Optimal tiling for a collection of partitions

Question

I'm interested in a possible generalization of Tiling relation on the set of partitions (the question has only been partially answered).

Let $x$ be an infinite set and let $\text{Part}(x)$ be the collection of all partitions of $x$. Moreover, let $P \in \text{Part}(x)$ and $t\subseteq x$. We set $$P_{[t]} = \{p\in P:p\cap t \neq \emptyset\}.$$ We define the tiling relation on $\text{Part}(x)$ by $$ P \triangleleft Q \textrm{ if and only if for all } S\subseteq P\textrm{ we have } \mathsf{card}(S) \leq \mathsf{card}(Q_{[\bigcup S]}).$$ In other words, the relation $P\triangleleft Q$ holds if no subset $S$ of $P$ is covered by a subset of $Q$ having a smaller cardinality than $S$.

Let $\mathcal{A}\subseteq \text{Part}(x)$ be a non-tempty collection of partitions on the set $x$ such that there is a set $M$ of subsets of $x$ so that every member of $\mathcal{A}$ refines $M$ (that is, for every $A\in\mathcal{A}$ and $a\in A$ there is $m\in M$ with $a\subseteq m$).

Question. Is there a partition $Z\in\text{Part}(x)$ such that

• $Z$ refines $M$ and
• $Z \triangleleft A$ for all $A\in\mathcal{A}$?

Answer 1

1) This question (as posed) has a simple negative answer. Just take any countable set $x$ and put $M$ be the family of finite subsets of $x$. Then for the family $\mathcal A$ of all possible partitions of $x$ into finite subsets there is no partition $Z$ of $x$ into finite sets such that $Z\triangleleft A$ for every $A\in\mathcal A$.

Indeed, any partition $Z$ of $x$ into finite sets is infinite. So, we can take any 2-element subfamily $S=\{z,z'\}\subset Z$ and take any partition $A\in\mathcal A$ such that $\{z,z'\}\in A$. Then $|S|=2>1=|A_{[\cup S]}|$, witnessing that $Z\not\triangleleft A$.

2) A bit more complicated counterexample can be constructed as follows. Consider the family $\mathcal A$ of all possible partitions of the set $\mathbb P:=\mathbb R\setminus\mathbb Q$ of irrational numbers into compact susbets. Let $M$ be the family of all compact subsets of $\mathbb P$, endowed with the Vietoris topology. It is well-known that the space $M$ is separable and metrizable.

Then there is no partition $Z$ of $\mathbb P$ into compact subsets such that $Z\triangleleft A$ for all $A\in\mathcal A$.

Indeed, since the space $\mathcal P$ is not $\sigma$-compact, each partition $Z$ of $\mathbb P$ into non-empty compact subsets of $\mathbb P$ is uncountable and considered as a subspace of the hyperspace $M$ is not discrete. Hence $Z$ contains a sequence $\{z_n\}_{n\in\omega}\subset Z$ of pairwise distinct elements that converges to a compact set $z_\omega\in Z$. Then the set $S:=\{z_n\}_{n\le\omega}$ has compact union in $\mathbb P$ and hence $\bigcup S\in A$ for some partition $A$ of $\mathbb P$ into compact sets. Taking into account that $|S|=\omega>|A_{[\cup S]}|=1$, we conclude that $Z\not\triangleleft A$.

3) In fact, the gap between $|S|$ and $|A_{[\cup S]}|$ can be arbitrarily high. Indeed, for any infinite regular cardinal $\kappa$ consider the set $x=\kappa$ and the family $M$ of subsets of $x$ of cardinality $<\kappa$. Let $\mathcal A$ be the family of all possible partitions of $x$ into subsets of cardinality $<\kappa$. Then for any partition $Z\in\mathcal A$ and any subfamily $S\subset Z$ of cardinality $|S|<\kappa$ there union $\bigcup S$ has cardinality $<\kappa$ and hence is an eleemnt of some partition $A\in\mathcal A$, for which $|A_{[\cup S]}|=1$.