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, \(S \subseteq T\), has an \(\omega\)-accumulation point, \(p \in T\), 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 \(\omega\)-accumulation point.
Let any infinite subset be \(S \subset T\). Let us choose any sequence of points from \(S\), \(p_1, p_2, ...\), which is possible because \(S\) is infinite. There is a subsequence, \(p'_1, p'_2, ...\), that converges to a point, \(p \in T\). \(p\) is an \(\omega\)-accumulation point of \(S\), because for any neighborhood, \(U_p \subseteq T\), there is an \(i_0\) such that \(p'_i \in U_p\) for each \(i\) such that \(i_0 \lt i\), which means that \(U_p \cap S\) 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.
