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".
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We study the deductive strength of the following statements: 𝖱𝖱: every set has a rigid binary relation, 𝖧𝖱𝖱: every set has a hereditarily rigid binary relation, 𝖲𝖱𝖱: every set has a strongly rigid binary relation, in set theory without the Axiom of Choice. 𝖱𝖱 was recently formulated by J. D. Hamkins and J. Palumbo, and 𝖲𝖱𝖱 is a classical (non-trivial) 𝖹𝖥𝖢-result by P. Vopěnka, A. Pultr and Z. Hedrlín.
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