Dense and locally compact subset of a Hausdorff space is open Let $X$ be a Hausdorff space and let $D \subseteq X$ be locally compact and dense in $X$. Why is $D$ open?
I can see that $D$ is regular but don't see why $D$ is in fact open. 

Here is a straightforward proof inspired by Theorem 2.70 in Aliprantis and Border's Infinite Dimensional Analysis (3rd ed.), p.56). Let $p \in D$. Since $D$ is locally compact, there is a neighborhood of $x$ in $D$ which is compact in $D$ and a neighborhood $V$ of $x$ in $D$ which is open in $D$ and $V \subset U$.
First, it is easy to see that $U$ is also compact in $X$. Since $X$ is Hausdorff, it implies that $U$ is closed (see, for example, Proposition 4.24 in Folland's Real Analysis (2nd ed.), p.128), and consequently $\overline{U}=U$.
By definition, there is an open set $W$ in the topology of $X$ such that $V = W\cap D$. Since $D$ is dense in $X$, it follows that $$W \subset \overline{W} = \overline{W \cap E} = \overline{V} \subset \overline{U} = U \subset D.$$ Hence for every  $p \in D$ there is a neighborhood of $p$ open in $X$ which is included in $D$, i.e., $D$ is open in $X$.
