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

### Volume 58 / 2010

#### 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$^{\rm{WO}}$) does not imply “the Tychonoff product $2^\mathbb R$, 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 $2^\mathbb{R}$ is countably compact in* ZF?

2. Assuming the Countable Axiom of Multiple Choice (CMC), the statements “every infinite subset of $2^{\mathbb{R}}$ has an accumulation point”, “every countably infinite subset of $2^{\mathbb{R}}$ has an accumulation point”, “$2^{\mathbb{R}}$ is countably compact”, and UF($\omega$) = “there is a free ultrafilter on $\omega$” 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 $2^{\mathbb{R}}$ has an accumulation point”, and UF($\omega$) are, in ZF, pairwise equivalent. Hence, in ZF, the statement “$2^{\mathbb{R}}$ is countably compact” implies UF($\omega$).

4. The statement “every infinite subset of $2^{\mathbb{R}}$ has an accumulation point” implies “every countable family of 2-element subsets of the powerset $\mathcal{P}(\mathbb{R})$ of $\mathbb{R}$ has a choice function”.

5. The Countable Axiom of Choice restricted to non-empty finite sets, (CAC$_{\rm{fin}}$), is, in ZF, strictly weaker than the statement “for every infinite set $X$, $2^{X}$ is countably compact”.