$\mathbb{R}/\mathbb{Q}$ has the indiscrete topology

Let \(U \subset \RR/\QQ\) be open and nonempty, show \(U = \RR/\QQ\).

Let \([x] \in U\), then \(x \in \inverseof{\pi}(U) \definedas V \subset\RR\) is open.

Then \(V\) contains an interval \((a, b)\)

Every \(y\in V\) satisfies \(y+q \in V\) for all \(q\in \QQ\), since \(y+q-y \in \QQ \implies [y+q] = [y]\).

So \((a-q, b+q) \in V\) for all \(q\in \QQ\).

So \(\union_{q\in \QQ}(a-q, b+q) \in V \implies \RR \subset V\).

So \(\pi(V) = \RR/\QQ = U\), and thus the only open sets are the entire space and the empty set.