Is it true that every normal countable topological space is metrizable?

I’ve been reading about and working on various proofs about metrizabililty. I’m having trouble answering the following question: Is it true that every normal countable topological space is metrizable? Then it asks to explain your reasoning.

Any help would be greatly appreciated.

Solutions Collecting From Web of "Is it true that every normal countable topological space is metrizable?"

No. An example is the Arens-Fort Space.

Let $X = \omega \times \omega$. For each $A \subseteq X$ and $n \in \omega$ we let
$$A_n = \{ m \in \omega : (n,m) \in A \}$$ denote the $n$th section of $A$.

Topologise $X$ by taking each point of $X \setminus \{ (0,0) \}$ to be isolated, and let $U \subseteq X$ be a neighbourhood of $(0,0)$ iff it contains $(0,0)$ and all but finitely many sections of $U$ are co-finite.

Clearly this space is T$_1$. If $F, E \subseteq X$ are disjoint closed sets, then one, say $E$, does not contain $(0,0)$. So $E$ is clopen, and $X \setminus E$ is an open set including $F$ which is disjoint from $E$. Thus $X$ is normal.

To show that $X$ is not metrizable, it suffices to show that it is not first-countable. For this we will show that there is no countable base at $(0,0)$. (Obviously, every other point has a countable base.) If $\{ U^{(i)} \}_{i \in \omega}$ is any family of open neighbourhoods of $(0,0)$, we inductively construct a sequence of pairs of natural numbers $\{ (n_i,m_i) \}_{i \in \omega}$ so that:

  • $n_i > n_{i-1}$ is such that $U^{(i)}_{n_i}$ is non-empty (say $n_{-1} = 0$); and
  • $m_i \in U^{(i)}_{n_i}$.

Then $V = X \setminus \{ ( n_i , m_i ) : i \in \omega \}$ is an open neighbourhood of $(0,0)$, and by construction $U^{(i)} \not\subseteq V$ for all $i$.

Off the top of my head, I think $\mathbb{Q}$ (or any countable set) with the topology in which only $\mathbb{Q}$ and $\varnothing$ are open is a counterexample. It’s definitely countable, and it’s not metrizable as in any metric there must exist an open ball around any point which is non-empty but not all of $\mathbb{Q}$, so this open ball wouldn’t be open in the topology.

The only point I’m a little uncomfortable with is the normality, but my feeling here is that the only two disjoint closed sets ($\mathbb{Q}$ are $\varnothing$) are their own disjoint open neighbourhoods. I guess this has to be true, otherwise $X$ and $\varnothing$ throw up problems in describing any topological space $X$ as normal.

I wonder if you want Hausdorff as an assumption – the wiki page
doesn’t include this in the definition of normal, but maybe you do.