Tychonoff Products of Two-Element Sets and Some Weakenings of the Boolean Prime Ideal Theorem
Tom 53 / 2005
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 $).
