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Consider the space $\mathbb{R}/\mathbb{Q}$ with metric $d([x],[y])=\inf_{r\in\mathbb{Q}}|x-y+r|$, where $[x]$ and $[y]$ are elements in the space. Since the quotient map $q:\mathbb{R}\rightarrow\mathbb{R}/\mathbb{Q}$ is continuous, any convergent sequence in $\mathbb{R}$ must also converge in the quotient space.

However if we consider a rational sequence $r_{n}$ converging to a irrational number $r$.In the quotient space, the sequence is mapped to $[0]$, so the limit of sequence is $[0]$, but this sequence converge to a irrational number, which does not get mapped to $[0]$. So the quotient sequence $q(r_{n})$ does not converge to $q(r)$, which is absurd. So what is wrong with my example?

Ken.Wong
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    That metric doesn't make sense. For example, $0$ and $\pi$ are not in the same equivalence class mod $\mathbb Q$, but $d([0],[\pi]) \le |\pi - r|$ for any $r \in \mathbb Q$. Now take a sequence $r$ of rational convergents of $\pi$ so that the distance goes to zero, contradiction. – Brevan Ellefsen Feb 24 '23 at 04:54
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    It seems like $\inf_{r\in\mathbb{Q}}|x-y+r| = 0$ for any $x, y$. – Maximal Ideal Feb 24 '23 at 04:55
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    The quotient topology on $\mathbb R/\mathbb Q$ is the indiscrete topology. Hence, it is not metrizable. See this for example. – azif00 Feb 24 '23 at 05:04

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