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).
Asked
Active
Viewed 352 times
2 Answers
10
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.
Brian M. Scott
- 631,399
-
Is cub a variant I haven't seen, or do you mean club? – Chris Eagle Jan 01 '13 at 21:01
-
@Chris: The former; that’s how I learnt it. – Brian M. Scott Jan 01 '13 at 21:03
-
1Hi Brian, I abusively updated one of your links. – Andrés E. Caicedo Jan 01 '13 at 23:24
-
@Andres: Thanks; nothing like going to the source. (I think that I’ve pointed more people to your blog than to any other except Dan Ma’s topology blog.) – Brian M. Scott Jan 02 '13 at 15:31
6
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.
Asaf Karagila
- 405,794