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

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Any circle around a real number will contain a rational, so... – Alex B. Apr 4 '11 at 8:56

2 Answers 2

Expanded version of user9077's answer:

Let $[0]$ be the class of $0$ (and thus of any rational), and let $[x]$, $x \in \mathbb{R}\setminus\mathbb{Q}$ be any other class in $Y = \mathbb{R}/\sim$. Let $O$ be any open set in $Y$ that contains $[x]$. This means that $O' = q^{-1}[O]$ (where $q$ is the quotient map from $\mathbb{R}$ onto $Y$) is open in $\mathbb{R}$ and contains $x$, so $O'$ contains a point $y \in \mathbb{Q}$, as $\mathbb{Q}$ is dense in $\mathbb{R}$. But this means that $[0] = [y] = q(y) \in q[O'] = O$. So every open neighbourhood of $[x]$ contains $[0]$, as claimed, and so $Y$ is not $T_1$.

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It is not even T1. As Alex mentioned above, just take two points $x,y$ in $\mathbb{R}\backslash \sim$ where $x\in \mathbb{Q}$ and $y\not\in \mathbb{Q}$. Then any open neighborhood containing $y$ contains $x$ as well.

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Pedantic note: if $x$ and $y$ are in $\mathbb{R}/\sim$, these are equivalence classes, hence the statements $x\in\mathbb{Q}$ and $y\notin\mathbb{Q}$ mean nothing and should be replaced by $x=\mathbb{Q}$ (because $\mathbb{Q}$ is obviously an equivalence class for the relation $\sim$) and $y\ne\mathbb{Q}$ respectively. – Did Apr 4 '11 at 9:32
@Didier: I wouldn't go so far as to call that pedantic :-) – joriki Apr 4 '11 at 9:34

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