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Theorem. Every regular space with a countable basis is normal.

Proof. Let X be a regular space with a countable basis $\mathcal{B}$. Let $A$ and $B$ be disjoint closed subsets of $X$. Each point $x$ of $A$ has a neighborhood $U$ disjoint from $B$. Using regularity, choose a neighborhood $V$ of $x$ whose closure lies in $U \dots$

I can't follow the last step made until this point; Using regularity, choose a neighborhood $V$ of $x$ whose closure lies in $U$.

I do know that $X$ is a regular space if, given any nonempty closed set $F$ and any point $x$ that does not belong to $F$, there exists a neighbourhood $U$ of $x$ and a neighbourhood $V$ of $F$ that are disjoint.

Could anyone help me with some insights?

Thanks in advance.

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How do you define regular? In some places the definition of regular is "separate a point and a disjoint closed set by disjoint open sets" (I may be forgetting something like an additional $T_1$ assumption, but the point is the same). – Asaf Karagila Jan 31 '13 at 12:13
Apply the regularity property to the point $x$ and the closed set $X\setminus U$ will give you that last step. – Willie Wong Jan 31 '13 at 12:28
@WillieWong. So that would mean that $\overline{X\setminus{U}} \subset U$. But how do I justify that the closure lies in $U$? Is this always true for any space and any two closed subsets? Or does the closure operation act proper to the regularity of the space? – omar Jan 31 '13 at 13:33
up vote 2 down vote accepted

$x$ is a point. $U\ni x$ is open. $X\setminus U$ is closed. Regularity implies that there are disjoint neighborhoods $V \ni x$ and $V' \supset X\setminus U$.

We have $$ V\cap V' = \emptyset \implies V \subseteq X \setminus V' $$ Note that the right hand side is the complement of an open set and so is closed. Take the closure on both sides we have $$ \bar{V} \subseteq \overline{X\setminus V'} = X\setminus V'$$

Now since $$ X\setminus U \subseteq V'$$ from de Morgan's laws we have $$ X\setminus V' \subseteq U $$ and so $$ \bar{V} \subseteq U $$ as desired.

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