A description/proof of that for product topological space, projection of compact subset is compact
Topics
About: topological space
The table of contents of this article
- Starting Context
- Target Context
- Orientation
- Main Body
- 1: Description 1
- 2: Proof 1
- 3: Description 2
- 4: Proof 2
Starting Context
- The reader knows a definition of product topology.
- The reader knows a definition of compact subset.
Target Context
- The reader will have a description and a proof of the proposition that for any product topological space, the projection of any compact subset into any constituent topological space is a compact subset of the constituent topological space.
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 1
For any possibly uncountable number of indexed topological spaces, \(\{T_\alpha \vert \alpha \in A\}\) where \(A\) is any possibly uncountable indices set, the product, \(T := \times_{\alpha \in A} T_\alpha\), and any compact subset, \(S \subseteq T\), the projection, \(\pi_{\alpha'} (S) \subseteq T_{\alpha'}\) where \(\pi_{\alpha'}: T \to T_{\alpha'}\), is a compact subset of \(T_{\alpha'}\).
2: Proof 1
For any \(p \in T\), let the \(\alpha\) component of \(p\) be denoted as \(p^\alpha\), which is \(\pi_\alpha (p) = p^\alpha\).
Let \(\{U_\beta \subseteq T_{\alpha'} \vert \beta \in B\}\) be any open cover of \(\pi_{\alpha'} (S)\) where \(B\) is any possibly uncountable indices set.
\(\{\times_{\alpha \in A} U_{\alpha, \beta} \subseteq T \vert \beta \in B\}\) where \(U_{\alpha', \beta} = U_\beta\) and \(U_{\alpha, \beta} = T_\alpha\) when \(\alpha \neq \alpha'\) is an open cover of \(S\), because for any point, \(p \in S\), \(p^{\alpha'} \in U_\beta = U_{\alpha', \beta}\) for a \(\beta \in B\), and \(p^\alpha \in T_\alpha = U_{\alpha, \beta}\) for any \(\alpha \neq \alpha'\) and the \(\beta\), so, \(p \in \times_{\alpha \in A} U_{\alpha, \beta}\) for the \(\beta\).
As \(S\) is a compact subset of \(T\), there is a finite open cover, \(\{\times_{\alpha \in A} U_{\alpha, \beta} \vert \beta \in J\}\) where \(J \subseteq B\) is a finite indices set.
\(\{U_\beta \subseteq T_{\alpha'} \vert \beta \in J\}\) is a finite subcover, because for any point, \(p^{\alpha'} \in \pi_{\alpha'} (S)\), it is the \(\alpha'\) component of a point, \(p \in S\), and \(p \in \times_{\alpha \in A} U_{\alpha, \beta}\) for a \(\beta \in J\), which means that \(p^{\alpha'} \in U_{\alpha', \beta} = U_\beta\).
3: Description 2
For any finite number of topological spaces, \(T_1, T_2, . . ., T_n\), the product, \(T = T_1 \times T_2 \times . . . \times T_n\), and any compact subset, \(S \subseteq T\), the projection, \(\pi_{j'} (S) \subseteq T_{j'}\) where \(\pi_{j'}: T \to T_{j'}\), is a compact subset of \(T_{j'}\).
4: Proof 2
For any \(p \in T\), let the \(j\) component of \(p\) be denoted as \(p^j\), which is \(\pi_j (p) = p^j\).
Let \(\{U_\beta \subseteq T_{j'} \vert \beta \in B\}\) be any open cover of \(\pi_{j'} (S)\) where \(B\) is any possibly uncountable indices set.
\(\{T_1 \times ... \times T_{j' - 1} \times U_\beta \times T_{j' + 1} \times ... \times T_n \subseteq T \vert \beta \in B\}\) is an open cover of \(S\), because for any point, \(p \in S\), \(p^{j'} \in U_\beta\) for a \(\beta \in B\), and \(p^j \in T_j\) for any \(j \neq j'\) and the \(\beta\), so, \(p \in T_1 \times ... \times T_{j' - 1} \times U_\beta \times T_{j' + 1} \times ... \times T_n\) for the \(\beta\).
As \(S\) is a compact subset of \(T\), there is a finite open cover, \(\{T_1 \times ... \times T_{j' - 1} \times U_\beta \times T_{j' + 1} \times ... \times T_n \vert \beta \in J\}\) where \(J \subseteq B\) is a finite indices set.
\(\{U_\beta \subseteq T_{j'} \vert \beta \in J\}\) is a finite subcover, because for any point, \(p^{j'} \in \pi_{j'} (S)\), it is the \(j'\) component of a point, \(p \in S\), and \(p \in T_1 \times ... \times T_{j' - 1} \times U_\beta \times T_{j' + 1} \times ... \times T_n\) for a \(\beta \in J\), which means that \(p^{j'} \in U_\beta\).
