Indiscreet topology on quotient space

Question

I'm studying basical topology and I can't figure out something. In a syllabus I read, it is written "Let $\mathbb{R}$ be a topological space with the euclidian topology. We can define an equivalence relation such that $ x \sim y \Leftrightarrow x - y \in \mathbb{Q}$". Moreover it is said that "the quotient topology on $\mathbb{R}/\sim$ is the indiscreet topology" but I can' figure out how to prove it.

I tried to prove that any subset (except the empty set) of $\mathbb{R}/\sim$ is close but without information about the topology it seems hard.

Can anyone have an hint?

Thanks!

Answer 1

Following my comment above:

Let $S$ be some nonempty proper subset of $\mathbb{R}$ that is a union of equivalence classes. Since it is non-empty it contains at least one point, call this $x$. Since it is a proper subset, there is an element $y\in \mathbb{R}\setminus S$.

In order to show that $S$ is not open, it suffices to show that every open ball of radius $r>0$ centred at $x$ is not contained in $S$. Can you show this?

Answer 2

Let $V = \{x_\alpha, \alpha < \mathfrak{c}\}$, be a set of representatives for the equivalence relation (one point for each class of $\sim$), a so-called Vitali set. So $\mathbb{R}/\mathbb{Q}$ can be seen as $ V$, essentially. Now let $O$ be any set that is non-empty open in this quotient space. Then $q^{-1}[O]$ is non-empty open in $\mathbb{R}$. And for every $\alpha$, $x_\alpha + \mathbb{Q}$ is dense in the reals as all shifts are autohomeomorphisms of the reals, so this set intersects $q^{-1}[O]$. But if $x\in q^{-1}[O]$ can be written as $x = x_\alpha + q, q \in \mathbb{Q}$, then $x \sim x_\alpha$, showing that $x_\alpha = q(x) \in O$. As this holds for all $\alpha$, $O$ must contain all classes of $\sim$, so the only non-empty open set is the whole space, ergo it's indiscrete.