Let $\kappa$ be an uncountable cardinal.
Given a set of S of cardinality $\kappa$, I want to construct a chain {$S_\lambda : \lambda \in \kappa$ } such that:
1) Each $S_\lambda$ is a proper subset of S with cardinality $\kappa$ and
2) For any two ordinals $\alpha < \beta$, $S_\alpha - S_\beta$ has cardinality $\kappa$

Now, using axiom of choice this is easy.
Simply take $\kappa$ many copies of S and index them using some well ordering of $\kappa$.
There is a bijection between this collection and S.
Now just remove one copy at a time and use the bijection to get the corresponding subsets of S. This gives us the chain we want.

My question is, can we always obtain such a chain in ZF without choice ?

  • 6
    $\begingroup$ Are we using here the usual ZF definition of "cardinal" as "an ordinal not in bijection with any lesser ordinal"? If so, then $\kappa$ automatically has a well ordering. $\endgroup$ – Nate Eldredge Sep 30 '19 at 3:05
  • 2
    $\begingroup$ @Nate: That is not the usual definition. However, in order to make sense of $\lambda\in\kappa$ we pretty much have to assume that $\kappa$ is indeed a well-ordered cardinal. $\endgroup$ – Asaf Karagila Sep 30 '19 at 5:27
  • 3
    $\begingroup$ Note that if we intend for $\kappa$ to not necessarily be well-orderable (and taking into account the other comments to make the statement make sense) this is definitely false in general, as $S$ may not have any proper subsets of the same cardinality. $\endgroup$ – Wojowu Sep 30 '19 at 8:09
  • 3
    $\begingroup$ If, as previous comments indicate, $\kappa$ is supposed to be a well-ordered cardinal, then the answer is yes, because it's provable in ZF (without choice) that every infinite ordinal $\alpha$ admits a bijection to $\alpha\times\alpha$. Indeed, if you know this with choice, then it follows without choice, because you can find the required bijection in Gödel's constructible universe $L$. $\endgroup$ – Andreas Blass Sep 30 '19 at 12:25
  • 1
    $\begingroup$ Thanks for the answers in the comments above. Why the votes to close, though ? It's not a super-advanced question, but from the comments , it looks like there are some non-trivial issues here. $\endgroup$ – Anindya Oct 1 '19 at 21:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.