description/proof of that for infinite product topological space and closed subset, point on product space whose each finite-components-projection belongs to corresponding projection of subset belongs to subset
Topics
About: topological space
The table of contents of this article
- Starting Context
- Target Context
- Orientation
- Main Body
- 1: Structured Description
- 2: Natural Language Description
- 3: Note 1
- 4: Proof
- 5: Note 2
Starting Context
- The reader knows a definition of product topological space.
Target Context
- The reader will have a description and a proof of the proposition that for any infinite product topological space and any closed subset, any point on the product space whose each finite-components-projection belongs to the corresponding projection of the subset belongs to the subset.
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:
\(A\): \(\in \{\text{ the possibly uncountable infinite index sets }\}\)
\(\{T_\alpha \vert \alpha \in A\}\): \(T_\alpha \in \{\text{ the topological spaces }\}\)
\(T\): \(= \times_{\alpha \in A} T_\alpha\) with the product topology
\(C\): \(\subseteq T\), \(\in \{\text{ the closed subsets }\}\)
\(p\): \(\in T\)
//
Statements:
\(\forall J \subset A, J \in \{\text{ the finite index sets }\} (\pi_{J} (p) \in \pi_{J} (C))\), where \(\pi_{J}: T \to \times_{j \in J} T_j\) is the projection
\(\implies\)
\(p \in C\)
//
2: Natural Language Description
For any possibly uncountable infinite index set, \(A\), any topological spaces, \(\{T_\alpha \vert \alpha \in A\}\), the product topological space, \(T := \times_{\alpha \in A} T_\alpha\), any closed \(C \subseteq T\), and any \(p \in T\), \(\pi_{J} (p) \in \pi_{J} (C)\) for each \(J \subset A\), where \(J\) is a finite index set and \(\pi_{J}: T \to \times_{j \in J} T_j\) is the projection, implies that \(p \in C\).
3: Note 1
A typical case is \(T = T_1 \times T_2 \times ...\), a infinitely countable product, and \(\pi_{J} (p) \in \pi_{J} (C)\) where \(J = \{1, ..., k\}\) for each \(k\) implies \(p \in C\): the requirement for such \(J\)s implies the requirement for each \(J' \subseteq J\), so, implies the requirement for each finite \(J' \subseteq A\).
4: Proof
Let us suppose that \(\pi_{J} (p) \in \pi_{J} (C)\) for each \(J\).
\(T \setminus C \subseteq T\) is open.
Let us suppose that \(p \notin C\).
\(p \in T \setminus C\). There would be an open neighborhood, \(U_p \subseteq T \setminus C\), of \(p\). \(U_p = \cup_{\beta \in B} \times_{\alpha \in A} U_{\beta, \alpha}\), where \(B\) would be a possibly uncountable index set and \(U_{\beta, \alpha} \subseteq T_\alpha\) would be open while only finite number of \(U_{\beta, \alpha}\) s would not be \(T_\alpha\) s for each \(\beta\) (see Note of the definition of product topology).
There would be a \(\beta\) such that \(p \in \times_{\alpha \in A} U_{\beta, \alpha} \subseteq T \setminus C\). As only finite of \(U_{\beta, \alpha}\) s would not be \(T_\alpha\) s, let \(J\) be of the finite components. \(\pi_J (p) \in \pi_J (\times_{\alpha \in A} U_{\beta, \alpha}) = \times_{j \in J} U_{\beta, j}\). Then, there would be a \(p' \in C\) such that \(\pi_J (p') = \pi_J (p) \in \times_{j \in J} U_{\beta, j}\). But \(p' \in \times_{\alpha \in A} U_{\beta, \alpha}\), because \(U_{\beta, \alpha} = T_\alpha\) for each \(\alpha \notin J\) and so, \(p'^\alpha \in U_{\beta, \alpha}\) when \(\alpha \in J\) and when \(\alpha \in A \setminus J\).
So, \(p' \in \times_{\alpha \in A} U_{\beta, \alpha} \subseteq \cup_{\beta \in B} \times_{\alpha \in A} U_{\beta, \alpha} = U_p \subseteq T \setminus C\), which would mean \(p' \notin C\), a contradiction.
So, \(p \in C\).
5: Note 2
When \(C \subseteq T\) is not closed, \(p \in C\) is not necessarily implied as is proved in another article.
