nyuuzyou/wikis huggingface dataset:
Equivalence of Definitions of Closed Set in Normed Vector Space

== Theorem ==
Let $V = \struct {X, \norm {\, \cdot \,}}$ be a normed vector space.
Let $F \subseteq X$.
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
Let $\sequence {x_n}_{n \mathop \in \N}$ be a sequence in $F$ with a limit point $x$ in $X$.
=== Definition 1 implies Definition 2 ===
Let $F$ be closed in $V$ according to Definition 1:
:$F$ is closed in $V$ its complement $X \setminus F$ is open in $V$.
 $x \notin F$.
Since $F$ is closed, $X \setminus F$ is open.
Let $\map {B_\epsilon} x := \set {y \in X : \norm {y - x} < \epsilon}$ be an open ball.
Then there exists $\map {B_\epsilon} x$, which belongs to $X \setminus F$:
:$\forall x \in X \setminus F: \exists \epsilon \in \R_{>0}: \map {B_\epsilon} x \subset X \setminus F$
$\sequence {x_n}_{n \mathop \in \N}$ converges to $x$.
Therefore:
:$\forall \epsilon \in \R_{>0} : \exists N \in \N : \forall n \in N : n > N \implies \norm {x_n - x} < \epsilon$
Hence:
:$\norm {x_{N + 1} - x} < \epsilon$
Then we have:
:$x_{N + 1} \in \map {B_\epsilon} x \subset X \setminus F$
But at the same time:
:$x_{N + 1} \in F$
and we have reached a contradiction.
Hence there is no such $x$ which is not in $F$.
It follows that every limit point of $F$ is also a point of $F$.
That is, $F$ is closed in $V$ according to Definition 2.
=== Definition 2 implies Definition 1 ===
Let $F$ be closed in $V$ according to Definition 2:
:$F$ is closed (in $V$) every limit point of $F$ is also a point of $F$.
We are given that $x$ is a limit point of $F$.
Let $x \in F$.
 $X \setminus F$ is not open.
Then:
:$\neg \paren {\forall x \in X \setminus F : \exists \epsilon \in \R_{> 0} : \map {B_\epsilon} x \subseteq X \setminus F}$
By Assertion of Existence, this is equivalent to:
:$(1): \quad \exists x \in X \setminus F : \forall \epsilon \in \R_{> 0} : \map {B_\epsilon} x \cap F \ne \O$
Set $\epsilon = \dfrac 1 n$ with $n \in \N$.
Then:
:$\exists n \in \N : x_n \in \map {B_\epsilon} x \cap F$
So there is a sequence $\sequence {x_n}_{n \mathop \in \N}$ in $F$ such that:
:$\forall n \in \N : \norm {x - x_n} < \dfrac 1 n$
and $x_n$ converges to $x$ in $F$.
By $(1)$:
:$x \in X \setminus F$
That is:
:$x \notin F$
This contradicts the condition in the definition that $x \in F$.
Hence, $X \setminus F$ has to be open.
That is, $F$ is closed in $V$ according to Definition 1.
== Sources ==
* : Chapter $\S 1.4$: Normed and Banach spaces. Sequences in a normed space; Banach spaces
Category:Closed Sets
Equivalence of Definitions of Closure Operator

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
Let $\struct {S, \preceq}$ be an ordered set.
Let $\cl: S \to S$ be a mapping.
=== Definition 1 implies Definition 2 ===
Let $\cl$ be an inflationary, increasing and idempotent mapping.
It is necessary to show that for all $x, y \in S$:
:$x \preceq \map \cl y \iff \map \cl x \preceq \map \cl y$
==== Necessary Condition ====
==== Sufficient Condition ====
Suppose that $\map \cl x \preceq \map \cl y$.
=== Definition 2 implies Definition 1 ===
Suppose that:
:$x \preceq \map \cl y \iff \map \cl x \preceq \map \cl y$
It is necessary to show that $\cl$ is inflationary, increasing and idempotent.
==== Inflationary ====
Let $x \in S$.
That is, $\cl$ is inflationary.
==== Increasing ====
It has been demonstrated that $\preceq$ is inflationary.
Let $x, y \in S$ such that $x \preceq y$.
Then:
That is, $\cl$ is increasing.
==== Idempotent ====
It has been demonstrated that $\preceq$ is inflationary.
Let $x \in S$.
Then:
Then as $\cl$ is inflationary:
:$\map \cl x \preceq \map \cl {\map \cl x}$
As $\preceq$ is antisymmetric by dint of being an ordering:
:$\map \cl {\map \cl x} = \map \cl x$
That is, $\cl$ is idempotent.
Thus $\cl$ has been shown to be inflationary, increasing and idempotent as required.
Category:Closure Operators
Equivalence of Definitions of Closure of Topological Subspace

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $H \subseteq S$.
=== Definition 1 ===
=== Definition 2 ===
=== Definition 3 ===
=== Definition 4 ===
=== Definition 5 ===
=== Definition 6 ===
== Proof ==
=== Definition $1$ is equivalent to Definition $2$ ===
This is proved in Set Closure as Intersection of Closed Sets.
=== Definition $2$ is equivalent to Definition $3$ ===
This is proved in Set Closure is Smallest Closed Set in Topological Space.
=== Definition $1$ is equivalent to Definition $4$ ===
By the definition of the interior, and Set is Subset of its Topological Closure, it easily follows that
:$H^\circ \subseteq H \subseteq H^-$
Then:
=== Definition $1$ is equivalent to Definition $5$ ===
Every isolated point of $H$ is a point of $H$.
So by Set Union Preserves Subsets:
:$(1): \quad H^i \cup H' \subseteq H \cup H'$
If $S \setminus \paren {H^i \cup H'} = \O$, then $(1)$ yields $H^i \cup H' = H \cup H'$ and so the proof is complete.
Otherwise, let $x \in S \setminus \paren {H^i \cup H'} \ne \O$.
From De Morgan's Laws:
:$S \setminus \paren {H^i \cup H'} = \paren {S \setminus H^i} \cap \paren {S \setminus H'}$
Thus $x$ is a point of $S$ that is neither an isolated point of $H$ nor a limit point of $H$.
Then there exists an open set $U$ that contains $x$ such that:
:$U \cap H \ne \set x$
and:
:$H \cap \paren {U \setminus \set x} = \O$
This implies that
:$U \cap H = \O$
By Condition for Point being in Closure:
:$x \notin H \cup H'$
It has been shown that:
:$x \notin H^i \cup H' \implies x \notin H \cup H'$
From Rule of Transposition:
:$x \in H \cup H' \implies x \in H^i \cup H'$
And so
:$(2): \quad H \cup H' \subseteq H^i \cup H'$
Combining $(1)$ and $(2)$:
:$H \cup H' = H^i \cup H'$
Hence the result.
=== Definition $1$ is equivalent to Definition $6$ ===
By one of the equivalent definitions of an adherent point:
:A point $x \in S$ is an adherent point of $H$ every open neighborhood $U$ of $x$ satisfies $H \cap U \ne \O$
From Condition for Point being in Closure:
:$x \in H^-$ every open set of $T$ which contains $x$ contains a point in $H$
where:
:$H^-$ is the union of $H$ and all the limit points of $H$ in $T$
The equivalence follows.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $1$: General Introduction: Closures and Interiors
Category:Set Closures
Equivalence of Definitions of Commutative Local Ring

== Theorem ==
Let $A$ be a commutative ring with unity.
=== Definition 1 ===
=== Definition 2 ===
=== Definition 3 ===
== Proof ==
=== Definition 1 implies Definition 2 ===
Let $\mathfrak m \subsetneq A$ be the unique maximal ideal.
First, $A$ is nontrivial, since $1 \notin \mathfrak m$.
Secondly, let $x, y \in A$ be non-units.
Let $\ideal x$ and $\ideal y$ be the principal ideals generated by $x$ and $y$, respectively.
In view of the unique maximality, $\ideal x \subseteq \mathfrak m$ and $\ideal y \subseteq \mathfrak m$.
In particular:
:$x, y \in \mathfrak m$
which implies:
:$x + y \in \mathfrak m$
Therefore $x + y$ is a non-unit.
=== Definition 2 implies Definition 3 ===
First, $0 \in M$, since $A$ is nontrivial.
Thus $M$ is non-empty.
Secondly, by the assumption:
:$x, y \in M \implies x + y \in M$
Thirdly, let $a \in A$ and $x \in M$.
We shall show that $a x \in M$.
 there exists a $u \in A$ such that:
:$u \paren {a x} = 1$
Thus:
:$\paren {u a} x = 1$
which means that $x$ is a unit.
This contradicts the fact that $x \in M$.
Therefore $a x \in M$.
Altogether $M$ is an ideal.
Finally, $1 \notin M$, since:
:$1 \cdot 1 = 1$
by .
Therefore $M$ is a proper ideal.
=== Definition 3 implies Definition 1 ===
Let $I$ be an arbitrary ideal such that $M \subsetneq I$.
Let $u \in I \setminus M$.
Then $u$ is a unit by the definition of $M$.
Therefore $M = A$.
Furthermore, $M$ is assumed to be a proper ideal.
Therefore $M$ is a maximal ideal.
On the other hand, let $N$ be an arbitrary maximal ideal.
As $N$ does not contain a unit:
:$N \subseteq M$
which implies:
:$N = M$
Therefore $M$ is the unique maximal ideal.
Category:Definitions/Local Rings
Equivalence of Definitions of Commutator of Group Elements

== Theorem ==
=== Definition $1$ ===
=== Definition $2$ ===
== Proof ==
Throughout the following, is assumed implicitly.
Let $g, h \in G$ be arbitrary elements of $G$.
Let $c$ be the commutator of $g$ and $h$ by definition $2$.
Then we have:
Category:Group Commutators
Equivalence of Definitions of Compact Topological Space

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
=== Definition 3 ===
=== Definition 4 ===
=== Definition 5 ===
== Proof ==
=== $(1) \iff (2)$: Compact Space satisfies Finite Intersection Axiom ===
=== $(1) \iff (3)$: Alexander's Compactness Theorem ===
=== $(4) \implies (5)$: Every Filter has Limit Point implies Every Ultrafilter Converges ===
=== $(5) \implies (4)$: Every Ultrafilter Converges implies Every Filter has Limit Point ===
=== $(2) \implies (4)$ ===
Let $T = \struct {S, \tau}$ satisfy the Finite Intersection Axiom.
Let $\FF$ be a filter on $X$.
 that $\FF$ has no limit point.
Thus:
: $\ds \bigcap_{F \mathop \in \FF} \overline F = \O$
By hypothesis there are therefore sets $F_1, \ldots, F_n \in \FF$ such that:
:$\overline F_1 \cap \ldots \cap \overline F_n = \O$
Because for any set $M$ we have $M \subseteq \overline M$:
:$\overline F_1, \ldots, \overline F_n \in \FF$
But by definition of a filter, $\FF$ must not contain the empty set.
Thus $\FF$ has a limit point.
=== $(4) \implies (2)$ ===
Let $\AA \subset \powerset S$ be a set of closed subsets of $S$.
Let:
: $\bigcap \tilde \AA \ne \O$
for all finite subsets $\tilde \AA$ of $\AA$.
We show that this implies $\bigcap \AA \ne \O$.
From our assumption, $\BB := \set {\bigcap \tilde \AA : \tilde \AA \subseteq \AA \text{ finite} }$ is a filter basis.
Let $\FF$ be the corresponding generated filter.
By hypothesis $\FF$ has a limit point.
Thus:
: $\ds \O \ne \bigcap_{F \mathop \in \FF} \overline F \subseteq \bigcap \BB \subseteq \bigcap \AA$.
Thus $\bigcap \AA \ne \O$.
Hence $T = \struct {S, \tau}$ satisfies the Finite Intersection Axiom.
Category:Compact Spaces
Equivalence of Definitions of Compact Topological Subspace

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $T_H = \struct {H, \tau_H}$ be a topological subspace of $T$, where $H \subseteq S$.
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
=== $1$ implies $2$ ===
Suppose $T_H$ is compact in the sense of Definition 1.
Let $\CC$ be a cover of $H$ by open sets of $T$.
Then for each $U \in \CC$, $U \cap H$ is open in $T_H$ by definition of the subspace topology.
Since $\CC$ is a cover of $H$ it follows that $\CC' = \set{U \cap H: U \in \CC}$ is also a cover of $H$.
By hypothesis, there exists a finite subcover $\FF'$ of $\CC'$ for $H$.
Thus let $\FF' = \set {U_1 \cap H, U_2 \cap H, \ldots, U_n \cap H}$ where $U_1, U_2, \ldots, U_n \in \CC$.
It follows that $\FF = \set {U_1, U_2, \ldots, U_n}$ is a finite subcover of $H$.
That is, $T_H$ is compact in the sense of Definition 2.
=== $2$ implies $1$ ===
Suppose $T_H$ is compact in the sense of Definition 2.
Let $\CC$ be a cover of $H$ by open sets of $H$.
By definition of the subspace topology, each $V \in \CC$ is of the form $U \cap H$ for some $U \in \tau$.
It follows that the collection of all such $U$ is a cover of $H$ by open sets of $T$.
By hypothesis, there exists a finite subcover $\FF = \set {U_1, U_2, \ldots, U_n}$ of $\CC$ for $H$.
Thus $\FF' = \set {U_1 \cap H, U_2 \cap H, \ldots, U_n \cap H}$ is also a finite subcover of $H$
That is, $T_H$ is compact in the sense of Definition 1.
== Also see ==
* Definition:Compact Subspace
* Definition:Compact Space
== Sources ==
* : $2.33$
* : $5$: Compact spaces: $5.2$: Definition of compactness
Category:Compact Spaces
Equivalence of Definitions of Compatible Atlases

== Theorem ==
Let $M$ be a topological space.
Let $\mathscr F, \mathscr G$ be $d$-dimensional atlases of class $C^k$ on $M$.
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
=== Definition $1$ implies Definition $2$ ===
Follows immediately from the definition of $C^k$-atlas.
=== Definition $2$ implies Definition $1$ ===
Let $\struct {U, \phi}$ and $\struct {V, \psi}$ be charts in $\mathscr F \cup \mathscr G$.
If they are both in $\mathscr F$, they are $C^k$-compatible because $\mathscr F$ is a $C^k$-atlas.
If they are both in $\mathscr G$, they are $C^k$-compatible because $\mathscr G$ is a $C^k$-atlas.
If $\struct {U, \phi} \in \mathscr F$ and $\struct {V, \psi} \in \mathscr G$, they are $C^k$-compatible .
Thus $\mathscr F \cup \mathscr G$ is a $C^k$-atlas.
Category:Compatible Atlases
Equivalence of Definitions of Complement of Subgroup

== Theorem ==
Let $G$ be a group with identity $e$.
Let $H$ and $K$ be subgroups.
=== Definition $1$ ===
=== Definition $2$ ===
== Proof ==
=== Definition $1$ implies Definition $2$ ===
Let $G = H K$.
Then $H K$ is a group.
By Subset Product of Subgroups:
:$H K = K H$
Thus $K H = G$.
=== Definition $2$ implies Definition $1$ ===
Let $G = K H$.
Then $K H$ is a group.
By Subset Product of Subgroups:
: $H K = K H$
Thus $H K = G$.
Category:Subgroup Complements
Equivalence of Definitions of Complete Elliptic Integral of the First Kind

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
Let $\map K k$ be the complete elliptic integral of the first kind by definition $1$.
Let $v := \sin \phi$.
Then we have:
Hence:
 
Thus $\map K k$ is the complete elliptic integral of the first kind by definition $2$.
Category:Complete Elliptic Integral of the First Kind
Equivalence of Definitions of Complete Elliptic Integral of the Second Kind

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
Let $\map E k$ be the complete elliptic integral of the second kind by definition $1$.
Let $v := \sin \phi$.
Then we have:
Hence:
 
Thus $\map E k$ is the complete elliptic integral of the second kind by definition $2$.
Category:Complete Elliptic Integral of the Second Kind
Equivalence of Definitions of Complete Elliptic Integral of the Third Kind

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
Let $\map \Pi {k, n}$ be the complete elliptic integral of the third kind by definition $1$.
Let $v := \sin \phi$.
Then we have:
Hence:
 
Thus $\map \Pi {k, n}$ is the complete elliptic integral of the third kind by definition $2$.
Category:Complete Elliptic Integral of the Third Kind
Equivalence of Definitions of Complete Metric Space

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
Definition 1 implies definition 2 by the Nested Sphere Theorem.
Let $M = \struct {A, d}$ be a complete metric space by definition 2.
Let $\sequence {x_n}$ be a Cauchy sequence in $M$.
By the definition of a Cauchy sequence, for each $k \in \N_{>0}$, there is an $N_k \in \N_{>0}$ such that for all $n, m \in \N$:
:$n, m \ge N_k \implies \map d {x_n, x_m} < \dfrac 1 {2^k}$
For each $k \in \N_{>0}$, let $C_k = \map { {B_{1/2^k} }^-} {x_{N_k} }$.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $5$: Metric Spaces: Complete Metric Spaces
Category:Complete Metric Spaces
Equivalence of Definitions of Completely Hausdorff Space

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
=== Definition 1 implies Definition 2 ===
Let $\struct {S, \tau}$ satisfy:
:$\forall x, y \in S, x \ne y: \exists U, V \in \tau: x \in U, y \in V: U^- \cap V^- = \O$ 
Let $x, y \subseteq S , x \ne y $ be arbitrary.
Then:
:$\exists U, V \in \tau: x \in U, y \in V: U^- \cap V^- = \O$ 
Let $N_x = U$ and $N_y = V$.
From Set is Subset of Itself then:
:$\exists N_x, N_y \subseteq S: \exists U, V \in \tau: x \in U \subseteq N_x, y \in V \subseteq N_y: N_x^- \cap N_y^- = \O$
Because $x, y \in S$ were arbitrary:
:$\forall x, y \in S, x \ne y : \exists N_x, N_y \subseteq S: \exists U, V \in \tau: x \subseteq U \subseteq N_x, y \subseteq V \subseteq N_y: N_x^- \cap N_y^- = \O$
=== Definition 2 implies Definition 1 ===
Let $\struct {S, \tau}$ satisfy:
:$\forall x, y \in S, x \ne y : \exists N_x, N_y \subseteq S: \exists U, V \in \tau: x \subseteq U \subseteq N_x, y \subseteq V \subseteq N_y: N_x^- \cap N_y^- = \O$
Let $x, y \subseteq S , x \ne y $ be arbitrary.
Then:
:$\exists N_x, N_y \subseteq S: \exists U, V \in \tau: x \in U \subseteq N_x, y \in V \subseteq N_y: N_x^- \cap N_y^- = \O$
From Topological Closure of Subset is Subset of Topological Closure:
:$U^- \subseteq N_x^-, V^- \subseteq N_y^-$
From Subsets of Disjoint Sets are Disjoint then:
:$U^- \cap V^- = \O$
Thus:
:$\exists U, V \in \tau: x \in U, y \in V: U^- \cap V^- = \O$
Because $x, y \in S$ were arbitrary:
:$\forall x, y \in S, x \ne y: \exists U, V \in \tau: x \in U, y \in V: U^- \cap V^- = \O$ 
Category:Separation Axioms
Equivalence of Definitions of Complex Cross Product

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
=== Definition 3 ===
=== Definition 4 ===
== Proof ==
=== Definition 1 equivalent to Definition 3 ===
=== Definition 2 equivalent to Definition 3 ===
=== Definition 1 equivalent to Definition 4 ===
== Sources ==
* : $1$: Complex Numbers: Dot and Cross Product
Category:Complex Cross Product
Equivalence of Definitions of Complex Dot Product

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
=== Definition 3 ===
=== Definition 4 ===
== Proof ==
=== Definition 1 equivalent to Definition 3 ===
=== Definition 2 equivalent to Definition 3 ===
=== Definition 1 equivalent to Definition 4 ===
== Sources ==
* : $1$: Complex Numbers: Dot and Cross Product
Category:Complex Dot Product
Equivalence of Definitions of Complex Exponential Function

== Theorem ==
=== As a Power Series Expansion ===
=== By Real Functions ===
=== As a Limit of a Sequence ===
=== As the Solution of a Differential Equation ===
== Proof ==
From Radius of Convergence of Power Series over Factorial: Complex Case, it follows that the power series $\ds \sum_{n \mathop = 0}^\infty \dfrac {z^n} {n!}$ is absolutely convergent over the entirety of $\C$.
Hence, the definition of $\exp z$ as a power series is valid.
It remains to demonstrate the logical equivalence of all the definitions.
=== Power Series Expansion equivalent to Solution of Differential Equation ===
=== Power Series Expansion equivalent to Definition by Real Functions ===
=== Power Series Expansion equivalent to Limit of Sequence ===
Category:Exponential Function
Equivalence of Definitions of Complex Exponential Function/Power Series Expansion equivalent to Definition by Real Functions

== Theorem ==
=== As a Power Series Expansion ===
=== By Real Functions ===
== Proof ==
We have the result:
:Power Series Expansion equivalent to Solution of Differential Equation
which gives that the definition of $\exp z$ as the power series:
:$\exp z := \ds \sum_{n \mathop = 0}^\infty \frac {z^n} {n!}$
is equivalent to the definition of $\exp z$ as the solution of the differential equation:
:$\dfrac {\d y} {\d z} = y$
satisfying the initial condition $y \paren 0 = 1$.
Let:
:$e: \R \to \R$ denote the real exponential function
:$\sin: \R \to \R$ denote the real sine function
:$\cos: \R \to \R$ denote the real cosine function.
Then:
== Sources ==
* : $\S 4.5$. The Functions $e^z$, $\cos z$, $\sin z$
Category:Equivalence of Definitions of Exponential Function
Equivalence of Definitions of Complex Exponential Function/Power Series Expansion equivalent to Differential Equation

== Theorem ==
=== As a Power Series Expansion ===
=== As the Solution of a Differential Equation ===
== Proof ==
==== Power Series Expansion implies Solution of Differential Equation ====
Let $\exp z$ be the complex function defined as the power series:
:$\exp z := \ds \sum_{n \mathop = 0}^\infty \frac {z^n} {n!}$
Let $y = \ds \sum_{n \mathop = 0}^\infty \frac {z^n} {n!}$.
Then:
We show that $\ds \sum_{n \mathop = 0}^\infty \dfrac {z^n} {n!}$ satisfies the initial condition:
:$\exp \paren 0 = 1$.
Setting $z = 0$ we find:
That is:
$\exp z$ is the particular solution of the differential equation:
:$\dfrac {\d y} {\d z} = y$
satisfying the initial condition $\map y 0 = 1$.
==== Solution of Differential Equation implies Power Series Expansion ====
Let $\exp z$ be the complex function defined as the particular solution of the differential equation:
:$\dfrac {\d y} {\d z} = y$
satisfying the initial condition $\map y 0 = 1$.
Let $f: \C \to \C$ be a solution to the differential equation $\dfrac {\d f} {\d z} = f$ with $f \paren 0 = 1$.
Then Holomorphic Function is Analytic shows that $f$ can be expressed as a power series:
:$\ds \sum_{n \mathop = 0}^\infty a_n \paren {z - \xi}^n$
about any $\xi \in \C$.
When $\xi = 0$, we have for all $n \in \N_{\ge 1}$:
As $a_0 = \dfrac {\map {f^{\paren 0} } 0} {0!} = 1$ by the initial condition, it follows inductively that:
:$a_n = \dfrac 1 {n!}$
Hence:
:$\ds \map f z = \sum_{n \mathop = 0}^\infty \dfrac 1 {n!} z^n$
Category:Equivalence of Definitions of Exponential Function
Equivalence of Definitions of Complex Exponential Function/Power Series Expansion equivalent to Limit of Sequence/Proof 1

== Theorem ==
=== As a Power Series Expansion ===
=== As a Limit of a Sequence ===
== Proof ==
Fix $z \in \C$.
For every $r, n \in \N$, define:
:$\map {v_r} n = \dbinom n r \paren {\dfrac z n}^r$
Also, for every $r \in \N$, define:
:$M_r = \dfrac {\size z^r} {r!}$
Then:
Furthermore, by Radius of Convergence of Power Series over Factorial:
:$\ds \sum_{r \mathop = 0}^\infty M_r$ converges
Now, define:
:$w_r = \dfrac {z^r} {r!}$
By Limit to Infinity of Binomial Coefficient over Power:
:$\ds \lim_{n \mathop \to \infty} \map {v_r} n = w_r$
Hence:
Category:Equivalence of Definitions of Exponential Function
Equivalence of Definitions of Complex Exponential Function/Power Series Expansion equivalent to Limit of Sequence/Proof 2

== Theorem ==
=== As a Power Series Expansion ===
=== As a Limit of a Sequence ===
== Proof ==
Let:
:$\ds s_n = \sum_{k \mathop = 0}^n \dfrac {z^k} {k!}$
:$a_n = \paren {1 + \dfrac z n}^n$
Then we can express $a_n$ as follows:
The limit of the difference between the $k$th terms of $a_n$ and $s_n$ is:
To show that $s_n$ and $a_n$ have the same limit, let $\epsilon \in \R_{>0}$.
From Tail of Convergent Series tends to Zero, it follows that we can find $M \in \N$ such that for all $m \ge M$:
:$\ds \sum_{k \mathop = m}^n \cmod {\dfrac {z^k} {k!} } < \dfrac \epsilon 2$
For all $k \in \left\{ {0, 1, \ldots, M - 1}\right\}$, we can find $N_k \in \N$ such that for all $n \ge N_k$:
:$\ds \cmod {\dfrac {z^k} {k!} - \dfrac {z^k} {k!} \prod_{j \mathop = 1}^{k - 1} \paren {1 - \dfrac j n} } < \dfrac \epsilon {2 M}$
Then for all $n \ge \max \paren {M, N_0, N_1, \ldots, N_{M - 1} }$, we have:
As an Absolutely Convergent Series is Convergent, $\sequence {s_n}$ converges.
Then:
The result follows.
Category:Equivalence of Definitions of Exponential Function