Equivalent statements to compactness of a metric space

Theorem statement

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

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]

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1. Using every sequence in a compact space is a lingering sequence and
2. 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

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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.
• 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.

Q.E.D.

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

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TODO: Do this

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Q.E.D.

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

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