1182: For Topological Space and Subset of Subspace, Closure of Subset on Subspace Is Contained in Closure of Subset on Base Space

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description/proof of that for topological space and subset of subspace, closure of subset on subspace is contained in closure of subset 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: Note
• 3: Proof

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Starting Context

• The reader knows a definition of closure of subset of topological space.
• The reader admits the proposition that any subset on any topological subspace is closed if and only if there is a closed set on the base space whose intersection with the subspace is the subset.

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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, the closure of the subset on the subspace is contained in 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:
${\bar{S}}^T\subseteq {\bar{S}}^{T^{\prime }}$, where ${\bar{S}}^T$ is the closure of $S$ on $T$ and ${\bar{S}}^{T^{\prime }}$ is the closure of $S$ on $T^{\prime }$
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2: Note

The equality does not necessarily hold.

For example, let $T^{\prime }=\mathbb{R}$ with the Euclidean topology, $T=(-1,1)$, and $S=(-1,1)$, then, ${\bar{S}}^T=(-1,1)\subset [-1,1]={\bar{S}}^{T^{\prime }}$.

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3: Proof

Whole Strategy: Step 1: see that $S\subseteq {\bar{S}}^{T^{\prime }}\cap T$ and ${\bar{S}}^{T^{\prime }}\cap T$ is closed on $T$, and see that ${\bar{S}}^T\subseteq {\bar{S}}^{T^{\prime }}\cap T\subseteq {\bar{S}}^{T^{\prime }}$.

Step 1:

$S\subseteq {\bar{S}}^{T^{\prime }}$.

As $S\subseteq T$, $S\subseteq {\bar{S}}^{T^{\prime }}\cap T$.

As ${\bar{S}}^{T^{\prime }}$ is closed on $T^{\prime }$, ${\bar{S}}^{T^{\prime }}\cap T$ is closed on $T$, by the proposition that any subset on any topological subspace is closed if and only if there is a closed set on the base space whose intersection with the subspace is the subset.

As ${\bar{S}}^T$ is the intersection of all the closed subsets of $T$ that contain $S$, ${\bar{S}}^T\subseteq {\bar{S}}^{T^{\prime }}\cap T$.

But ${\bar{S}}^T\subseteq {\bar{S}}^{T^{\prime }}\cap T\subseteq {\bar{S}}^{T^{\prime }}$.

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References

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