Let $\kappa < \lambda < \mu$ be infinite cardinals. Is there a collection ${\cal U}\subseteq {\cal P}(\mu)$ of subsets of $\mu$ with the following properties?

  1. for all $U\in {\cal U}$ we have $|U| = \lambda$;
  2. every $S\subseteq \mu$ with $|S| = \kappa$ is contained in exactly one member of ${\cal U}$; and
  3. for all $\alpha,\beta \in \mu$ we have $|{\cal U}_\alpha| = |{\cal U}_\beta|$, where ${\cal U}_\alpha = \{U\in {\cal U}: \alpha\in U\}$, and ${\cal U}_\beta$ is defined similarly.
  • 4
    $\begingroup$ If $\kappa$ is infinite, then no. Let $V$ be a $\kappa$ sized subset of $U$, and let $W$ have one more element than $V$ with that element outside of $U$. If there is $U'$ containing $W$, then both $U$ and $U'$ contain $V$, violating condition $2$. Gerhard "Not Quite A Projective Plane" Paseman, 2018.12.18. $\endgroup$ – Gerhard Paseman Dec 18 '18 at 15:49
  • 1
    $\begingroup$ Isn't 3) implied by 2)? Just enumerate all $\kappa$-sets containing $\alpha$... $\endgroup$ – Ilya Bogdanov Dec 18 '18 at 18:32

Extending the problem, admit the possibility of equality of cardinals, so consider that $\kappa \leq \mu$. If indeed $\kappa=\mu$, then the collection of subsets with one member, namely $\mu$, suffices. But that is the only case.

For we borrow the argument from the comments. Let $U$ be a proper subset of $\mu$ that is part of a suitable collection, and let $V$ be $\kappa$-sized and contained in $U$. If $\kappa$ is infinite, pick some element of $\mu \setminus U$ and consider a set $U'$ from the collection that contains the $\kappa$-sized set which is $V$ union (the singleton set containing) this element. Then both $U$ and $U'$ contain $V$, which means condition two does not hold, and so a suitable collection cannot have a proper subset of $\mu$.

For finite $\kappa$ and infinite $\lambda \lt \mu$, here is an idea which may provide a suitable collection, but some work needs to be done. Namely, well order all of the $\lambda$-sized subsets of $\mu$, and pick the least allowed $\lambda$ sized subset $U$, and then "throw out" the $\lambda$ sets that intersect $U$ in a subset of size at least $\kappa$. Conditions 1) and 2) are partly satisfied during the construction, but it remains to show that every $\kappa$ subset is uniquely covered. Since $\mu$ is strictly larger than $\lambda$, this should be possible to show.

Gerhard "Infinitely Elementary, My Dear Dominic" Paseman, 2018.12.20.

| cite | improve this answer | |
  • $\begingroup$ Thanks for the answer @gerhardpaseman! - Dominic "I always enjoy your signatures!" van der Zypen $\endgroup$ – Dominic van der Zypen Dec 20 '18 at 16:06
  • $\begingroup$ If we consider $\kappa=1$, the picture changes, and any equipartition of $\mu$ works. Things get interesting for larger $\kappa$, and it is unclear to me what a suitable collection would be even for $\kappa=2$. Probably this problem was considered by Erdos, but I have no references for you. (And, you are welcome!) Gerhard "Finitely Unclear, My Dear Dominic" Paseman, 2018.12.20. $\endgroup$ – Gerhard Paseman Dec 20 '18 at 16:48

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.