Show that the right half-open topology on $\mathbb R$ is not metrisable. The right half-open topology on the real line $\mathbb R$ is the topology generated by the right half-open intervals $[a,b)$ for $ -\infty < a < b < \infty$.
How to show it does not arise from a metric? I know it's a Hausdorff space so this approach doesn't work.
Any idea is appreciated.
 A: A separable metric space is second countable (that is, its topology has a countable base).
$\mathbb{R}$ in the right half-open topology is separable ($\mathbb{Q}$ is dense), but not second countable (proof left as an exercise).
A: This can be proven using separability, as Daniel Fischer pointed out, or the fact that $\Bbb{R}$ is Lindelöf with this topology. 
A space is Lindelöf if every open covering contains a countable subcovering. To show that this topology on $\Bbb{R}$ is Lindelöf we show that every covering by basis elements has a countable subcovering. Let $\Bbb{R} = \cup_{i\in I} C_i,$ where $C_i = [a_i,b_i)$. Set $C = \cup_{i\in I} (a_i,b_i)$.  We have that $\Bbb{R}  - C$ is countable, since it contains at most those points $a_i$ and for each point in this there is a rational point in its associated basis element, giving an injection from $\Bbb{R} - C$ into $\Bbb{Q}$. The set $\Bbb{R}$ is covered by the basis elements in the open cover assocated with the points in $\Bbb{R} - C$ and this is therefore a countable subcover, since there are a countable number of these.
We use the fact that every metrizable Lindelöf space is second countable (an easy exercise).
All that's left to show is that $\Bbb{R}$ with this topology is non second countable. For a basis $B$ of $\Bbb{R}$ pick an element of this basis $B_x$ for each $x\in \Bbb{R}$ such that $x \in B_x \subset [x,x+1)$. The function which takes $x$ to $B_x$ is injective, and thus $B$ cannot be countable. 
A: Its square is not normal, so it cannot be metrizable, which implies that $X$ is not metrizable.
A: In a metrizable space, Lindelof $\iff$ separable $\iff$ second countable.
