2010
DOI: 10.1007/s10587-010-0042-2
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On sequential properties of Banach spaces, spaces of measures and densities

Abstract: We answer in negative the problem if the existence of a Pmeasure implies the existence of a P-point. Namely, we show that if we add random reals to a certain 'unique P-point' model, then in the resulting model we will have a P-measure but not P-points. Also, we investigate the question if there is a P-measure in the Silver model. We show that rapid filters cannot be extended to a P-measure in the extension by ω product of Silver forcings and that in the model obtained by the countable support ω 2 -iteration of… Show more

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Cited by 7 publications
(6 citation statements)
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“…This property simplifies the form of functionals on the space of measures on A and in this way it can be used to achieve Mazur property. In [3] it was also proved that if p K (Z) > h, then there is a Boolean algebra as described above.…”
Section: Problem 37mentioning
confidence: 94%
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“…This property simplifies the form of functionals on the space of measures on A and in this way it can be used to achieve Mazur property. In [3] it was also proved that if p K (Z) > h, then there is a Boolean algebra as described above.…”
Section: Problem 37mentioning
confidence: 94%
“…The idea of the cardinal invariant p K (I) came from certain analytic considerations contained in [3], where authors were exploring a problem if there is a Mazur space without the Gelfand-Phillips property. The Gelfand-Phillips condition is widely used in functional analysis.…”
Section: Theorem 32 P K (I) Is the Smallest Weight Of A Countable Spmentioning
confidence: 99%
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“…The space K = 2 c enjoys that property as well [8, 491Q], where c stands for the cardinality of the continuum. The Stone space K of a minimally generated Boolean algebra satisfies Seq(co∆ K ) = P (K) (see [2]) and, in fact, this result can be strengthen to saying that equality (1.1) holds. Under the Continuum Hypothesis, we present in Section 2 a construction of a compact 0-dimensional space K such that Seq 1 (co∆ K ) = Seq(co∆ K ) = P (K) (see Theorem 2.7).…”
mentioning
confidence: 99%