Let $X\subset \mathbb{R}$ be Borel measurable. Can it be that $\aleph_0 <|X|<2^{\aleph_0}$?

Question

I want to know if every Borel measurable set in the real line has cardinality either that of the naturals or of the reals. Of course the Continuum Hypothesis is not assumed.

It is clear that every open set has cardinality $2^{\aleph_0}$ (think of a ball of positive radius). I already feel lost when it comes to the closed sets.

Answer 1

It is not a trivial theorem, and it certainly needs the axiom of choice (well, after a certain number of unions and complements have been applied anyway). So the proof is not very obvious.

First we show that given a Polish space $X$ (separable and completely metrizable space), then every uncountable closed set has size continuum, this is done by essentially constructing a copy of the Cantor set inside the closed set.

Next we show that given any Borel set $A$, there is a topology extending the given Polish topology such that:

1. $A$ is clopen in the new topology.
2. The new topology is Polish.
3. The Borel sets are the same Borel sets.

If $A$ is uncountable, then it has to have size $2^{\aleph_0}$ since it must have a copy of the Cantor set inside of it.

You can find the proof in Kechris "Classical Descriptive Set Theory" on pp. 82-83.