Difference between revisions of "Sequential compactness"

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(Created page with "The Bolzano-Weierstrass theorem states that every bounded sequence has a convergent subsequence. Sequential compactness extends this notion to general topolo...")
 
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==Definition==
 
==Definition==
 
A [[Topological space|topological space]] {{M|(X,\mathcal{J})}} is sequentially compact if every (infinite) [[Sequence]] has a [[Convergence of a sequence|convergent]] subsequence.
 
A [[Topological space|topological space]] {{M|(X,\mathcal{J})}} is sequentially compact if every (infinite) [[Sequence]] has a [[Convergence of a sequence|convergent]] subsequence.
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===Common forms===
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====Functional Analysis====
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A subset {{M|S}} of a [[Norm|normed]] [[Vector space|vector space]] <math>(V,\|\cdot\|,F)</math> is sequentially compact if any sequence <math>(a_n)^\infty_{n=1}\subset k</math> has a convergent subsequence <math>(a_{n_i})_{i=1}^\infty</math>, that is <math>(a_{n_i})_{i=1}^\infty\rightarrow a\in K</math>
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Like with compactness, we consider the [[Subspace topology|subspace topology]] on a subset, then see if that is compact to define "compact subsets" - we do the same here. As warned below a [[Topological space|topological space]] is not sufficient for sequentially compact <math>\iff</math> compact, so one ought to use [[Mertric subspace|a metric subspace]] instead. Recalling that a [[Norm|norm]] can give rise to the metric <math>d(x,y)=\|x-y\|</math>
 
==Warning==
 
==Warning==
 
Sequential compactness and [[Compactness|compactness]] are not the same for a general [[Topological space|topology]]
 
Sequential compactness and [[Compactness|compactness]] are not the same for a general [[Topological space|topology]]

Revision as of 22:40, 8 March 2015

The Bolzano-Weierstrass theorem states that every bounded sequence has a convergent subsequence.

Sequential compactness extends this notion to general topological spaces.

Definition

A topological space [ilmath](X,\mathcal{J})[/ilmath] is sequentially compact if every (infinite) Sequence has a convergent subsequence.

Common forms

Functional Analysis

A subset [ilmath]S[/ilmath] of a normed vector space [math](V,\|\cdot\|,F)[/math] is sequentially compact if any sequence [math](a_n)^\infty_{n=1}\subset k[/math] has a convergent subsequence [math](a_{n_i})_{i=1}^\infty[/math], that is [math](a_{n_i})_{i=1}^\infty\rightarrow a\in K[/math]

Like with compactness, we consider the subspace topology on a subset, then see if that is compact to define "compact subsets" - we do the same here. As warned below a topological space is not sufficient for sequentially compact [math]\iff[/math] compact, so one ought to use a metric subspace instead. Recalling that a norm can give rise to the metric [math]d(x,y)=\|x-y\|[/math]

Warning

Sequential compactness and compactness are not the same for a general topology

Uses

  • A metric space is compact if and only if it is sequentially compact, a theorem found here