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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.

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  • $\begingroup$ 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? $\endgroup$
    – David Roberts ♦
    Commented Oct 6, 2010 at 4:33
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    $\begingroup$ 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). $\endgroup$
    – Todd Trimble
    Commented Oct 6, 2010 at 5:13
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    $\begingroup$ Well, there's the opposite of Fields, if that's natural enough? $\endgroup$
    – George Lowther
    Commented Oct 6, 2010 at 6:58
  • $\begingroup$ 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. $\endgroup$
    – David Roberts ♦
    Commented Oct 6, 2010 at 8:00
  • $\begingroup$ 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. $\endgroup$
    – David Roberts ♦
    Commented Oct 6, 2010 at 8:02

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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.

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  • $\begingroup$ HI Todd - took away the 'accepted answer' tick, as more answers are rolling in, and I really need simple examples (really simple!) $\endgroup$
    – David Roberts ♦
    Commented Oct 6, 2010 at 8:53
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How about the category of sets with injective maps as morphisms? As an ad hoc example, this may not count as "natural", but it's simple enough.

[EDIT] following Martin's comment: take the dual, or replace "injective" by "surjective".

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  • $\begingroup$ I think you mean "surjective" (with "injective", the empty set is initial). Then this is almost the same example as the dual category of Fields. $\endgroup$
    – Martin Brandenburg
    Commented Oct 6, 2010 at 8:50
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    $\begingroup$ Sorry, "surjective" indeed. Actually I had in mind the dual category, in analogy with the dual of Fields. I agree about your other comment; my point is only that this example is in a sense more straightforwrd. $\endgroup$
    – Laurent Moret-Bailly
    Commented Oct 6, 2010 at 10:54

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