The base theory is $\mathsf{ZF}$.
The Boolean Prime Ideal Theorem ($\mathsf{BPI}$) states
Every Boolean algebra contains a prime ideal.
The Countable Axiom of Choice ($\mathsf{AC}^{\omega}$) states
Every countable collection of non-empty sets has a choice function.
The Infinite Ramsey Theorem ($\mathsf{RT}$) states
Let $A$ be an infinite set. For any $r,p\in\mathbb{N}$, and for any function $c:[A]^{p}\rightarrow [r]$, there exists an infinite subset $B\subseteq A$ such that $c|_{[B]^{p}}$ is constant.
Where $[X]^{p}$ is the set of $p$-subsets of $X$ and $[r]=\{1,\ldots,r\}$.
In Ramsey's theorem in the hierarchy of choice principles, by Andreas Blass, it is stated that
- Neither $\mathsf{BPI}\not\Rightarrow\mathsf{RT}$ nor $\mathsf{RT}\not\Rightarrow\mathsf{BPI}$. Although, $\mathsf{BPI}\Rightarrow\mathsf{AC}_{\mathrm{fin}}$ (the axiom of choice for families of finite sets) and this implication is not reversible.
- $\mathsf{AC}^{\omega}\Rightarrow \mathsf{RT}$, and this implication is not reversible. Although, $\mathsf{RT}\Rightarrow\mathsf{AC}^{\omega}_{\mathrm{fin}}$ (the countable axiom of choice for non-empty finite sets, which is $\mathsf{ZF}$-equivalent to Kőnig's Infinity Lemma) and this implication is not reversible.
Hence, $\mathsf{BPI}\not\Rightarrow\mathsf{AC}^{\omega}$, for otherwise $\mathsf{BPI}$ would imply $\mathrm{RT}$. Also, $\mathrm{AC}^{\omega}\not\Rightarrow\mathsf{BPI}$, for otherwise $\mathsf{AC}^{\omega}$ would imply $\mathsf{AC}_{\mathrm{fin}}$.
It appears then $\mathsf{BPI}$ and $\mathsf{AC}^{\omega}$ have no logical relation (in the sense described no one is weaker/stronger than the other). Is this correct? Are they related in any other way?
The reason why I wanted to find out how they relate to each other is that they both potentially involve the uncountable. On the one hand, for example, $\mathsf{BPI}$ is $\mathsf{ZF}$-equivalent to full propositional (or first-order) compactness, which may involve an uncountable set of formulas. On the other, the sets involved in the statement of $\mathsf{AC}^{\omega}$ may be uncountable. I guess, in a sense, this is a relationship between them.
Edit: This answer has related, and interesting results.
Best Answer
Yes. That is correct.
In the Cohen model the Boolean Prime Ideal theorem holds, but there is a Dedekind finite set, which is a contradiction to countable choice.
On the other hand, $L(\Bbb R)$ of the Cohen model satisfy Dependent Choice, which is stronger than countable choice, and there are no free ultrafilters on $\omega$, so the Boolean Prime Ideal theorem fails.