Also asked at MO.
It's a fun exercise to show in $\mathsf{ZF}$ that "the free complete lattice on $3$ generators" doesn't actually exist. The punchline, unsurprisingly, is size: a putative free complete lattice on $3$ generators would surject onto the class of ordinals.
This obstacle isn't a problem however in the context of $\mathsf{NFU}$; on the other hand, recursive constructions (which were utterly unproblematic in $\mathsf{ZF}$) now become more complicated. So the situation is unclear to me: is it consistent with $\mathsf{NFU}$ that there is a free complete lattice on $3$ generators?
- I don't have much experience with $\mathsf{NFU}$, so as far as I know it's possible that in in this context the phrase "free complete lattice on $3$ generators" is ambiguous - specifically, I'm a bit worried about the word "free" in this context. To make things precise, I'll use the homomorphism-based notion of freeness, that is, I'm looking for a complete lattice $L$ with three distinguished elements $a,b,c$ such that for every other complete lattice $M$ and distinguished elements $u,v,w\in M$ there is exactly one complete lattice homomorphism $L\rightarrow M$ sending $a$ to $u$, $b$ to $v$, and $c$ to $w$.
I would also be interseted in the answer to the same question for other set theories admitting a universal set, such as $\mathsf{GPK}_\infty^+$. However, my current impression is that $\mathsf{NFU}$ is by far the best-understood such theory, so it seems like a good starting point.
