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I'm trying to prove that the class $\mathbb{Z}$ has no countable neighbourhood basis in the quotient space $\mathbb{R} / \mathbb{Z}$. I started with taking a neighbourhood basis $\mathcal{U} = \{ U_{i} : i \in I \}$ of $\mathbb{Z}$ in $\mathbb{R} / \mathbb{Z}$. I'd like to prove that $\mathcal{U}$ is uncountable but I'm stucked at this point. I'll appreciate it if someone could give a hint on how to proceed/prove this. Thank you.

$\mathbb{R} / \mathbb{Z}$ here is the partition $\{ \{ x \} : x \in \mathbb{R} \setminus \mathbb{Z} \} \cup \{ \mathbb{Z} \}$ on $\mathbb{R}$.

Shuichi
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    it has a countable neighborhoood basis: $\mathbb{R}/\mathbb{Z} \cong S^1$ – psl2Z Sep 14 '22 at 11:43
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    @psl2Z no, $\mathbb{R} / \mathbb{Z}$ means the set $\mathbb{Z}$ is collapsed to a point. Hence the class where a non-integer $x$ belongs to is a singleton, namely ${ x }$. – Shuichi Sep 14 '22 at 11:48
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    Take a look at our guidelines for how to avoid "I'm not sure where to start" questions. For example, I'm sure you were able to write down the definitions and convert them into a proof strategy, so adding that to your question would be a good start. – Lee Mosher Sep 14 '22 at 11:50
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    Please edit your question to explain what $\mathbb R/\mathbb Z$ is. Since $\mathbb Z$ is a subgroup of $\mathbb R$, the reader is tempted to understand it as the quotient group. – Paul Frost Sep 14 '22 at 11:55
  • Suggestion: Can you be more explicit about what a typical neighborhood $U_i$ of $\mathbf{Z}$ looks like? – Andrew D. Hwang Sep 14 '22 at 12:19
  • See here https://math.stackexchange.com/a/1417425/1016538 – psl2Z Sep 14 '22 at 12:26

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