# Equivalent statements to compactness of a metric space

## Theorem statement

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

1. [ilmath]X[/ilmath] is compact
2. Every sequence in [ilmath]X[/ilmath] has a subsequence that converges (AKA: having a convergent subsequence)
3. [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]

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:

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

TODO: Do this

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

## Notes

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