions of Equivalent Division Ring Norms/Norm is Power of Other Norm implies Topologically Equivalent

== Theorem ==
Let $R$ be a division ring.
Let $\norm {\, \cdot \,}_1: R \to \R_{\ge 0}$ and $\norm {\, \cdot \,}_2: R \to \R_{\ge 0}$ be norms on $R$.
Let $d_1$ and $d_2$ be the metrics induced by the norms $\norm {\, \cdot \,}_1$ and $\norm {\, \cdot \,}_2$ respectively.
Let $\norm {\, \cdot \,}_1$ and $\norm {\, \cdot \,}_2$ satisfy:
:$\exists \alpha \in \R_{\gt 0}: \forall x \in R: \norm x_1 = \norm x_2^\alpha$
Then $d_1$ and $d_2$ are topologically equivalent metrics.
== Proof ==
Let $x \in R$ and $\epsilon \in \R_{\gt 0}$
Then for $y \in R$:
Hence:
:$\map {B^1_\epsilon} x = \map {B^2_{\epsilon^{1 / \alpha} } } x$
where:
:$\map {B^1_\epsilon} x$ is the open ball in $d_1$ centered on $x$ with radius $\epsilon$
:$\map {B^2_{\epsilon^{1 / \alpha} } } x$ is the open ball in $d_2$ centered on $x$ with radius $\epsilon^{1 / \alpha}$
Since $x$ and $\epsilon$ were arbitrary then:
:every $d_1$-open ball is a $d_2$-open ball.
Similarly, for $y \in R$:
So:
:every $d_2$-open ball is a $d_1$-open ball.
By the definition of an open set of a metric space it follows that $d_1$ and $d_2$ are topologically equivalent metrics,
== Sources ==
* : $\S 3.1$ Absolute Values on $\Q$, Lemma $3.1.2$ and Problem $66$
Category:Equivalence of Definitions of Equivalent Division Ring Norms
Equivalence of Definitions of Equivalent Division Ring Norms/Null Sequence Equivalent implies Open Unit Ball Equivalent

== Theorem ==
Let $R$ be a division ring.
Let $\norm {\, \cdot \,}_1: R \to \R_{\ge 0}$ and $\norm {\, \cdot \,}_2: R \to \R_{\ge 0}$ be norms on $R$.
Let $\norm {\, \cdot \,}_1$ and $\norm {\, \cdot \,}_2$ satisfy:
:for all sequences $\sequence {x_n}$ in $R:\sequence {x_n}$ is a null sequence in $\norm {\, \cdot \,}_1 \iff \sequence {x_n}$ is a null sequence in $\norm {\, \cdot \,}_2$ 
Then $\forall x \in R$:
:$\norm x_1 < 1 \iff \norm x_2 < 1$ 
== Proof ==
Let $x \in R$.
Let $\sequence {x_n}$ be the sequence defined by: $\forall n: x_n = x^n$.
== Sources ==
* : $\S 3.1$ Absolute Values on $\Q$, Lemma 3.1.2 and Problem 66
Category:Equivalence of Definitions of Equivalent Division Ring Norms
Equivalence of Definitions of Equivalent Division Ring Norms/Open Unit Ball Equivalent implies Norm is Power of Other Norm

== Theorem ==
Let $R$ be a division ring.
Let $\norm {\, \cdot \,}_1: R \to \R_{\ge 0}$ and $\norm {\, \cdot \,}_2: R \to \R_{\ge 0}$ be norms on $R$.
Let $\norm {\, \cdot \,}_1$ and $\norm {\, \cdot \,}_2$ satisfy:
:$\forall x \in R: \norm x_1 < 1 \iff \norm x_2 < 1$ 
Then:
:$\exists \alpha \in \R_{> 0}: \forall x \in R: \norm x_1 = \norm x_2^\alpha$
== Proof ==
==== Case 1 ====
For all $x \in R: x \ne 0_R$, let $x$ satisfy $\norm x_1 \ge 1$.
===== Lemma 1 =====
By assumption, for all $x \in R, x \ne 0_R$, then $\norm x_2 \ge 1$.
Similarly $\norm {\, \cdot \,}_2$ is the trivial norm.
Hence $\norm {\, \cdot \,}_1$ and $\norm {\, \cdot \,}_2$ are equal.
For $\alpha = 1$ the result follows.
==== Case 2 ====
Let $x_0 \in R$ such that $x_0 \ne 0_R$ and $\norm {x_0}_1 < 1$.
By assumption then $\norm {x_0}_2 < 1$.
Let $\alpha = \dfrac {\log \norm {x_0}_1 } {\log \norm {x_0}_2 }$.
Then $\norm {x_0}_1 = \norm {x_0}_2^\alpha$.
As $\norm {x_0}_1, \norm {x_0}_2 < 1$:
:$\log \norm {x_0}_1 < 0$
:$\log \norm {x_0}_2 < 0$
So $\alpha > 0$.
====== Lemma 2 ======
== Sources ==
* : $\S 3.1$ Absolute Values on $\Q$, Lemma $3.1.2$ and Problem $66$
* : $\S 1.2$ Normed Fields, Proposition $1.10$ and Exercise $13$
Category:Equivalence of Definitions of Equivalent Division Ring Norms
Equivalence of Definitions of Equivalent Division Ring Norms/Open Unit Ball Equivalent implies Norm is Power of Other Norm/Lemma 1

== Theorem ==
Let $R$ be a division ring.
Let $\norm {\, \cdot \,}_1: R \to \R_{\ge 0}$ be a norm on $R$.
Let $\norm {\, \cdot \,}_1$ satisfy:
:$\forall x \in R, x \ne 0_R:\norm x_1 \ge 1$ 
Then:
:$\norm {\, \cdot \,}_1$ is the trivial norm.
== Proof ==
We prove the contrapositive.
Let $\norm {\, \cdot \,}_1$ be a nontrivial norm.
Then:
:$\exists y \in R: \norm y_1 \ne 0, \norm y_1 \ne 1$.
By Real Numbers form Totally Ordered Field either $\norm y_1 < 1$ or $\norm y_1 > 1$.
Suppose $\norm y_1 > 1$.
By Norm axiom $(\text N 1)$: Positive Definiteness:
:$y \ne 0_R$
By Norm of Inverse in Division Ring:
:$\norm {y^{-1} }_1 = \dfrac 1 {\norm y_1} < 1$
So either $\norm y_1 < 1$ or $\norm {y^{-1} }_1 < 1$
So:
:$\exists x \in R, x \ne 0_R:\norm x_1 < 1$
The theorem now follows by the Rule of Transposition. 
Category:Equivalence of Definitions of Equivalent Division Ring Norms
Equivalence of Definitions of Equivalent Division Ring Norms/Open Unit Ball Equivalent implies Norm is Power of Other Norm/Lemma 2

== Theorem ==
Let $R$ be a division ring.
Let $\norm {\, \cdot \,}_1: R \to \R_{\ge 0}$ and $\norm {\, \cdot \,}_2: R \to \R_{\ge 0}$ be norms on $R$.
Let $\norm {\, \cdot \,}_1$ and $\norm {\, \cdot \,}_2$ satisfy:
:$\forall x \in R:\norm x_1 < 1 \iff \norm x_2 < 1$ 
Let $x_0 \in R$ such that $x_0 \ne 0_R$ and $\norm {x_0}_1 < 1$.
Let $\alpha = \dfrac {\log \norm {x_0}_1 } {\log \norm {x_0}_2 }$.
Then:
:$\forall x \in R: \norm x_1 = \norm x_2^\alpha$
== Proof ==
Since $\norm {x_0}_1 < 1$ then $\norm {x_0}_2 < 1$ and:
:$\log \norm {x_0}_1 < 0$
:$\log \norm {x_0}_2 < 0$
Hence $\alpha > 0$
Since $\forall x \in R: \norm x_1 < 1 \iff \norm x_2 < 1$: 
===== Lemma 1 =====
===== Lemma 2 =====
Let $x \in R, x \ne 0_R$.
===== Case 1 =====
Let $\norm x_1 = 1$.
By Lemma 2 then:
:$\norm x_2 = 1$
Hence:
:$\norm x_1 = 1 = 1^\alpha = \norm x_2^\alpha$
 
===== Case 2 =====
Let $\norm x_1 \ne 1$.
By Lemma 2 then:
:$\norm x_2 \ne 1$
Let $\beta = \dfrac {\log \norm x_1 } {\log \norm x_2 }$.
Then $\norm x_1 = \norm x_2^\beta$.
====== Lemma 3 ======
Hence $\norm x_1 = \norm x_2^\alpha$.
</onlyinclude>
Category:Equivalence of Definitions of Equivalent Division Ring Norms
Equivalence of Definitions of Equivalent Division Ring Norms/Open Unit Ball Equivalent implies Norm is Power of Other Norm/Lemma 2/Lemma 2.1

== Theorem ==
Let $R$ be a division ring.
Let $\norm {\, \cdot \,}_1: R \to \R_{\ge 0}$ and $\norm {\, \cdot \,}_2: R \to \R_{\ge 0}$ be norms on $R$.
Let $\norm {\, \cdot \,}_1$ and $\norm {\, \cdot \,}_2$ satisfy:
:$\forall y \in R: \norm y_1 < 1 \iff \norm y_2 < 1$ 
Then:
:$\forall y \in R: \norm y_1 > 1 \iff \norm y_2 > 1$ 
== Proof ==
For $y \in R$ then:
Category:Equivalence of Definitions of Equivalent Division Ring Norms
Equivalence of Definitions of Equivalent Division Ring Norms/Open Unit Ball Equivalent implies Norm is Power of Other Norm/Lemma 2/Lemma 2.2

== Theorem ==
Let $R$ be a division ring.
Let $\norm {\, \cdot \,}_1: R \to \R_{\ge 0}$ and $\norm {\, \cdot \,}_2: R \to \R_{\ge 0}$ be norms on $R$.
Let $\norm {\, \cdot \,}_1$ and $\norm {\, \cdot \,}_2$ satisfy:
:$\forall y \in R:\norm y_1 \lt 1 \iff \norm y_2 \lt 1$ 
Then:
:$\forall y \in R:\norm y_1 = 1 \iff \norm y_2 = 1$ 
== Proof ==
By assumption:
:$\forall y \in R:\norm y_1 \ge 1 \iff \norm y_2 \ge 1$
By Lemma 1:
:$\forall y \in R:\norm y_1 \le 1 \iff \norm y_2 \le 1$
Hence $\forall y \in R$:
 
Category:Equivalence of Definitions of Equivalent Division Ring Norms
Equivalence of Definitions of Equivalent Division Ring Norms/Open Unit Ball Equivalent implies Norm is Power of Other Norm/Lemma 2/Lemma 2.3

== Theorem ==
Let $R$ be a division ring.
Let $\norm {\, \cdot \,}_1: R \to \R_{\ge 0}$ and $\norm {\, \cdot \,}_2: R \to \R_{\ge 0}$ be norms on $R$.
Let $\norm {\, \cdot \,}_1$ and $\norm {\, \cdot \,}_2$ satisfy:
:$\forall x \in R:\norm x_1 < 1 \iff \norm x_2 < 1$ 
Let $x, y \in R \setminus 0_R$ such that $\norm x_1, \norm y_1 \ne 1$.
Let $\alpha = \dfrac {\log \norm x_1} {\log \norm x_2}$ and $\beta = \dfrac {\log \norm y_1} {\log \norm y_2}$.
Then:
:$\alpha = \beta$
== Proof ==
Because $x, y \in R \setminus 0_R$:
:$\norm x_1 , \norm y_1, \norm x_2 , \norm y_2 > 0$.
Because $\norm{x}_1 , \norm {y}_1 \ne 1$, by Lemma 2:
:$\norm x_2 , \norm y_2 \ne 1$.
Hence:
:$\log \norm x_1 , \log \norm y_1, \log \norm x_2, \log \norm y_2 \ne 0$ 
and $\alpha, \beta$ are well-defined.
Let $r = \dfrac n m \in \Q$ be any rational number where $n, m \in \Z$ are integers and $m \ne 0$.
Then:
By Logarithm is Strictly Increasing:
:$\log \norm y_1^n < \log \norm x_1^m \iff \log \norm y_2^n < \log \norm x_2^m$
By Sum of Logarithms:
:$n \log \norm y_1 < m \log \norm x_1 \iff n \log \norm y_2 < m \log \norm x_2$
Because $m, \log \norm x_1, \log \norm x_2 \ne 0$:
:$r = \dfrac n m < \dfrac {\log \norm x_1} {\log \norm y_1} \iff r = \dfrac n m < \dfrac {\log \norm x_2} {\log \norm y_2}$
By Between two Real Numbers exists Rational Number:
:$\dfrac {\log \norm x_1} {\log \norm y_1} = \dfrac {\log \norm x_2} {\log \norm y_2}$
Because $\log \norm x_2, \log \norm y_2 \ne 0$:
:$\dfrac {\log \norm x_1} {\log \norm x_2} = \dfrac {\log \norm y_1} {\log \norm y_2}$
That is:
:$\alpha = \beta$
Category:Equivalence of Definitions of Equivalent Division Ring Norms
Equivalence of Definitions of Equivalent Division Ring Norms/Topologically Equivalent implies Convergently Equivalent

== Theorem ==
Let $R$ be a division ring.
Let $\norm {\, \cdot \,}_1: R \to \R_{\ge 0}$ and $\norm {\, \cdot \,}_2: R \to \R_{\ge 0}$ be norms on $R$.
Let $d_1$ and $d_2$ be the metrics induced by the norms $\norm {\, \cdot \,}_1$ and $\norm {\, \cdot \,}_2$ respectively.
Let $d_1$ and $d_2$ be topologically equivalent metrics.
Then:
:$d_1$ and $d_2$ are convergently equivalent metrics.
== Proof ==
Let $\sequence {x_n}$ converge to $l$ in $\norm {\, \cdot \,}_1$.
Let $\epsilon \in \R_{> 0}$ be given.
Let $\map {B_\epsilon^2} i$ denote the open ball centered on $l$ of radius $\epsilon$ in $\struct {R, \norm {\, \cdot \,}_2}$.
By Open Ball of Metric Space is Open Set then $\map {B_\epsilon^2} l$ is open set in $\struct {R, d_2}$.
Since $d_1$ and $d_2$ are topologically equivalent metrics then $\map {B_\epsilon^2} l$ is open set in $\struct {R, d_1}$.
By the definition of an open set in a metric space then:
:$\exists \delta \in \R_{> 0}: \map {B_\delta^1} l \subseteq \map {B_\epsilon^2} l$
Hence:
:$\forall x \in R: \norm {x - l}_1 < \delta \implies \norm {x - l}_2 < \epsilon$
Since $\sequence {x_n}$ converges to $l$ in $\norm {\, \cdot \,}_1$ then:
:$\exists N \in \N: \forall n \ge N: \norm {x_n - l}_1 < \delta$
Hence:
:$\exists N \in \N: \forall n \ge N: \norm {x_n - l}_2 < \epsilon$
Since $\sequence {x_n}$ and $\epsilon > 0$ were arbitrary then it has been shown that for all sequences $\sequence {x_n}$ in $R$: 
:$\sequence {x_n}$ converges to $l$ in $\norm {\, \cdot \,}_1 \implies \sequence {x_n}$ converges to $l$ in $\norm {\, \cdot \,}_2$. 
By a similar argument it is shown that for all sequences $\sequence {x_n}$ in $R$:
:$\sequence {x_n}$ converges to $l$ in $\norm {\, \cdot \,}_2 \implies \sequence {x_n}$ converges to $l$ in $\norm {\, \cdot \,}_1$. 
The result follows. 
== Sources ==
* : $\S 3.1$ Absolute Values on $\Q$, Lemma $3.1.2$ and Problem $66$
Category:Equivalence of Definitions of Equivalent Division Ring Norms
Equivalence of Definitions of Euler's Number

== Theorem ==
=== Limit of Series ===
=== Limit of Sequence ===
=== Base of Logarithm ===
=== Exponential Function ===
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Equivalence of Definitions of Exponential Function
Category:Equivalence of Definitions of Euler's Number
Category:Euler's Number
Equivalence of Definitions of Euler's Number/Proof 1

== Theorem ==
== Proof ==
See Equivalence of Definitions of Real Exponential Function: Inverse of Natural Logarithm implies Limit of Sequence for how $\ds \lim_{n \mathop \to \infty} \paren {1 + \frac 1 n}^n = e$ follows from the definition of $e$ as the number satisfied by $\ln e = 1$.
See Euler's Number: Limit of Sequence implies Limit of Series for how $\ds e = \sum_{n \mathop = 0}^\infty \frac 1 {n!}$ follows from $\ds \lim_{n \mathop \to \infty} \paren {1 + \frac 1 n}^n = e$.
Now suppose $e$ is defined as $\ds e = \sum_{n \mathop = 0}^\infty \frac 1 {n!}$.
Let us consider the series $\ds \map f x = \sum_{n \mathop = 0}^\infty \frac {x^n} {n!}$.
From Series of Power over Factorial Converges, this is convergent for all $x$.
We differentiate $\map f x$ $x$ term by term (justified by Power Series is Differentiable on Interval of Convergence), and get:
Thus we have:
:$\map {D_x} {\map f x} = \map f x$
From Derivative of Exponential Function:
:$\map f x = e^x$
From Derivative of Inverse Function:
:$\map {D_x} {\map {f^{-1} } x} = \dfrac 1 {\map {f^{-1} } x}$
Hence from Derivative of Natural Logarithm Function:
:$\map {f^{-1} } x = \ln x$
It follows that $e$ can be defined as that number such that $\ln e = 1$.
Hence all the definitions of $e$ as given here are equivalent.
Category:Equivalence of Definitions of Euler's Number
Equivalence of Definitions of Euler's Number/Proof 2

== Theorem ==
== Proof ==
=== 1 implies 2 ===
This is proved in Euler's Number: Limit of Sequence implies Limit of Series.
=== 2 implies 3 ===
This is proved in Euler's Number: Limit of Sequence implies Base of Logarithm.
=== 3 implies 4 ===
Let $e$ be the unique solution to the equation $\ln x = 1$.
We want to show that $\exp 1 = e$, where $\exp$ is the exponential function.
where the final equation holds .
Hence the result.
=== 4 implies 1 ===
Let $e = \exp 1$, where $\exp$ denotes the exponential function.
We want to show that:
:$\ds \sum_{n \mathop = 0}^\infty \frac 1 {n!} = e$
By definition of $\exp$:
:$\ds \sum_{n \mathop = 0}^\infty \frac 1 {n!} = \exp 1$
We also have that $\exp 1 = e$ .
Hence the result.
Category:Equivalence of Definitions of Euler's Number
Equivalence of Definitions of Euler-Gompertz Constant

== Theorem ==
=== Integral Form ===
=== As a Continued Fraction ===
== Proof ==
=== Integral Form implies Continued Fraction Form ===
Let $G$ be the integral form of the Euler-Gompertz constant.
Then by definition:
:$G = \ds \int_0^\infty \dfrac {e^{-u} } {1 + u} \rd u$
Thus $G$ equals the continued fraction form of the Euler-Gompertz constant.
=== Continued Fraction Form implies Integral Form ===
Let $G$ be the continued fraction form of the Euler-Gompertz constant.
Then by definition:
:$G = \cfrac 1 {2 - \cfrac {1^2} {4 - \cfrac {2^2} {6 - \cfrac {3^2} {8 - \cfrac {4^2} {10 - \dotsb} } } } }$
Thus $G$ equals the integral form of the Euler-Gompertz constant.
== Sources ==
* : $0,59634 7355 \ldots$
Category:Euler-Gompertz Constant
Equivalence of Definitions of Euler Lucky Number

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
Let $p$ be an Euler lucky number.
Let $f_p: \Z \to \Z$ be the mapping defined as:
:$\forall n \in \Z: f_p \left({n}\right) = n^2 + n + p$
Let $m = n - 1$.
Then:
We have that $f_p \left({n}\right)$ is prime for $0 \le n < p - 1$.
Thus $f_p \left({m}\right)$ is prime for $0 \le \left({n - 1}\right) < p - 1$.
and so $f_p \left({m}\right)$ is prime for $1 \le n < p$.
Category:Euler Lucky Numbers
Equivalence of Definitions of Even Integer

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
=== Definition 3 ===
== Proof ==
=== $(1)$ $(2)$ ===
By definition of divisor, $n$ is divisible by $2$ :
:$n = 2 r$
where $r \in \Z$.
Thus definition 1 is logically equivalent to definition 2.
=== $(2)$ $(3)$ ===
By definition of congruence modulo $2$:
:$x \equiv y \pmod 2 \iff \exists r \in \Z: x - y = 2 r$
Setting $y = 0$:
:$x \equiv 0 \pmod 2 \iff \exists r \in \Z: x = 2 r$
Thus definition 2 is logically equivalent to definition 3.
Category:Even Integers
Equivalence of Definitions of Even Permutation

== Theorem ==
Let $n \in \N$ be a natural number.
Let $S_n$ denote the symmetric group on $n$ letters.
Let $\rho \in S_n$ be a permutation in $S_n$.
=== Definition $1$ ===
=== Definition $2$ ===
== Proof ==
The sign of $\rho$ is defined as:
:$\map \sgn \rho = \begin {cases} 1 & : \text {$k$ even} \\ -1 & : \text {$k$ odd} \\ \end {cases}$
The result follows.
Category:Even Permutations
Equivalence of Definitions of Everywhere Dense

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $H \subseteq S$ be a subset.
=== Definition 1 ===
=== Definition 2 ===
=== Definition 3 ===
== Proof ==
=== $(1)$ implies $(2)$ ===
Let $H$ be a subset of $S$ which is everywhere dense in $T$ by definition $1$.
Then by definition:
:$H^- = S$
where $H^-$ is the closure of $H$.
 there exists $U \in \tau$ such that $U \cap H = \O$.
Let $x \in S$ such that $x \in U$.
Thus $U$ is an open set of $T$ which does not contain an element of $H$ which is distinct from $x$.
Indeed, $U$ is an open set of $T$ which does not contain any elements of $H$ at all.
Hence, by definition, $x$ is not a limit point of $H$.
By definition, the closure of $H$ is the union of $H$ with its limit points.
But $x$ is not in $H$, nor is it a limit point of $H$.
That is:
:$x \notin H^-$
So $H^- \ne S$.
But this contradicts our definition of everywhere dense in $T$ by definition $1$.
Hence our assertion that there exists $U \in \tau$ such that $U \cap H = \O$ must be false.
So, by Proof by Contradiction:
:$\forall U \in \tau: H \cap U \ne \O$
Thus $H$ is a subset of $S$ which is everywhere dense in $T$ by definition $2$.
=== $(2)$ implies $(1)$ ===
Let $H$ be a subset of $S$ which is everywhere dense in $T$ by definition $2$.
Then by definition:
:$\forall U \in \tau \setminus \set \O: H \cap U \ne \O$
It is taken for granted that $H^- \subseteq S$.
 $H^- \ne S$.
Then:
:$\exists x \in S: x \notin H^-$
Thus:
:$x \notin H$
and also, $x$ is not a limit point of $H$.
Hence, by definition of limit point:
:$\exists U \in \tau: \not \exists y \in U: y \in H$
Hence by definition of set intersection:
:$U \cap H = \O$
But this contradicts our definition of everywhere dense in $T$ by definition $2$.
So, by Proof by Contradiction:
:$H^- = S$
Thus $H$ is a subset of $S$ which is everywhere dense in $T$ by definition $1$.
Category:Everywhere Dense
Equivalence of Definitions of Exterior

== Theorem ==
Let $T$ be a topological space.
Let $H \subseteq T$.
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
Let $H^e$ be defined as:
:$H^e$ is the complement of the closure of $H$ in $T$.
Then:
Thus:
:$H^e$ is the interior of the complement of $H$ in $T$.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $1$: General Introduction: Closures and Interiors
Category:Set Exteriors