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

I'm asking merely out of curiosity, but I'll mention that this would imply $2^{\aleph_1}=2^{\aleph_0}$. This is a hopefully more interesting adaption of a recent too easy question.


1 Answer 1


Yes! Martin's Axiom implies that if $S \subseteq \mathbb R$ and $|S| < \mathfrak{c}$, then every subset $D$ of $S$ is a relative $G_\delta$ in $S$: i.e., there is a $G_\delta$ set $X \subseteq \mathbb R$ with $X \cap S = D$. (And let me note that $2^{\aleph_0} = 2^{\aleph_1}$ is another consequence of Martin's Axiom.)

  • $\begingroup$ Nice, this applies to Kathleen Romanik's testing dimension as well. $\endgroup$ Sep 14, 2021 at 17:29

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