Let $\mathbb{R}$ be the set of reals with a topology $\mathcal{T}$ such, that each $x\in\mathbb{R}$ has a base of open regions $$\mathcal{B}_x=\{\{x\}\cup\{y\in\mathbb{Q}:|x-y|<1/n\}\}_{n\in\mathbb{N}}$$
- Show the space is seperable and $T_2$
- Show that for each $x\in\mathbb{R}$ and $U$ open subset of $(\mathbb{R}, \mathcal{T})$ which contains $x$, there is an open set $V$ such that $x\in V\subseteq\overline{V}\subseteq U$. Then conclude that the space is $T_3$.
- Show that the singleton $\{q\}$, where $q\in\mathbb{Q}$, and the set of irrationals $\mathbf{Irr}$ are closed and not separable by two disjoint open sets. Conclude that the space is not $T_4$.
My question is
How can this space be $T_3$ when, according to (3), for $q\in\mathbb{Q}$ and $\mathbf{Irr}$ (which is closed and does not contain $q$) there is no open set separating them? Isn't this exercise contradictory?
You are right, the exercise is contradictory. And the space is not $T_3$. For rational $x$, the closed neighbourhoods don't form a neighbourhood base.
If $x\in\mathbb{Q}$, take $U = (x-1,x+1) \cap \mathbb{Q}$. This is an open set, and every open set $x \in V \subset U$ contains $W_\varepsilon := (x-\varepsilon,x+\varepsilon)\cap\mathbb{Q}$ for some $\varepsilon > 0$. But $(x-\varepsilon,x+\varepsilon) \subset \overline{W}_\varepsilon$, whence $\overline{W}_\varepsilon \not\subset U$.