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In ${\sf ZFC}$ it can be easily proved that we cannot have infinitely descending sequences of cardinalities, that is, the following statement does not hold:

(DescSeq) There is a set $A$ a map $\alpha: \omega \to {\cal P}(A)$ such that for all $n\in \omega$ we have $\alpha(n+1) \subseteq \alpha(n)$, and there is no injective function from $\alpha(n)$ into $\alpha(n+1)$.

It seems to be unknown whether $\neg$(DescSeq) implies the Axiom of Choice.

Question. In ${\sf ZF}$, are there any implications between $\neg$(DescSeq), the Boolean Prime Ideal Theorem, and the Partition Principle?

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    $\begingroup$ I would find it more natural, in both of your questions, to consider whether the cardinals are well-founded, that is, whether every set of sets has a member of minimal cardinality. This prevents a descending $\omega$-sequence, but without AC (specifically, without DC), the failure of well-foundedness doesn't seem to imply that there must be a descending $\omega$-sequence, since you'd have to pick the sets. So well-foundedness appears to be stronger than the nonexistence of a descending $\omega$-sequence. $\endgroup$
    – Joel David Hamkins
    Commented Jul 14, 2021 at 9:29
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    $\begingroup$ Throwing a stone into a well is a lot easier than digging it out. $\endgroup$
    – Asaf Karagila ♦
    Commented Jul 14, 2021 at 11:18
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    $\begingroup$ Asking questions about PP is almost disingenuous. We don't know of any models of ZF where it holds and choice fails. $\endgroup$
    – Asaf Karagila ♦
    Commented Jul 14, 2021 at 11:19
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    $\begingroup$ @AsafKaragila: I don't think it's fair to call a question "disingenuous," at least not without strong evidence of the asker's motivations. $\endgroup$
    – Sam Hopkins
    Commented Jul 14, 2021 at 12:01
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    $\begingroup$ @Sam: I said almost. In every place it is mentioned, it is also mentioned that we know practically nothing about it, except the obvious implications. So this is the choice principle equivalent of asking what is dark matter. $\endgroup$
    – Asaf Karagila ♦
    Commented Jul 14, 2021 at 12:07

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You are asking about three choice principles, two of which have practically no research around them. Mainly due to the lack of tools we have for dealing with them.

The most you can find is the following paper,

Howard, Paul; Tachtsis, Eleftherios, No decreasing sequence of cardinals, Arch. Math. Logic 55, No. 3-4, 415-429 (2016). ZBL1339.03038.

Where the authors show that $\sf AC_{WO}$ does not imply that there are no decreasing sequences of cardinals (which they take to mean actual sets, rather than the cardinals of these sets), at least in $\sf ZF$ (in $\sf ZFA$ we can say ever so slightly more).

This is really not a whole lot. As for $\sf BPI$, we know that it holds in Cohen's model, where there is a decreasing sequence of sets, simply because there are infinite Dedekind-finite sets. But since we don't have any models of $\sf ZF+\lnot AC$ where we know that the cardinals are well-founded, we cannot prove it will not imply $\sf BPI$, mainly because it may very well imply $\sf AC$ for all we know.

So, you either need to come up with a completely new proof that somehow $\sf BPI$ follows from the well-foundedness of the cardinals, or prove that it implies choice outright, or better yet: develop new tools for positive results about cardinal structure in $\sf ZF$ that will let you deal with this sort of choice principle.

Appendix A.

One can ask about decreasing chains of sets/cardinals versus well-foundedness of sets/cardinals.

In $\sf ZF+DC$ the four questions are equivalent, and given a sequence of cardinals we can choose (using $\sf DC$) the sets to match. So it all falls back to be the same.

If there is a Dedekind-finite set, then there is a decreasing sequence of cardinals, but there is also one where we cannot match a decreasing sequence of sets. However, there are other sequences of cardinals where we do have a decreasing sequence of sets to match.

So the real question about this is really "Suppose that $\sf ZF+\lnot DC+DF=F$ holds, are the four principles still equivalent?"

Again, due to the lack of tools, this is frustratingly difficult to answer. Quite possibly, the assumption of $\sf DC$ can be weakened to $\sf AC_\omega$, which closes the gap a bit further, but still not enough to conclude the answer is positive.

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    $\begingroup$ Thanks Asaf for writing this (very nice) answer despite not liking my question too much. I agree it certainly does not belong to my better questions here on MO (and I have my share of very bad questions.) $\endgroup$
    – Dominic van der Zypen
    Commented Jul 14, 2021 at 15:52

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