2023-10-22

391: Topological Space Is Countably Compact if It Is Sequentially Compact

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A description/proof of that topological space is countably compact if it is sequentially compact

Topics


About: topological space

The table of contents of this article


Starting Context



Target Context


  • The reader will have a description and a proof of the proposition that any topological space is countably compact if the space is sequentially compact.

Orientation


There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.


Main Body


1: Description


Any topological space, T, is countably compact if T is sequentially compact.


2: Proof


Let suppose that T is sequentially compact. We will prove that each infinite subset, ST, has an ω-accumulation point, pT, which will imply that T is countably compact, by the proposition that any topological space is countably compact if and only if each infinite subset has an ω-accumulation point.

Let any infinite subset be ST. Let us choose any sequence of points from S, p1,p2,..., which is possible because S is infinite. There is a subsequence, p1,p2,..., that converges to a point, pT. p is an ω-accumulation point of S, because for any neighborhood, UpT, there is an i0 such that piUp for each i such that i0<i, which means that UpS is infinite.


3: Note


The converse is not true in general. The converse with the requirement that T is 1st-countable is true, as is proved in another article.


References


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