1993
DOI: 10.1007/bf01874222
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On the product of twob f-spaces

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Cited by 12 publications
(3 citation statements)
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“…Proof. Since E 1 is a Tychonoff space, (1)=⇒(2) is a consequence of Lemma 8 in [7]. Thus, we only need to prove (2)=⇒(1).…”
Section: Preliminaries and Notationmentioning
confidence: 97%
“…Proof. Since E 1 is a Tychonoff space, (1)=⇒(2) is a consequence of Lemma 8 in [7]. Thus, we only need to prove (2)=⇒(1).…”
Section: Preliminaries and Notationmentioning
confidence: 97%
“…It is easy to see that B is a proper subclass of the class of all pseudocompact spaces. The spaces X ∈ B admit the following nice characterization [13,5,42]: for every sequence {U n : n ∈ ω} of disjoint non-empty open subsets of X, there is an infinite subset J ⊆ ω such that for any filter ξ on J, the set F ∈ξ n∈F U n is not empty. By [38, Corollary 2.14], every pseudocompact topological group is in B.…”
Section: Corollary 16 a Countably Compact Tikhonov Space With A Conmentioning
confidence: 99%
“…Define now a function g μ B from G × cl μX B into μX as g μ B (g, x) = g A,B (g, x) whenever g ∈ A with A a bounded subset of G. It is straightforward that g μ B is well-defined and b f -continuous. Since cl μX B is compact, G × cl μX B is a b f -space[7, Theorem 9], so that g μ B is continuous. Next define a function h on G × μX as h(g, x) = g μ B (g, x) whenever x ∈ cl μX B for some bounded subset B of X.…”
mentioning
confidence: 99%