Difference between revisions of "Equivalent statements to compactness of a metric space"
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| − | {{M|1)\implies 2)}}: {{M|X}} is [[compact]] {{M|\implies}} {{M|1=\forall(a_n)_{n=1}^\infty\ \exists}} a [[sub-sequence]] {{M|1=(a_{k_n})_{n=1}^\infty}} that [[Convergence (sequence)|coverges]] in {{M|X}} | + | {{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}} | {{Begin Inline Proof}} | ||
{{:Equivalent statements to compactness of a metric space/1-implies-2 proof}} | {{:Equivalent statements to compactness of a metric space/1-implies-2 proof}} | ||
{{End Proof}}{{End Theorem}} | {{End Proof}}{{End Theorem}} | ||
{{Todo|Rest}} | {{Todo|Rest}} | ||
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==Notes== | ==Notes== | ||
<references group="Note"/> | <references group="Note"/> | ||
Revision as of 16:29, 1 December 2015
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]
TODO: Rest
Notes
- ↑ To say statements are equivalent means we have one [ilmath]\iff[/ilmath] one of the other(s)
References
