202303011803

Type :Note Tags : Topology

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Countably Compact

Note

A space $X$ is called countably compact if every open cover of $X$ has a finite subcover.

Proposition

Note

A first countable space is countably compact iff it is sequentially compact.

Proof:

If it is sequentially compact, then take a minimal countable cover, take points $x_1,x_2,\dots$ such that each $x_i$ is in exactly one of the open sets of the cover, then there is a point $x$ such that the element of the cover containing it has infinitely many of these $x_i$‘s. Contradiction.

If it is countably compact, then take a sequence $x_1,x_2,\dots$, assume this is a sequence without any convergent subsequence. Then for any $x_i$, there is some $V_i$ around it which contains only finitely many of the other $x_j^{\prime }s$. Then let $U=(\bigcup V_i{)}^c$, so the open cover

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Related Problems

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References