# Showing a space is regular by knowing each point has a nbd whose closure is regular

Can you please let me know if my "proof" is correct: ?

Let $Y$ be a Hausdorff space and assume that each $y \in Y$ has a neighborhood $V$ such that $\overline{V}$ is regular. Prove $Y$ is regular.

Proof:

In order to show that $Y$ is regular we are going to show that for each $y \in Y$ and each open set $U$ which contains $y$ there exists an open neighborhood $W$ of $y$ such that $y \in W \subseteq \overline{W} \subseteq U$.

First observation: since regularity is hereditary then $V$ is regular.

Now let $y \in Y$ and $U$ a neighborhood of $y$. By assumption there exists a neighborhood $V$ of $y$ such that $\overline{V}$ is regular and hence $V$ is regular. Observe $U \cap V$ is a non-empty subset of $V$ and this set is open in $V$ as well. Since $V$ is regular we can find a non-empty subset $S$ of $V$ such that:

$S \subseteq \overline{S} \subseteq U \cap V \subseteq U$

But since $V$ is open then $S$ is open in $X$ so the same set $S$ given by regularity of $V$ shows that $U$ is regular as well.

Is the above incorrect?

Thank you.

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There is a problem with your argument. The closure of $S$ in $V$ might not be closed in $X$. I can see why you went to $V$, because you wanted $S$ to be open. However, if you work instead in $\overline V$, then the closure of $S$ will remain closed in $X$, and since $x$ is in the interior of $\overline V$, you'll still be able to find an open set that does what you want.
You have to use the regularity of $\overline V$, and not just of $V$, because there are examples of nonregular spaces in which every point has a regular neighborhood, like the line with 2 origins.
@Jonas Meyer: Thank you. I still have trouble and get confused using subspace topology of $\overline V$. Can you please help some more? (not homeowork, it is for fun).
Similar to your argument, suppose you have $S$ open in $\overline V$ such that $y\in S\subset \overline S\subset U\cap\overline V\subset U$. Because $y$ is in the interior of $\overline V$, you can show that it is in the interior of $S$, $S^\circ$, and since $\overline{S^\circ}\subset\overline S$, $S^\circ$ will work as the $W$ you want. –  Jonas Meyer Dec 30 '10 at 5:18