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\subseteq X\ \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}}
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{{:Equivalent statements to compactness of a metric space/1-implies-2 proof}}
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# Using [[every sequence in a compact space is a lingering sequence]] and
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# [[every lingering sequence has a convergent subsequence]]
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We see that every sequence in a compact space has a convergent subsequence.
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{{End Proof}}{{End Theorem}}
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{{Begin Inline Theorem}}
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{{M|2)\implies 3)}}: Suppose for all sequences {{M|1=(x_n)_{n=1}^\infty\subseteq X}} then {{M|(X,d)}} is a [[complete metric space]] and is [[totally bounded]]
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{{Begin Inline Proof}}
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'''Proof of completeness:'''
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: 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}}
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:** Recall that [[If a subsequence of a Cauchy sequence converges then the sequence converges also converges]]
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:*** 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.
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:* As our choice of Cauchy sequence was arbitrary we conclude that all Cauchy sequences converge in {{M|X}}, which is the definition of completeness.
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{{QED}}
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'''Proof that {{M|(X,d)}} is [[totally bounded]]'''
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{{Todo|Do this}}
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{{QED}}
 
{{End Proof}}{{End Theorem}}
 
{{End Proof}}{{End Theorem}}
 
{{Todo|Rest}}
 
{{Todo|Rest}}

Revision as of 00:39, 8 December 2015

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]

  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] then [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:

Q.E.D.

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

TODO: Do this

Q.E.D.


TODO: Rest


Notes

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

References

  1. ↑ Introduction to Topology - Theodore W. Gamelin & Robert Everist Greene



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