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Possible Duplicate:
Quotient Space $\mathbb{R} / \mathbb{Q}$
For $x, y \in \mathbb{R}$, define $x \sim y $ if $x-y \in \mathbb{Q}$. Is $\mathbb{R}/\!\!\sim$ Hausdorff?

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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    Just saw that in the related sidebar, question answered!2012-06-06
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    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!2012-06-06
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    I agree it's kind of trivial, but it's good practice for working with quotient spaces. Something other than the usual examples given.2012-06-06
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    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.2012-06-06

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