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: Structured Description
Here is the rules of Structured Description.
Entities:
\(T\): \(\in \{\text{ the topological spaces }\}\)
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Statements:
\(T \in \{\text{ the sequentially compact topological spaces }\}\)
\(\implies\)
\(T \in \{\text{ the countably compact topological spaces }\}\)
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2: Proof
Whole Strategy: apply the proposition that any topological space is countably compact if and only if each infinite subset has an \(\omega\)-accumulation point; Step 1: suppose that \(T\) is sequentially compact; Step 2: see that each infinite subset has an \(\omega\)-accumulation point; Step 3: conclude the proposition.
Step 1:
Let suppose that \(T\) is sequentially compact.
Step 2:
Let \(S \subseteq T\) be any infinite subset.
Let us choose any sequence of distinct points from \(S\), \((s_1, s_2, ...)\), which is possible because \(S\) is infinite.
There is a subsequence, \((s^`_1, s^`_2, ...)\), that converges to a point, \(t \in T\), because \(T\) is sequentially compact.
\(t\) is an \(\omega\)-accumulation point of \(S\), because for any neighborhood of \(t\), \(U_t \subseteq T\), there is an \(N \in \mathbb{N} \setminus \{0\}\) such that for each \(n \in \mathbb{N} \setminus \{0\}\) such that \(N \lt n\), \(s^`_n \in U_t\), which means that \(U_t \cap S\) is infinite.
So, each infinite subset of \(T\) has an \(\omega\)-accumulation point.
Step 3:
By the proposition that any topological space is countably compact if and only if each infinite subset has an \(\omega\)-accumulation point, \(T\) is countably compact.
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.
