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.

Solution 1:

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 $i$.

Solution 2:

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.