description/proof of that for product topological space and constituent s.t. other constituents are compact, projection onto constituent is closed
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 topological space.
- The reader knows a definition of projection from product set onto subproduct set.
- The reader knows a definition of closed map.
- The reader admits the proposition that for any product topological space and any partition of the index set, the product space is homeomorphic to the product of the subproduct topological spaces with the divided index sets.
- The reader admits the proposition that for any product topological space and any neighborhood of any point, there is an open neighborhood of the point contained in the neighborhood as the product of some open neighborhoods of the components of the point.
- The reader admits the proposition that the product of any possibly uncountable number of compact topological spaces is compact (the Tychonoff theorem).
- The reader admits the local criterion for openness.
Target Context
- The reader will have a description and a proof of the proposition that for any product topological space and any constituent such that the other constituents are compact, the projection onto the constituent is closed.
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:
\(J\): \(\in \{\text{ the possibly uncountable index sets }\}\)
\(\{T_j \in \{\text{ the topological spaces }\} \vert j \in J\}\):
\(\times_{j \in J} T_j\): \(= \text{ the product topological space }\)
\(j_0\): \(\in J\)
\(\pi^{j_0}\): \(: \times_{j \in J} T_j \to T_{j_0}\), \(= \text{ the projection }\)
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Statements:
\(\forall j \in J \setminus \{j_0\} (T_j \in \{\text{ the compact topological spaces }\})\)
\(\implies\)
\(\pi^{j_0} \in \{\text{ the closed maps }\}\)
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2: Proof
Whole Strategy: Step 1: take any closed \(C \subseteq \times_{j \in J} T_j\) and see that \(T_{j_0} \setminus \pi^{j_0} (C)\) is open using homeomorphic \(f: \times_{j \in J} T_j \to (\times_{j \in J \setminus \{j_0\}} T_j) \times T_{j_0}\).
Step 1:
Let \(C \subseteq \times_{j \in J} T_j\) be any closed subset.
Let us see that \(T_{j_0} \setminus \pi^{j_0} (C) \subseteq T_{j_0}\) is open.
Let \(t_{j_0} \in T_{j_0} \setminus \pi^{j_0} (C)\) be any.
Let us take \(f: \times_{j \in J} T_j \to (\times_{j \in J \setminus \{j_0\}} T_j) \times T_{j_0}, \times_{j \in J} t_j \mapsto (\times_{j \in J \setminus \{j_0\}} t_j) \times t_{j_0}\), which is a homeomorphism, by the proposition that for any product topological space and any partition of the index set, the product space is homeomorphic to the product of the subproduct topological spaces with the divided index sets.
For each \(s \in \times_{j \in J \setminus \{j_0\}} T_j\), \((s, t_{j_0}) \notin f (C)\), and so, \((s, t_{j_0}) \in ((\times_{j \in J \setminus \{j_0\}} T_j) \times T_{j_0}) \setminus f (C)\), because if there was a \((s, t_{j_0}) \in f (C)\), \(f^{-1} ((s, t_{j_0})) \in C\), and \(\pi^{j_0} (f^{-1} ((s, t_{j_0}))) = t_{j_0} \in \pi^{j_0} (C)\), a contradiction.
\(f (C) \subseteq (\times_{j \in J \setminus \{j_0\}} T_j) \times T_{j_0}\) is closed, because \(f\) is a homeomorphism.
So, \(((\times_{j \in J \setminus \{j_0\}} T_j) \times T_{j_0}) \setminus f (C) \subseteq (\times_{j \in J \setminus \{j_0\}} T_j) \times T_{j_0}\) is open.
For each \(s \in \times_{j \in J \setminus \{j_0\}} T_j\), there is an open neighborhood of \((s, t_{j_0})\), \(U_{(s, t_{j_0})} \subseteq (\times_{j \in J \setminus \{j_0\}} T_j) \times T_{j_0}\), such that \(U_{(s, t_{j_0})} \subseteq ((\times_{j \in J \setminus \{j_0\}} T_j) \times T_{j_0}) \setminus f (C)\), by the local criterion for openness, and there are an open neighborhood of \(s\), \(U_s \subseteq \times_{j \in J \setminus \{j_0\}} T_j\) and an open neighborhood of \(t_{j_0}\), \(U_{t_{j_0}, s} \subseteq T_{j_0}\), such that \(U_s \times U_{t_{j_0}, s} \subseteq U_{(s, t_{j_0})} \subseteq ((\times_{j \in J \setminus \{j_0\}} T_j) \times T_{j_0}) \setminus f (C)\), by the proposition that for any product topological space and any neighborhood of any point, there is an open neighborhood of the point contained in the neighborhood as the product of some open neighborhoods of the components of the point.
\(\times_{j \in J \setminus \{j_0\}} T_j\) is compact, by the proposition that the product of any possibly uncountable number of compact topological spaces is compact (the Tychonoff theorem).
\(\{U_s \vert s \in \times_{j \in J \setminus \{j_0\}} T_j\}\) is an open cover of \(\times_{j \in J \setminus \{j_0\}} T_j\).
There is a finite subcover, \(\{U_{s_l} \vert l \in L\}\).
Let \(U_{t_{j_0}} := \cap_{l \in L} U_{j_0, s_l}\), which is an open neighborhood of \(t_{j_0}\) as a finite intersection.
\(U_{t_{j_0}} \subseteq T_{j_0} \setminus \pi^2 (f (C))\), where \(\pi^2: (\times_{j \in J \setminus \{j_0\}} T_j) \times T_{j_0} \to T_{j_0}\) is the projection, because for each \(u \in U_{t_{j_0}}\), for each \(s \in \times_{j \in J \setminus \{j_0\}} T_j\), \(s \in U_{s_l}\) for an \(l \in L\), \(u \in U_{t_{j_0}} \subseteq U_{t_{j_0}, s_l}\), and \(U_{s_l} \times U_{t_{j_0}, s_l} \subseteq ((\times_{j \in J \setminus \{j_0\}} T_j) \times T_{j_0}) \setminus f (C)\), so, \(U_{s_l} \cap U_{t_{j_0}} \subseteq ((\times_{j \in J \setminus \{j_0\}} T_j) \times T_{j_0}) \setminus f (C)\), so, \((s, u) \in ((\times_{j \in J \setminus \{j_0\}} T_j) \times T_{j_0}) \setminus f (C)\), which means that \(u \notin \pi^2 (f (C))\), so, \(u \in T_{j_0} \setminus \pi^2 (f (C))\).
But \(\pi^2 (f (C)) = \pi^{j_0} (C)\), because for each \(\pi^2 (f (c)) = c^{j_0} \in \pi^2 (f (C))\), \(c^{j_0} = \pi^{j_0} (c) \in \pi^{j_0} (C)\); for each \(\pi^{j_0} (c) = c^{j_0} \in \pi^{j_0} (C)\), \(c^{j_0} = \pi^2 (f (c)) \in \pi^2 (f (C))\).
So, \(U_{t_{j_0}} \subseteq T_{j_0} \setminus \pi^{j_0} (C)\).
So, \(T_{j_0} \setminus \pi^{j_0} (C) \subseteq T_{j_0}\) is open, by the local criterion for openness.
So, \(\pi^{j_0} (C) \subseteq T_{j_0}\) is closed.
So, \(\pi^{j_0}\) is closed.
