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

Eleftherios Tachtsis

Bulletin of the Polish Academy of Sciences. Mathematics, Tome 58 (2010), p. 91-107 / Harvested from The Polish Digital Mathematics Library

Résumé

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 (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”.

Détail

Publié le : 2010-01-01
EUDML-ID : urn:eudml:doc:281170