description/proof of that for topological space and subset of subspace, closure of subset on base space contained in subspace is closure on 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.
- The reader admits 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.
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 base space contained in the subspace is the closure of the subset on 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:
\(T'\): \(\in \{\text{ the topological spaces }\}\)
\(T\): \(\in \{\text{ the topological subspaces of } T'\}\)
\(S\): \(\subseteq T\)
\(\overline{S}^{T'}\): \(= \text{ the closure of } S \text{ on } T'\) such that \(\overline{S}^{T'} \subseteq T\)
\(\overline{S}^T\): \(= \text{ the closure of } S \text{ on } T\)
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Statements:
\(\overline{S}^{T'} = \overline{S}^T\)
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2: Proof
Whole Strategy: Step 1: apply 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.
Step 1:
By 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, \(\overline{S}^T = \overline{S}^{T'} \cap T = \overline{S}^{T'}\).
