A locally compact and dense subset of a Hausdorff space is open

Question

Following a reference from "Elementos de Topología General" by Angel Tamariz and Fidel Casarrubias.

Definition

A topological space is locally compact if for any its point there exist a compact neighborhood.

Theorem

Let be $X$ a Hausdorff locally compact space and let be $Y\subseteq X$ a dense set: so if $Y$ is locally compact, then $Y$ is open in $X$.

proof. Let be $y\in Y$. Since $Y$ is locally compact there exist a open set $A$ in $Y$ and a compact $K$ in $Y$ such that $y\in A\subseteq K\subseteq Y$. So we choose an open set $V$ in $X$ such that $A=Y\cap V$ and we prove that $y\in V\subseteq Y$.

Clearly $y\in V$; then we observe that

$$\mathscr{cl}_X(Y\cap V)\cap Y=\mathscr{cl}_X(A)\cap Y=\mathscr{cl}_Y(A)$$

and moreover since $\mathscr{cl}_Y(A)$ is compact, then $\mathscr{cl}_X(Y\cap V)\cap Y$ is compact and so this set is a closed set in $X$. Then $\mathscr{cl}_X(Y\cap V)\cap Y$ contains $Y\cap V$ and so

$$\mathscr{cl}_X(Y\cap V)\subseteq\mathscr{cl}_X(Y\cap V)\cap Y$$

that is $\mathscr{cl}_X(Y\cap V)\subseteq Y$. Howewer it result that

$$\mathscr{cl}_X(Y)\cap V\subseteq \mathscr{cl}_X(Y\cap V)$$

and so by density of $Y$ it is $V\subseteq\mathscr{cl}_X(Y\cap V)\subseteq Y$.

Unfortunately I don't understand why $\mathscr{cl}_X(Y)\cap V\subseteq\mathscr{cl}_X(Y\cap V)$. If someone know another proof he could show it.

Could someone help me, please?

Answer 1

This follows straightforwardly from the definition of closure.

Take $x\in cl_X(Y)\cap V$. Then $Y\cap U\neq \emptyset$ for all neighborhoods $U$ of $x$

Next, note that if $U$ is a neighborhood of $x$, then $U \cap V$ is also a neighborhood of $x$, as $V$ is a neighborhood of $x$. Thus by the above we get $\emptyset \neq (U\cap V)\cap Y = U\cap (V\cap Y)$ showing that $x\in cl_X(V\cap Y)$.

Hence, we have shown the desired inclusion.

Answer 2

It is crucial that $V$ is open in $X$:

Let $a \in \overline{Y} \cap V$. Then $a \in V$ and there exists a net $(x_i)_{i \in I} \subseteq Y$ s.t. $x_i \rightarrow a$. Since $V$ is open, there exists an open neighborhood $a \in V' \subseteq V$ and since $(x_i)_{i \in I}$ converges, the tail $\{x_i: i \geq i_0\}$ for a large neugh $i$ is contained in $V'$ and thus in $V$. Thus this tail is a net in $Y \cap V$ which converges to $a$ and thus $a \in \overline{Y \cap V}$.