Weakly initial sets - examples and nonexamples

A weakly initial set in a category C is a set of objects I of C such that every object a of C has at least one arrow from an object contained in I.

The question is then, does Fields have a weakly initial set? This is equivalent to the collection of prime fields being a set.

The converse is, is there a (fairly natural) example of a category without a weakly initial set? Aside from obvious things like the discrete category on the objects of a large category.

-
Actually, this is a bit of a folly. My guess is that the fields Q, F_p are enough, but I wonder if there are crazy model theory type things that go on at large cardinals. Or am I worrying about nothing? –  David Roberts Oct 6 '10 at 4:33
Sorry, David, I'm not sure what you're worried about. Any field contains a smallest field (the smallest field containing 1) where 1 is either torsion (making the smallest field a finite field F_p) or not (making it Q). –  Todd Trimble Oct 6 '10 at 5:13
Well, there's the opposite of Fields, if that's natural enough? –  George Lowther Oct 6 '10 at 6:58
Ah, that's nice. So Fields has a weakly initial set, and Fields^op doesn't. Thanks. For the amount of space I want to talk about these two examples, that should be better than Todd's answer. –  David Roberts Oct 6 '10 at 8:00
And I realised how stupid my question regarding Fields was, but the second part of the question was less stupid, so I left it up. –  David Roberts Oct 6 '10 at 8:02

Regarding the second question: I'm not sure what should count as "natural", but couldn't you just work with examples where the solution set condition in an adjoint functor theorem fails? The solution set is a weakly initial set in a comma category.

For example, there is no left adjoint to the underlying-set functor $U$ from complete Boolean algebras to sets, and in particular no free complete Boolean algebra on a countably infinite set. But the category of complete Boolean algebras is small-complete and $U$ preserves all small limits. So it's the solution set condition that fails, and therefore the comma category

$$\mathbb{N} \downarrow U$$

has no weakly initial set.

Edit: After reading David's request for really simple, I offer instead $Ord^{op}$, where $Ord$ is the class of ordinals ordered by inclusion. I acknowledge the influence of Laurent's answer.

-
HI Todd - took away the 'accepted answer' tick, as more answers are rolling in, and I really need simple examples (really simple!) –  David Roberts Oct 6 '10 at 8:53