Co-absolutes of \(U(\omega_ 1)\) (Q1313931)
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scientific article; zbMATH DE number 500694
| Language | Label | Description | Also known as |
|---|---|---|---|
| English | Co-absolutes of \(U(\omega_ 1)\) |
scientific article; zbMATH DE number 500694 |
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Co-absolutes of \(U(\omega_ 1)\) (English)
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25 March 1997
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The Stone space of the regular open algebra of a space is known as the absolute of a space. Two spaces which have isomorphic regular open algebras are called co-absolute. The symbol \(U(\omega_1)\) denotes the space of uniform ultrafilters on \(\omega_1\). Miller has shown that if \(\kappa\) many Cohen reals \((\kappa\leq 2^{\omega_2})\) are added to a model of \(2^{\omega_1}= \omega_2\), then the density of \(U(\omega_1)\) is still \(\omega_2\). This paper removes Miller's restriction that \(\kappa\leq 2^{\omega_2}\). Specifically, it shows that after adding any number of Cohen reals to a model of GCH, the space \(U(\omega_1)\) is co-absolute with a power of \(\omega_2\) from which it can be concluded that the density of \(U(\omega_1)\) is \(\omega_2\) (co-absolute compact spaces have the same density). This result generalizes a result of Balcar and Vopěnka which states that if \(2^{\omega_1}= \omega_2\), then \(U(\omega_1)\) is co-absolute with \(\omega_2^\omega\). A second result is that there is a model in which \(U(\omega_1)\) is not co-absolute with any product of discrete spaces, disproving what might be a natural conjecture based on the first theorem. A final result shows that after adding \(\kappa\geq 2^{\omega_1}\) many Cohen reals, there is a point \(p\) in \(U(\omega_1)\) whose type has density equal to the \(\pi\) weight of \(U(\omega_1)\) (which is \(2^{\omega_1}\)) and that every subset of the type of \(p\) of size less than \(2^{\omega_1}\) is relatively discrete.
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Suslin trees
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co-absolute compact spaces
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absolute
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uniform ultrafilters
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density
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Cohen reals
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