A description/proof of that topological space is countably compact iff each infinite subset has \(\omega\)-accumulation point
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of countably compact topological space.
- The reader knows a definition of \(\omega\)-accumulation point of subset.
Target Context
- The reader will have a description and a proof of the proposition that any topological space is countably compact if and only if each infinite subset has an \(\omega\)-accumulation point.
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 and only if each infinite subset, \(S\), has an \(\omega\)-accumulation point, \(p \in T\).
2: Proof
Let us suppose that each \(S\) has a \(p\). Let us suppose that there was an open countable cover of \(T\), \(O = \{U_i\vert i \in I\}\), where \(I\) was a countable indices set, without any finite subcover. Let us suppose that \(U_i \neq \emptyset\), because any empty set could be removed and the elements could be re-indexed. Let us suppose that \(I = \{1, 2, ...\}\) without loss of generality. There would be a point, \(p_1 \in U_1\), and let us define \(i_1 := 1\) and \(U'_1 := U_1\). There would be a point, \(p_2 \notin U'_1\), so, \(p_2 \in U_{i_2}\) for an \(i_1 \lt i_2 \in I\), and let us define \(U'_2 := \cup_{1 \le i \le i_2} U_i\). There would be a point, \(p_3 \notin U'_2\), so, \(p_3 \in U_{i_3}\) for an \(i_2 \lt i_3 \in I\), and so on, which would go on infinitely because there would be no finite subcover of \(O\). \(O' := \{U'_i\vert i \in I\}\) would be an open cover of \(T\). Let us choose \(S = \{p_1, p_2, ...\}\) and let \(p\) be an \(\omega\)-accumulation point of \(S\). \(p \in U'_j\) for an \(j \in I\), because \(O'\) would be a cover of \(T\). But while \(U'_j\) would be a neighborhood of \(p\), \(U'_j \cap S = \{p_1, p_2, ..., p_j\}\), which is a contradiction against \(p\)'s being an \(\omega\)-accumulation point.
Let us suppose that \(T\) is countably compact. Let us suppose that there was an infinite subset, \(S\), that had no \(\omega\)-accumulation point. Also any countable subset, \(S' \subseteq S\), would have no \(\omega\)-accumulation point, because for any point, \(p \in T\), there would be a neighborhood, \(U_p\), such that \(U_p \cap S\) would have only finite points, and also \(U_p \cap S'\) would have only finite points, because \(S'\) would be smaller than \(S\). For each finite (can be empty) \(S'' \subseteq S'\), let us define \(f (S'') := \cup \{U_p\vert U_p \cap S' = S''\}\), which might be empty for an \(S''\), but that would not matter. \(O := \{f (S'') \vert S'' \subseteq S'\}\) would be an open cover of \(T\), because as \(S'\) would have no \(\omega\)-accumulation point, any point, \(p \in T\), would have at least 1 \(U_p\) such that \(U_p \cap S'\) would have only finite points, and in fact, countable, because the set of the finite subsets of any countable set would be countable: for any countable set, \(S' = \{1, 2, ...\}\), choose the finite sets in this order: 1st take \(\emptyset\); 2nd take \(\{1\}\) and choose \(\{1\}\); 3rd take \(\{1, 2\}\) and choose \(\{2\}\) and \(\{1, 2\}\); 4th take \(\{1, 2, 3\}\) and choose \(\{3\}\), \(\{2, 3\}\), and \(\{1, 2, 3\}\), for example.
There would be a finite subcover, \(O' := \{f (S''_i)\} \subseteq O\). There would be a point, \(p \in S' \setminus \cup_i S''_i\), because \(S'\) would be infinite while \(\cup_i S''_i\) would be finite. But \(p \notin f (S''_i)\) for any \(i\), because \(f (S''_i) \cap S' = S''_i\), a contradiction against \(O'\)'s being a cover of \(T\).
