Difference between revisions of "Sequential compactness"

Revision as of 22:40, 8 March 2015 (view source) Alec (Talk | contribs) m ← Older edit		Latest revision as of 15:37, 24 November 2015 (view source) Alec (Talk | contribs) m (Fixed category typo)
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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]]
−	{{Definition|Topology|Metric Spaces}}	+	{{Definition|Topology|Metric Space}}

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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 $(X,\mathcal{J})$ is sequentially compact if every (infinite) Sequence has a convergent subsequence.

Common forms

Functional Analysis

A subset $S$ of a normed vector space

$$(V,\|\cdot\|,F)$$ is sequentially compact if any sequence $$(a_n)^\infty_{n=1}\subset k$$ has a convergent subsequence $$(a_{n_i})_{i=1}^\infty$$ , that is $$(a_{n_i})_{i=1}^\infty\rightarrow a\in K$$

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

$$\iff$$ compact, so one ought to use a metric subspace instead. Recalling that a norm can give rise to the metric $$d(x,y)=\|x-y\|$$

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