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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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    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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How do I take an arbitrary open set in such an equivalence class?

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.