2025-06-29

1179: For Disjoint Union Topological Space, Closure of Disjoint Union of Subsets Is Disjoint Union of Closures of Subsets

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description/proof of that for disjoint union topological space, closure of disjoint union of subsets is disjoint union of closures of subsets

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 disjoint union topological space, the closure of the disjoint union of any subsets is the disjoint union of the closures of the subsets.

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:
J: { the possibly uncountable index sets }
{Tj|jJ}: Tj{ the topological spaces }
T: =jJTj
{SjTj|jJ}:
//

Statements:
jJSj=jJSj, where the 1st overline is the closure on T and the 2nd overline is the closure on Tj
//


2: Proof


Whole Strategy: Step 1: see that jJSjjJSj; Step 2: see that jJSjjJSj.

Step 1:

jJSjjJSj, and the right hand side is closed on T, by the definition of disjoint union topology.

So, jJSjjJSj, by the definition of closure of subset of topological space.

Step 2:

Let pjJSj be any.

pSj for a j.

Let UpT be any open neighborhood of p.

UpTj is open on Tj, by the definition of disjoint union topology, and pUpTj.

So, UpTj is an open neighborhood of p on Tj.

UpTjSj, because pSj, by the proposition that the closure of any subset is the union of the subset and the accumulation points set of the subset.

So, UpjJSj.

So, pjJSj, by the proposition that the closure of any subset is the union of the subset and the accumulation points set of the subset.

So, jJSjjJSj.

So, αASα=αASα.


References


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