Can we define Urysohn function here?

Question

Let $X$ be a topological space such that any two distinct points $x,y$ in $X$ can be separated by open subsets $U,V$ of $X$ such that $x\in U$, $y\in V$ and $cl(U), cl(V)$ are disjoint. For any distinct $x,y$, is it possible to define a continuous function $f:X\rightarrow [0,1]$ on $X$ such that $f(x)=0$ and $f(y)=0$?

I think this is related to the Urysohn function, but from what I understand, for Urysohn function to exist, it is required that $X$ to be normal first and I can't prove that $X$ is normal.

Thanks in advance.

Answer 1

A space $X$ satisfying the first condition is is called a Uryso(h)n space. A space satisfying the second condition is called a completely Hausdorff or functionally Hausdorff space. Every completely Hausdorff space is Uryson, but not every Uryson space is completely Hausdorff.

One Uryson space that is not completely Hausdorff is the space called the Arens square in Steen & Seebach, Counterexamples in Topology. It’s defined as follows.

Let $Q=\Bbb Q\cap(0,1)$, and let $S=Q\times Q$. Let $p_0=\langle 0,0\rangle$ and $p_1=\langle 1,0\rangle$. For each $r\in\Bbb Q\cap\left(0,\frac12\sqrt2\right)$ let $p_r=\left\langle\frac12,r\sqrt2\right\rangle$, and let $L=\left\{p_r:r\in\Bbb Q\cap\left(0,\frac12\sqrt2\right)\right\}$. Finally, let $X=S\cup L\cup\{p_0,p_1\}$. Points of $S$ have the nbhds that they inherit from the usual topology of $\Bbb R^2$. For $n\in\Bbb Z^+$ let $$\begin{align*}B(0,n)&=\{p_0\}\cup\left\{\langle x,y\rangle\in S:x<\frac14\text{ and }y<\frac1n\right\}\text{ and}\\B(1,n)&=\{p_1\}\cup\left\{\langle x,y\rangle\in S:x>\frac34\text{ and }y<\frac1n\right\}\;,\end{align*}$$ and for $n\in\Bbb Z^+$ and $r\in\Bbb Q\cap\left(0,\frac12\sqrt2\right)$ let $$B(r,n)=\left\{\langle x,y\rangle\in S\cup L:\frac14<x<\frac34\text{ and }\left|y-r\sqrt2\right|<\frac1n\right\}\;.$$ Then $\{B(0,n):n\in\Bbb Z^+\}$ is a local base at $p_0$, $\{B(1,n):n\in\Bbb Z^+\}$ is a local base at $p_1$, and $\{B(r,n):n\in\Bbb Z^+\}$ is a local base at $p_r$ for each $r\in\Bbb Q\cap\left(0,\frac12\sqrt2\right)$.

It’s not hard to verify that $X$ is Uryson; the key observation is that no point of $X\setminus L$ has the same $y$-coordinate as any point of $L$.

To show that $X$ is not completely Hausdorff, we show that there is no continuous $f:X\to[0,1]$ such that $f(p_0)=0$ and $f(p_1)=1$. Suppose that $f$ is such a function. The sets $\left[0,\frac14\right)$ and $\left(\frac34,1\right]$ are open in $[0,1]$, so their inverse images under $f$ are open in $X$. This means that there are $m,n\in\Bbb Z^+$ such that $f\big[B(0,m)\big]\subseteq\left[0,\frac14\right)$ and $f\big[B(1,n)\big]\subseteq\left(\frac34,1\right]$. Now choose any $r\in\Bbb Q\cap\left(0,\frac12\sqrt2\right)$ such that $r\sqrt2<\min\left\{\frac1m,\frac1n\right\}$.

Now $f(p_r)$ cannot be in both $\left[0,\frac14\right)$ and $\left(\frac34,1\right]$, so without loss of generality suppose that $f(p_r)\notin\left[0,\frac14\right)$. Then there are $a,b\in(0,1)$ such that $\frac14<a<f(p_r)<b$. Clearly $\left[0,\frac14\right]$ and $[a,b]$ are disjoint closed subsets of $[0,1]$, so their inverse images under $f$ are disjoint closed sets in $X$ that contain open nbhds of $p_0$ and $p_r$, respectively. In fact, $f^{-1}\left[\left[0,\frac14\right]\right]\supseteq B(0,m)$, and I’ll leave it to you to show that the choice of $r$ ensures that for each $k\in\Bbb Z^+$,

$$\big(\operatorname{cl}B(0,m)\big)\cap\operatorname{cl}B(r,k)\ne\varnothing\;,$$

contradicting the disjointness of the inverse images of $\left[0,\frac14\right)$ and $\left(\frac34,1\right]$.