Difference between revisions of "Equivalent statements to compactness of a metric space"
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{{End Proof}}{{End Theorem}} | {{End Proof}}{{End Theorem}} | ||
{{Begin Inline Theorem}} | {{Begin Inline Theorem}} | ||
| − | {{M|2)\implies 3)}}: Suppose for all sequences {{M|1=(x_n)_{n=1}^\infty\subseteq X}} | + | {{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}} | {{Begin Inline Proof}} | ||
'''Proof of completeness:''' | '''Proof of completeness:''' | ||
: To show {{M|(X,d)}} is complete we must show that every [[Cauchy sequence]] converges. To do this: | : 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}} | :* 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 sequence | + | :** 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. | :*** 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. | :* As our choice of Cauchy sequence was arbitrary we conclude that all Cauchy sequences converge in {{M|X}}, which is the definition of completeness. | ||
| Line 24: | Line 24: | ||
{{QED}} | {{QED}} | ||
{{End Proof}}{{End Theorem}} | {{End Proof}}{{End Theorem}} | ||
| − | {{Todo|Rest}} | + | {{Todo|Rest, namely: {{M|3\implies 1}}}} |
==Notes== | ==Notes== | ||
Revision as of 11:37, 27 May 2016
Contents
[hide]Theorem statement
Given a metric space (X,d), the following are equivalent[1][Note 1]:
- X is compact
- Every sequence in X has a subsequence that converges (AKA: having a convergent subsequence)
- X is totally bounded and complete
Proof
[Expand]
1)⟹2): X is compact ⟹ ∀(an)∞n=1⊆X ∃ a sub-sequence (akn)∞n=1 that coverges in X
[Expand]
2)⟹3): Suppose for all sequences (xn)∞n=1⊆X that (xn)∞n=1 has a convergent subsequence ⟹ (X,d) is a complete metric space and is totally bounded
TODO: Rest, namely: 3⟹1
Notes
- Jump up ↑ To say statements are equivalent means we have one ⟺ one of the other(s)
References
