# Well-foundedness of cardinals and the axiom of choice

Without axiom of choice it is not generally true that the class of all cardinal (in this question we consider Scott cardinal rather than cardinals as ordinals) is not well-founded under the ordinary cardinality comparison. However, we also know that assuming well-foundedness of such ordering causes no consistency problem.

Under ZF with the assumption, we can prove every infinite set is Dedekind-infinite as follows: for infinite set $X$, consider the collection of cardinals $$\mathcal{A} = \{|A| : A\subseteq X \text{ and A is infinite}\}.$$ From assumption, $\mathcal{A}$ is well-founded. If $|B|$ is a minimal element, then $B$ should be Dedekind-infinite, since $|B|-1 = |B|$. (where $|B|-1$ is a cardinality of the set $B$ except one element in $B$.)

I wonder we can prove more stronger result; for example, axiom of choice follows from that the class of cardinals are well-founded? I would appreciate your answer.

• In ZF the assertion that "there is no strictly decreasing infinite sequence of cardinals" seems to be weaker than "every nonempty set of cardinals has a minimal element" (is it really weaker?), and the weaker assertion is enough to prove that every infinite set is Dedekind-infinite. In other words, given a Dedekind-finite infinite set, you can easily (and choicelessly) define a strictly decreasing infinite sequence of cardinals. – bof Feb 22 '16 at 8:36
• This question is not clear. What do you mean by "The class of cardinals are well-founded?" That sentence is not even describable in ZF because ZF does not even mention proper classes but it might be describable in NBG. Did you really mean a different statement which ZF can describe? – Timothy Jan 6 '18 at 0:17
• @Timothy I didn't consider it in detail when I make this question. However formulating my question on NBG works. Moreover, as the axiom of regularity proves every class has $\in$-minimal element, formulating my statement for set-sized collection of cardinals also seem to work. – Hanul Jeon Jan 6 '18 at 8:48
• @Hanul Jeon I still don't know what you mean by "The class of all cardinals is well founded." If you edit your question to clarify it, I'll have a better idea of what you're asking. Do you mean "There is no descending sequence of cardinal numbers?" Do you mean "For any nonempty class of cardinal numbers, there is a cardinal number in that class such that no cardinal number in it is strictly smaller?" – Timothy Jan 7 '18 at 1:06
• @Timothy I think conditions you have presented are equivalent, though the well-foundness usually denotes the latter. – Hanul Jeon Jan 7 '18 at 2:17

## 1 Answer

This is an open problem.

It was shown that for every $\kappa$, $\sf DC_\kappa$ cannot prove that the cardinals are well-founded. While not enough to conclude the principle is equivalent to the axiom of choice ($\sf BPI$ does not follow from $\sf DC_\kappa$ either), it is worth remarking that we really don't know much about this principle.

A very recent paper gave a nice survey of this problem and related results:

Paul Howard, Eleftherios Tachtsis, "No decreasing sequence of cardinals", Archive for Mathematical Logic, First online: 28 December 2015.

Let me finish by stating that generally speaking the structure of the cardinals is a bit of a wild beast when it comes to the axiom of choice. We don't have good techniques to control it very well in order to produce separating models for much awaited-results (e.g. the Partition Principle is a statement about the structure of the cardinals). So we mainly know how to violate things wildly (e.g. embed partial orders into the cardinals of a model), but not how to fine tune this in order to produce nice results.