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It seems that the answer is no, but I'm struggling to think of a proper example.

Would I be best trying to think of a closed set whose complement is open?

Dazzler95
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2 Answers2

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$ S:=\left\{\frac{1}{n}: n\ge 1\right\} $ is neither closed nor open (in the usual topology of $\mathbf{R}$).

Paolo Leonetti
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$\mathbb{Q}$ is not closed in $\mathbb{R}$ (with the usual metric)

Netchaiev
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