# VC dimension of standard topology on the reals

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

• 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.
– bof
Sep 14 at 0:11
• See also a more interesting question replacing open by Borel sets: mathoverflow.net/q/403888/4600 Sep 14 at 6:30

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