Difference between revisions of "Sequential compactness"
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*A [[Metric space|metric space]] is compact if and only if it is sequentially compact, a theorem found [[Metric space is compact iff sequentially compact|here]] | *A [[Metric space|metric space]] is compact if and only if it is sequentially compact, a theorem found [[Metric space is compact iff sequentially compact|here]] | ||
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Latest revision as of 15:37, 24 November 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
