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 of topological space.
- The reader admits the proposition that for any set, the intersection of the union of any possibly uncountable number of subsets and any subset is the union of the intersections of each of the subsets and the latter 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: Structured Description
Here is the rules of Structured Description.
Entities:
\(T\): \(\in \{\text{ the topological spaces }\}\)
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Statements:
\(T \in \{\text{ the countably compact topological spaces }\}\)
\(\iff\)
\(\forall S \in \{\text{ the infinite subsets of } T\} (\exists t \in \{\text{ the } \omega \text{ -accumulation points of } S\})\)
//
2: Proof
Whole Strategy: Step 1: suppose that each \(S\) has a \(t\); Step 2: see that \(T\) is countably compact; Step 3: suppose that \(T\) is countably compact; Step 4: see that each \(S\) has a \(t\).
Step 1:
Let us suppose that each \(S\) has a \(t\).
Step 2:
Let us suppose that \(T\) was not countably compact.
There would be a countable open cover of \(T\), \(\{U_j \vert j \in J\}\) where \(J\) was a countable index set, without any finite subcover.
The open cover would be inevitably infinite, because otherwise, it would have the finite subcover as itself.
Let us suppose that \(U_j \neq \emptyset\) for each \(j \in J\), which would be possible, because any empty set could be removed and the remaining elements could be re-indexed, still covering \(T\) (removing the empty subsets would not influence covering), still infinitely countable (otherwise, the set of the remaining elements would be a finite subcover), and still having no finite subcover (if the set of the remaining elements had a finite subcover, that would be a finite subcover of the original cover).
Let us suppose that \(J = \{1, 2, ...\}\) without loss of generality, just for convenience of expressions.
There would be a point, \(t_1 \in U_1\), and let us define \(j_1 := 1\) and \(U'_1 := U_1\).
There would be a point, \(t_2 \in T\) such that \(t_2 \notin U'_1\), because \(U'_1\) did not cover \(T\), so, \(t_2 \in U_{j_2}\) for a \(j_2 \in J\) such that \(j_1 \lt j_2\), and let us define \(U'_2 := \cup_{1 \le j \le j_2} U_j\).
Note that \(2 \le j_2\) and \(U_2 \subseteq U'_2\).
There would be a point, \(t_3 \in T\), such that \(t_3 \notin U'_2\), so, \(t_3 \in U_{j_3}\) for an \(j_3 \in J\) such that \(j_2 \lt j_3\), and let us define \(U'_3 := \cup_{1 \le j \le j_3} U_j\).
Note that \(3 \le j_3\) and \(U_3 \subseteq U'_3\).
And so on, which would go on infinitely because there would be no finite subcover of \(\{U_j \vert j \in J\}\).
Note that for each \(l \in J\), \(l \le j_l\) and \(U_l \subseteq U'_l\).
\(\{U'_j \vert j \in J\}\) would be an open cover of \(T\), because for each \(t \in T\), \(t \in U_j \subseteq U'_j\) for a \(j \in J\).
Let us choose \(S = \{t_1, t_2, ...\}\) and let \(t\) be an \(\omega\)-accumulation point of \(S\).
\(t \in U'_j\) for a \(j \in J\), so, \(U'_j\) would be an open neighborhood of \(t\).
But \(U'_j \cap S = \{t_1, t_2, ..., t_j\}\), a contradiction against that \(t\) was an \(\omega\)-accumulation point of \(S\).
So, there is no countable open cover of \(T\) without any finite subcover, which means that each countable open cover of \(T\) has a finite subcover.
So, \(T\) is countably compact.
Step 3:
Let us suppose that \(T\) is countably compact.
Step 4:
Let us suppose that there was an infinite subset, \(S \subseteq T\), that had no \(\omega\)-accumulation point.
Also any countable subset, \(S^` \subseteq S\), would have no \(\omega\)-accumulation point, because for each point, \(t \in T\), there would be a neighborhood of \(t\), \(U_t \subseteq T\), such that \(U_t \cap S\) would have only some finite points, and also \(U_t \cap S^`\) would have only some finite points, because \(S^`\) would be smaller than \(S\).
For each finite (can be empty) \(S^{``} \subseteq S^`\), let us define \(f (S^{``}) := \cup \{U \in \{\text{ the open subsets of } T\} \vert U \cap S^` = S^{``}\}\), which might be empty for an \(S^{``}\), but that would not matter.
\(\{f (S^{``}) \vert S^{``} \subseteq S^`\}\) would be an open cover of \(T\), because as \(S^`\) would have no \(\omega\)-accumulation point, each point, \(t \in T\), would have at least 1 open neighborhood of \(t\), \(U_t \subseteq T\), such that \(U_t \cap S^`\) would have only finite points, so, for \(S^{``} := U_t \cap S^`\), \(t \in U_t \subseteq f (S^{``})\).
\(\{f (S^{``}) \vert S^{``} \subseteq S^`\}\) would be a countable open cover of \(T\), because the set of the finite subsets of any countable set would be countable: for any countable set, \(S^` = \{1, 2, ...\}\), choose the finite subsets 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 \(\{3\}\), \(\{1, 3\}\), \(\{2, 3\}\), and \(\{1, 2, 3\}\), and so on, for example.
There would be a finite subcover, \(\{f (S^{``}_j) \vert j \in J\}\), where \(J\) would be a finite index set, because \(T\) was countably compact.
There would be a point, \(t \in S^` \setminus \cup_{j \in J} S^{``}_j\), because \(S^`\) would be infinite while \(\cup_{j \in J} S^{``}_j\) would be finite.
But \(t \notin f (S^{``}_j)\) for each \(j \in J\), because \(f (S^{``}_j) \cap S^` = \cup \{U \in \{\text{ the open subsets of } T\} \vert U \cap S^` = S^{``}_j\} \cap S^` = \cup \{U \cap S^` \vert U \in \{\text{ the open subsets of } T\} \text{ such that } U \cap S^` = S^{``}_j\}\), by the proposition that for any set, the intersection of the union of any possibly uncountable number of subsets and any subset is the union of the intersections of each of the subsets and the latter subset, \(= \cup \{S^{``}_j \vert U \in \{\text{ the open subsets of } T\} \text{ such that } U \cap S^` = S^{``}_j\} = S^{``}_j\), a contradiction against \(\{f (S^{``}_j) \vert j \in J\}\)'s being a cover of \(T\).
So, there is no infinite subset that has no \(\omega\)-accumulation point, so, each infinite subset has a \(\omega\)-accumulation point.
