Is it conceivable that [Frankl's union closed sets conjecture](http://en.wikipedia.org/wiki/Union-closed_sets_conjecture) is undecidable in $\mathsf{ZFC}$, or is this quite implausible, perhaps due to the "finitistic" nature of the statement, or for some other reason?



Decidability of Frankl's union-closed sets conjecture
The statement has [complexity $\Pi^0\_1$](http://en.wikipedia.org/wiki/Arithmetical_hierarchy), which means that it has a
single universal quantifier, quantifying over the possible
union-closed sets, and then making a simple assertion about those
objects.


Although this is a very simple level of complexity, it is the same
complexity as consistency assertions, and these admit of a robust
independence phenomenon. For example, even ZFC, if consistent,
admits of independent statements of this level of complexity, such as the statement Con(ZFC), which asserts that ZFC is consistent.


Meanwhile, if we can prove in ZFC (or some other theory) that the
union-closed set conjecture was independent of PA, then it would
follow that the statement was true in the standard model. The
reason is that if a $\Pi^0\_1$ statement is independent of PA, then
it can admit of no standard counterexample, since PA would prove
that this was a counterexample. So, the situation is that if we
can prove in ZFC that the union-closed set conjecture is
independent of PA, then we would be able to prove in ZFC that it
is true.


