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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Sign up to join this communityCan 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 $2^{\aleph_1}=2^{\aleph_0}$.
An uncountable $S\subseteq \mathbb{R}$ has an accumulation point $x\in S$. Then for $D=\{x\}$ there is no such open set $U$.