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

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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 if and only if:

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.

$\Box$

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:

\(\ds \map d {x_{n_j}, y}\) \(\le\) \(\ds \map d {x_{n_j}, y_{n_j} } + \map d {y_{n_j}, y}\) Metric Space Axiom $(\text M 2)$: Triangle Inequality
\(\ds \) \(=\) \(\ds \frac 1 {n_j} + \map d {y_{n_j}, y}\) from the construction of $\sequence {y_n}_{n \mathop \in \N}$

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.

$\blacksquare$


Sources

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