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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.

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Renan Mezabarba
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2 Answers2

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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.

Asaf Karagila
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  • Well, I actually had something else to add to this answer, but I forgot it while I was typing this part in. Maybe tomorrow I'll remember it. – Asaf Karagila Oct 30 '16 at 21:47
  • Both the answers are obviously right, but I was wondering of a statement that is not "obviously related to UF", in some reasonable sense. It would be against the rules to add this to my question now? – Renan Mezabarba Oct 30 '16 at 21:48
  • Well, I don't see who $\sf AC_\kappa$ is related to the Ultrafilter Lemma. – Asaf Karagila Oct 30 '16 at 21:49
  • AC$\kappa$ is not related, but $UF+AC\kappa$ is, because it is explicitly part of $UF+AC_\kappa$. – Renan Mezabarba Oct 30 '16 at 21:51
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    Does "Banach-Alaoglu+..." work for you? I could add that as an option. – Asaf Karagila Oct 30 '16 at 21:52
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    As for the edit, I'd rather that you think through before you post this edit. I know, it's hard, because you can never know when your question is going to have a trivial answer. I don't mind improvements to the question after I post an answer, but I do rather have them final, and not incremental where each time someone gives you a "trivial answer" and you "improve". So maybe take a day to think about it, then come back and see what's what. – Asaf Karagila Oct 30 '16 at 21:54
  • "Banach-Alaoglu+..." I see what you did there :) Anyway, if nothing "less trivial" appear, I'll accept your answer. – Renan Mezabarba Oct 30 '16 at 21:59
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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.

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.

Noah Schweber
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