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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

  1. 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.
  2. $\mathsf{AC}^{\omega}\Rightarrow \mathsf{RT}$, 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.

John
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1 Answers1

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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.

Asaf Karagila
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  • I see, right Ramsey's Theorem is false in the Cohen model. Then Dependent Choice and the Boolean Prime Ideal Theorem have no logical relationship either. A follow-up question: the statement that every Dedekind finite set is finite follows from countable of choice, does it not? Do you know if it implies implies Ramsey's Theorem (in ZF)? – John May 26 '23 at 15:20
  • I'm not sure that RT as you expressed it is false in the Cohen model, as it may very well follow from compactness. To the other things, DC and BPI are indeed independent, as I explained in my answer. Countable choice is enough to show that every infinite set is Dedekind-infinite, yes. To the final question, I'm not sure. Ramsey stuff was never my strong suit. – Asaf Karagila May 26 '23 at 16:02
  • Thank you. It is from Blass' paper, which in turn references Jech's The Axiom of Choice. I checked and it is on p.164 problem 20: "The following version of Ramsey’s Theorem is not provable without the Axiom of Choice: If $A$ is an infnite set and $[A]^2=W_1 \cup W_2$ is a partition of $[A]^2={{a,b}:a,b\in A, a\neq b}$ then there is an infnite homogeneous subset $H\subseteq A$. [Hint: The second Fraenkel (or Cohen) model. Let $A=\cup_{n=0}^{\infty}P_{n}$, and let ${a, b}\in W_1$ just in case $a, b$ are in the same pair $P_{n}$. A homogeneous set would be an infinite choice function.]" – John May 26 '23 at 16:18
  • I do not think RT would follow from compactness alone without more choice, since BPI does not imply RT. Although, I am aware there is some way to use ultrafilters (hence compactness) to prove RT. But it uses more choice if I am not mistaken. Sorry, I don't know it from the top of my head. But after using a non-principal ultrafilter we have to pick an element somewhere. – John May 26 '23 at 16:45
  • But you asked about Cohen's model, and Cohen's model is not Fraenkel's second model. – Asaf Karagila May 26 '23 at 16:52
  • My apologies, I rushed and misunderstood what Blass meant. The reference from Jech is correctly sated but has no bearing on what is proven in Blass' paper. There it is stated (p.288 - Theorem 1) "Ramsey's Theorem is false in the basic Cohen model" (Theorem 2 asserts that Ramsey's Theorem is true in the basic Fraenkel model). I hope that clears things up. – John May 26 '23 at 17:14