Non-measurable subset of $\omega_1$

Question

Consider $\omega_1$ equipped with the order topology. Then Borel subsets of $\omega_1$ are precisely those which contain a closed and unbounded set or the complement contains such a set. There must be (in ZFC) sets which lack this property, as otherwise $\omega_1$ would be measurable. Can one please tell how to "construct" such sets (using some form of choice, of course).

Answer 1

A set $S\subseteq\omega_1$ is stationary if it has non-empty intersection with every cub set, so it suffices to construct two disjoint stationary sets: neither can contain or be disjoint from any cub set. In fact this post from Andres Caicedo’s blog shows how to construct $\omega_1$ pairwise disjoint stationary subsets of $\omega_1$.

You may also find the two references in Joel David Hamkins’s answer to this MathOverflow question helpful.

Answer 2

It requires the axiom of choice, but there are sets which are stationary and co-stationary. Namely they do not contain a club not they a disjoint from one.

For example using Solovay's theorem we can partition $\omega_1$ into $\omega_1$ disjoint stationary sets.

In some models where the axiom if choice fails every subset of $\omega_1$ contains a club or is disjoint from one.