Borel measures in consonant spaces

Bouziad, Ahmed

doi:10.1016/0166-8641(95)00089-5

Cited by 25 publications

(17 citation statements)

References 6 publications

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“…Since Q is not consonant [1] (see also Remark 9 below) and consonance is hereditary with respect to closed subsets, in this model every analytic metrizable consonant space is complete.…”

Section: Where Y Is Compact Every Open Section Of ϕ Contains a Compamentioning

confidence: 99%

“…Answering a question from [2], the nonconsonance of Q was established in [1] by using a measure theoretical argument. The proof is as follows: Let C denote the Cantor set and, following [6], let A ⊂ C × C be such that the projection map π : A → C is open, onto and not compact covering.…”

Section: Remarkmentioning

confidence: 99%

“…In [3], among other things, it is proved that everyČech-complete space is consonant. It is also known that metrizable consonant spaces are hereditarily Baire [1] (i.e. every closed subspace of a metrizable consonant metric space is a Baire space).…”

mentioning

confidence: 99%

“…This remained one of the central open problems in the theory of consonance. For coanalytic metrizable spaces, this problem is entirely solved: A co-analytic metrizable space X is consonant if and only if X is Polish (see [1]). In this note we show that the statement "every analytic metrizable consonant space is complete" is independent of the usual axioms of set theory (Theorem 6).…”

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confidence: 99%

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Consonance and topological completeness in analytic spaces

Bouziad¹

1999

Proc. Amer. Math. Soc.

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Abstract. We give a set-valued criterion for a topological space X to be consonant, i.e. the upper Kuratowski topology on the family of all closed subsets of X coincides with the co-compact topology. This characterization of consonance is then used to show that the statement "every analytic metrizable consonant space is complete" is independent of the usual axioms of set theory. This answers a question by Nogura and Shakhmatov. It is also proved that continuous open surjections defined on a consonant space are compact covering.A topological space X is said to be consonant if the co-compact topology on the set of all closed subsets of X coincides with the upper Kuratowski topology. The class of consonant spaces was introduced by Dolecki, Greco and Lechicki in [2], [3] and was recently studied rather intensively. It was noticed by Nogura and Shakhmatov in [8] that two other classical topologies (automatically) coincide on the set of closed subsets of a consonant space, namely, the Fell topology and the Kuratowski topology. In [3], among other things, it is proved that everyČech-complete space is consonant. It is also known that metrizable consonant spaces are hereditarily Baire [1] (i.e. every closed subspace of a metrizable consonant metric space is a Baire space). These results prompted Nogura and Shakhmatov in [8, Problem 11.4] to ask whether consonant metrizable spaces are (Čech-)complete. This remained one of the central open problems in the theory of consonance. For coanalytic metrizable spaces, this problem is entirely solved: A co-analytic metrizable space X is consonant if and only if X is Polish (see [1]). In this note we show that the statement "every analytic metrizable consonant space is complete" is independent of the usual axioms of set theory (Theorem 6). It is also proved that continuous open surjections defined on a consonant space are compact covering (Corollary 8), which gives a generalization of the classical theorem of Pasynkov stated forČech-complete spaces. Corollary 8, along with a construction of Michael [6], allows us to give a new and topological proof of the nonconsonance of the usual space of rationals.This note is based on a characterization of consonance in terms of a special property of lower semicontinuous set-valued maps (Theorem 2). Concerning topological completeness, an important and recent result of this kind, achieved by van Mill,

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“…Since Q is not consonant [1] (see also Remark 9 below) and consonance is hereditary with respect to closed subsets, in this model every analytic metrizable consonant space is complete.…”

Section: Where Y Is Compact Every Open Section Of ϕ Contains a Compamentioning

confidence: 99%

Section: Remarkmentioning

confidence: 99%

mentioning

confidence: 99%

mentioning

confidence: 99%

See 2 more Smart Citations

Consonance and topological completeness in analytic spaces

Bouziad¹

1999

Proc. Amer. Math. Soc.

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show abstract

“…It is also known that metrizable consonant spaces are hereditarily Baire [4]. These results prompted Nogura and Shakhmatov to ask whether consonant metrizable spaces arě Cech-complete [25,Problem 11.4].…”

Section: Introductionmentioning

confidence: 99%