Difference between revisions of "Equivalent statements to compactness of a metric space"
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| − | ==Theorem statement== | + | ==[[:Equivalent statements to compactness of a metric space/Statement|Theorem statement]]== |
{{:Equivalent statements to compactness of a metric space/Statement}} | {{:Equivalent statements to compactness of a metric space/Statement}} | ||
| + | |||
==Proof== | ==Proof== | ||
{{Begin Inline Theorem}} | {{Begin Inline Theorem}} | ||
| − | {{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}} | ||
| − | + | # 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}} | {{End Proof}}{{End Theorem}} | ||
| − | {{Todo|Rest}} | + | {{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== | ==Notes== | ||
<references group="Note"/> | <references group="Note"/> | ||
Latest revision as of 11:37, 27 May 2016
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
