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 $2^{\aleph_1}=2^{\aleph_0}$.

  • 4
    $\begingroup$ 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
  • 1
    $\begingroup$ 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 $U$.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.