Can there be an uncountable set $S\subseteq\mathbb R$ such that for each subset $D\subseteq S$, there is an open set $U$ with $D=S\cap U$?

I'm asking merely out of curiosity, but I'll mention that this would imply $2^{\aleph_1}=2^{\aleph_0}$.

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    $\begingroup$ You are asking whether the real line has an uncountable discrete subspace. The answer is no, because the real line is a separable metric space, and every subspace is separable and second countable. $\endgroup$
    – bof
    Sep 14, 2021 at 0:11
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    $\begingroup$ See also a more interesting question replacing open by Borel sets: mathoverflow.net/q/403888/4600 $\endgroup$ Sep 14, 2021 at 6:30

1 Answer 1


An uncountable $S\subseteq \mathbb{R}$ has an accumulation point $x\in S$. Then for $D=\{x\}$ there is no such open set $U$.

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    $\begingroup$ Nice... I should have tried harder but hopefully you enjoyed writing that :) $\endgroup$ Sep 14, 2021 at 2:31

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