2000
DOI: 10.1002/(sici)1521-3870(200001)46:1<3::aid-malq3>3.0.co;2-e
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Compactness in Countable Tychonoff Products and Choice

Abstract: Abstract. We study the relationship between the countable axiom of choice and the Tychonoff product theorem for countable families of topological spaces.Mathematics Subject Classification: 03E25, 04A25, 54B10.

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Cited by 15 publications
(23 citation statements)
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“…It is well known that the countable version of Tychonoff's compactness theorem, i. e., countable products of compact topological spaces are compact, implies the axiom of countable choice CAC, see [11]. The status of the reverse implication is still an open problem, see [1,5,8]. However, it is known, see [12], that if we restrict to the class of compact metric spaces, then Tychonoff's theorem for countable products of compact metric spaces, CPM(C,C), is a theorem of ZF + CAC.…”
Section: Introduction and Some Known Resultsmentioning
confidence: 99%
“…It is well known that the countable version of Tychonoff's compactness theorem, i. e., countable products of compact topological spaces are compact, implies the axiom of countable choice CAC, see [11]. The status of the reverse implication is still an open problem, see [1,5,8]. However, it is known, see [12], that if we restrict to the class of compact metric spaces, then Tychonoff's theorem for countable products of compact metric spaces, CPM(C,C), is a theorem of ZF + CAC.…”
Section: Introduction and Some Known Resultsmentioning
confidence: 99%
“…(ii)→(i) Let A = (A n ) n∈N be a disjoint family of non-empty sets such that no infinite subfamily of A has a choice function and consider the pseudometric d on X = {A n : n ∈ N} given by (5). Clearly, X is complete.…”
Section: Clearly Ac Implies the Statement: (H)mentioning
confidence: 99%
“…) 3. CFE(TP) (Closed Filter Extendability in Tychonoff Products): If {(X i , T i ) : i ∈ I} is a family of compact topological spaces, then every family G ⊂ K(X), X = i∈I X i , with the fip extends to a closed ultrafilter F. 4 Kelley's proof of the implication "TCT → AC" can be adopted to show: Theorem 2.2 CTCT implies CAC. However, one cannot adopt his proof of the implication "AC → TCT" to establish "CAC → CTCT".…”
Section: An Open Set O Of X Is Called Restricted Open If There Is a Fmentioning
confidence: 99%
“…Several partial answers to Question 2.3 have been given in [4], where, in addition, the following characterization of CTCT has been established:…”
Section: Let X Be a Non Empty Set And E ⊂ P(x) \ {∅} A Non Empty Colmentioning
confidence: 99%
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