description/proof of that for topological sum, constituent is topological subspace of topological sum
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of topological sum.
- The reader knows a definition of topological subspace.
Target Context
- The reader will have a description and a proof of the proposition that for any topological sum, each constituent is the topological subspace of the topological sum.
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:
\(J\): \(\in \{\text{ the possibly uncountable index sets }\}\)
\(\{T_j \in \{\text{ the topological spaces }\} \vert j \in J\}\):
\(\coprod_{j \in J} T_j\): \(= \text{ the topological sum }\)
//
Statements:
\(\forall j \in J (T_j \in \{\text{ the topological subspaces of } \coprod_{j \in J} T_j\})\)
//
2: Proof
Whole Strategy: Step 1: see that each \(S \subseteq T_j\) is open if and only if \(S\) is the intersection of an open \(U' \subseteq \coprod_{j \in J} T_j\) and \(T_j\).
Step 1:
Let \(S \subseteq T_j\) be any.
Let us suppose that \(S\) is open.
Let us take \(U' := S \subseteq \coprod_{j \in J} T_j\).
\(U'\) is open, because \(U' \cap T_j = S \subseteq T_j\) is open and \(U' \cap T_l = \emptyset \subseteq T_l\) for each \(l \in J \setminus \{j\}\) is open, by the definition of topological sum.
\(S = U' \cap T_j\).
So, each subset of \(T_j\) is open only if it is the intersection of an open subset of \(\coprod_{j \in J} T_j\) and \(T_j\).
Let \(U' \subseteq \coprod_{j \in J} T_j\) be any open subset.
\(U' \cap T_j \subseteq T_j\) is open, by the definition of topological sum.
So, each subset of \(T_j\) is open if it is the intersection of an open subset of \(\coprod_{j \in J} T_j\) and \(T_j\).
So, each subset of \(T_j\) is open if and only if it is the intersection of an open subset of \(\coprod_{j \in J} T_j\) and \(T_j\).
That means that \(T_j\) is the topological subspace of \(\coprod_{j \in J} T_j\).
