9
$\begingroup$

Motivated by this question I'm looking for a pair of Baire topological spaces whose product is not Baire and whose construction does not need the axiom of choice.

The example of such spaces I'm familiar with, which is due to Cohen, is to consider $\Bbb P_{S_1}\times\Bbb P_{S_2}$, where $S_1,S_2\subseteq\omega_1$ are disjoint stationary subsets of $\omega_1$ and $\Bbb P_{S_i}$ is the usual $\sigma$-distributive forcing that adds to $\omega_1$ a club contained in $S_i$, but this construction needs some choice.

I'm not too familiar with other examples of Baire spaces which are not productively Baire, but those that I could find in the literature all seemed to need some form of choice. Is there a construction in $\mathsf{ZF}$ of two Baire spaces $X,Y$ such that $X\times Y$ is not Baire? Or at the opposite, is it consistent with $\mathsf{ZF}$ that every finite product of Baire spaces is Baire?

Improve this question
$\endgroup$
9
  • $\begingroup$ I think that your example can be stretched even more. If we assume DC, then for all regular $\kappa$, the club filter restricted to $S^{\kappa}_{\omega}$ is an ultrafilter (if $S_1, S_2 \subseteq S^{\kappa}_{\omega}$ are both stationary and disjoint, then the forcings $Col(\omega_1, \kappa) * \mathbb{P}_{S_1}$ and $Col(\omega_1, \kappa) * \mathbb{P}_{S_2}$ will have non-$\sigma$-distributive product). This looks like a ZF variant of the HOD-conjecture... $\endgroup$
    – Yair Hayut
    Commented Jan 22, 2020 at 19:26
  • $\begingroup$ @Yair: If you assume DC, might as well just $\operatorname{Add}(\omega_1,2)$ and then take $S_i$ to be the $i$th Cohen subset (or some other configuration of easily definable set from each Cohen subset). $\endgroup$
    – Asaf Karagila ♦
    Commented Jan 22, 2020 at 21:02
  • $\begingroup$ @Asaf: I don't understand your example. I think that the product that you describe is going to be $\sigma$-distributive (under DC), because those two Cohen sets are going to have a large intersection. In the example that I gave, I'm using the fact that $S_1$, $S_2$ are from the ground model. $\endgroup$
    – Yair Hayut
    Commented Jan 23, 2020 at 8:02
  • $\begingroup$ @YairHayut: Okay, then consider the almost disjoint forcing for $\omega_1$ (so a condition is a countable initial segment, and a countable family of NS subsets), then iterate it twice. Since we have DC, and these are $\sigma$-closed, each step also preserves DC. And yes yes, I know what you're going to say, why are these stationary? Well, they are meeting every club in the ground model, and I think there is enough structure there to ensure that new clubs are not too many, and are also met. This can probably be significantly tweaked and improved, too. $\endgroup$
    – Asaf Karagila ♦
    Commented Jan 23, 2020 at 8:09
  • $\begingroup$ Alessandro, the real question is whether or not it is consistent that there are no $\sigma$-distributive forcings to begin with. For example, what is the situation in Gitik's model where every set is a countable union of smaller sets? Unclear. This somewhat relates to Foreman's maximality principle as well (without choice), that every forcing adds a real or collapses cardinals. But without choice. $\endgroup$
    – Asaf Karagila ♦
    Commented Jan 23, 2020 at 14:16

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .