Subspace of Complete Metric Space is Relatively Compact iff Every Sequence has Cauchy Subsequence

== Theorem ==
Let $\struct {X, d}$ be a complete metric space.
Let $\struct {H, d_H}$ be a metric subspace of $\struct {X, d}$. 
Then $\struct {H, d_H}$ is relatively compact :
:every sequence in $H$ has a Cauchy subsequence.
== Proof ==
=== Necessary Condition ===
Suppose that: 
:$\struct {H, d_H}$ is relatively compact
Let $\sequence {x_n}_{n \mathop \in \N}$ be a sequence in $H$.
Then $\sequence {x_n}_{n \mathop \in \N}$ is a sequence in $\map \cl H$, where $\map \cl H$ is the closure of $H$ in $\struct {X, d}$.
Since $\struct {H, d_H}$ is relatively compact, we have: 
:$\map \cl H$ is compact.
From Compact Subspace of Metric Space is Sequentially Compact in Itself, we have: 
:$\map \cl H$ is sequentially compact.
So:
:$\sequence {x_n}_{n \mathop \in \N}$ has a subsequence $\sequence {x_{n_j} }_{j \mathop \in \N}$ that converges to a point in $\map \cl H$.
From Convergent Sequence in Metric Space is Cauchy Sequence, we have: 
:$\sequence {x_{n_j} }_{j \mathop \in \N}$ is a Cauchy sequence.
So:
:$\sequence {x_n}_{n \mathop \in \N}$ has a Cauchy subsequence $\sequence {x_{n_j} }_{j \mathop \in \N}$.
Since $\sequence {x_n}_{n \mathop \in \N}$ was an arbitrary sequence in $H$, we have: 
:every sequence in $H$ has a Cauchy subsequence.
=== Sufficient Condition ===
Suppose that: 
:every sequence in $H$ has a Cauchy subsequence.
We show that $\map \cl H$ is sequentially compact.
Let $\sequence {x_n}_{n \mathop \in \N}$ be a sequence in $\map \cl H$. 
From Condition for Point being in Closure, for each $n \in \N$ we can pick $y_n \in H$ such that: 
:$\ds \map d {x_n, y_n} < \frac 1 n$
From hypothesis, there exists a Cauchy subsequence $\sequence {y_{n_j} }_{j \mathop \in \N}$ of $\sequence {y_n}_{n \mathop \in \N}$.
Since $\struct {X, d}$ is complete, $\sequence {y_{n_j} }_{j \mathop \in \N}$ converges to a point $y$.
We show that $\sequence {x_{n_j} }_{j \mathop \in \N}$ converges to $y$.
We have:
Let $\epsilon > 0$.
Since $\sequence {y_{n_j} }_{j \mathop \in \N}$ converges to $y$, there exists $N \in \N$ such that: 
:$\ds \map d {y_{n_j}, y} < \frac \epsilon 2$
for $j \ge N$. 
Now, let $N' \in \N$ be such that: 
:$\ds N' \ge \max \set {\frac 2 \epsilon, N}$
Then: 
:$\ds \frac 1 {n_j} < \frac \epsilon 2$
for $j \ge N'$, giving:
:$\map d {x_{n_j}, y} < \epsilon$
for $j \ge N'$.
Since $\epsilon > 0$ was arbitrary, we have that $\sequence {x_{n_j} }_{j \mathop \in \N}$ converges to $y$.
Since $\sequence {x_{n_j} }_{j \mathop \in \N}$ is a sequence in $\map \cl H$, we have that $y$ is a limit point of $\map \cl H$. 
From Topological Closure is Closed, we have that $\map \cl H$ is a closed set.
So $y \in \map \cl H$. 
So $\sequence {x_n}_{n \mathop \in \N}$ has a subsequence that converges to a point in $\map \cl H$.
Since $\sequence {x_n}_{n \mathop \in \N}$ was an arbitrary sequence in $\map \cl H$, we have: 
:every sequence in $\map \cl H$ has a subsequence that converges to a point in $\map \cl H$.
So:
:$\map \cl H$ is sequentially compact.
So, from Sequentially Compact Metric Space is Compact:
:$\map \cl H$ is compact.
So we conclude:
:$\struct {H, d_H}$ is relatively compact.
== Sources ==
* $6.3$: Arzelà-Ascoli Theorem
Category:Relatively Compact Subspaces
Category:Complete Metric Spaces
Category:Cauchy Sequences