How to show that normality is equivalent to the intermediate open set property? (Step-1 of Urysohn Lemma in Munkres)

Question

In the process of a proof of the Urysohn Lemma, Munkres' Topology says two things that I can't prove them:

1- Let $X$ be a normal space. Let $U_p$ and $U_q$ are two open sets such that $\bar U_p \subset U_q$. Using normality of $X$, we can always find an open set $U_r$ of $X$ such that $\ \bar U_p \subset U_r \ $ and $\ \bar U_r \subset U_q \ $. Why? I can't see any connection to the definition of normality.

2- Now let $X$ be a regular space. Considering the same things said in "1" we may not find an open set $U_r$ of $X$ such that $\ \bar U_p \subset U_r \ $ and $\ \bar U_r \subset U_q \ $. Why? I can't see any connection to the definition of regularity, as well!

Answer 1

Apply the definition of normality to the disjoint closed sets $\bar{U}_p$ and $X -U_q$: there exists some open $A$ containing $\bar{U}_p$ and some open $B$ containing $X-U_q$ such that $A$ and $B$ are disjoint. Note that $\bar{A}$ and $B$ are also disjoint: every neighborhood of a point in $\bar{A}$ contains points in $A$ (which are not in $B$), so this point cannot lie in $B$. Then take $U_r:= A$; it contains $\bar{U}_p$ by construction, and also $\bar{U}_r \subset (X-B) \subset U_q$.

Answer 2

I’ll address your second question. If $\operatorname{cl}U_p\subseteq U_q$, let $H=\operatorname{cl}U_p$ and $K=X\setminus\operatorname{cl}U_q$; $H$ and $K$ are disjoint closed sets in $X$. If there is an open set $U_r$ such that $\operatorname{cl}U_p\subseteq U_r\subseteq\operatorname{cl}U_r\subseteq U_q$, let $V=X\setminus\operatorname{cl}U_r$. Then $H\subseteq U_r$, $K\subseteq V$, and $U_r\cap V=\varnothing$. In other words, $U_r$ and $V$ are disjoint open nbhds of $H$ and $K$, respectively. If $X$ is normal, by definition such open nbhds exist. If $X$ is only regular, however, they need not exist: regularity only ensures that a point and a closed set not containing it have disjoint open nbhds.

You might wonder whether this problem can actually arise when one of the closed sets is the closure of an open set. The answer is yes; here’s an example.

Let $Q=(0,1)\cap\Bbb Q$, $Y=(-1,1)\times Q$, $V=\{0\}\times Q$, $H=(0,1)\times\{0\}$, and $X=Y\cup H$. We topologize $X$ as follows.

• Points of $Y\setminus V$ are isolated.
• For each $x\in(0,1)$ and $n\in\Bbb Z^+$ the sets $B_n(x)=\{x\}\times\left(\left[0,\frac1n\right)\cap Q\right)$ are a local base at the point $\langle x,0\rangle\in H$.
• For each $y\in Q$ and finite $F\subseteq(-1,0)\cup(0,1)$ the sets $B(y,F)=\big((-1,1)\setminus F\big)\times\{y\}$ are a local base at the point $\langle 0,y\rangle\in V$.

It’s straightforward to check that this topology is $T_3$ (i.e., regular and $T_1$).

Let $U_p=(-1,0)\times Q$ and $U_q=X\setminus H$; $U_p$ and $U_q$ are open sets, and $\operatorname{cl}U_p=U_p\cup V\subseteq U_q$. Suppose that $U_r$ is an open set such that $\operatorname{cl}U_p\subseteq U_r\subseteq\operatorname{cl}U_r$. $V\subseteq\operatorname{cl}U_p$, so for each $q\in Q$ there is a finite $F_q\subseteq(-1,0)\cup(0,1)$ such that $B(q,F_q)\subseteq U_r$. Let $C=\bigcup_{q\in Q}F_q$; $C$ is the union of countably many finite sets, so $C$ is countable, and there is some $x\in(0,1)\setminus C$. Let $q\in Q$ be arbitrary; $x\notin F_q$, so $\langle x,q\rangle\in B(q,F_q)\subseteq U_r$, and therefore $B_n(x)\cap U_r\ne\varnothing$ for each $n\in\Bbb Z^+$. But then $\langle x,0\rangle\in\operatorname{cl}U_r\setminus U_q$, so $\operatorname{cl}U_r\nsubseteq U_q$. That is, there is no open set $U_r$ such that $\operatorname{cl}U_p\subseteq U_r\subseteq\operatorname{cl}U_r\subseteq U_q$.