In this post I will discuss some cardinal characteristic of the continuum, related to partitions of $\omega$ and would like to know if it is equal to some known cardinal characteristic.

By a partition of $\omega$ I understand a cover of $\omega$ by pairwise disjoint nonempty subsets. A partition $\mathcal P$ is called finitary if $\sup_{P\in\mathcal P}|P|$ is finite.

A family $\mathfrak P$ of partitions of $\omega$ is called directed if for any two partitions $\mathcal A,\mathcal B\in\mathfrak P$ there exists a partition $\mathcal C\in\mathfrak P$ such that each set $S\in\mathcal A\cup\mathcal B$ is contained in some set $C\in\mathcal C$.

Let $\mathfrak P$ is a family of partitions of $\omega$. An infinite subset $D\subset\omega$ is called $\mathfrak P$-discrete if for any partition $\mathcal P\in\mathfrak P$ there exists a finite set $F\subset D$ such that for any $P\in\mathcal P$ the intersection $P\cap (D\setminus F)$ contains at most one point.

Let $\kappa$ be the smallest cardinality of a directed family $\mathfrak P$ of finitary partitions of $\omega$ admitting no infinite $\mathfrak P$-discrete set $D\subset\omega$.

It can be shown that $\mathfrak b\le\kappa\le\mathfrak c$ (the upper bound follows from the observation that any maximal directed family of finitary partitions has no infinite discrete set, see Proposition 6.5 in this preprint).

Problem 1. Is $\kappa$ equal to some known cardinal characteristic of the continuum?

Problem 2. Is $\kappa=\mathfrak c$ in ZFC?

Problem 3. Find lower and upper bounds on $\kappa$ (which are better than $\mathfrak b\le\kappa\le\mathfrak c$).

Added in Edit. The lower bound $\sup_{U\in\beta\omega}\pi(U)\le\kappa$, suggested by Todd Eisworth can be improved to $\mathfrak s\le \kappa$. One can also prove that $\max\{\mathfrak b,\mathfrak s,\}\le\mathfrak j\le\kappa\le\mathrm{non}(\mathcal M)$ and hence $\kappa$ is not equal to $\mathfrak c$. The cardinal $\mathfrak j$ is discussed in this MO-post.

  • $\begingroup$ Possibly a stupid sub-question. For any finitary partition ${\cal P}$ let $\mu({\cal P})$ be the smallest $k\in \omega$ such that for all $P\in{\cal P}$ we have $|P|\leq k$. Let ${\frak P}$ be a directed family of finitary partitions of $\omega$ with the property that there is $N\in \omega$ with $\mu({\cal P}) \leq N$ for all ${\cal P}\in {\frak P}$. (In other words, there is a global "block size bound" for every partition in ${\frak P}$.) Does this imply that ${\frak P}$ admint a ${\frak P}$-discrete set $D\in[\omega]^\omega$? $\endgroup$ Feb 6, 2020 at 20:28
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    $\begingroup$ @DominicvanderZypen Very good question! The answer is "yes". If $\mathfrak P$ is directed and the cardinality of cells is bounded from above, then there exists an increasing sequence of partitions $(\mathcal P_n)_{n\in\omega}$ in $\mathfrak P$ that converges (in a pointwise sense) to some partition $\mathcal P_\infty$. Take a $\{\mathcal P_\infty\}$-discrete set and notice that it is also $\mathfrak P$-discrete. $\endgroup$ Feb 6, 2020 at 21:43
  • $\begingroup$ Thanks for your answer, Taras! $\endgroup$ Feb 7, 2020 at 13:05

1 Answer 1


This is not an answer, but hopefully it's a helpful observation:

(1) If $U$ is an ultrafilter on $\omega$ and $\mathcal{P}$ is a finitary partition of $\omega$, then there is $A\in U$ such that $A\cap P$ contains at most one element for each $P\in\mathcal{P}$.

(As if each piece of the partition has cardinality at most $n$, then there is a $k\leq n$ such that the union of pieces with size exactly $k$ is in $U$. Now split this union up into $k$ pieces in the obvious way, and one of these is in $U$.)

(2) Given an ultrafilter $U$, let $\tau(U)$ be the least cardinal $\tau$ such that some subfamily of $U$ of cardinality $\tau$ fails to have an infinite pseudo-intersection. (We do not require the pseudo-intersection to be in $U$, so $\aleph_1\leq\tau(U)\leq\mathfrak{c}$.)

If $U$ is an ultrafilter on $\omega$, then $\tau(U)\leq\kappa$.

Proof. Given a family $\mathfrak{P}$ of finitary partitions of $\omega$ (directed or not), we fix for each $P\in\mathfrak{P}$ a set $A_P\in U$ meeting each element of $P$ in at most one point. If $|\mathfrak{P}|<\tau(U)$ then we can find an infinite pseudo-intersection $X$ for the collection $\{A_P:P\in\mathfrak{P}\}$, and $X$ is $\mathfrak{P}$-discrete.$_\square$

I don't know anything about the cardinals $\tau(U)$. I note that at one point Blass and Shelah claimed to have model containing both simple $P_{\aleph_1}$ and simple $P_{\aleph_2}$ points, but Alan Dow discovered an error in the paper, and I'm not sure if it has ever been repaired. (The existence of a simple $P_{\aleph_1}$-point implies $\mathfrak{b}=\mathfrak{u}=\aleph_1$, while the simple $P_{\aleph_2}$ point is an ultrafilter $U$ with $\tau(U)=\aleph_2$. In such a model, $\kappa$ would be strictly greater than $\mathfrak{b}$.)

Clearly this is all tied up with the topology of $\beta\omega$, so I suspect much more is known by the experts.

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    $\begingroup$ Indeed, very interesting! Especially the cardinal $\sup_{U\in\omega^*}\tau(U)$. Maybe @Andreas Blass could comment on the consistency of $\mathfrak b<\sup_{U\in\omega^*}\tau(U)$? $\endgroup$ Feb 3, 2020 at 20:13
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    $\begingroup$ And what about the consistency of $\sup_{U\in\omega^*}\tau(U)<\mathfrak b$? Is there any model of ZFC where this strict inequality is true? $\endgroup$ Feb 3, 2020 at 20:20
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    $\begingroup$ As far as I know, the error about simple P-points for two different cardinals has not yet been repaired. Shelah and Mildenberger each had arguments to repair it, but neither argument seems to have survived. $\endgroup$ Feb 3, 2020 at 22:57
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    $\begingroup$ I think the second comment by @TarasBanakh can be answered by starting with a model of MA + $\neg$CH and then adjoining $\aleph_1$ random reals. The random reals don't affect $\mathfrak b$, which stays large. But any ultrafilter will have to contain each of the random reals or its complement, and I don't see any chance for a pseudo-intersection (which would have to depend on only countably many of the random reals, by ccc). (Unfortunately, I have no answer yet for the first comment, the one relevant to the actual question.) $\endgroup$ Feb 3, 2020 at 23:02
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    $\begingroup$ It is consistent that $\mathfrak{b}<\kappa$. See the answer to mathoverflow.net/questions/352034/… $\endgroup$ Feb 14, 2020 at 0:56

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