A description/proof of that finite product of compact topological spaces is compact
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of compact topological space.
- The reader knows a definition of product topology.
- The reader admits the proposition that the product of any finite number of topological spaces equals the sequential products of the topological spaces.
Target Context
- The reader will have a description and a proof of the proposition that the product of any finite number of compact topological spaces is 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
For any compact topological spaces, \(T_1, T_2, . . ., T_n\), the product topological space, \(T = T_1 \times T_2 \times . . . \times T_n\), is compact.
2: Proof
1st, let us think of the case, \(n = 2\).
For any open cover, \(\{U_\alpha \subseteq T \vert \alpha \in A\}\) where \(A\) is a possibly uncountable indices set, of \(T\), \(U_\alpha = \cup_{\beta \in B_\alpha} U_{1, \alpha, \beta} \times U_{2, \alpha, \beta}\) where \(B_\alpha\) is a possibly uncountable indices set that depends on \(\alpha\).
For each point, \(p \in T_1\), there is the subset, \(S_p := \{U_{1, \alpha, \beta} \times U_{2, \alpha, \beta} \vert (\alpha, \beta) \in C_p\}\} \subseteq \{U_{1, \alpha, \beta} \times U_{2, \alpha, \beta} \vert \alpha \in A, \beta \in B_\alpha\}\) where \(C_p\) is the possibly uncountable indices set, such that \(p \in U_{1, \alpha, \beta}\) for each \((\alpha, \beta) \in C_p\). \(S_{2, p} := \{U_{2, \alpha, \beta} \vert (\alpha, \beta) \in C_p\}\) covers \(T_2\).
For \(S_{2, p}\), there is a finite subcover, \(\{U_{2, \alpha, \beta} \vert (\alpha, \beta) \in D_p\}\) where \(D_p \subseteq C_p\) is a finite indices set, and there is corresponding \(\{U_{1, \alpha, \beta} \times U_{2, \alpha, \beta} \vert (\alpha, \beta) \in D_p\} \subseteq S_p\) for each \(p\).
Let us take \(U'_{1, p} := \cap_{(\alpha, \beta) \in D_p} U_{1, \alpha, \beta}\), open on \(T_1\) as a finite intersection of open sets. \(\{U'_{1, p} \vert p \in T_1\}\) covers \(T_1\) and there is a finite subcover, \(\{U'_{1, p} \vert p \in E\}\) where \(E \subseteq T_1\) is a finite subset.
Let us take \(F := \{\alpha \in A \vert \exists \beta \in B_\alpha ((\alpha, \beta) \in D_p, p \in E)\}\), which is a finite indices set, because \(E\) is finite and \(D_p\) is finite for each \(p\).
\(\{U_\alpha \vert \alpha \in F\}\) covers \(T_1 \times T_2\), because for any \((p^1, p^2) \in T_1 \times T_2\), \(p^1 \in U'_{1, p} \subseteq U_{1, \alpha, \beta}\) for a \(p \in E\) and each \((\alpha, \beta) \in D_p\) and \(p_2 \in U_{2, \alpha, \beta}\) for a \((\alpha, \beta) \in D_p\), because \(\{U_{2, \alpha, \beta} \vert (\alpha, \beta) \in D_p\}\) covers \(T_2\), so, \((p^1, p^2) \in U_{1, \alpha, \beta} \times U_{2, \alpha, \beta}\) for a \((\alpha, \beta) \in D_p\), while \(U_{1, \alpha, \beta} \times U_{2, \alpha, \beta} \subseteq U_\alpha\) where \(\alpha \in F\).
So, \(\{U_\alpha \vert \alpha \in F\}\) is a finite subcover of \(\{U_\alpha \vert \alpha \in A\}\).
So, \(T_1 \times T_2\) is compact.
For any general \(n\), \(T_1 \times T_2 \times . . . \times T_n = ( . . . (T_1 \times T_2) \times . . . ) \times T_n\), by the proposition that the product of any finite number of topological spaces equals the sequential products of the topological spaces, but \(T_1 \times T_2\) is compact by the above paragraph, then, \((T_1 \times T_2) \times T_3\) is compact by the above paragraph, and so on (we can use the induction principle to be exact). So, \(T_1 \times T_2 \times . . . \times T_n\) is compact.
