Difference between revisions of "Sequential compactness"
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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. | ||
| + | ===Common forms=== | ||
| + | ====Functional Analysis==== | ||
| + | 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> | ||
| + | |||
| + | 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
