I stumbled upon this [link](https://math.stackexchange.com/questions/50044/quotient-space-of-hausdorff-space) 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 $[q]$.