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: Structured Description
Here is the rules of Structured Description.
Entities:
\(T\): \(\in \{\text{ the 1st-countable topological spaces }\}\)
//
Statements:
\(T \in \{\text{ the countably compact topological spaces }\}\)
\(\implies\)
\(T \in \{\text{ the sequentially compact topological spaces }\}\)
//
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 countably compact; Step 2: see that \(T\) is sequentially compact.
Step 1:
Let \(s: \mathbb{N} \setminus \{0\} \to T\) be any.
Let us define the subset, \(S := s (\mathbb{N} \setminus \{0\}) \subseteq T\).
If \(S\) is finite, at least an element of \(S\) appears infinite times, so, the subsequence that chooses only the element converges to the element.
Let us suppose that \(S\) is infinite hereafter.
\(S\) has an \(\omega\)-accumulation point, \(t \in T\), 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 neighborhoods basis at \(t\), \(\{U_{t, j} \vert j \in J\}\).
There is a point, \(t'_1 \in U_{t, 1} \cap S\), because \(t\) is an \(\omega\)-accumulation point of \(S\).
There is a point, \(t'_2 \in U_{t, 1} \cap U_{t, 2} \cap S\), which can be chosen to be later than \(t'_1\) in the sequence, because \(U_{t, 1} \cap U_{t, 2} \cap S\) has some infinite points.
And so on, after all, for each \(j \in \mathbb{N} \setminus \{0\}\), there is a point, \(t'_j \in U_{t, 1} \cap ... \cap U_{t, j} \cap S\), which is later than \(t'_1, ..., t'_{j - 1}\) in the sequence.
\((t'_1, t'_2, ...)\) is a subsequence of the original sequence.
For each open neighborhood of \(t\), \(U_t \subseteq T\), there is a \(U_{t, N} \subseteq U_t\) for an \(N\), and for each \(j \in \mathbb{N} \setminus \{0\}\) such that \(N \lt j\), \(t'_j \in U_{t, 1} \cap ... \cap U_{t, j} \cap S \subseteq U_{t, N} \subseteq U_t\), which means that \((t'_1, t'_2, ...)\) converges to \(t\).
