# Equivalent statements to compactness of a metric space

## Contents

## Theorem statement

Given a metric space [ilmath](X,d)[/ilmath], the following are equivalent^{[1]}^{[Note 1]}:

- [ilmath]X[/ilmath] is compact
- Every sequence in [ilmath]X[/ilmath] has a subsequence that converges (AKA: having a
*convergent subsequence*) - [ilmath]X[/ilmath] is totally bounded and complete

## Proof

[ilmath]1)\implies 2)[/ilmath]: [ilmath]X[/ilmath] is compact [ilmath]\implies[/ilmath] [ilmath]\forall(a_n)_{n=1}^\infty\subseteq X\ \exists[/ilmath] a sub-sequence [ilmath](a_{k_n})_{n=1}^\infty[/ilmath] that coverges in [ilmath]X[/ilmath]

- Using every sequence in a compact space is a lingering sequence and
- every lingering sequence has a convergent subsequence

We see that every sequence in a compact space has a convergent subsequence.

[ilmath]2)\implies 3)[/ilmath]: Suppose for all sequences [ilmath](x_n)_{n=1}^\infty\subseteq X[/ilmath] that [ilmath] ({ x_n })_{ n = 1 }^{ \infty } [/ilmath] has a convergent subsequence [ilmath]\implies[/ilmath] [ilmath](X,d)[/ilmath] is a complete metric space and is totally bounded

**Proof of completeness:**

- To show [ilmath](X,d)[/ilmath] is complete we must show that every Cauchy sequence converges. To do this:
- Let [ilmath](x_n)_{n=1}^\infty\subseteq X[/ilmath] be any Cauchy sequence in [ilmath]X[/ilmath]
- Recall that If a subsequence of a Cauchy sequence converges then the Cauchy sequence itself also converges
- By hypothesis all sequences in [ilmath]X[/ilmath] have a convergent subsequence. By this theorem our [ilmath](x_n)_{n=1}^\infty[/ilmath] converges to the same thing.

- Recall that If a subsequence of a Cauchy sequence converges then the Cauchy sequence itself also converges
- As our choice of Cauchy sequence was arbitrary we conclude that all Cauchy sequences converge in [ilmath]X[/ilmath], which is the definition of completeness.

- Let [ilmath](x_n)_{n=1}^\infty\subseteq X[/ilmath] be any Cauchy sequence in [ilmath]X[/ilmath]

**Proof that [ilmath](X,d)[/ilmath] is totally bounded**

TODO: Do this

TODO: Rest, namely: [ilmath]3\implies 1[/ilmath]

## Notes

- ↑ To say statements are equivalent means we have one [ilmath]\iff[/ilmath] one of the other(s)

## References