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Tychonoff Products of Two-Element Sets and Some Weakenings of the Boolean Prime Ideal Theorem

Tom 53 / 2005

Bulletin Polish Acad. Sci. Math. 53 (2005), 349-359 MSC: 03G05, 03E25, 54B10, 54D30. DOI: 10.4064/ba53-4-1

Streszczenie

Let $X$ be an infinite set, and $\mathcal{P}(X)$ the Boolean algebra of subsets of $X$. We consider the following statements:

BPI($X$): Every proper filter of $\mathcal{P}(X)$ can be extended to an ultrafilter.

UF($X$): $\mathcal{P}(X)$ 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($\omega$).

(ii) The Tychonoff product $2^{\mathbb{R}}$, where $2$ is the discrete space $\{0,1\}$, is compact.

(iii) The Tychonoff product $[0,1]^{\mathbb{R}}$ is compact.

(iv) In a Boolean algebra of size $\leq|\mathbb{R}|$ every filter can be extended to an ultrafilter.

We will also show that in ZF, UF($\mathbb{R}$) does not imply BPI($\mathbb{R}%$). Hence, BPI($\mathbb{R}$) is strictly stronger than UF($\mathbb{R}$). We do not know if UF($\omega$) implies BPI($\omega$) in ZF.

Furthermore, we will prove that the axiom of choice for sets of subsets of $\mathbb{R}$ does not imply BPI($\mathbb{R}$) and, in addition, the axiom of choice for well orderable sets of non-empty sets does not imply BPI($\omega$).

Autorzy

• Kyriakos KeremedisDepartment of Mathematics
University of the Aegean
83 200 Karlovassi (Samos), Greece
e-mail

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