I'm studying the Ostaszewski's article "On Countably Compact, Perfectly Normal Spaces". I'll add some context. Lemma 1.2 says the following:

Let $X$ be a locally compact, zero-dimensional and metrizable space. Suppose $A, B \subset X$ are infinite countable sets. Suppose no point of $X$ is a point of accumulation of $A$ or $B$. Then there exists a space $\kappa(X, A, B)$ such that:

  1. $X$ is a subspace of $\kappa(X, A, B)$
  2. $\kappa(X, A, B)\setminus X$ is countable and nas no points of accumulation
  3. $\kappa(X, A, B)$ is locally compact, zero dimensional and metrizable
  4. If $p \in \kappa(X, A, B)\setminus X$ then $p$ is a point of accumulation of both $A$ and $B$.

The proof consists of first reducing the case for $A$ and $B$ disjoint infinite sets such that $X \setminus A \cup B$ is infinite then writing $A \cup B=\langle a_n: n \in \omega\rangle$ where the even elements enumerates $A$ and odd enumerate $B$. Then for each $n$ we inductively choose sets $G_{2n}$, $G_{2n+1}$ such that:

$$a_{2n} \in G_{2n} \subset X \setminus \left[\bigcup_{m<n} G_{2m} \cup G_{2m+1}\cup (B \cup A \setminus \{a_{2n}\})\right],$$ $$a_{2n+1} \in G_{2n+1} \subset X \setminus \left[G_{2n}\cup \bigcup_{m<n} G_{2m} \cup G_{2m+1}\cup (B \cup A \setminus \{a_{2n+1}\})\right],$$ $$\text{diam}(G_{2n+1}), \, \text{diam}(G_{2n}) \leq 2^{-2n}.$$

We partition $\omega$ into mutually disjoint, infinite sets $\mathbb N_m$ ($m \in \omega$)Then we pick a countable set disjoint from $X$ consisting of some $p_n$'s and then define $\kappa(X, A, B)=X\cup\{p_n: n \in \omega\}$. Then we topologise each point of $x$ with it's original topology and each $p_n$ with a base consisting of the sets $U_k(p_n)=\{p_n\}\cup \bigcup_{m>k, \, m \in \mathbb N_n}(G_{2m+1}\cup G_{2m})$. Then he proves some of the assertions and some others are left for the reader.

Question: I'm ok with all the assertions but the one that says $\kappa(X, A, B)$ is metrizable. He says it's obvious.

Well, I tried to use the following theorem:

A space $X$ is metrizable if, and only if, it's regular, Hausdorff and has a $\sigma$-locally finite base $\mathscr B$.

I tried to proceed as following (at this point, I already knew the space is regular and hausdorff)

Let $\mathscr B$ be a $\sigma$-locally finite base of $X$. Write $\mathscr B=\bigcup_{l \in \omega}F_l$, where $F_n$ is locally finite. Let $\mathscr C=\bigcup_{l \in \omega}F_l \cup \bigcup_{n, k \in \omega}\{U_k(p_n)\}$.

$\mathscr C$ is clearly a base of $\kappa(X, A, B)$. If $x$ is any point of $X$ it's easy to see that for each of those sets $x$ has a neighborhood that only intersects a finite number of their elements. But it remains to show the same for each $p_n$ and I don't know how to proceed.

Can someone help me?

Edit: I think this approach might fail. Pick $X=\omega$. Let $A=\{n \in \omega: n \mod 3 = 0\}$, $B=\{n \in \omega: n \mod 3 = 1\}$.

Now for each $n$, let $G_n=\{n\}$. Now notice that $\{{n}: n \in \omega\}$ is a locally finite base and notice that every neighborhood of each $p_n$ intersects this set $\omega$ times. Therefore I must try another approach.

  • 2
    $\begingroup$ You don't need to use one of the special metrization theorems since we are dealing with locally compact spaces. In particular, a locally compact space is metrizable if and only if it can be written as a disjoint union of second countable spaces (this little result follows easily from the fact that metric spaces are paracompact and every paracompact locally compact space is a disjoint union of $\sigma$-compact spaces). Also, it does not seem too difficult to show that this space is a disjoint union of second countable spaces using the fact that $X$ space is locally compact and metrizable. $\endgroup$ – Joseph Van Name Aug 22 '14 at 1:43

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