# Quotient Space $\mathbb{R} / \mathbb{Q}$

I've just learned about topological quotient spaces and was wondering if anyone can help me with this example I thought of.

Let $(\mathbb{Q}, +)$ be the usual group of rational numbers for addition, likewise $(\mathbb{R}, +)$. Set $S$ to be the set of all cosets, t.i. $S=\mathbb{R}/\mathbb{Q}=\{x + \mathbb{Q} \mid x \in \mathbb{R} \}$. What is the quotient space $\mathbb{R} / S$ like? ($\mathbb{R}$ is equipped with the regular euclidian topology) What is it homeomorphic to? What does a typical open set look like?

Thanks.

• Hint: the topology on this space is a "standard" topology that one learns as an example, upon first learning what a topological space is. And you can determine which topology it is by knowing that between every two irrational numbers there is a rational number, together with the definition of the quotient topology. – Ryan Budney Feb 27 '11 at 23:12
• I think it is easier simply to use directly the fact that open sets in $\mathbb R/\mathbb Q$ are in bijections with the saturated open subsets of $\mathbb R$, that is, the open subsets $U\subseteq\mathbb R$ such that $$x\in U,q\in\mathbb Q\implies x+q\in U.$$ Can you describe all the saturated open subsets of $\mathbb R$? – Mariano Suárez-Álvarez Feb 28 '11 at 0:23
• How do you want to take the quotient of $\mathbb{R}^2$ with respect to $S$? Since $S$ is not a subset of $\mathbb{R}^2$, I don't see a canonical way here. – Nemo Mar 27 '19 at 13:51

I'm thinking the topology is trivial on the set $S$. Since if the set $U$ is open in $\mathbb{R} / S$ then it's preimage of $q$ (where $q$ is quotient mapping) must be open in $\mathbb{R}$, meaning there exists an open interval $J \subseteq q^{-1}(U)$. But $q(J)$ equals all of the cosets in $\mathbb{R} / S$. Am I right?
• Do you really mean $\mathbb{R}/S=\mathbb{R}/(\mathbb{R}/\mathbb{Q})$ and not $S=\mathbb{R}/\mathbb{Q}$? If so, how do you define $\mathbb{R}/(\mathbb{R}/\mathbb{Q})$? – joriki Jun 17 '11 at 5:50