Is the Alexandroff extension of a locally compact, second-countable space second-countable

Question

If $X$ is a locally compact, second-countable topological space, then is its Alexandroff extension $X^*$ also second-countable?

Our definition of locally compact is that for every $x$ in $X$, we have an open subset $U$ of $X$ such that $\operatorname{cl}_X U$ is compact and $x \in U$. I was able to prove that if $\mathcal{B}$ is a base for $X$, then the subset

$$
\bigl\{U \in \mathcal{B} : \text{$\operatorname{cl}_X U$ is compact}\bigr\}
$$

is also a base for $X$. So we can assume without loss of generality that our (countably many) basic open sets have compact closure.

Since the set of finite subsets of a countable set is also countable, we have that the set $\mathcal{B}_\mathrm{f}$ of finite unions of sets in $\mathcal{B}$ is also countable. Define

$$
\mathcal{C} := \mathcal{B} \cup \bigl\{X^* \setminus \operatorname{cl}_X U : U \in \mathcal{B}_\mathrm{f}\bigr\}
$$

which is a countable collection of open subsets of $X^*$.

Is the following proof that every open subset of $X^*$ is a union of sets in $\mathcal{C}$ correct?

Let $x \in U$ which is an open subset of $X^*$. If $x$ is a point in $X$, then $U \cap X$ is open in $X$, and so there is a $V$ in $\mathcal{B} \subset \mathcal{C}$ such that $x \in V$ and $V \subset U \cap X \subset U$. Otherwise $x$ is the point at infinity, so $U = X^* \setminus K$ where $K$ is a compact, closed subset of $X$. We can cover $K$ by finitely many sets $U_n$ in $\mathcal{B}$, so that $K \subset \bigcup_n U_n \in \mathcal{B}_\mathrm{f}$. It follows that $x \in X^*\setminus \operatorname{cl}_X \bigl(\bigcup_n U_n\bigr)$ which is contained in $X^* \setminus K = U$.

In both cases, we have found a set in $\mathcal{C}$ which is a neighborhood of the point $x$ and is contained in $U$. Hence, $\mathcal{C}$ is a base for $X^*$.

Answer 1

Your definition of "locally compact" is unusual. I know two variants:

1. Each point has a compact neighborhood.

2. Each point has a neighborhood base consisting of compact sets.

They are equivalent for Hausdorff spaces, but you do not require spaces to be Hausdorff. As you say, you use

3. Each point has a compact closed neighborhood.

This property also seems to be denoted as strongly locally compact.

In the context of the Alexandroff extension this variant is in fact extremely useful because the open neigborhoods of $\infty$ in $X^*$ are precisely the complements $X^* \setminus K$, where $K \subset X$ is compact and closed in $X$.

Your proof is correct. Its essence is that $X$ has a countable base $\mathcal B$ with the following properties:

• For each $U \in \mathcal B$ the closure $\operatorname{cl}_X U$ is compact.
• Each compact $K \subset X$ is contained is some $U \in \mathcal B$.

You start with a countable base $\mathcal B_0$ of $X$ and observe that $\mathcal B_1 = \{ U \in \mathcal B \mid \operatorname{cl}_X U \in \mathcal B \}$ is also a base for $X$ (which is clearly countable). Then trivially the set $\mathcal B$ of all finite unions of elements of $\mathcal B_1$ is again a basis. Clearly $\mathcal B$ is countable and for each $U \in \mathcal B$ the closure $\operatorname{cl}_X U$ is compact. This holds because if $U = \bigcup_{i=1}^n U_i$ with $U_i \in \mathcal B_1$, then $U \subset C := \bigcup_{i=1}^n \operatorname{cl}_X U_i$, thus $\operatorname{cl}_X U \subset C$. Because $C$ is compact and $\operatorname{cl}_X U$ is closed in $C$, it is compact. Moreover each compact $K \subset X$ can be covered by finitely many $U_i \in \mathcal B_0$, thus it is contained in some $U \in \mathcal B$.

Therefore $\mathcal{C} := \mathcal{B} \cup \bigl\{X^* \setminus \operatorname{cl}_X U : U \in \mathcal{B}\bigr\} $ is a countable base of $X^*$.