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

Eleftherios Tachtsis. "On the Set-Theoretic Strength of Countable Compactness of the Tychonoff Product $2^{ℝ}$." Bulletin of the Polish Academy of Sciences. Mathematics 58.2 (2010): 91-107. <http://eudml.org/doc/281170>.

@article{EleftheriosTachtsis2010,
abstract = {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”.},
author = {Eleftherios Tachtsis},
journal = {Bulletin of the Polish Academy of Sciences. Mathematics},
language = {eng},
number = {2},
pages = {91-107},
title = {On the Set-Theoretic Strength of Countable Compactness of the Tychonoff Product $2^\{ℝ\}$},
url = {http://eudml.org/doc/281170},
volume = {58},
year = {2010},
}

TY - JOUR
AU - Eleftherios Tachtsis
TI - On the Set-Theoretic Strength of Countable Compactness of the Tychonoff Product $2^{ℝ}$
JO - Bulletin of the Polish Academy of Sciences. Mathematics
PY - 2010
VL - 58
IS - 2
SP - 91
EP - 107
AB - 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”.
LA - eng
UR - http://eudml.org/doc/281170
ER -