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9.39 (1984)

Open

Let $\Omega$ be a countable set and $\mathfrak{m}$ a cardinal number such that $\aleph_0 \leqslant \mathfrak{m} \leqslant 2^{\aleph_0}$ (we assume Axiom of Choice but not Continuum Hypothesis). Does there exist a permutation group $G$ on $\Omega$ that has exactly $\mathfrak{m}$ orbits on the power set $\mathcal{P}(\Omega)$?

Contributor: P. M. Neumann

Progress

Comment of 2001: It is proved (S. Shelah, S. Thomas, Bull. London Math. Soc., 20, no. 4 (1988), 313–318) that the answer is positive in set theory with Martin’s Axiom. The question is still open in ZFC.

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