Right closed-ray topology on $\mathbb{R}$ is not second countable

Question

I am taking the collection $\mathcal B$ of $\{ [a,\infty), a\in\mathbb{R}\}$, then this collection makes basis set on $\mathbb{R}$. Now i am claiming that this not a second countable topology.

$\text{Proof}$: Claim: if there exist basis set $\mathcal B'$ then $\mathcal B \subset \mathcal B'$.

Let there is $[x, \infty) \not\in \mathcal B'$, then there is no basis element $B\in \mathcal B'$ such that $x\in B\subset [x,\infty)$.

I think my idea is correct, what do you think?

when i said this is basis for usual topology, this a new topology which generate by $\mathcal B$ — Texas, Feb 25 '25 at 14:02
@Maxime please go through whole post, maybe there is misunderstanding — Texas, Feb 25 '25 at 14:07
https://math.stackexchange.com/questions/3983378/why-is-the-sorgenfrey-line-not-second-countable/3983403#3983403 — Evangelopoulos Foivos, Feb 25 '25 at 14:49
@Texas yes indeed, I thought that you were talking about a basis of $\mathbb R$ not $T_{ray}$. Sorry :)` — Iq-n-dI, Feb 25 '25 at 15:06
@EvangelopoulosFoivos you're overthinking this. You just say that every open set of the form $[x, \infty)$ needs to be in any basis for $X$. The OP has correct idea — Jakobian, Feb 25 '25 at 19:16
@EvangelopoulosFoivos: I disagreee that you are not overthinking this. The OP has the correct idea, but hasn't given enough detail. I will add a comment to this effect to Jakobian's answer. — Rob Arthan, Feb 25 '25 at 20:02

Jakobian · Accepted Answer · 2025-02-25T20:33:04.023

Yes, you're correct. Given a basis $\mathcal{B}'$ on $X = \mathbb{R}$ with topology generated by $\mathcal{B} = \{[x, \infty) : x\in X\}$, we need to have $\mathcal{B}\subseteq \mathcal{B}'$, so any basis for $X$ must have at least size continuum.

To continue from where you left over, if $x\in B\subseteq [x, \infty)$ where $B\in\mathcal{B}'$, then $B$ is a union of sets of the form $[y, \infty)$. Since $B\subseteq [x, \infty)$ we need to have $x\leq y$, and one of the sets $[y, \infty)$ needs to be of the form $[x, \infty)$ since otherwise $B\subseteq (x, \infty)$ so $x\notin B$. It follows that $B = [x, \infty)$.

yes thank you, i appreciate you – Texas Feb 26 '25 at 08:39 — Texas, Feb 26 '25 at 08:39

Right closed-ray topology on $\mathbb{R}$ is not second countable

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