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I want to show that if $X$ is regular then for any $x \in X $ and any open set $U$ containing $x$, there is an open set $V$ s.t. $x \in \overline{V} \subseteq U$.

From the regularity definition it follows: Given $x \in X$, since $X$ is regular, there exist a closed set C s.t. the open neighborhood of $x$ and $C$ are disjoint.

But I don't understand how I can make use of this, since I want $x$ to be in the closed set.

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    Also your phrasing "the open neighborhood of $x$ and $C$ are disjoint" is incorrect, because *there are*, using *the* implies some uniqueness, which is not assured here. Since the union of two open environments of $x$ and of $C$ would be an environment of $x$ which is not disjoint of any environment of $C$.2011-10-19

1 Answers 1

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If you apply the definition of “regular” to the point $x$ and the closed set $U^{c}$ you get disjoint open sets $V \ni x$ and $W \supset U^c$. It follows that $\overline{V}$ is contained in the closed set $W^c \subset U$.

Added:

It may be worth noting that the following properties of a topological space $X$ are equivalent:

  1. For every point $x$ and every open set $U \ni x$ there is an open neighborhood $V$ of $x$ with $\overline{V} \subset U$.
  2. The space is regular in the sense that for every point $x$ and every closed set $C$ with $x \notin C$ there are disjoint open sets $U \ni x$ and $V \supset C$.

1. implies 2.:

If $x \in X$ and $C \subset X$ is closed with $x \notin C$ put $U = C^{c}$. Then $x \in U$ and $U$ is open. By 1., we can find an open neighborhood $V$ of $x$ with $\overline{V} \subset U$. Then $W = \overline{V}^c \supset C$ and $V \cap W = \emptyset$ are disjoint open sets separating $x$ and $C$.

2. implies 1.:

That's the argument given at the beginning of this answer.