description/proof of that finite product of open maps is open
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of product map.
- The reader knows a definition of product topology.
- The reader knows a definition of open map.
- The reader admits the proposition that for any map, the map image of any union of sets is the union of the map images of the sets.
- The reader admits the proposition that for any product map, the image of any product subset is the product of the images of the component subsets under the component maps.
Target Context
- The reader will have a description and a proof of the proposition that any finite product of open maps is open.
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:
\(J\): \(\in \{\text{ the finite index sets }\}\)
\(\{T_{1, j} \vert j \in J\}\): \(T_{1, j} \in \{\text{ the topological spaces }\}\)
\(\{T_{2, j} \vert j \in J\}\): \(T_{2, j} \in \{\text{ the topological spaces }\}\)
\(\{f_j \vert j \in J\}\): \(f_j: T_{1, j} \to T_{2, j} \in \{\text{ the open maps }\}\)
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Statements:
\(\times_{j \in J} f_j \in \{\text{ the open maps }\}\)
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2: Note
This proposition requires that \(J\) is finite, because of the reason mentioned in Proof.
3: Proof
Whole Strategy: Step 1: take any open subset, \(U \subseteq \times_{j \in J} T_{1, j}\), and see that \(U = \cup_{j' \in J'} \times_{j \in J} U_{1, j, j'}\); Step 2: see that \((\times_{j \in J} f_j) (U) = \cup_{j' \in J'} \times_{j \in J} f_j (U_{1, j, j'})\).
Step 1:
Let \(U \subseteq \times_{j \in J} T_{1, j}\) be any open subset.
\(U = \cup_{j' \in J'} \times_{j \in J} U_{1, j, j'}\) where \(J'\) is a possibly uncountable index set and \(U_{1, j, j'} \subseteq T_{1, j} \) is open: refer to Note for the definition of product topology.
Step 2:
\((\times_{j \in J} f_j) (U) = (\times_{j \in J} f_j) (\cup_{j' \in J'} \times_{j \in J} U_{1, j, j'}) = \cup_{j' \in J'} (\times_{j \in J} f_j) (\times_{j \in J} U_{1, j, j'})\), by the proposition that for any map, the map image of any union of sets is the union of the map images of the sets, \(= \cup_{j' \in J'} \times_{j \in J} f_j (U_{1, j, j'})\), by the proposition that for any product map, the image of any product subset is the product of the images of the component subsets under the component maps.
As each \(f_j\) is open, each \(f_j (U_{1, j, j'}) \subseteq T_{2, j}\) is open.
So, \(\cup_{j' \in J'} \times_{j \in J} f_j (U_{1, j, j'})\) is open: refer to Note for the definition of product topology.
Note that this proposition requires \(J\) to be finite, because otherwise, for each fixed \(j'\), while only finite of \(f_j (U_{1, j, j'})\) s were allowed to be not \(T_{2, j}\), which would not be guaranteed: the fact that only finite of \(U_{1, j, j'}\) s were not \(T_{1, j}\) would not guaranteed the requirement, because \(f_j\) s were not necessarily surjective.
So, \(\times_{j \in J} f_j\) is open.
