description/proof of that for topological space and disjoint subsets, union of subsets as subspace is not necessarily 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.
Target Context
- The reader will have a description and a proof of 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.
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:
Not necessarily \(T = \coprod_{j \in J} S_j\)
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The reason is not that \(J\) is not necessarily finite.
2: Note
Compare with 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.
3: Proof
Whole Strategy: Step 1: see a counter example.
Step 1:
Let us see a counterexample.
Let \(T' = \mathbb{R}\) as the Euclidean topological space and \(\{S_j \subseteq T' \vert j \in J\} = \{S_1 = [-1, 0), S_2 = [0, 1]\}\).
\(T = [-1, 1]\).
Let us think of \(S = [0, 1)\).
\(S \cap S_1 = \emptyset\), open on \(S_1\); \(S \cap S_2 = [0, 1)\), open on \(S_2\), because \(S \cap S_2 = (-1, 1) \cap S_2\), where \((-1, 1)\) is open on \(T'\).
So, \(S\) is open on \(\coprod_{j \in J} S_j\).
But \(S\) is not open on \(T\), because there is no open neighborhood of \(0 \in S\) on \(T\) that is contained in \(S\).
So, \(T \neq \coprod_{j \in J} S_j\).
