1176: For Topological Space and Subset of Subspace, if Its Closure on Subspace Is Closed on Base Space, the Closure Is Closure on Base Space

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

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Topics

About: topological space

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Proof

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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 knows a definition of closed subset of topological space.

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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 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.

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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.

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Main Body

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1: Structured Description

Here is the rules of Structured Description.

Entities:
$T^{\prime }$: $\in \{\text{ the topological spaces }\}$
$T$: $\in \{\text{ the topological subspaces of }{T}^{\prime }\}$
$S$: $\subseteq T$
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Statements:
${\bar{S}}^T\in \{\text{ the closed subsets of }{T}^{\prime }\}$, where ${\bar{S}}^T$ denotes the closure of $S$ on $T$
$⟹$
${\bar{S}}^T={\bar{S}}^{T^{\prime }}$, where ${\bar{S}}^{T^{\prime }}$ denotes the closure of $S$ on $T^{\prime }$
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2: Proof

Whole Strategy: Step 1: see that ${\bar{S}}^{T^{\prime }}\subseteq {\bar{S}}^T$; Step 2: see that ${\bar{S}}^T\subseteq {\bar{S}}^{T^{\prime }}$; Step 3: conclude the proposition.

Step 1:

${\bar{S}}^{T^{\prime }}\subseteq {\bar{S}}^T$, because ${\bar{S}}^{T^{\prime }}$ is the smallest closed subset of $T^{\prime }$ that (the closed subset) contains $S$ while ${\bar{S}}^T$ is a closed subset of $T^{\prime }$ that (the closed subset) contains $S$, by the supposition.

Step 2:

$S\subseteq {\bar{S}}^{T^{\prime }}\cap T$, which is closed on $T$, so, ${\bar{S}}^T\subseteq {\bar{S}}^{T^{\prime }}\cap T$, because ${\bar{S}}^T$ is the smallest closed subset of $T$ that (the closed subset) contains $S$ while ${\bar{S}}^{T^{\prime }}\cap T$ is a closed subset of $T$ that (the closed subset) contains $S$, so, ${\bar{S}}^T\subseteq {\bar{S}}^{T^{\prime }}$.

Step 3:

So, ${\bar{S}}^T={\bar{S}}^{T^{\prime }}$.

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References

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