2025-06-29

1178: For Topological Space and Subset of Subspace, if Subspace Is Closed, Closure of Subset on Subspace Is Closure on Base Space

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description/proof of that for topological space and subset of subspace, if subspace is closed, closure of subset on subspace is closure on base space

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, if the subspace is closed on the base space, the closure of the subset on the subspace is the closure of the subset 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: { the topological spaces }
T: { the topological subspaces of T}
S: T
//

Statements:
T{ the closed subspaces of T}

ST=ST, 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=STT; Step 2: see that ST=ST.

Step 1:

ST=STT, 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.

STT is closed on T as an intersection of closed subsets.

Step 2:

ST=ST, by the proposition that for any topological space and any subset of any subspace, if the closure of the subset on the subspace is closed on the base space, the closure is the closure of the subset on the base space.


References


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