VC dimension of Borel sets [duplicate]

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.

• FYI, $2^{\aleph_1}=2^{\aleph_0}$ is sometimes called Lusin's second continuum hypothesis or Lusin's hypothesis. Sep 14, 2021 at 6:09
• Sep 14, 2021 at 12:38
• @Joseph ah yes, the answer is the same although the question is asking for less... Sep 14, 2021 at 16:21
• This question is originally due to Keith Ramsay. Sep 14, 2021 at 17:31
• For every uncountable complete separable metric space X, there is a bijection $f:X\rightarrow\mathbb{R}$ such that $A\subseteq X$ is Borel if and only if $f[A]$ is Borel. With this Borel equivalence in mind, we know that my old question is equivalent to this question. Sep 14, 2021 at 19:55

1 Answer

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.)

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