# Indiscreet topology on quotient space

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!

• That a set is not open does not mean that it is closed. Consider the definition of open set in Euclidean topology. It is that for every element in $U$, there exists some ball (open interval) with positive radius around it, that lies entirely in $U$. – JSchoone Dec 30 '16 at 16:15
• Hint: have a look at MSE here, e.g., this question; so indiscrete topology just means trivial topology. – Dietrich Burde Dec 30 '16 at 16:16
• You want to prove that the only set of equivalence classes whose union is an open subset of $\mathbb{R}$ is the empty set and the set of all equivalence classes. If you take any other set of equivalence classes you can see it's not open by a direct argument. – AnonymousCoward Dec 30 '16 at 16:17
• For topology the word is indiscrete. Indiscreet is a real word, it means not being able to keep something secret. – Will Jagy Dec 30 '16 at 17:44
• You only need the definition that $O$ in the quotient is open iff $q^{-1}[O]$ is open in the original space. – Henno Brandsma Dec 31 '16 at 15:57

## 2 Answers

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?

• At first sight I would say no. We have $[x] = \{ z \in \mathbb{R} | x - z \in \mathbb{Q}$ so if we take an open ball with sufficient small radius (for example if we have $x = 2.73$ and we take $r = 0.01$), there is an open ball include in $S$ no? – Jacobi Dec 30 '16 at 16:36
• Hint: the rationals are dense in the reals. – AnonymousCoward Dec 30 '16 at 17:00
• I am asking you to prove that every ball around $x$ must contain an element in the equivalence class of $y$. – AnonymousCoward Dec 30 '16 at 17:01
• Ok, I got it! Thanks a lot! – Jacobi Dec 31 '16 at 16:07

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.