# Showing $\mathbb{R}/\mathbb{Q}$ is not Hausdorff

I stumbled upon this link while googling for an answer to the same question.

Under point number 3, the sole answer mentions an example where it is said $\mathbb{R}/\mathbb{Q}$ is not Hausdorff, because the topology generated is trivial, assuming $\mathbb{R}$ has the standard topology. I agree with the conclusion but I think the reason given is wrong. I just wanted to cross-check my argument.

If I take $(\mathbb{R}/\mathbb{Q}) \backslash [e]$ where e is the equivalence class of the Euler number $e$,

then the inverse of the projection map $\pi: \mathbb {R}\rightarrow \mathbb{R}/\mathbb{Q}$ is $\mathbb{R}\backslash \{e\}$ which is an open set in $\mathbb{R}$.

And so the topology consists of open sets other than $X$ and $\phi$, which means the topology is not trivial.

Is there something wrong with the above argument?

NOTE: I agree with the conclusion though. Let $U \neq \phi$ be an open set in $\mathbb{R}/\mathbb{Q}$ . Let $q\in \mathbb{Q}$ then if $[q] \notin U$, $\pi^{-1}(U)$ will consist of only irrationals which cannot be open, and so the space is not hausdorff, since any two non-empty open sets have non-empty intersection containing atleast $[q]$.

• $\pi^{-1}(\mathbb{R/Q} \setminus [e]) = \mathbb{R} \setminus (e + \mathbb{Q})$, which contains no intervals. – Anthony Carapetis Aug 28 '13 at 1:55
• The inverse of the projection map seems to also contain $e+0.0123$. – André Nicolas Aug 28 '13 at 2:11

Let's get really broad, and also hope I'm not making any major errors. Forgive (or fix) the awful variable names.

Let $D$ be a dense set in a space $S$.

Let $f\colon S \to X$ be any quotient map that is constant on $D$, and suppose that $f$ maps every element of $D$ to $q$.

Let $y \in X\setminus\{q\}$.

Let $U$ be an open set about $y$, so $f^{-1}[U]$ is a non-empty open set.

Then $f^{-1}[U]$ contains an element, $b$, of $D$, so $q\in U$.

So we see that every non-empty open set in $X$ contains $q$, so $X$ cannot even be $T_1$, let alone Hausdorff.

The topology is trivial: every open set in $\mathbb{R}$ contains a complete set of coset representatives for $\mathbb{R}/\mathbb{Q}$. So if $U\subseteq\mathbb{R}/\mathbb{Q}$ is open and nonempty, $\pi^{-1}(U)$ is open in $\mathbb{R}$ and $\pi(\pi^{-1}(U))$ is all of $\mathbb{R}/\mathbb{Q}$.

• Oh I see where my confusion is occuring. The answer uses $\mathbb{R}/\mathbb{Q}$ as the collection of cosets of $\mathbb{Q}$ where $\mathbb{R}$ is considered as a group, but the book I am using uses this notation for the concept of adjunction space, as mentioned in this link on Wikipedia en.wikipedia.org/wiki/Quotient_space under the tag adjunction space. Well, anyway the topologies in both senses are non-Hausdorff. :-) – smilingbuddha Aug 28 '13 at 2:27