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Tychonoff products of two-element sets and some weakenings of the Boolean prime ideal theorem

Keremedis K.

Bulletin of the Polish Academy of Sciences. Mathematics

2005

Vol. 53, no 4

349-359

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).