5
$\begingroup$

Working in $ZFC$ every cardinal is either finite or in bijection with a proper subset of itself (Dedekind infinite). Without Choice it is consistent that there are infinite sets which can't be partitioned into two infinite subsets (amorphous sets), so the above statement no longer holds since a bijection to a proper subset implies a partition into two disjoint infinite subsets as proven on the wiki -- all of this is discussed in the question and answers here much more succinctly.

Is it consistent in $MK$ without Global Choice that there are amorphous proper classes, meaning proper classes which can't be partitioned into two proper class sized subclasses?

Directly generalizing the argument given on the wiki article for amorphous sets seems to require a notion of transfinite function composition which can be defined in good categorical generality using colimits, but it is not immediately apparent how to generalize the recursive definition of the $S_i$'s for limit ordinal $i$ since the given definitions depend on immediate predecessor steps.

$\endgroup$
7
$\begingroup$

Unless I'm missing something, the answer is no: we have a surjection $s$ from a given proper class to the class of ordinals - sending each element to its rank and then "collapsing" appropriately - and this lets us partition the original class into two proper classes, for example $s^{-1}(limits)$ versus $s^{-1}(successors)$.

$\endgroup$
  • $\begingroup$ Very nice, you haven't missed anything -- if I create a moving target and ask about $MK-Foundation$ is the answer still trivially no? $\endgroup$ – Alec Rhea Mar 19 at 21:09
  • 1
    $\begingroup$ @Alec: In that case the answer is positive. Just do Fraenkel's model over a proper class of atoms. $\endgroup$ – Asaf Karagila Mar 19 at 21:29
  • $\begingroup$ @AsafKaragila Much appreciated Asaf and thanks again Noah, I'm not familiar with Fraenkel's model but I'll look into it. $\endgroup$ – Alec Rhea Mar 19 at 21:32
  • 1
    $\begingroup$ @Alec: That's the OG model for amorphous sets. Just remember that ZFA (or ZFU) is equivalent to ZF-Foundation with Quine atoms for the atoms. $\endgroup$ – Asaf Karagila Mar 19 at 21:33
  • 2
    $\begingroup$ Hahah, it’s an abbreviation for the american colloquialism “original gangster” meaning a member of the original older generation of badasses in a given gang/discipline of mathematics. $\endgroup$ – Alec Rhea Mar 20 at 0:25

Your Answer

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

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