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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.

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  • $\begingroup$ 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. $\endgroup$ – bof Feb 22 '16 at 8:36
  • $\begingroup$ 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? $\endgroup$ – Timothy Jan 6 '18 at 0:17
  • $\begingroup$ @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. $\endgroup$ – Hanul Jeon Jan 6 '18 at 8:48
  • $\begingroup$ @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?" $\endgroup$ – Timothy Jan 7 '18 at 1:06
  • $\begingroup$ @Timothy I think conditions you have presented are equivalent, though the well-foundness usually denotes the latter. $\endgroup$ – Hanul Jeon Jan 7 '18 at 2:17
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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.

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