Prove that exist subset B of reals, with properties as in body (too long too write in title).

Question

Prove that there exists subset B of reals, such that B is linearly independent over $Q$ and nor B neither $R$\B contain any uncountable closed subset of $R$.

I had such exercise on set theory exam, and I was unsuccessful to obtain nonzero amount of points from that.

It is exercise for induction construction. What we know:

• base of $R$ over $Q$ has power continuum;

• there are continuum uncountable closed subsets of $R$;

• from Zermelo Theorem, we can create well order on set of indices of base of $R$ over $Q$ and a set of indices of uncountable closed sets;

There's no point in me writing what more I wrote, bcs it was bad. Could You please help me what to do next?

Answer 1

This is very close to the construction of a Bernstein set - we just need to fold in linear independence requirements.

In the standard construction of a Bernstein set, at stage $\alpha$ we have disjoint sets $In_\alpha, Out_\alpha$ of reals we've decided are in and out of our Bernstein set respectively, each of which has cardinality $<2^{\aleph_0}$, and we add an element to each of $In_\alpha$ and $Out_\alpha$ to ensure that the $\alpha$th uncountable closed set will not be a subset of either our Bernstein set or its complement. (And at limit stages we take unions.)

The only thing we need to be careful of here is the additional requirement that at each stage $In_\alpha$ should be linearly independent over $\mathbb{Q}$. So the key thing to prove is:

If $I,O$ are disjoint sets of reals each of cardinality $<2^{\aleph_0}$ with $I$ linearly independent, and $C$ is an uncountable closed set of reals, then there is some $x\in C$ such that $I\cup \{x\}$ is still linearly independent.

To prove this, just think about cardinality ...