In set theory without the Axiom of Choice (AC), we investigate the deductive strength and mutual relationships of the following statements: 1) Every infinite set X has an almost disjoint family A of infinite subsets of X with [formula]. (2) Every infinite set X has an almost disjoint family A of infinite subsets of X with [formula]. (3) For every infinite set X, every almost disjoint family in X can be extended to a maximal almost disjoint family in X. (4) For every infinite set X, no infinite maximal almost disjoint family in X has cardinality [formula]. (5) For every infinite set A, there is a continuum sized almost disjoint family A ⊆ Aω. (6) For every free ultrafilter U on ω and every infinite set A, the ultrapower Aω/U has cardinality at least [formula].
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In the framework of ZF (Zermelo-Fraenkel set theory without the Axiom of Choice) we provide topological and Boolean-algebraic characterizations of the statements "2R is countably compact" and "2R is compact".
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Let X be an infinite set, and P(X) the Boolean algebra of subsets of X. We consider the following statements: BPI(X): Every proper filter of P(X) can be extended to an ultrafilter. UF(X): P(X) has a free ultrafilter. We will show in ZF (i.e., Zermelo–Fraenkel set theory without the Axiom of Choice) that the following four statements are equivalent: (i) BPI(ω). (ii) The Tychonoff product 2R, where 2 is the discrete space {0, 1}, is compact. (iii) The Tychonoff product [0, 1] R is compact. (iv) In a Boolean algebra of size ≤ |R| every filter can be extended to an ultrafilter. We will also show that in ZF, UF(R) does not imply BPI(R). Hence, BPI(R) is strictly stronger than UF(R). We do not know if UF(ω) implies BPI(ω) in ZF. Furthermore, we will prove that the axiom of choice for sets of subsets of R does not imply BPI(R) and, in addition, the axiom of choice for well orderable sets of non-empty sets does not imply BPI(ω).
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