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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 $[q]$.

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

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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 $\mathbb{R}/\mathbb{Q}$.

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  • $\begingroup$ 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. :-) $\endgroup$ – smilingbuddha Aug 28 '13 at 2:27
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The topology $\Bbb R\backslash \Bbb Q$ is trivial topology and hence not Hausdorff. It is trivial as follows:
Let $x \in \Bbb R$. If $x \in \Bbb Q'$, then, equivalence class of $x$ is $$[x] = x + \Bbb Q \\= \{x+p\,|\; p \in \Bbb Q\}$$ Similarly, if $x \in \Bbb Q$, then the equivalence class of $x$ is $\Bbb Q$ (easy to see).

Also note that all of these are closed sets and there is no open set in $\Bbb R$ which is saturated for if it were open, then there should be an open ball contained in the set, around every point of the set. But clearly, we can always find a point in these open balls which is not in the set. This can be prevented only if we take the set $\Bbb R$ and thus, $\phi$ and $\Bbb R$ are the only saturated sets in $\Bbb R$.

Since open sets in quotient topology are in one-one correspondence with the saturated open sets in original space, the topology on $\Bbb R\backslash \Bbb Q$ is trivial.

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