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. 
 A: If $D$ is not open, let $p \in D$ such that $D$ contains no neighbourhood (in $X$) of $p$.
Since $D$ is locally compact, there is a compact $K \subseteq D$ that is a neighbourhood (in $D$) of $p$.  That is, there is an open set $U$ of $X$ containing $p$ such that  $U \cap D \subseteq K$.  Now since $D$ is dense in $X$, if the open set $U \backslash K$ was nonempty it would contain a member of $D$, which contradicts $U \cap D \subseteq K$.  So in fact
$U \backslash K = \emptyset$, i.e. $U \subseteq K \subseteq D$, i.e. $D$ does contain a neighbourhood of $p$.
A: Whenever I write $\overline{A}$ this will mean the closure of $A$ in $X$.
Take a point $d \in D$ and choose an open neighborhood $U$ of $d$ in $D$ such that $F = \overline{U} \cap D$ is compact (by local compactness of $D$). Then $F$ is  also closed in $X$ since the inclusion $D \subset X$ is continuous. Because $U \subset F$ and $F \subset X$ is closed we have $\overline{U} \subset F \subset D$.
Now $U$ is open in $D$, so there is an open $V \subset X$ such that $U = D \cap V$. As $D \subset X$ is dense and $V$ is open, $\overline{D \cap V} = \overline V$. Therefore $V \subset \overline{V} = \overline{D\cap V} = \overline{ U} \subset F \subset D$. But by assumption $d \in V$ and $V$ is open in $X$ and we've just argued that  $V \subset D$. Therefore $D$ is open in $X$.
A: Let $x \in D$. Let $U$ be an open neighborhood of $x$ such that $\overline{U \cap D}^D$ is a compact subset of $D$. $D$ is dense in $X$, so $\overline{U \cap D}^X = \overline{U}^X$. Let $y \in \overline{U \cap D}^X$. Let $\left\langle y_\alpha \right\rangle_{\alpha \in A}$ be a net in $U \cap D$ such that $y_\alpha \to y$. $\overline{U \cap D}^D$ a compact subset of $D$, so $y=\lim y_\alpha \in \overline{U \cap D}^D \subset D$. Therefore, $\overline{U} \subset D$.
A: 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 $X$.
