586: For Topological Space, Subspace, and Subset of Superspace, Subspace Minus Subset as Subspace of Subspace Is Subspace of Superspace Minus Subset

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description/proof of that for topological space, subspace, and subset of superspace, subspace minus subset as subspace of subspace is subspace of superspace minus subset

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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: Natural Language Description
• 3: Proof

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Starting Context

• The reader knows a definition of topological subspace.

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Target Context

• The reader will have a description and a proof of the proposition that for any topological space, its any subspace, and any subset of the superspace, the subspace minus the subset as the subspace of the subspace is the subspace of the superspace minus the subset.

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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:
$T^{\prime }$: $\in \{\text{ the topological spaces }\}$
$T$: $\subseteq T^{\prime }$, with the subspace topology
$S$: $\subseteq T^{\prime }$
$T^{\prime }\setminus S\subseteq T^{\prime }$: with the subspace topology
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Statements:
$T\setminus S$ as the subspace of $T$ is $T\setminus S$ as the subspace of $T^{\prime }\setminus S$.
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2: Natural Language Description

For any topological space, $T^{\prime }$, any subspace, $T\subseteq T^{\prime }$, any $S\subseteq T^{\prime }$, and $T^{\prime }\setminus S\subseteq T^{\prime }$ with the subspace topology, $T\setminus S$ as the subspace of $T$ is $T\setminus S$ as the subspace of $T^{\prime }\setminus S$.

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3: Proof

Let $U\subseteq T\setminus S$ be any open subset of $T\setminus S$ as the subspace of $T$.

$U=U^{\prime }\cap (T\setminus S)$, where $U^{\prime }\subseteq T$ is an open subset of $T$. $U^{\prime }=U^{″}\cap T$ where $U^{″}\subseteq T^{\prime }$ is an open subset of $T^{\prime }$. $U=U^{″}\cap T\cap (T\setminus S)=U^{″}\cap (T\setminus S)=U^{″}\cap (T^{\prime }\setminus S)\cap (T\setminus S)$, but $U^{″}\cap (T^{\prime }\setminus S)\subseteq T^{\prime }\setminus S$ is an open subset of $T^{\prime }\setminus S$, and $U^{″}\cap (T^{\prime }\setminus S)\cap (T\setminus S)$ is an open subset of $T\setminus S$ as the subspace of $T^{\prime }\setminus S$.

Let $U\subseteq T\setminus S$ be any open subset of $T\setminus S$ as the subspace of $T^{\prime }\setminus S$.

$U=U^{\prime }\cap (T\setminus S)$, where $U^{\prime }\subseteq T^{\prime }\setminus S$ is an open subset of $T^{\prime }\setminus S$. $U^{\prime }=U^{″}\cap (T^{\prime }\setminus S)$, where $U^{″}\subseteq T^{\prime }$ is an open subset of $T^{\prime }$. $U=U^{″}\cap (T^{\prime }\setminus S)\cap (T\setminus S)=U^{″}\cap (T\setminus S)=U^{″}\cap T\cap (T\setminus S)$, but $U^{″}\cap T\subseteq T$ is an open subset of $T$, and $U^{″}\cap T\cap (T\setminus S)\subseteq T\setminus S$ is an open subset of $T\setminus S$ as the subspace of $T$.

So, $T\setminus S$ as the subspace of $T$ is $T\setminus S$ as the subspace of $T^{\prime }\setminus S$.

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References

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