Is this a normal topology?

Question

Let $\tau$ be the usual topology on $\Bbb R$. I'm studying the following topology on $\Bbb R$:

$$\alpha = \{O \cup U: O \in \tau \text{ and } \Bbb Q \subset \Bbb R \setminus U\}$$

So far, I found that:

• $(\Bbb R,\alpha)$ is regular. Indeed, for $a \in \Bbb R$ and $A = (\Bbb R \setminus O) \cap (\Bbb R \setminus U)$ a closed set for $\alpha$ with $a \notin A$, we have two possibilities: if $a \in \Bbb Q$, then $a \in \Bbb R \setminus U$, so $a \notin \Bbb R \setminus O$, which is closed, so by regularity of $(\Bbb R, \tau)$, there exist disjoint $\tau$-open neighbourhoods $G$ and $H$ for $x$ and $\Bbb R \setminus O$ respectively, but since $\alpha$ is finer than $\tau$, these are in $\alpha$, and $A \subset H$ since $A \subset \Bbb R \setminus O$. If $a \notin \Bbb Q$, then $\{a\}$ is open and so $\{a\}$ and $\Bbb R \setminus \{a\}$ are disjoint $\alpha$-open neighbourhoods of $x$ and $A$.

• $(\Bbb R, \alpha)$ is not separable (since any dense set must contain $\Bbb R \setminus \Bbb Q$)

• $(\Bbb R, \alpha)$ is first countable: if $x \in \Bbb R \setminus \Bbb Q$, $\{\{x\}\}$ is a local base. Otherwise, $\{]x-\frac1n, x+\frac1n[: n \in \Bbb N\}$ is a local base.

I still don't understand this topology well enough. So, I'm investigating whether it's normal, and if not, whether it's completely regular. I guess it's not normal because it seems that it has "too many open sets", but I'm not sure how to proceed. Can someone help me figure this out? Thanks.

Answer 1

It has a base of clopen sets, so it’s automatically completely regular; since it’s also $T_1$, it’s actually Tikhonov. But in fact it’s hereditarily normal.

Let

$$X=\{\langle x,n\rangle\in\Bbb R\times\Bbb Z:n=0\text{ or }x\in\Bbb R\setminus\Bbb Q\}\;,$$

and let $\preceq$ be the lexicographic order on $X$. Endow $X$ with the order topology generated by $\preceq$, i.e., the topology generated by the subbase of sets of the forms $L_x=\{y\in X:y\prec x\}$ and $R_x=\{y\in X:x\prec y\}$ for $x\in X$.) Every space whose topology is generated in this way by a linear order is $T_5$ (i.e., hereditarily normal and $T_1$); there’s a proof here.

Let $Y=\Bbb R\times\{0\}$, with the relative topology that it inherits from $X$; $Y$ is $T_4$ (i.e., normal and $T_1$). To complete the proof that $\langle\Bbb R,\alpha\rangle$ is $T_5$, show that it is homeomorphic to $Y$.

This space is called the Michael line, after Ernie Michael, who first used it as a counterexample and studied it in some detail; its properties are covered at some length in Dan Ma’s Topology Blog.