The topology $\mathcal{C}$ is given by $[a,b)$ where $a,b\in\mathbb{Q}$ . We noted in the solution for Exercise 8 of §13 that this topology is strictly finer than the standard topology and strictly coarser than the lower limit topology. The finer is the topology, the (weakly) smaller is the closure of any set, as there are more neighborhoods of points not in the set. So, we expect the closures of the two sets in both topologies to be subsets of their closures in $\mathbb{R}$ , i.e. $[0,\sqrt{2}]$ and $[\sqrt{2},3]$ .

In the lower limit topology, $\overline{A}=[0,\sqrt{2})$ and $\overline{B}=[\sqrt{2},3)$ , as the closure of any interval $(a,b)$ is $[a,b)$ (point $b$ has the neighborhood $[b,+\infty)$ not intersecting the interval). So, additionally, in $\mathcal{C}$ we would expect the closures to be also supersets of these half-open intervals.

In $\mathcal{C}$ , $\overline{A}=[0,\sqrt{2}]$ and $\overline{B}=[\sqrt{2},3)$ . Indeed, for the set $A$ the argument is close to the one described in the solution of Exercise 8 of §13: the set $[\sqrt{2},2)$ is open in the lower limit topology, but not in $\mathcal{C}$ , where every open interval containing $\sqrt{2}$ has a rational lower bound and, hence, a point below $\sqrt{2}$ from the interval $A$ .