Are there models of ZFC in which $\mathfrak{r}$ is strictly less than $\mathfrak{s}$? I've not been able to find any forcings that end up with this result.

Here $\mathfrak{r}$ is the reaping number $\lvert\lvert([\omega]^\omega,[\omega]^\omega,\text{does not split})\rvert\rvert$ and $\mathfrak{s}$ is the splitting number, its dual.


1 Answer 1


The inequality $\mathfrak{r} \leq \mathfrak{u}$ is provable in ZFC (because every base for an ultrafilter is a reaping family). Blass and Shelah proved the consistency of $\mathfrak{u} < \mathfrak{s}$ in

Blass, Andreas; Shelah, Saharon, There may be simple $P_{\aleph_1}$- and $P_{\aleph_2}$-points and the Rudin-Keisler ordering may be downward directed, Ann. Pure Appl. Logic 33, 213-243 (1987). ZBL0634.03047.

As far as I am aware, this was the first proof of the consistency of $\mathfrak{r} < \mathfrak{s}$. Other models for this have been constructed since, for example in this recent paper by Guzman and Kalajdzievski.

Guzmán, Osvaldo; Kalajdzievski, Damjan, The ultrafilter and almost disjointness numbers, Adv. Math. 386, Article ID 107805, 41 p. (2021). ZBL1493.03009.

  • 2
    $\begingroup$ We have an "add citation button". $\endgroup$
    – Asaf Karagila
    Nov 19, 2022 at 15:36
  • $\begingroup$ @AsafKaragila: I've been using MO for over 7 years and somehow never noticed :) Thanks for letting me know! $\endgroup$
    – Will Brian
    Nov 19, 2022 at 15:44
  • $\begingroup$ Roughly 9 years now, or even more, meta.mathoverflow.net/questions/1485/… $\endgroup$
    – Asaf Karagila
    Nov 19, 2022 at 15:47

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