5
$\begingroup$

This is a variation on this question with amorphous cardinals replaced with dedekind finite sets.

Dedekind finite sets are sets that have no countable subset, and it is well known that this is a weaker assertion than a set not having a countable chain of subsets with respect to the strict inclusion relation. For example as Joel pointed out in a comment, there can be infinite Dedekind finite sets of reals, and one can construct a sequence of slices of such a set. A set like that would also have an uncountable chain of subsets, but I am interested only in well-ordered chains.

I want to know how long such well-ordered chains can be, and I have the same 3 questions as in the last question (we work in $\sf ZF$, and a "chain" means a chain with respect to the strict inclusion relation):

  1. Given an ordinal $\alpha$, is it consistent that there exists a chain of infinite Dedekind finite sets of length $\alpha$?

  2. Is it consistent that for every ordinal $\alpha$ there exists a chain of infinite Dedekind finite sets of length $\alpha$?

  3. Is it consistent that there is a class chain of length $\sf Ord$ of Dedekind finite sets?

Also, if the answer to any of these is no, does it change when we consider Dedekind finite cardinals instead of actual sets (with strict inclusion replaced by strict inequality of cardinals)?

$\endgroup$

1 Answer 1

6
$\begingroup$

The answer to all of these is yes. Monro constructed a model where for every ordinal $\alpha$ there is a Dedekind-finite set mapped onto $\alpha$.

This alone gives us arbitrarily long chains. Take such $A$ that is mapped onto $\alpha$ and $f\colon A\to\alpha$ witnessing that, and let $A_\beta=f^{-1}\beta$, the preimage of the initial segment not the element, then this is a chain of length $\alpha$.

Monro's construction has the benefit that there is a canonical Dedekind-finite set that is mapped onto $\alpha$, and the class of these is amenable to the model, which allows us to add it as a predicate to get a chain of length $\rm Ord$.

Monro, G.P., Independence results concerning Dedekind finite sets.

$\endgroup$
6
  • $\begingroup$ Apologies for the bad citation formatting, I'm typing this on my phone. $\endgroup$
    – Asaf Karagila
    Mar 30 at 10:52
  • $\begingroup$ You say into $\alpha$, in first paragraph, but I think you mean onto. $\endgroup$ Mar 30 at 11:29
  • 4
    $\begingroup$ Ducking auto portrait. $\endgroup$
    – Asaf Karagila
    Mar 30 at 13:43
  • $\begingroup$ I don't understand the $\sf Ord$ chain construction. Assuming $F(\alpha)$ maps onto $\alpha$ for all $\alpha$, then what exactly is the sequence? $\endgroup$
    – Ynir Paz
    Mar 30 at 14:14
  • $\begingroup$ @YnirPaz: The point is that $\alpha\mapsto A_\alpha$ is an amenable class function, and we can add a predicate for $\{A_\alpha\mid\alpha\in\rm Ord\}$. And moreover, the union of these sets is itself Dedekind-finite. $\endgroup$
    – Asaf Karagila
    Mar 30 at 15:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.