1175: For Topological Space, if Intersection of Subset and Open Subset Is Closed on Open Subset Subspace, Intersection Equals Intersection of Closure of Subset and Open Subset

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description/proof of that for topological space, if intersection of subset and open subset is closed on open subset subspace, intersection equals intersection of closure of subset and open 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: Proof
• 3: Note

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Starting Context

• The reader knows a definition of subspace topology of subset of topological space.
• The reader knows a definition of closed subset of topological space.
• The reader knows a definition of closure of subset of topological space.
• The reader admits the proposition that for any topological space, the intersection of the closure of any subset and any open subset is contained in the closure of the intersection of the subset and the open subset.

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Target Context

• The reader will have a description and a proof of the proposition that for any topological space, any subset, and any open subset, if the intersection of the subset and the open subset is closed on the open subset subspace, the intersection equals the intersection of the closure of the subset and the open 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:
$T$: $\in \{\text{ the topological spaces }\}$
$S$: $\subseteq T$
$U$: $\in \{\text{ the open subsets of }T\}$ with the subspace topology
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Statements:
$S\cap U\in \{\text{ the closed subsets of }U\}$
$⟹$
$S\cap U=\bar{S}\cap U$, where $\bar{S}$ is the closure of $S$ on $T$
//

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2: Proof

Whole Strategy: Step 1: see that $S\cap U=\overline{S\cap U}\cap U$; Step 2: see that $S\cap U\subseteq \bar{S}\cap U\subseteq \overline{S\cap U}$; Step 3: see that $S\cap U\subseteq \bar{S}\cap U\subseteq \overline{S\cap U}\cap U$.

Step 1:

There is a closed subset, $C\subset T$, such that $S\cap U=C\cap U$, by the definition of subspace topology.

$S\cap U\subseteq C$ and $\overline{S\cap U}\subseteq C$ where the closure is on $T$, because the closure is the smallest closed subset that contains the concerned subset.

Then, $C$ can be taken to be $\overline{S\cap U}$, which is a closed subset of $T$, contains $S\cap U$, so, the replacement does not lose any point on $S\cap U$, and is contained in $C$, so the replacement does not add any extra point. So, $S\cap U=\overline{S\cap U}\cap U$.

Step 2:

On the other hand, $S\cap U\subseteq \bar{S}\cap U\subseteq \overline{S\cap U}$, by the proposition that for any topological space, the intersection of the closure of any subset and any open subset is contained in the closure of the intersection of the subset and the open subset.

Step 3:

By taking the intersection of it with $U$, $S\cap U\cap U=S\cap U\subseteq \bar{S}\cap U\cap U=\bar{S}\cap U\subseteq \overline{S\cap U}\cap U$. But as the 1st term and the last term equal, the middle term equals them, so, $S\cap U=\bar{S}\cap U$.

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3: Note

$U$ has to be open for this proposition: for example, if $U$ is not open, here is a counterexample: $T=\mathbb{R}$ with the Euclidean topology, $U=[0,1]$, and $S=(-1,0)$, then $S\cap U=\mathrm{\varnothing }$, closed, but $\bar{S}\cap U=[-1,0]\cap [0,1]=\{0\}$. Such a counterexample does not work for when $U$ is open, because if $U=(0,1)$, $S\cap U=(-1,0)\cap (0,1)=\mathrm{\varnothing }$ and $\bar{S}\cap U=[-1,0]\cap (0,1)=\mathrm{\varnothing }$, and if $U=(-p,1)$ for any $0<p<1$, $S\cap U=(-1,0)\cap (-p,1)=(-p,0)$, which is not closed.

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References

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