avewright/proofwiki-math huggingface dataset:
Theorem: Metric Defines Norm iff it Preserves Linear Structure

Let $\struct {k, \norm {\,\cdot\,}_k}$ be a valued field.
Let $V$ be a vector space over the valued field $\struct {k, \norm {\,\cdot\,}_k}$.
Let $d: V \times V \to k$ be a metric on $V$.
Then the function $\norm v := \map d {v, 0}$ is a norm on $V$ {{iff}} for all $x, y, z \in V$, $\lambda \in k$:
:$(1): \quad \map d {x + z, y + z} = \map d {x, y}$ (homogeneity or translation invariance)
:$(2): \quad \map d {\lambda x, \lambda y} = \norm \lambda_k \map d {x, y}$ (the enlargement property)

Proof:
Suppose first that $d$ satisfies the hypotheses $(1)$ and $(2)$.
From {{Metric-space-axiom|4}}:
:$\forall u, v \in V: \map d {u, v} \ge 0$
Hence:
:$\forall u \in V: \norm u = \map d {u, 0} \ge 0$
Moreover, from {{Metric-space-axiom|1}}:
:$\norm u = 0 \implies \map d {u, 0} = 0$
and hence:
:$u = 0$
Now let $\lambda \in K$, $u \in V$.
Then, using the enlargement property of $d$:
{{begin-eqn}}
{{eqn | l = \norm {\lambda v}
      | r = \map d {\lambda v, 0}
      | c = Definition of $\norm {\,\cdot\,}$
}}
{{eqn | r = \map d {\lambda v, \lambda \cdot 0}
      | c = 
}}
{{eqn | r = \norm \lambda_k \map d {v, 0}
      | c = Enlargement Property
}}
{{eqn | r = \norm \lambda_k \norm v
      | c = Definition of $\norm{\,\cdot\,}$
}}
{{end-eqn}}
Finally if $u,v \in V$, then we have
{{begin-eqn}}
{{eqn | l = \norm {u + v}
      | r = \map d {u + v, 0}
      | c = Definition of $\norm {\,\cdot\,}$
}}
{{eqn | r = \map d {u, -v}
      | c = Translation Invariance of $d$
}}
{{eqn | r = \map d {u, 0} + d(0,-v)
      | o = \le
      | c = {{Metric-space-axiom|2}}
}}
{{eqn | r = \map d {u, 0} + \map d {v, 0}
      | c = Enlargement Property
}}
{{eqn | r = \norm u + \norm v
      | c = Definition of $\norm {\,\cdot\,}$
}}
{{end-eqn}}
{{stub|converse to add}}
{{qed}}
Category:Metric Spaces
7w5xuv3wbl5c608navp8f84nbkqteoz
Theorem: Real Numbers form Perfect Set

Consider the set of real numbers $\R$ as a (complete) metric space with the usual (Euclidean) metric.
Then $\R$ forms a perfect set.

Proof:
{{Recall|Perfect Set|perfect set}}
{{:Definition:Perfect Set/Definition 1}}
Let $x \in \R$.
Consider the sequence:
:$\sequence {y_k} = x + \dfrac 1 k$
Then as $\sequence {z_k} = \dfrac 1 k$ is a basic null sequence it follows that:
:$\ds \lim_{n \mathop \to \infty} \sequence {y_k} = x$
Thus we see that $x$ is a limit point of $S$.
{{qed}}
Category:Real Numbers
Category:Examples of Perfect Sets
e4y5sx1f1ko0vgj9o70qnuvti5q46lf
Theorem: Closed Interval in Reals is Uncountable

Let $a, b$ be extended real numbers such that $a < b$.
Then the closed interval $\set {x \in \R : a \le x \le b} \subseteq \R$ is uncountable.

Proof:
First suppose that $a, b \in \R$.
We have that the unit interval is uncountable.
{{MissingLinks|Unit Interval is Uncountable (page does not exist at time of editing)}}
Let $f: \closedint 0 1 \to \closedint a b$ such that $\map f x = a + \paren {b - a} x$.
Then if $\map f {x_1} = \map f {x_2}$, we have:
:$a + \paren {b - a} x_1 = a + \paren {b - a} x_2$
Because $b > a$:
:$b - a > 0$
so this implies:
:$x_1 = x_2$
Therefore $f$ is injective.
Now if $\closedint a b$ were countable, there would be an injective function $g : \closedint a b \to \N$.
By Composite of Injections is Injection, this implies that $g \circ f$ is an injection, so $\closedint 0 1$ is countable, a contradiction.
If one or both of $a, b$ is (are) not real, then we can pick a closed interval $\closedint c d \subseteq \closedint a b$ with $c, d \in \R, c < d$.
We know by the above that $\closedint c d$ is uncountable.
Let $S = \set {x \in \R : a \le x \le b}$. 
The identity function $I_\R : \closedint c d \to S$ is trivially injective.
If $S$ were uncountable, we would have an injective function $g : S \to \N$.
Then by Composite of Injections is Injection we would have an injection $\closedint c d \to \N$ given by $g \circ I_\R$.
This was shown above to be false, so $S$ is uncountable.
{{qed}}
Theorem: Metric Space is T5

A metric space $M = \struct {A, d}$ is a $T_5$ space.

Proof:
{{Recall|T5 Space|$T_5$ space}}
{{:Definition:T5 Space/Definition 1}}
Let $S, T \subseteq A$ such that $S$ and $T$ are separated in $A$.
Then:
:each point $x \in S$ has an open $\epsilon$-ball $\map {B_{\epsilon_x} } x$ which is disjoint from $T$
:each point $y \in T$ has an open $\epsilon$-ball $\map {B_{\epsilon_y} } y$ which is disjoint from $S$.
Then:
:$U_S = \ds \bigcup_{x \mathop \in S} \map {B_{\epsilon_x / 2} } x$
:$U_T = \ds \bigcup_{y \mathop \in T} \map {B_{\epsilon_y / 2} } y$
are disjoint open neighborhoods of $S$ and $T$ respectively.
{{finish|that's cutting the corner}}
Hence the result by the definition of $T_5$ space.
{{qed}}
Theorem: Metric Space is Perfectly T4

A metric space $M = \struct {A, d}$ is a perfectly $T_4$ space.

Proof:
We have that a metric space is $T_4$.
We also have that every closed set in a metric space is a $G_\delta$ set.
Hence the result, by definition of a perfectly $T_4$ space.
{{qed}}
Theorem: Metric Space is Fully T4

A metric space $M = \struct {A, d}$ is a fully $T_4$ space.

Proof:
{{Recall|Fully T4 Space|fully $T_4$ space}}
{{:Definition:Fully T4 Space}}
{{proof wanted}}
Theorem: Metric Space is Fully Normal

A metric space $M = \struct {A, d}$ is a fully normal space.

Proof:
{{Recall|Fully Normal Space|fully normal space}}
{{:Definition:Fully Normal Space/Definition 2}}
From Metric Space is Fully $T_4$
:$M$ is a fully $T_4$ space.
From Metric Space is $T_1$
:$M$ is a  $T_1$ space.
The result follows.
{{qed}}
Theorem: Metric Space is Paracompact

Let $M = \struct {A, d}$ be a metric space.
Then $M$ is a paracompact space.

Proof:
We have that a Metric Space is Fully Normal.
Then we have that a Fully Normal Space is Paracompact.
{{qed}}
Theorem: Metric Space is Paracompact

Let $M = \struct {A, d}$ be a metric space.
Then $M$ is a paracompact space.

Proof:
Let $M = \struct {A, d}$ be a metric space.
Let $\sequence {C_\alpha}$ be an open cover of $M$ indexed by ordinals.
Let $\map {B_r} x$ be the open $r$-ball in $M$ around $x$.
For $n \in \N$, let $D_{\alpha n}$ denote the union of all the open $r$-balls $\map {B_{2^{-n} } } x$ such that:
:$(1): \quad \alpha$ is the smallest ordinal such that $x \in C_\alpha$
:$(2): \quad x \notin D_{\beta j}$ for $j < n$
:$(3): \quad \map {B_{3 \times 2^{-n} } } x \subseteq C_\alpha$
It is to be shown that $\sequence {D_{\alpha n} }$ is a locally finite open refinement of $\sequence {C_\alpha}$ which covers $M$.
By construction, $D_{\alpha n}$ is a refinement of $\sequence {C_\alpha}$.
Let $x \in A$.
Then there exists a smallest ordinal $\alpha$ such that $x \in C_\alpha$.
Then there exists $n$ sufficiently large such that $(3)$ holds.
Then by $(2)$:
:$x \in D_{\beta j}$
for some $j \le n$.
Thus $\sequence {D_{\alpha n} }$ covers $M$.
Now assume $x \in A$.
Let $\alpha$ be the smallest ordinal such that $x \in D_{\alpha n}$ for some $n$.
Choose $j \in \N$ such that $\map {B_{2^{-j} } } x \subseteq D_{\alpha n}$.
Let $i \ge n + j$.
As $i > n$, every one of the open balls of radius $2^{-i}$ used to define $D_{\beta i}$ has its center $y$ outside $D_{\alpha n}$
And because $\map {B_{2^{-j} } } x \subseteq D_{\alpha n}$, we have that $\map d {x, y} \ge 2^{-j}$.
But $i \ge j + 1$ and $n + j \ge j + 1$.
So:
:$\map {B_{2^{-n - j} } } x \cap \map {B_{2^{-i} } } y = \O$
Thus if $i \ge n$, then $\map {B_{2^{-n - j} } } x$ intersects no $D_{\beta i}$.
Let $i < n+j$.
Let us choose:
:$p \in D_{\beta i}$
:$q \in D_{\gamma i}$
:$\beta < \gamma$
There exist points $y$ and $z$ such that:
:$p \in \map {B_{2^{-i} } } y \subseteq D_{\beta i}$
:$q \in \map {B_{2^{-i} } } z \subseteq D_{\gamma i}$
By $(3)$:
:$\map {B_{3 \times 2^{-i} } } y \subseteq C_\beta$
But by $(1)$:
:$z \notin C_\beta$
So:
:$\map d {y, z} \ge 3 \times 2^{-i}$
and:
:$\map d {p, q} > 2^{-i} \ge 2^{-n - j + 1}$
Thus if $i < n + j$, then $\map {B_{2^{-n - j} } } x$ intersects $D_{\beta i}$ for at most one $\beta$.
Thus $D_{\alpha n}$ is locally finite.
Then $D_{\alpha n}$ is a locally finite open refinement of $\sequence {C_\alpha}$ which covers $M$.
Hence by definition $M$ is paracompact.
{{qed}}
Theorem: Metric Space is First-Countable

Let $M = \struct {A, d}$ be a metric space.
Then $M$ is first-countable.

Proof:
{{Recall|First-Countable Space|first-countable space}}
{{:Definition:First-Countable Space}}
Let $x \in A$ be arbitrary.
Let:
:$\BB = \set {\map {B_{1/n} } x: n \in \N_{>0} }$
where $\map {B_\epsilon} x$ denotes the open $\epsilon$-ball of $x$ in $M$.
By Surjection from Natural Numbers iff Countable, we have that $\BB$ is countable.
By the definition of a first-countable space, it suffices to show that $\BB$ is a local basis at $x$.
By Open Ball of Metric Space is Open Set, every element of $\BB$ is an open neighborhood of $x$.
Let $U$ be an open neighborhood of $x$.
By the definition of an open set, there exists a strictly positive real number $\epsilon$ such that $\map {B_\epsilon} x \subseteq U$.
By the Axiom of Archimedes, there exists a natural number $n > \dfrac 1 \epsilon$.
That is, $\dfrac 1 n < \epsilon$, and so:
:$\map {B_{1/n} } x \subseteq \map {B_\epsilon} x \subseteq U$
The result follows from Subset Relation is Transitive.
{{qed}}
Theorem: Metric Space is Separable iff Second-Countable

A metric space is separable {{iff}} it is second-countable.

Proof:
Follows directly from:
:Separable Metric Space is Second-Countable
:Second-Countable Space is Separable
{{qed}}
{{ACC|Second-Countable Space is Separable}}
Theorem: Metric Space is Lindelöf iff Second-Countable

Let $M = \struct {A, d}$ be a metric space.
Then $M$ is Lindelöf {{iff}} $M$ is second-countable.

Proof:
=== Sufficient Condition ===
We have from Second-Countable Space is Lindelöf that second-countability implies Lindelöf in all topological spaces, regardless of whether they are metric spaces or not.
{{qed|lemma}}
Theorem: Metric Space is Countably Compact iff Sequentially Compact

Let $M$ be a metric space.
Then $M$ is countably compact {{iff}} $M$ is sequentially compact.

Proof:
This follows directly from the results:
:Countably Compact Metric Space is Sequentially Compact
:Sequentially Compact Space is Countably Compact
{{qed}}
{{ACC|Sequentially Compact Space is Countably Compact}}
Theorem: Metric Space is Weakly Countably Compact iff Countably Compact

A metric space $M = \struct {A, d}$ is weakly countably compact {{iff}} $M$ is countably compact.

Proof:
From Metric Space is $T_1$, $M$ is a $T_1$ space.
The result follows from $T_1$ Space is Weakly Countably Compact iff Countably Compact.
{{qed}}
Theorem: Countably Compact Metric Space is Compact

Let $M = \struct {A, d}$ be a metric space.
Let $M$ be countably compact.
Then $M$ is compact.

Proof:
This follows directly from:
:Metric Space is First-Countable
:Countably Compact First-Countable Space is Sequentially Compact
:Sequentially Compact Metric Space is Second-Countable
:Second-Countable Space is Compact iff Countably Compact
{{qed}}
{{ACC|Sequentially Compact Metric Space is Second-Countable}}
Theorem: Metric Space is Weakly Locally Compact iff Strongly Locally Compact

Let $M = \struct {A, d}$ be a metric space.
Then $M$ is weakly locally compact {{iff}} $M$ is strongly locally compact.

Proof:
Let $M = \struct {A, d}$ be a metric space.
Theorem: Sequence of Implications of Metric Space Compactness Properties

Let $P_1$ and $P_2$ be compactness properties.
Let:
:$P_1 \implies P_2$
mean:
:If a metric space $M$ satisfies property $P_1$, then $M$ also satisfies property $P_2$.
Then the following sequence of implications holds:
{|
|-
| align="center" | Sequentially Compact ||
| align="center" | $\implies$ ||
| align="center" | Weakly $\sigma$-Locally Compact ||
| align="center" | $\implies$ ||
| align="center" | Weakly Locally Compact ||
|-
| align="center" | $\Big \Updownarrow$ ||
| align="center" | ||
| align="center" | $\Big \Downarrow$ ||
| align="center" | ||
| align="center" | $\Big \Updownarrow$ ||
|-
| align="center" | Countably Compact ||
| align="center" | ||
| align="center" | $\sigma$-Compact ||
| align="center" | ||
| align="center" | Strongly Locally Compact ||
|-
| align="center" | $\Big \Updownarrow$ ||
| align="center" | ||
| align="center" | $\Big \Downarrow$ ||
|-
| align="center" | Compact ||
| align="center" | ||
| align="center" | Separable ||
|-
| align="center" | $\Big \Updownarrow$ ||
| align="center" | ||
| align="center" | $\Big \Updownarrow$ ||
|-
| align="center" | Weakly Countably Compact ||
| align="center" | ||
| align="center" | Lindelöf ||
|-
| align="center" | ||
| align="center" | ||
| align="center" | $\Big \Updownarrow$ ||
|-
| align="center" | ||
| align="center" | ||
| align="center" | Second-Countable ||
|}
If there is no implication indicated, then there exists a counterexample demonstrating that no such implication exists.

Proof:
The relevant justifications are listed as follows:
:Metric Space is Compact iff Countably Compact.
:Metric Space is Countably Compact iff Sequentially Compact.
:Metric Space is Weakly Countably Compact iff Countably Compact.
:Compact Space is Weakly $\sigma$-Locally Compact.
:By definition, a weakly $\sigma$-locally compact space is both weakly locally compact and $\sigma$-compact.
:$\sigma$-Compact Space is Lindelöf.
:Metric Space is Lindelöf iff Second-Countable.
:Metric Space is Separable iff Second-Countable.
:Metric Space is Weakly Locally Compact iff Strongly Locally Compact.
{{WIP|Determine which results serve as counterexamples}}
{{qed}}
{{ACC}}
{{explain|Indicate which of the above theorems uses ACC in its proof}}
Theorem: Extremally Disconnected Metric Space is Discrete

Let $M = \struct {A, d}$ be a metric space which is extremally disconnected.
Then $M$ is the discrete topology.

Proof:
Let $M = \struct {A, d}$ be extremally disconnected.
Let $p \in A$.
As $M$ is a metric space, $\set p$ can be expressed as:
:$\set p = \ds \bigcap_{n \mathop \in \N_{>0} } \paren {\map {B_{1 / n} } p}^-$
where:
:$\map {B_{1 / n} } p$ denotes the open $1 / n$-ball of $p$
:$\paren {\map {B_{1 / n} } p}^-$ denotes the closure of $\map {B_{1 / n} } p$
That is, as the intersection of the closures of the open $1 / n$-ball of $p$ for all non-zero natural numbers.
Now let:
:$\ds U = \bigcup_{n \mathop \in \N_{>0} } \map {B_{1 / 2 n} } p \setminus \paren {\map {B_{1 / \paren {2 n + 1} } } p}^-$
Then either $U$ or the complementary set of annuli is an open set which has $p$ as a non-interior limit point provided $\set p$ is not open.
{{explain|Take another run at that last sentence, it needs amplification.}}
So if $M$ is not the discrete topology, it cannot be extremally disconnected.
{{qed}}
Theorem: Range of Characters

Let $G$ be a finite abelian group of order $m$.
Let $\chi: G \to \C^\times$ be a character on $G$.
Then for any $g \in G$, $\map \chi g$ is an $m$th root of unity.
If $e$ is the identity of $G$ then $\map \chi g = 1$.

Proof:
The claim that $\map \chi e = 1$ is shown by Group Homomorphism Preserves Identity.
Let $g \in G$ be arbitrary.
Let $k$ be the order of $g$.
By Order of Element Divides Order of Finite Group:
:$\exists l \in \Z: m = k l$
Therefore:
:$g^m = \paren {g^k}^l = e^l = e$
By the homomorphism property:
:$1 = \map \chi e = \map \chi {g^m} = \map \chi g^m$
Hence the result.
{{Qed}}
Category:Analytic Number Theory
0ze9r5yuv8ntst78141gfauk9wn6afm
Theorem: Indicator is Well-Defined

Let $G$ be a finite group, and $a \in G$.
Let $H$ be a subgroup of $G$.
Then the indicator of $a$ in $H$ is well defined.

Proof:
If $a \in H$, then for $n = 1$, $a^n \in H$.
Suppose that $a \notin H$. By Identity of Subgroup, we have that the identity $e$ of $G$ is in $H$.
Moreover, by Element of Finite Group is of Finite Order, there is $n \in \N$ such that $a^n = e \in H$.
Therefore, for any $g \in G$, there is ''some'' $n \in \N$ such that $a^n \in H$.
That is, the set of positive integers $n$ such that $a^n \in H$ is non-empty.
So by the Well-Ordering Principle, this set has a least element, and the indicator is well defined.
{{qed}}
Category:Subgroups
9ellxqergqaoxuoyn896eccdklzha7o
Theorem: Compact Metric Space is Totally Bounded

Let $M = \struct {A, d}$ be a metric space which is compact.
Then $M$ is totally bounded.

Proof:
Let $M = \struct {A, d}$ be compact.
Let $\epsilon > 0$.
Then the family $\set {\map {B_\epsilon} x: x \in A}$ of open $\epsilon$-balls forms an open cover of $A$.
By the definition of compact, every open cover for $A$ has a finite subcover.
That is, there are points $x_0, \ldots, x_n$ such that:
:$\ds A = \bigcup_{0 \mathop \le i \mathop \le n} \map {B_\epsilon} {x_i}$
as required. 
{{qed}}
Theorem: Compact Metric Space is Totally Bounded

Let $M = \struct {A, d}$ be a metric space which is compact.
Then $M$ is totally bounded.

Proof:
We have {{hypothesis}} that $M$ is a sequentially compact space.
So {{afortiori}} every infinite sequence in $M$ has a subsequence which converges to a point in $A$.
Let $\epsilon \in \R_{>0}$ be a strictly positive real number.
By definition, $M$ is totally bounded {{iff}} there exists a finite $\epsilon$-net for $M$.
{{AimForCont}} there exists no finite $\epsilon$-net for $M$.
The aim is to construct an infinite sequence $\sequence {x_n} _{n \mathop \in \N}$ in $A$ that has no convergent subsequence.
Suppose that $x_0, x_1, \ldots, x_r \in A$ have been defined such that:
:$\forall m, n \in \set {0, 1, \ldots, r} : m \ne n \implies \map d {x_m, x_n} \ge \epsilon$
That is, any two distinct elements of $\set {x_0, x_1, \ldots, x_r}$ are at least $\epsilon$ apart.
We have {{hypothesis}} that there exists no finite $\epsilon$-net for $M$.
Specifically, $\set {x_0, x_1, \ldots, x_r}$ is therefore not an $\epsilon$-net for $M$.
So, by definition of a $\epsilon$-net:
:$\ds A \nsubseteq \bigcup_{i \mathop = 0}^r \map {B_\epsilon} {x_i}$
where $\map {B_\epsilon} {x_i}$ denotes the open $\epsilon$-ball of $x_i$ in $M$.
Thus there must exist $x_{r + 1} \in A$ such that:
:$\ds x_{r + 1} \notin \bigcup_{i \mathop = 0}^r \map {B_\epsilon} {x_i}$
That is:
:$\exists x_{r + 1} \in A: \forall m, n \in \set {0, 1, \ldots, r, r + 1} : m \ne n \implies \map d {x_m, x_n} \ge \epsilon$
Thus, by induction, the infinite sequence $\sequence {x_n}$ in $A$ has been constructed such that:
:$\forall m, n \in \N: m \ne n \implies \map d {x_m, x_n} \ge \epsilon$
Thus $\sequence {x_n}$ has no Cauchy subsequence.
From Convergent Sequence in Metric Space is Cauchy Sequence, $\sequence {x_n}$ has no convergent subsequence either. 
Thus, by definition, $M$ is not sequentially compact.
But {{hypothesis}} $M$ is sequentially compact.
Hence by Proof by Contradiction there exists a finite $\epsilon$-net for $M$.
Hence {{afortiori}} $M$ is totally bounded.
{{qed}}
Theorem: Subgroup Generated by Subgroup and Element

Let $G$ be a finite abelian group.
Let $H$ be a proper subgroup of $G$.
Let $a \in G \setminus H$.
Let $n$ be the indicator of $a$ in $H$.
Then:
:$K = \left\{{x a^k: x \in H, \ 0 \le k < n}\right\}$
is a subgroup of $G$ such that $H \subseteq K$, and each element of $K$ has a unique representation in this form.
Moreover:
:$K = \left\langle{H, a}\right\rangle$
where $\left\langle{H, a}\right\rangle$ denotes the subgroup generated by $\left\{{H, a}\right\}$, and:
:$\left\vert{K}\right\vert = n \left\vert{H}\right\vert$
where $\left\vert{K}\right\vert$ denotes the order of $K$.

Proof:
=== $K$ is Subgroup of $G$ ===
We first show that $K$ is a subgroup of $G$ using the Two-Step Subgroup Test.
$(1): \quad K \ne \varnothing$:
From Identity of Subgroup:
:$e \in H$
From Indicator is Well-Defined, $n > 0$ and so $n - 1 \ge 0$.
Thus $k = 0$ fulfils the condition that $0 \le k < n$, and so:
:$e = e a^0 \in K$
Thus $K \ne \varnothing$.
$(2): \quad K$ is closed:
Let $r = x a^k \in K$ and $s = y a^l \in K$ where $x, y \in H$ and $0 \le k < n, 0 \le l < n$.
$G$ is abelian, so $r$, $s$ and $a$ commute.
So:
:$r s = x y a^{k + l}$.
Since $H$ is a group and $x, y \in H$ then also $x y \in H$.
Let $k + l \le n - 1$.
Then $r s \in K$ by definition of $K$.
Let $n \le k + l \le 2 n$.
Then:
:$r s = x y a^n a^{k + l - n}$
By definition of indicator, $a^n \in H$.
Thus as $x, y, a^n \in H$ it follows that $x y a^n \in H$.
As $0 \le k + l - n \le n - 1$ it follows that $\left({x y a^n}\right) a^{k + l - n} \in K$ by definition of $K$. 
So again $r s \in K$.
Thus $K$ is closed.
$(3): \quad K$ is closed under inversion:
Let $r = x a^k \in K$ where $x \in H$ and $0 \le k < n$.
Then $x^{-1} \in H$.
Let $k = 0$.
Then 
:$r^{-1} = x^{-1} e = x^{-1} a^0 \in K$
Let $k > 0$.
Then:
:$r^{-1} = x^{-1} a^{-k} = x^{-1} a^{-n} a^{n - k}$
By definition of indicator, $a^n \in H$.
We have that $x^{-1}, a^n \in H$.
As $G$ is abelian, $x$ and $a$ commute.
Thus from Inverse of Commuting Pair:
: $x^{-1} a^{-n} = \left({x a^n}\right)^{-1} \in H$
Also:
:$0 \le n - k < n -1$
so:
:$r^{-1} = \left({x a^n}\right)^{-1} a^{n - k} \in K$
Therefore by the Two-Step Subgroup Test we have shown that $K$ is a subgroup of $G$.
Theorem: Number of Characters on Finite Abelian Group

Let $G$ be a finite abelian group.
Then the number of characters $G \to \C^\times$ is $\order G$.

Proof:
=== Lemma ===
Let $H \le G$ be a subgroup.
Let $\chi: H \to \C^\times$ be a character on $G$.
Let $a \in G \divides H$ where $\divides$ denotes divisibility.
Then:
:$\chi$ extends to $\index G H$ distinct characters on $G$
where $\index G H$ denotes the index of $H$ in $G$.
Theorem: Sequentially Compact Metric Space is Complete

Let $M = \struct {A, d}$ be a metric space which is sequentially compact.
Then $M$ is complete.

Proof:
Let $\sequence {x_n}_{n \mathop \in \N}$ be a Cauchy sequence in $A$.
As $M$ is sequentially compact, $\sequence {x_n}$ has a convergent subsequence.
By Convergent Subsequence of Cauchy Sequence in Metric Space, this implies that the entire sequence $\sequence {x_n}$ is convergent.
Hence, $M$ is complete.
{{qed}}
Theorem: Complete Metrizability is Weakly Hereditary

Let $T = \struct {S, \tau}$ be a topological space which is completely metrizable.
Let $V \subseteq S$ be a closed subspace of $T$.
Then $V$ is also completely metrizable.
That is, complete metrizability is a weakly hereditary property.

Proof:
{{finish|Use Subspace of Complete Metric Space is Closed iff Complete}}
Theorem: Metric Space Completeness is Preserved by Isometry

Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be metric spaces.
Let $\phi: M_1 \to M_2$ be an isometry.
If $M_1$ is complete then so is $M_2$.

Proof:
Let $\tau_1$ be the topology on $A_1$ induced by $d_1$.
Let $\tau_2$ be the topology on $A_2$ induced by $d_2$.
Let $\sequence {x_n}$ be a Cauchy sequence in $A_2$.
From Inverse of Isometry of Metric Spaces is Isometry, $\phi^{-1}$ is an isometry.
Since Isometric Image of Cauchy Sequence is Cauchy Sequence, $\sequence {\map {\phi^{-1} } {x_n} }$ is a Cauchy sequence.
Since $M_1$ is a complete metric space, $\sequence {\map {\phi^{-1} } {x_n} }$ converges.
Since Isometry Preserves Sequence Convergence, $\sequence {\map \phi {\map {\phi^{-1} } {x_n} } }$ converges.
But:
:$\sequence {\map \phi {\map {\phi^{-1} } {x_n} } } = \sequence {x_n}$
so $\sequence {x_n}$ converges.
Thus each Cauchy sequence in $M_2$ converges.
It follows by definition that so $M_2$ is a complete metric space.
{{qed}}