description/proof of that for topological space and subset of subspace, closure of subset on subspace is intersection of closure of subset on base space and subspace
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of closure of subset of topological space.
- The reader knows a definition of subspace topology of subset of topological space.
Target Context
- The reader will have a description and a proof of the proposition that for any topological space and any subset of any subspace, the closure of the subset on the subspace is the intersection of the closure of the subset on the base space and the subspace.
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:
\(T'\): \(\in \{\text{ the topological spaces }\}\)
\(T\): \(\in \{\text{ the topological subspaces of } T'\}\)
\(S\): \(\subseteq T\)
//
Statements:
\(\overline{S}^T = \overline{S}^{T'} \cap T\), where \(\overline{S}^T\) is the closure of \(S\) on \(T\) and \(\overline{S}^{T'}\) is the closure of \(S\) on \(T'\)
//
2: Proof
Whole Strategy: Step 1: see that \(\overline{S}^{T'} = \cap_{j \in J} C'_j\) where \(\{C'_j \vert j \in J\}\) is the set of the closed subsets of \(T'\) that contain \(S\); Step 2: see that \(\overline{S}^{T'} \cap T = \cap_{j \in J} C'_j \cap T\) is the intersection of the closed subsets of \(T\) that contain \(S\).
Step 1:
\(\overline{S}^{T'}\) is the intersection of the closed subsets of \(T'\) that contain \(S\).
So, letting \(\{C'_j \vert j \in J\}\) be the set of the closed subsets of \(T'\) that contain \(S\) where \(J\) is a possibly uncountable index set, \(\overline{S}^{T'} = \cap_{j \in J} C'_j\).
Step 2:
\(\overline{S}^{T'} \cap T = (\cap_{j \in J} C'_j) \cap T = \cap_{j \in J} (C'_j \cap T)\).
\(C'_j \cap T\) is a closed subset of \(T\) by the definition of subspace topology and \(S \subseteq C'_j \cap T\), and any closed subset of \(T\) that contains \(S\), \(C\), appears as \(C = C'_j \cap T\) for a \(j\), because there is a closed subset of \(T'\), \(C' \subseteq T'\), such that \(C = C' \cap T\), but \(S \subseteq C \subseteq C'\), so, \(C'\) is a closed subset of \(T'\) that contains \(S\), so, \(C' = C'_j\) for a \(j\).
So, \(\cap_{j \in J} (C'_j \cap T)\) is the intersection of all the closed subsets of \(T\) that contain \(S\), which is \(\overline{S}^{T}\).
So, \(\overline{S}^{T} = \overline{S}^{T'} \cap T\).
