300: For Topological Space, Intersection of Closure of Subset and Open Subset Is Contained in Closure of Intersection of Subset and Open Subset

<The previous article in this series | The table of contents of this series | The next article in this series>

description/proof of that for topological space, intersection of closure of subset and open subset is contained in closure of intersection of subset and open subset

---

Topics

About: topological space

---

The table of contents of this article

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

---

Starting Context

• The reader knows a definition of topological space.
• The reader knows a definition of closure of subset of topological space.

---

Target Context

• The reader will have a description and a proof of 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.

---

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:
$T$: $\in \{\text{ the topological spaces }\}$
$S$: $\subseteq T$
$U$: $\in \{\text{ the open subsets of }T\}$
//

Statements:
$\bar{S}\cap U\subseteq \overline{S\cap U}$, where the over lines denote the closures on $T$
//

---

2: Proof

Whole Strategy: Step 1: let $p\in \bar{S}\cap U$ be any; Step 2: see that when $p\in S$, $p\in \overline{S\cap U}$; Step 3: see that when $p\notin S$, $p\in \overline{S\cap U}$.

Step 1:

For any point, $p\in \bar{S}\cap U$, $p\in \bar{S}$ and $p\in U$.

Step 2:

When $p\in S$, $p\in \overline{S\cap U}$.

Step 3:

When $p\notin S$, $p$ is an accumulation point of $S$, so, for any open neighborhood of $p$, $U_p\subseteq T$, $U_p\cap S\ne \mathrm{\varnothing }$.

$U_p\cap (S\cap U)\ne \mathrm{\varnothing }$?

As $p\in U$, $U_p\cap U\ne \mathrm{\varnothing }$, open, and is an open neighborhood of $p$, so, $U_p\cap (S\cap U)=(U_p\cap U)\cap S\ne \mathrm{\varnothing }$. So, $p$ is an accumulation point of $S\cap U$, so, $p\in \overline{S\cap U}$.

---

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 $\bar{S}\cap U=[-1,0]\cap [0,1]=\{0\}$ while $\overline{S\cap U}=\bar{\mathrm{\varnothing }}=\mathrm{\varnothing }$. Such a counterexample does not work for when $U$ is open, because if $U=(0,1)$, $\bar{S}\cap U=[-1,0]\cap (0,1)=\mathrm{\varnothing }$, and if $U=(-p,1)$ for any $0<p<1$, $\bar{S}\cap U=[-1,0]\cap (-p,1)=(-p,0]$ and $\overline{S\cap U}=\overline{(-p,0)}=[-p,0]$.

---

References

<The previous article in this series | The table of contents of this series | The next article in this series>