# Show that a quotient space of $\mathbb{R}$ isn't Hausdorff [duplicate]

I've got a fun question, which is somewhat testing my topology skills.

The space we're working with is $\mathbb{R} \rightarrow \mathbb{R}/\sim$, which sends $x$ to $[x] = \{y \in \mathbb{R}: x-y \in \mathbb{Q} \}$, and what I'm trying to show is that $\mathbb{R}/\sim$ isn't Hausdorff.

What I'm struggling with is proving that, for certain $[x],[y] \in \mathbb{R}/\sim$, that ALL open $U_{[x]}, U_{[y]}$ have a non-empty intersection. Intuition says that these open sets overlap, since in any open set around $[x]$ or $[y]$, there must be a rational, so this open set must also conatin all of $\mathbb{Q}$, so these open sets of $\mathbb{Q}$ in common.

Formalizing this is giving me trouble. How do I take an arbitrary open set in such an equivalence class? Is this just $[B_\varepsilon(x)]=\{[x] \in \mathbb{R}/\sim: x \in B_\varepsilon(x)\}$?

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## marked as duplicate by martini, azarel, t.b., Asaf Karagila, Zev ChonolesJun 6 '12 at 18:58

This question was marked as an exact duplicate of an existing question.

Just saw that in the related sidebar, question answered! – BallzofFury Jun 6 '12 at 18:35
It's a shame, because your question is much better-written than that one :P in any case, this is sort of a "trick" question inasmuch as the topology is actually trivial (i.e. the only nonempty open set is the whole set). No points are even topologically distinguishable, let alone separated by disjoint open sets! – Ben Millwood Jun 6 '12 at 18:37
I agree it's kind of trivial, but it's good practice for working with quotient spaces. Something other than the usual examples given. – BallzofFury Jun 6 '12 at 18:42
Oh yes. That's sort of what I meant - it's a "trick" question because the substance of the question isn't anything to do with Hausdorffness. – Ben Millwood Jun 6 '12 at 18:42

In answer to this, you don't: your open sets are sets of equivalence classes, and are roughly $$\left\{ [y] : y - x \in \mathbb Q, x \in B_\varepsilon(x_0) \right\}$$
It's now easier to see that for any $y \in \mathbb R$, you can subtract off some rational to land in $B_\varepsilon(x_0)$, so $[y]$ is in the open set above. The arbitrary open set above. Hence there is only one non-empty open set, namely the whole space.
Once you're accustomed to the definition, you'll stop thinking of them as equivalence classes and just think of it as "$\mathbb R$, but with points separated by a rational distance glued together". Then every point is "near" every other point, because every point is "near" a rational, and all the rationals are glued together. Hence the topology is trivial.