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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).

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2 Answers 2

up vote 9 down vote accepted

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.

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Is cub a variant I haven't seen, or do you mean club? –  Chris Eagle Jan 1 '13 at 21:01
    
@Chris: The former; that’s how I learnt it. –  Brian M. Scott Jan 1 '13 at 21:03
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Hi Brian, I abusively updated one of your links. –  Andres Caicedo Jan 1 '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 2 '13 at 15:31

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.

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