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

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?

• I didn't downvote, but I will underline the fact that this community appreciates the poster showing their own effort 🙂. Apr 16, 2021 at 18:25
• Thanks for the tip. I will edit with some ideas that I've thought. Apr 16, 2021 at 18:27

• 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 $$B_x=B_y$$?