Partition subalgebras for maximal almost disjoint families (Q1849859)

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scientific article; zbMATH DE number 1838863
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Partition subalgebras for maximal almost disjoint families
scientific article; zbMATH DE number 1838863

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    Partition subalgebras for maximal almost disjoint families (English)
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    2 December 2002
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    Let \({\mathcal M}\) be a maximal almost disjoint family on \(\omega\). A partitioner of \({\mathcal M}\) is a set \(A\subseteq\omega\) such that the sets \({\mathcal X}=\{M\in {\mathcal M}:M \subseteq^*A\}\) and \({\mathcal Y}=\{M\in {\mathcal M}:M\cap A=^* \emptyset\}\) form a partition of \({\mathcal M}\). The set of all partitioners of \({\mathcal M}\) forms a Boolean algebra \({\mathcal A}\). Let \(I_{\mathcal M}\) denote the ideal in \({\mathcal A}\) which is generated by \({\mathcal M}\) together with the finite subsets of \(\omega\). The partitioner algebra \(P_{\mathcal M}\) of \({\mathcal M}\) is the quotient algebra \({\mathcal A}/I_{\mathcal M}\). A Boolean algebra is representable if it is isomorphic to a partitioner algebra of some mad family on \(\omega\). Previously, \textit{J. E. Baumgartner} and \textit{M. Weese} [``Partition algebras for almost disjoint families'', Trans. Am. Math. Soc. 274, 619-630 (1982; Zbl 0515.03032)] asked if every subalgebra of a representable algebra is also representable. It is known that the answer is yes under CH. Here the authors show that the existence of a supercompact cardinal implies that it is consistent that the answer is no. The proof uses \(\text{PFA}^+\) and \(\diamondsuit (S^2_0)\).
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    consistency
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    maximal almost disjoint family
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    Boolean algebra
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    partitioner algebra
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    representable algebra
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