Is there anything between the Axiom of Choice and the Ultrafilter Lemma?

Question

It is well known that the Axiom of Choice (AC) implies the Ultrafilter Lemma (UF), and also that this implication is not reversible.

I was wondering if there exists a statement (S) strictly between them, in the following sense:

AC $\Rightarrow (S)\Rightarrow$ UF and UF $\not\Rightarrow (S)\not\Rightarrow$ AC, both in ZF.

What motivates this question is the fact that the statement "every open cover (of a topological space $X$) without finite subcovers is contained in a maximal open cover without finite subcovers" ($\dagger$) is enough to prove Alexander's Subbase Theorem (AST), and AST is equivalent to UF.

Since I don't know how to prove neither UF $\Rightarrow (\dagger)$ nor $(\dagger)\Rightarrow$ AC, I started to wonder if there exists a statement S as described above.

Answer 1

Of course. There are plenty!

For one, the Ultrafilter Lemma does not even prove the axiom of countable choice. Therefore $\sf UF+AC_\kappa$ is a strictly in between $\sf UF$ and $\sf AC$, for any $\kappa$. Therefore also $\sf UF+DC_\kappa$ is another intermediate statement.

Answer 2

Sure - let $\varphi$ be any sentence independent of $ZF+UL+\neg AC$ (say, "There is no uncountable set of reals not in bijection with all of $\mathbb{R}$" - the weak continuum hypothesis). Then let $S$ be the sentence $$AC\mbox{ }\vee \mbox{ }(UL\mbox{ }\wedge\mbox{ } \varphi).$$ We clearly have $AC$ implies $S$ over ZF, and this implication is strict since $ZF+UL+\neg AC+\varphi$ is consistent. On the other hand, clearly $S$ implies $UL$, and this is strict, since $ZF+UL+\neg\varphi$ is consistent.

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More generally, given any theory $T$ which is "nice" (here, I mean that Goedel's theorems apply to it) and sentences $\varphi,\psi$ such that $T\vdash \varphi\implies\psi$ strictly, we can find a $\theta$ such that $T\vdash\varphi\implies\theta$ and $\theta\implies\psi$, both strictly.