I really think the following one is just intuitive. However, not only is it consistent to be false in ZFC, it is actually not known to be consistent in ZFC (its consistency follows from the existence of a weakly compact cardinal).

The conjecture is second-order, and thus must be regarded in NBG rather than ZFC, though I think this isn't too much of a cheat for the question. Furthermore, the negation of this conjecture is only known to be consistent as long as there is a worldly cardinal. A meager assumption nonetheless, but still quite a jump from ZFC in consistency strength.

Let $D$ be the class of all doubletons of ordinals (i.e. $\{\alpha,\beta\}$ such that $\alpha<\beta$). Let a class $H\subseteq D$ be *containable* iff there is some class $h$ such that every element of $H$ is a subclass of $h$ and every doubleton of $h$ is an element of $H$. The *Doubleton Ordinal conjecture* is as follows:

**For any class proper class $X\subseteq D$, there is a containable proper class $H\subseteq D$ such that either $H\subseteq X$ or $H$ is disjoint from $X$.**

Although this seems quite reasonable, it does not hold in $V_{\kappa+1}$ (the successor of $\kappa$ is used for second-order logic) for the least worldly cardinal $\kappa$. This is quite a strange result indeed.

In fact, $V_{\kappa+1}$ satisfies the Doubleton Ordinal conjecture iff $\kappa$ is $0$, $\omega$, or weakly compact.

3I think this is a very good question, but the example given doesn't seem particularly outside of normal logic intuition. When you say the group is "free" it means "it can be proven that there exists a basis B such that..." so surely you need find that specific B --- that really can't follow from a description like you gave without some mechanism! – Ilya Nikokoshev – 2009-10-22T21:41:17.623

6@ilya: I don't understand your objection to the example. The statement is that Whitehead's problem is independent of ZFC (which allows you to do things like chose elements of sets, and lots of other constructions). I agree it would be silly to say that it's independent of the empty list of axioms. – Anton Geraschenko – 2009-10-22T21:49:26.020

1Mm, I think I misunderstood what ZFC is. I wanted to point out that axiom of choice (or something) should be necessary which was obvious to the asker from the start. Apologies. – Ilya Nikokoshev – 2009-10-22T22:10:09.803