If you look at the set of finite subsets of $\mathbb N$ (side question: Is there a standard notation for that?) partially ordered by the subset relation, you see that it can be partitioned into maximal antichains (that is, you have a set of maximal antichains so that any two of them are disjunct, and their union is the set of finite subsets of $\mathbb N). One possibility (but by far not the only one) is to have each antichain consist of all sets of the same cardinality.

Also for the set of all subsets of $\mathbb N$ that are either finite or cofinite, it's not hard to define such a partition; partition the finite sets according to their cardinality, and the cofinite sets according to the cardinality of the complement.

Of course such simple strategies won't work any more with the full $\mathcal P(\mathbb N)$. Indeed, I have no idea how one would either define such a partition, or prove that there doesn't exist one. Of course there's also the possibility that there exists such a partition, but you cannot explicitly specify it.

Therefore my question is: Does there exist a partition of $(\mathcal P(\mathbb N),\subseteq)$ into maximal antichains, and if so, is is possible to explicitly specify one?

  • 1
    $\begingroup$ One standard notation for the family of subsets of $A$ of cardinality $\kappa$ is $[A]^\kappa$, with $[A]^{<\kappa}$ for the family of subsets of $A$ of cardinality less than $\kappa$, and so on. The set of finite subsets of $\Bbb N$ is then $[\Bbb N]^{<\omega}$. I have also seen $\wp_{\text{fin}}(\Bbb N)$. $\endgroup$ – Brian M. Scott Nov 26 '16 at 19:50
  • 1
    $\begingroup$ I can second the notation $[\mathbb{N}]^{< \omega}$. Also, great question! $\endgroup$ – Dániel G. Nov 26 '16 at 20:00
  • $\begingroup$ @BrianM.Scott: Thank you. But, if the exponent is a cardinality, shouldn't it then be $[\mathbb N]^{<\aleph_0}$? $\endgroup$ – celtschk Nov 26 '16 at 20:26
  • 1
    $\begingroup$ @MitchellSpector: Actually, since we were talking about subsets if given cardinality, rather than of given order type, your argument is actually an argument against using ordinal notation for cardinality. In particular, since the cardinality of $\omega+\omega$ is still $\aleph_0$, the subsets of order type $\omega+\omega$ should definitely be in $[X]^{\aleph_0}$. If not distinguishing between $\omega$ and $\aleph_0$, you then also cannot distinguish in notation between the subsets of $X$ of order type $\omega$ ($[X]^\omega$) and the subsets of cardinality $\aleph_0$ ($[X]^{\aleph_0}$). $\endgroup$ – celtschk Nov 27 '16 at 6:36
  • 1
    $\begingroup$ @MitchellSpector: OK, that's a different argument for the specific notation $[\mathbb N]^{<\omega}$, but not the one Brian M. Scott used, and that was discussed later (in particular in the comment you directly responded to). $\endgroup$ – celtschk Nov 27 '16 at 7:04

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.