369: For Regular Topological Space, Neighborhood of Point Contains Closed Neighborhood

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

A description/proof of that for regular topological space, neighborhood of point contains closed neighborhood

---

Topics

About: topological space

---

The table of contents of this article

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

---

Starting Context

• The reader knows a definition of regular topological space.

---

Target Context

• The reader will have a description and a proof of the proposition that for any regular topological space, any neighborhood of any point contains a closed neighborhood.

---

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: Description

For any regular topological space, $T$, any point, $p\in T$, and any neighborhood of $p$, $p\in N_p\subseteq T$, $N_p$ contains a closed neighborhood.

---

2: Proof

There is an open set, $U_p\subseteq N_p,p\in U_p$. $T\setminus U_p$ is closed and does not contain $p$. So, by the definition of regular topological space, there are some disjoint neighborhoods, $N_1,N_2$, such that $p\in N_1\subseteq T$ and $T\setminus U_p\subseteq N_2\subseteq T$, $N_1\cap N_2=\mathrm{\varnothing }$ where we take $N_1$ and $N_2$ as open neighborhoods, which is always possible (see the Note of the definition article). In fact, $\overline{N_1}\cap N_2=\mathrm{\varnothing }$ where $\overline{N_1}$ is the closure of $N_1$, because $\overline{N_1}=N_1\cup ac(N_1)$ where $ac(N_1)$ is the set of all the accumulation points of $N_1$, but any accumulation point of $N_1$ does not belong to $N_2$, because for any point, $p^{\prime }\in N_2$, as $N_2$ is open, there is an open set, $U_{p^{\prime }}$, $p^{\prime }\in U_{p^{\prime }}\subseteq N_2$, which does not contain any point of $N_1$, so, $p^{\prime }$ is not any accumulation point of $N_1$. So, $\overline{N_1}\subseteq T\setminus N_2\subseteq T\setminus (T\setminus U_p)=U_p$, because $T\setminus U_p\subseteq N_2$. So, $\overline{N_1}$ is a closed neighborhood contained in $U_p$, which is contained in $N_p$.

---

References

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