# Does "$X \not\to (\omega)^\omega_2$ for every infinite $X$" imply ${\sf AC}$?

For any set $$X$$ and cardinal $$\mu \neq \emptyset$$, we denote by $$[X]^\mu$$ the collection of subsets of cardinality $$\mu$$. If $$\kappa, \mu \neq \emptyset$$ are cardinals and $$f: [X]^\mu\to \kappa$$ is a map, we say that $$H\subseteq X$$ is homogeneous with respect to $$f$$ if the restriction $$f|_{[H]^\mu}: [H]^\mu \to \kappa$$ is constant.

For cardinals $$\lambda, \mu, \kappa\neq \emptyset$$ and any set $$X\neq \emptyset$$ we write $$X \to (\lambda)^\mu_\kappa$$ if for every map $$f: [X]^\mu\to\kappa$$ there is $$H\subseteq X$$ such that $$H$$ is homogeneous with respect to $$f$$ and $$|H|=\lambda$$.

With the help of the Axiom of Choice $${\sf (AC)}$$ one can prove that $$X \not\to (\omega)^\omega_2$$ for every infinite $$X$$ (see Theorem 7, p. 5 of this recommended introduction to infinite combinatorics, thank you to Burak for writing it!).

Question. Does the statement "$$X \not\to (\omega)^\omega_2$$ for every infinite set $$X$$" imply $${\sf (AC)}$$?

• As the author of the notes, I tried to use the standard notation (known as the Erdös-Rado arrow notation) instead of inventing a new notation. Mar 21 at 17:46
• @ToddTrimble I have encountered the notation so many times that I am quite certain that it is standard in that field. For instance Saharon Shelah uses it in many articles. (By the way, I didn't find what "Egad" stands for, I took it to be an abbreviation like "WLG") Mar 21 at 18:18
• Why all this bad publicity? It is useful, and compact. No one who works in the field gets confused. Mar 21 at 22:13
• Compact it is, but for the life of me I can’t remember which number in the notation denotes which parameter. Mar 23 at 8:01
• If only cardinal and ordinal numbers were under consideration, we could just define $(b)^r_k$ to be the corresponding "Ramsey number" and then we could write $a\ge(b)^r_k$ or $a\lt(b)^r_k$ instead of $a\to(b)^r)_k$ or $a\not\to(b)^r_k$.
– bof
Mar 24 at 0:44

The answer is no, the statement that for every set $$X$$ we have $$X\not\to(\omega)^\omega_2$$ does not imply the axiom of choice.

This was shown by Kleinberg and Seiferas in 1973, see

MR0340025 (49 #4782) Kleinberg, E. M.; Seiferas, J. I. Infinite exponent partition relations and well-ordered choice. J. Symbolic Logic 38 (1973), 299–308. https://doi.org/10.2307/2272066

For $$\kappa$$ a (well-ordered) infinite cardinal, $$\kappa$$-well-ordered choice, $$\mathsf{AC}_\kappa$$, is the statement that every $$\kappa$$-sequence of nonempty sets admits a choice function.

The axiom of well-ordered choice $$\mathsf{WOC}$$ is the statement that $$\mathsf{AC}_\kappa$$ holds for all infinite well-ordered $$\kappa$$.

This statement is strictly weaker than the axiom of choice: it does not imply that $$\mathbb R$$ is well-orderable, and even if we add this assumption, the result is still weaker than choice. See for instance theorem 5.1 in

MR1351415 (96h:03087) Higasikawa, Masasi Partition principles and infinite sums of cardinal numbers. Notre Dame J. Formal Logic 36 (1995), no. 3, 425–434. https://doi.org/10.1305/ndjfl/1040149358

However, as shown in the paper by Kleinberg and Seiferas, $$\mathsf{WOC}$$ plus the existence of a well-ordering of $$[\omega]^\omega$$ rules out all infinite exponent partition relations. It is still open (as far as I know) whether $$\mathsf{WOC}$$ suffices for this result. What Kleinberg and Seiferas show is that, under $$\mathsf{WOC}$$, either all infinite exponent partition relations fail, or else $$\omega\to(\omega)^\omega_2$$. (And the latter fails if $$[\omega]^\omega$$ is well-orderable.)

• So, really it's just WOC + $\omega\nrightarrow(\omega)^\omega_2$, which is a consequence of WOC + well-orderable continuum. And the question remains whether or not the partition relation just follows from WOC, and can therefore be eliminated as an explicit assumption? If $\omega\to(\omega)^\omega_2$ has some LC strength to it, that means that we can arrange all kind of models where the continuum is not well-orderable, but the partition relation fails and WOC holds, just by taking symmetric extensions over some LC-challenged model. Mar 29 at 16:50
• Yes, that's the situation at the moment. And sure, if we knew that the Mathias relation gives you an inaccessible, we would be done. The alternative is to force $\mathsf{WOC}$ over, say, a Solovay model, while preserving the relation. Mar 29 at 17:55
• Do we even know that WOC is forceable over Solovay models (without forcing full AC, that is)? Mar 29 at 20:18