For $x, y \in \mathbb{R}$, define $x \sim y $ if $x-y \in \mathbb{Q}$. Is $\mathbb{R}/\!\!\sim$ Hausdorff?
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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4Any circle around a real number will contain a rational, so... – 2011-04-04
2 Answers
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 $T_1$.
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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1@Didier: I wouldn't go so far as to call that pedantic :-) – 2011-04-04