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In model theory, for the construction of an ultraproduct one needs the existence of an ultrafilter.

I heard that one can't construct an ultrafilter by hand, one cannot write it down, just merely knowing that one exists by using the axiom of choice. How is this meant? Even if I give you a concrete base set, say $X=\mathbb N$, is it not possible to write down an ultrafilter on $X$? (I just notice, that $U$ defined by $A\in U:\Longleftrightarrow 3\in A$ probably is an ultrafilter on $\mathbb N$, but it is a boring one, so my question should ask for real examples.)

Of course, up to now, my question is quite imprecise. Maybe one can formalize it like so: in ZF (without AC), is it possible to prove the existence of an ultrafilter on $\mathbb N$ which is non-principal?

What about other base sets, say finite ones? Can ZF prove the existence of an ultrafilter on a finite set?

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    This is an interesting situation. An object proved to exist by ZFC but cannot be written down explicitly. Example... a set that is not Lebesgue measurable. Or a discontinuous linear functional on a Banach space. A group isomorphism between additive groups $\mathbb R$ and $\mathbb R^2$. Many more... – GEdgar Nov 20 '19 at 14:37

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Let me first answer the last question. ZF can certainly prove existence of ultrafilters on (nonempty) finite sets, but it also proves all such ultrafilters are principal - indeed, such an ultrafilter $U$ must have finitely many elements, so we can take an intersection of all elements of $U$, and the result will be some one-element set $\{a\}$ and then $A\in U$ iff $a\in A$.

This means that we have to move to infinite sets to get interesting, nonprincipal ultrafilters. But it turns out that that ZF cannot prove that there is a nonprincipal ultrafilter on $\mathbb N$. Indeed, it is known that ZF cannot prove that there is a nonprincipal ultrafilter on any set, i.e. it is consistent that all ultrafilters on all sets are principal! You can find references and more details in answers to this question.

Wojowu
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