description/proof of that for topological space that is union of finite number of subspaces, subset of intersection of subspaces that is open on each subspace is open on base space
Topics
About: topological space
The table of contents of this article
Starting Context
- 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 space that is the union of any finite number of subspaces, any subset of the intersection of the subspaces that is open on each subspace is open on the base space.
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:
\(T\): \(\in \{\text{ the topological spaces }\}\)
\(J\): \(\in \{\text{ the finite index sets }\}\)
\(\{T_j \subseteq T \text{ with the subspace topology } \vert j \in J\}\): such that \(\cup_{j \in J} T_j = T\)
\(S\): \(\subseteq \cap_{j \in J} T_j\)
//
Statements:
\(\forall j \in J (S \in \{\text{ the open subsets of } T_j\})\)
\(\implies\)
\(S \in \{\text{ the open subsets of } T\}\)
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2: Proof
Whole Strategy: Step 1: see that for each \(j \in J\), \(S = U_j \cap T_j\) for an open \(U_j \subseteq T\), and see that \(S = \cap_{j \in J} U_j\).
Step 1:
For each \(j \in J\), \(S = U_j \cap T_j\) where \(U_j \subseteq T\) is open, by the definition of subspace topology.
Let us see that \(S = \cap_{j \in J} U_j\).
For each \(s \in S\), \(s \in U_j\) for each \(j \in J\), so, \(s \in \cap_{j \in J} U_j\).
So, \(S \subseteq \cap_{j \in J} U_j\).
For each \(t \in \cap_{j \in J} U_j\), as \(T = \cup_{j \in J} T_j\), \(t \in T_j\) for a \(j \in J\), so, \(t \in T_j \cap U_j\) for that \(j\), so, \(t \in U_j \cap T_j = S\).
So, \(\cap_{j \in J} U_j \subseteq S\).
So, \(S = \cap_{j \in J} U_j\).
\(S = \cap_{j \in J} U_j\) is open on \(T\) as a finite intersection of open subsets.
