2025-06-29

1177: For Topological Space and Subset of Subspace, Closure of Subset on Subspace Is Intersection of Closure of Subset on Base Space and Subspace

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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



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: { the topological spaces }
T: { the topological subspaces of T}
S: T
//

Statements:
ST=STT, where ST is the closure of S on T and ST is the closure of S on T
//


2: Proof


Whole Strategy: Step 1: see that ST=jJCj where {Cj|jJ} is the set of the closed subsets of T that contain S; Step 2: see that STT=jJCjT is the intersection of the closed subsets of T that contain S.

Step 1:

ST is the intersection of the closed subsets of T that contain S.

So, letting {Cj|jJ} be the set of the closed subsets of T that contain S where J is a possibly uncountable index set, ST=jJCj.

Step 2:

STT=(jJCj)T=jJ(CjT).

CjT is a closed subset of T by the definition of subspace topology and SCjT, and any closed subset of T that contains S, C, appears as C=CjT for a j, because there is a closed subset of T, CT, such that C=CT, but SCC, so, C is a closed subset of T that contains S, so, C=Cj for a j.

So, jJ(CjT) is the intersection of all the closed subsets of T that contain S, which is ST.

So, ST=STT.


References


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