Is this topology metrizable? Is the irrational sequence topology metrizable?
The irrational sequence topology is generated by the clopen basis 
$$
\big\{\{x\}\,:\,x\,\text{is irrational}\big\}\cup\big\{A_n(x) \cup \{x\}\,:\,n\in\mathbb{N}\,\text{if}\,x\,\text{is rational}\big\},
$$ 
where $A_n(x)$ is a tail of a sequence $x_n$ of irrational numbers that converge to $x$. (In this topology, we fix a convergent sequence for every rational number).
I proved that this space is $T_4$, but I couldn't prove that this space is not $T_5$ or $T_6$, so that it cannot be metrizable. Can someone give me a hint?
 A: The space is metrizable.  For this I will make use of the Nagata-Smirnov Metrization Theorem:

A topological space $X$ is metrizable iff it is regular and has a $\sigma$-locally finite base.

(A family of subsets of a topological space is called 


*

*locally finite if each point has an (open) neighbourhood which meets only finitely many sets in the family.  

*$\sigma$-locally finite if it is a countable union of locally finite subfamilies.)


First let $Y = \bigcup_{q \in \mathbb{Q}} A_0 ( q )$, and $Z = \mathbb{R} \setminus ( \mathbb{Q} \cup Y)$.  Note that $Z$ is a discrete clopen subseteq of $\mathbb{R}$, and $Y$ is countable.


*

*For each $q \in \mathbb{Q}$ and $n \in \mathbb{N}$ let $\mathcal{B}_{q,n} = \{ \{ q \} \cup A_n(q) \}$.

*For each $y \in Y$ let $\mathcal{B}_y = \{ \{ y \} \}$.

*Let $\mathcal{B}^\prime = \{ \{ z \} : z \in Z \}$.


It is clear that each $\mathcal{B}_{q,n}$ and each $\mathcal{B}_y$ ($y \in Y$) is (locally) finite.  Note that as for each $q \in \mathbb{Q}$ the set $\{ q \} \cup A_0(q)$ is an open neighbourhood disjoint from $Z$, it follows that $\mathcal{B}^\prime$ is locally finite.
Therefore $$\mathcal{B} = \bigcup_{q \in \mathbb{Q}} \bigcup_{n \in \mathbb{N}} \mathcal{B}_{q,n} \cup 
\bigcup_{y \in Y} \mathcal{B}_{y} \cup 
\mathcal{B}^\prime$$ is a $\sigma$-locally finite base for $\mathbb{R}$.
As you have shown that the space is normal (and hence regular) it follows that it is metrizable.
(Nagata-Smirnov is surely overkill, but I'm not seeing a more direct proof at the moment.)

Gaaack!!
There is a way to get this result using much weaker tools.  Note that the space is the topological sum of $Z$ and $Y \cup \mathbb{Q}$, so it suffices to show that each of these subspaces are metrizable.  Clearly $Z$ is metrizable since it is discrete, so all that remains is $Y \cup \mathbb{Q}$.  For this we make use of Urysohn's Metrization Theorem:

Every second-countable regular (Hausdorff) space is metrizable.

It is easy to see that the irrational singletons together with the basic open neighbourhoods of the rationals forms a countable base for the subspace $Y \cup \mathbb{Q}$, and as regularity (and Hausdorffness) was already established, we're done!
