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This comment appeals to a property of connected component in $\mathbb R^n$. I'm trying to see how it works. This property was also mentioned in this thread.

Theorem: A connected metric space $(E, d)$ with at least $2$ points has no isolated points.

Could you confirm if my below reasoning is correct?

Proof: Assume $a \in E$ is an isolated point of $E$. Fix $b\in E \setminus \{a\}$. There is $r>0$ such that $X :=B(a, r) = \{a\}$, which is both closed and open. On the other hand, $X \notin \{E, \emptyset \}$, which is a contradiction.

Didier
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Akira
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    This is good. Alternatively, you can say that both ${a}$ and $E\setminus {a}$ are open, so that $E={a}\cup(E\setminus {a})$ is a disconnection of $E$, and $E$ is thus not connected. – Didier Oct 02 '22 at 09:06

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