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• # Artykuł - szczegóły

## Bulletin of the Polish Academy of Sciences. Mathematics

2010 | 58 | 2 | 91-107

## On the Set-Theoretic Strength of Countable Compactness of the Tychonoff Product $2^{ℝ}$

EN

### Abstrakty

EN
We work in ZF set theory (i.e., Zermelo-Fraenkel set theory minus the Axiom of Choice AC) and show the following:
1. The Axiom of Choice for well-ordered families of non-empty sets ($AC^{WO}$) does not imply "the Tychonoff product $2^{ℝ}$, where 2 is the discrete space {0,1}, is countably compact" in ZF. This answers in the negative the following question from Keremedis, Felouzis, and Tachtsis [Bull. Polish Acad. Sci. Math. 55 (2007)]: Does the Countable Axiom of Choice for families of non-empty sets of reals imply $2^{ℝ}$ is countably compact in ZF?
2. Assuming the Countable Axiom of Multiple Choice (CMC), the statements "every infinite subset of $2^{ℝ}$ has an accumulation point", "every countably infinite subset of $2^{ℝ}$ has an accumulation point", "$2^{ℝ}$ is countably compact", and UF(ω) = "there is a free ultrafilter on ω" are pairwise equivalent.
3. The statements "for every infinite set X, every countably infinite subset of $2^{X}$ has an accumulation point", "every countably infinite subset of $2^{ℝ}$ has an accumulation point", and UF(ω) are, in ZF, pairwise equivalent. Hence, in ZF, the statement "$2^{ℝ}$ is countably compact" implies UF(ω).
4. The statement "every infinite subset of $2^{ℝ}$ has an accumulation point" implies "every countable family of 2-element subsets of the powerset 𝓟(ℝ) of ℝ has a choice function". 5. The Countable Axiom of Choice restricted to non-empty finite sets, ($CAC_{fin}$), is, in ZF, strictly weaker than the statement "for every infinite set X, $2^{X}$ is countably compact".

91-107

wydano
2010

### Twórcy

autor
• Department of Statistics and Actuarial-Financial Mathematics, University of the Aegean, Karlovassi, Samos, 83200, Greece

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