A description/proof of that 1st-countable topological space is sequentially compact if it is countably compact
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of 1st-countable topological space.
- The reader knows a definition of sequentially compact topological space.
- The reader knows a definition of countably compact topological space.
- The reader admits the proposition that any topological space is countably compact if and only if each infinite subset has an \(\omega\)-accumulation point.
Target Context
- The reader will have a description and a proof of the proposition that any 1st-countable topological space is sequentially compact if the space is countably 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 1st-countable topological space, \(T\), is sequentially compact if \(T\) is countably compact.
2: Proof
Let us suppose that \(T\) is countably compact. Let any infinite points sequence be \(p_1, p_2, ...\). Let us define the subset, \(S \subseteq T:= \{p_1, p_2, ...\}\). \(S\) has an \(\omega\)-accumulation point, \(p \in T\), by the proposition that any topological space is countably compact if and only if each infinite subset has an \(\omega\)-accumulation point. There is a countable neighborhood basis, \(\{B_{p, i}\}\) at \(p\). There is a point, \(p'_1 \in B_{p, 1} \cap S\). There is a point, \(p'_2 \in B_{p, 1} \cap B_{p, 2} \cap S\), which can be chosen to be later than \(p'_1\) in the sequence, because \(B_{p, 1} \cap B_{p, 2} \cap S\) has some infinite points, and so on. After all, there is a point, \(p'_i \in B_{p, 1} \cap B_{p, 2} \cap ... \cap B_{p, i} \cap S\), which is later than \(p'_1, p'_2, ..., p'_{i - 1}\) in the sequence. \(p'_1, p'_2, ...\) is a subsequence of the original sequence. The subsequence converges to \(p\), because for any neighborhood, \(U_p\), there is a \(B_{p, i} \subseteq U_p\), and \(p'_j \in B_{p, i} \subseteq U_p\) for any \(i \le j\).
