# $R_0$ and normal implies completely regular

We say a space ($$X, \tau$$) is $$R_0$$ if for each $$x,y \in X$$ such that $$\overline{\{x\}}\neq\overline{\{y\}}$$ there exists open sets with $$x\in U-V$$ and $$y \in V-U$$.

If $$F, K \subseteq X$$ disjoint closed sets there exists disjoints open sets $$U$$ and $$V$$ such that $$F \subseteq U, K \subset V$$ we say that the space is normal.

And if for each $$x \in X$$ and $$F \subset X$$ closed such that $$x \notin F$$ there exists $$f:X \rightarrow [0,1]$$ continuous with $$f(x)=0$$ and $$f[F]=\{1\}$$ we say that the space is completely regular.

I want to show that if a space is $$R_0$$ and normal then is completely regular, I've tried to show that with this conditions the space must be $$T_1$$.

If $$\overline{\{x\}} \neq \overline{\{y\}}$$ for any pair of distinct points then I'm done, this is equivalent to be $$T_0$$, but I don't know if this is always true?

• If $X$ is $R_0$ and also $T_0$ then it is $T_1$. But either of them by itself doesn’t imply $T_1$. – Henno Brandsma Jan 19 at 6:40

## 1 Answer

The space need not be $$T_1$$. Let $$X=\{0,1,2\}$$ and $$\tau=\big\{\varnothing,\{0\},\{1,2\},X\big\}$$; then $$\langle X,\tau\rangle$$ is $$R_0$$ and normal but not $$T_0$$ (and hence not $$T_1$$). And it is completely regular but not Tikhonov.

HINT: To prove the theorem, use Urysohn’s lemma: if $$F$$ and $$K$$ are disjoint closed set in $$X$$, there is a continuous $$f:X\to[0,1]$$ such that $$f[F]=\{0\}$$ and $$f[K]=\{1\}$$. Suppose that $$x\in X$$, $$F\subseteq X$$ is closed, and $$x\notin F$$. Let $$K=\operatorname{cl}\{x\}$$. If $$K\cap F=\varnothing$$, we can immediately apply Urysohn’s lemma to get the desired result.

If not, let $$y\in K\cap F$$. $$X\setminus F$$ is an open nbhd of $$x$$ that does not contain $$y$$, so $$x\notin\operatorname{cl}\{y\}$$, and therefore $$\operatorname{cl}\{x\}\ne\operatorname{cl}\{y\}$$. Now use the hypothesis that $$X$$ is $$R_0$$ to get a contradiction, showing that in fact $$K\cap F=\varnothing$$.