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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?

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

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