Separation Axiom in a Topology of $\mathbb{R}$

Question

In $\mathbb{R}$, let $\mathcal{B}=\{\emptyset,\mathbb{R}\}\cup\{[a,b); \, a,b \in \mathbb{R}, a < b\}.$ See that, besides $\emptyset,\mathbb{R} \in \mathcal{B}$, for all $B_1,B_2 \in \mathcal{B}$, we have $B_1\cap B_2 \in \mathcal{B}$. So, the collection of all possible unions of subsets of $\mathcal{B}$ is a topology in $\mathbb{R}$, that we will call $\tau_U$.

• Show that all intervals $(a,b)$, with $a,b \in \mathbb{R}, a< b$ are in $\tau_U$. Then, conclude that $(\mathbb{R},\tau_U)$ is a Hausdorff space.

• Show that $(\mathbb{R},\tau_U)$ is not a second-countable space.

For the first one, I was thinking of mount some union of subsets of $\mathcal{B}$ that would be equal to $(a,b)$, but I couldn't find such combination of subsets. Can I construct an union like that?

For the second one, I've tried by contradiction, but didn't got anywhere with it. Any leads?

Answer 1

HINT:

• What is $\bigcup\limits_{x\in(a,b)}[x,b)$? And remember, the usual topology on $\Bbb R$ is Hausdorff.
• If $\mathscr{B}$ is a base for $\tau_U$, then for each $x\in\Bbb R$ there must be a $B_x\in\mathscr{B}$ such that $x\in B_x\subseteq[x,x+1)$ (why?); if $x\ne y$, is it possible that $B_x=B_y$?