An improper arithmetically closed Borel subalgebra of \(\mathcal P(\omega)\) mod FIN
For a subalgebra \(\mathcal{A}\subseteq\mathcal{P}(\omega)\) let \(\mathbb{P}_\mathcal{A}\) be the partial order obtained by reducing \((\mathcal{A},\subseteq)\) modulo the ideal of finite sets. We say that a subalgebra \(\mathcal{A}\) is arithmetically closed when it is closed under the Turing jump and the Turing reducibility. Results obtained by \textit{V. Gitman} in [``Scott's problem for proper Scott sets'', J. Symb. Log. 73, No. 3, 845--860 (2008; Zbl 1161.03024); ``Proper and piecewise proper families of reals'', Math. Log. Q. 55, No. 5, 542--550 (2009; Zbl 1181.03050)] motivate the question whether there is an arithmetically closed \(\mathcal{A}\) for which the poset \(\mathbb{P}_\mathcal{A}\) is not proper. This question was answered in the affirmative by \textit{A. Enayat} in [``A standard model of Peano arithmetic with no conservative elementary extension'', Ann. Pure Appl. Logic 156, No. 2--3, 308--318 (2008; Zbl 1156.03035)], using a highly nonconstructive reasoning that establishes the existence of an arithmetically closed \(\mathcal{A}\) of power \(\aleph_1\) such that \(\mathbb{P}_\mathcal{A}\) collapses \(\aleph_1\) (and is therefore not proper). He also posed the question whether \(\mathbb{P}_\mathcal{A}\) must be proper if \(\mathcal{A}\) is both arithmetically closed and Borel. In the paper under review the authors provide a strong negative answer to this question, proving the existence of an arithmetically closed Borel subalgebra \(\mathcal{A}\subseteq\mathcal{P}(\omega)\) such that \(\mathbb{P}_\mathcal{A}\) is equivalent to \(((2^{\aleph_0})^{<\omega},\subseteq)\), the usual partial order that collapses \(2^{\aleph_0}\) to \(\aleph_0\).
scientific article; zbMATH DE number 5985547
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