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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.
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    $\begingroup$ 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
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    $\begingroup$ Isn't 3) implied by 2)? Just enumerate all $\kappa$-sets containing $\alpha$... $\endgroup$ – Ilya Bogdanov Dec 18 '18 at 18:32
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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.

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  • $\begingroup$ 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

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