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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 $[B_\varepsilon(x)]=\{[x] \in \mathbb{R}/\sim: x \in B_\varepsilon(x)\}$?