Consider the topology $\tau_1$, the standard topology on $\mathbb R$, and $\tau_2$, the cocountable topology (here we will consider a set open with respect to $\tau_2$ if its complement is either finite or countable). We define $\tau=\tau_1\lor \tau_2$ as the collection of subsets $W\subseteq \mathbb R$ such that for every $x \in W$, there exist subsets $U,V \subseteq \mathbb R$ with $x \in U$, $x \in V$, $U \in \tau_1$, $V \in \tau_2$, and $U \cap V \subseteq W$. Show that $(\mathbb R, \tau)$ is not regular.

I know that a regular topological space is defined as a $T_1$ topological space such that for every $x \in \mathbb R$ and closed set $E \subseteq \mathbb R$ with $x \notin E$, there are disjoint open sets $U,V \subseteq \mathbb R$ such that $x \in U$ and $E \subseteq V$. So, I want to show that there exists some $p \in \mathbb R$ and a closed set $E \subseteq \mathbb R$ such that $p \notin E$ and $U$ and $V$ are not disjoint for every $U,V \subseteq \mathbb R$ with $p \in U$ and $E \subseteq V$.

I'm not really sure how to set this proof up. I have a proposition that says a regular topological space is Hausdorff, so one idea I had was to show that $(\mathbb R, \tau)$ is not Hausdorff, but I'm not sure if it would be better to attempt a direct proof or not. For a direct proof, I want to consider a set $E$ which is countable (or finite). Then, $E$ will be closed with respect to $\tau$. I'm not sure how to find a $p$ to fit my needs, though.

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    $\begingroup$ $(X,\tau)$ is Hausdorff as $\tau_1 \subseteq \tau$, so that route is out. $\endgroup$ – Henno Brandsma Apr 15 '18 at 21:17

Note that $\mathbb{Q}$ is closed in $\tau$, as $\mathbb{Q}$ is countable and so $\mathbb{R} \setminus \mathbb{Q} \in \tau_2 \subseteq \tau$ and so its complement is closed.

Also, $\sqrt{2} \notin \mathbb{Q}$. Suppose we have $U$ and $V$ in $\tau$ such that $\sqrt{2} \in U$, $\mathbb{Q} \subseteq V$.

By definition of $\tau$ we can find a Euclidean open set $O_1$ and an at most countable set $N_1 \subseteq \mathbb{R}$ such that $$\sqrt{2} \in O_1 \setminus N_1 \subseteq U$$

We can find a rational number $q \in O_1$, as $\mathbb{Q}$ is dense in the usual topology on the reals.

As $V$ is open and $q \in V$ we again by definition of $\tau$ find a Euclidean open set $O_2$ and an at most countable set $N_2 \subseteq \mathbb{R}$ such that $$q \in O_2 \setminus N_2 \subseteq V$$

But then note that $O_1 \cap O_2$ is a non-empty Euclidean subset of $\mathbb{R}$ (it contains $q$ e.g.) and so it contains uncountably many irrational numbers. We can thus find an irrational $p$ such that

$$p \in O_1 \cap O_2 , p \notin N_1 \cup N_2$$

And we can immediately check that $p \in U \cap V \neq \emptyset$. This implies that $(X,\tau)$ is not regular.

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