Michael Line Topology on $\mathbb{R}$

Question

Is the Michael Line topology on $\mathbb{R}$ second-countable?

In order to be second-countable, we need to consider a countable base for $\mathbb{R}$. Can $\{(a,b) : a,b \in \mathbb{R}\} \cup \{\{x\} : x \in \mathbb{P}\}$ be used as a countable base for $\mathbb{R}$ or this uncountable?

You do know that $\mathbb P$ is uncountable? All singleton irrationals must be in any base for the Michael line. — Henno Brandsma, Jun 08 '19 at 07:00
For those of us who have never heard of the Michael line, would someone please post a link? — Lee Mosher, Jun 08 '19 at 13:54

score 3 · Answer 1 · answered Jun 08 '19 at 00:00

3

Every irrational number $x$ is an isolated point and there are uncountable many irrationals so this topology is not second countable.

answered Jun 08 '19 at 00:00

Kavi Rama Murthy

359,332

To the proposer: We have $B\supset {{x}:x\in \Bbb R \setminus \Bbb Q}$ for any base (basis) $B$ for the Michael line. – DanielWainfleet Jun 08 '19 at 03:30

score 0 · Answer 2 · answered Jun 08 '19 at 13:33

Let $\mathcal{B}$ be any base for the Michael line.

For each $x \in \Bbb P$, $\{x\}$ is open and there must be some $B_x \in \mathcal{B}$ such that $x \in B_x \subseteq \{x\}$, and it follows that $\{x\} = B_x \in \mathcal{B}$. So (as the map $\Bbb P \ni x \to \{x\} \in \mathcal{B}$ is an injection) $|\mathcal{B}| \ge |\Bbb P| = |\Bbb R|$ and the base $\mathcal{B}$ is uncountable.

Hence $X$ is not second countable.

Michael Line Topology on $\mathbb{R}$

2 Answers2