description/proof of that for topological space and disjoint open subsets, union of subsets as subspace is disjoint union topological space of subset subspaces
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of topological subspace.
- The reader knows a definition of disjoint union topology.
- The reader admits the proposition that for any set, the intersection of the union of any possibly uncountable number of subsets and any subset is the union of the intersections of each of the subsets and the latter subset.
- The reader admits the proposition that for any topological space and any topological subspace that is not necessarily open on the basespace, any subset of the subspace is open on the subspace if it is open on the basespace.
Target Context
- The reader will have a description and a proof of the proposition that for any topological space and any disjoint open subsets, the union of the subsets as the subspace is the disjoint union topological space of the subset subspaces.
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:
\(T'\): \(\in \{\text{ the topological spaces }\}\)
\(J\): \(\in \{\text{ the possibly uncountable index sets }\}\)
\(\{S_j \subseteq T' \vert j \in J\}\): such that for each \(j, j' \in J\) such that \(j \neq j'\), \(S_j \cap S_{j'} = \emptyset\)
\(T\): \(= \cup_{j \in J} S_j \subseteq T'\) as the topological subspace
\(\coprod_{j \in J} S_j\): \(= \text{ the disjoint union topological space }\)
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Statements:
\(\forall j \in J (S_j \in \{\text{ the open subsets of } T'\})\)
\(\implies\)
\(T = \coprod_{j \in J} S_j\)
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2: Note
Compare with the proposition that for any topological space and any disjoint subsets, the union of the subsets as the subspace is not necessarily the disjoint union topological space of the subset subspaces.
3: Proof
Whole Strategy: Step 1: let \(S \subseteq T\) be any open subset of \(T\), and see that \(S\) is open on \(\coprod_{j \in J} S_j\); Step 2: let \(S \subseteq \coprod_{j \in J} S_j\) be any open subset of \(\coprod_{j \in J} S_j\), and see that \(S\) is open on \(T\).
Step 1:
Let \(S \subseteq T\) be any open subset of \(T\).
\(S = U' \cap T\) where \(U' \subseteq T'\) is an open subset of \(T'\).
For each \(j \in J\), \(S \cap S_j = U' \cap T \cap S_j = U' \cap S_j\), which is open on \(S_j\).
So, \(S\) is open on \(\coprod_{j \in J} S_j\).
Note that this Step does not need \(S_j\) to be open on \(T'\).
Step 2:
Let \(S \subseteq \coprod_{j \in J} S_j\) be any open subset of \(\coprod_{j \in J} S_j\).
For each \(j \in J\), \(S \cap S_j\) is open on \(S_j\).
So, \(S \cap S_j = U'_j \cap S_j\) for an open \(U'_j \subseteq T'\).
\(S = S \cap T = S \cap (\cup_{j \in J} S_j) = \cup_{j \in J} (S \cap S_j)\), by the proposition that for any set, the intersection of the union of any possibly uncountable number of subsets and any subset is the union of the intersections of each of the subsets and the latter subset, \(= \cup_{j \in J} (U'_j \cap S_j)\), which is open on \(T'\) because \(S_j\) is open on \(T'\).
So, \(S\) is open on \(T\), by the proposition that for any topological space and any topological subspace that is not necessarily open on the basespace, any subset of the subspace is open on the subspace if it is open on the basespace.
Note that this Step needs \(S_j\) to be open on \(T'\).
