View source for Equivalent statements to compactness of a metric space

==[[:Equivalent statements to compactness of a metric space/Statement|Theorem statement]]==
{{:Equivalent statements to compactness of a metric space/Statement}}

==Proof==
{{Begin Inline Theorem}}
{{M|1)\implies 2)}}: {{M|X}} is [[compact]] {{M|\implies}} {{M|1=\forall(a_n)_{n=1}^\infty\subseteq X\ \exists}} a [[sub-sequence]] {{M|1=(a_{k_n})_{n=1}^\infty}} that [[Convergence (sequence)|coverges]] in {{M|X}}
{{Begin Inline Proof}}
# 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.
{{End Proof}}{{End Theorem}}
{{Begin Inline Theorem}}
{{M|2)\implies 3)}}: Suppose for all sequences {{M|1=(x_n)_{n=1}^\infty\subseteq X}} that {{MSeq|x_n}} has a convergent subsequence {{M|\implies}} {{M|(X,d)}} is a [[complete metric space]] and is [[totally bounded]]
{{Begin Inline Proof}}
'''Proof of completeness:'''
: To show {{M|(X,d)}} is complete we must show that every [[Cauchy sequence]] converges. To do this:
:* Let {{M|1=(x_n)_{n=1}^\infty\subseteq X}} be any [[Cauchy sequence]] in {{M|X}}
:** Recall that [[If a subsequence of a Cauchy sequence converges then the Cauchy sequence itself also converges]]
:*** By hypothesis all sequences in {{M|X}} have a convergent subsequence. By this theorem our {{M|1=(x_n)_{n=1}^\infty}} converges to the same thing.
:* As our choice of Cauchy sequence was arbitrary we conclude that all Cauchy sequences converge in {{M|X}}, which is the definition of completeness.
{{QED}}

'''Proof that {{M|(X,d)}} is [[totally bounded]]'''
{{Todo|Do this}}
{{QED}}
{{End Proof}}{{End Theorem}}
{{Todo|Rest, namely: {{M|3\implies 1}}}}

==Notes==
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==References==
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{{Theorem Of|Metric Space}}