description/proof of that for infinite product topological space and subset, point on product space whose each finite-components-projection belongs to corresponding projection of subset does not necessarily belong 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 an infinite product topological space and a subset, a point on the product space whose each finite-components-projection belongs to the corresponding projection of the subset does not necessarily belong 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
\(S\): \(\subseteq T\)
\(p\): \(\in T\)
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Statements:
\(\forall J \subset A, J \in \{\text{ the finite index sets }\} (\pi_{J} (p) \in \pi_{J} (S))\) does not necessarily imply \(p \in S\), where \(\pi_{J}: T \to \times_{j \in J} T_j\) is the projection.
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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 \(S \subseteq T\), and any \(p \in T\), \(\pi_{J} (p) \in \pi_{J} (S)\) 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, does not necessarily imply that \(p \in S\).
3: Note 1
A typical case is \(T = T_1 \times T_2 \times ...\), a infinitely countable product, and \(\pi_{J} (p) \in \pi_{J} (S)\) where \(J = \{1, ..., k\}\) for each \(k\) does not imply \(p \in S\).
4: Proof
A counterexample suffices.
Let each \(T_\alpha\) have at least 2 points, \(p_{\alpha, 1}, p_{\alpha, 2} \in T_\alpha\), let \(S = T \setminus \times_{\alpha \in A} \{p_{\alpha, 1}\}\), and let each \(\alpha\) component of \(p\) be \(p_{\alpha, 1}\).
\(p \in T\). \(\pi_{J} (p) \in \pi_{J} (S)\) for each \(J\), because \(S\) has at least 1 point, \(p'\), whose \(J\) components are \(p_{j, 1}\) s (\(j \in J\)) but whose \(\alpha\) component is \(p_{\alpha, 2}\) where \(\alpha \notin J\); \(p' \in S\), because \(p_{\alpha, 2} \notin \{p_{\alpha, 1}\}\).
But \(p \notin S\), because \(p \in \times_{\alpha \in A} \{p_{\alpha, 1}\}\).
Letting \(A\) be countable does not help. Let \(A = \{1, 2, ...\}\), let each \(T_j\) have at least 2 points, \(p_{j, 1}, p_{j, 2} \in T_j\), let \(S = T \setminus (\{p_{1, 1}\} \times \{p_{2, 1}\} \times ...)\), and let \(p\) be \((p_{1, 1}, p_{2, 1}, ...)\).
\(p \in T\). \(\pi_{J} (p) \in \pi_{J} (S))\) for each \(J\), because, for example, when \(J = \{2, 4\}\), \((p_{2, 2}, p_{2, 1}, p_{3, 1}, p_{4, 1}, ...), (p_{1, 1}, p_{2, 1}, p_{3, 2}, p_{4, 1}, ...), ... \in S\), while \(\pi_{J} (p) = \pi_{J} ((p_{2, 2}, p_{2, 1}, p_{3, 1}, p_{4, 1}, ...))) = (p_{2, 1}, p_{4, 1})\).
But \(p \notin S\), because \(p \in \{p_{1, 1}\} \times \{p_{2, 1}\} \times ...\).
5: Note 2
When \(S \subseteq T\) is closed, \(p \in S\) is implied as is proved in another article.
