The Galvin-Prikry theorem says that Borel sets are Ramsey. This means that for every Borel set $S\subseteq[\omega]^\omega$, there is an $A\in[\omega]^\omega$ such that either $[A]^\omega \subseteq S$ or $[A]^\omega \cap S=\emptyset$.

My question is, for which $\kappa>\omega$ does the analogous statement hold (call it the $\kappa$-Galvin-Prikry theorem):

For every open set $S\subseteq [\kappa]^\omega$, there is an $A\in[\kappa]^\kappa$ such that $[A]^\omega\subseteq S$ or $[A]^\omega \cap S=\emptyset$.

With the open sets given by the usual topology on $[\kappa]^\omega$, with basic open sets $[\eta,\omega]=\{s\in[\omega]^\omega:\eta \prec s\}$, where $\prec$ denotes initial segment.

At $\omega$, the Galvin-Prikry theorem is equivalent to Nash-Williams' "every block contains a barrier". (See page 644 in http://www.cs.elte.hu/~kope/bqoint.pdf ). Ie, for every Block $B\subseteq[\omega]^{<\omega}$, there is an $A\in[\omega]^\omega$ such that $B\cap[A]^{<\omega}$ is a barrier.

The $\kappa$-version is equivalent to "every $\kappa$-block contains a $\kappa$-barrier", with the definitions given in Shelah's paper:

Better quasi-orders for uncountable cardinals

Israel Journal of Mathematics September 1982, Volume 42, Issue 3, pp 177-226 http://link.springer.com/article/10.1007%2FBF02802723

The important implication here can be seen because the $\kappa$-Galvin-Prikry theorem is equivalent to $\kappa\stackrel{\mbox{open}}{\longrightarrow}(\kappa)^\omega$ which means $\forall f:[\kappa]^\omega\rightarrow 2$ with $f$ continuous, $\exists A\in[\kappa]^\kappa$ such that $|f"[A]^\omega|=1$. Which in turn is equivalent to $\forall f:B\rightarrow 2$ with $B\subset[\kappa]^{<\omega}$ a $\kappa$-block, $\exists A\in[\kappa]^\kappa$ such that $|f"[A]^{<\omega}\cap B|=1$. From this "every $\kappa$-block contains a $\kappa$-barrier" quickly follows.

On page 2 of the quoted paper, Shelah says that for "every $\kappa$-block contains a $\kappa$-barrier" to hold, $\kappa$ must be Ramsey. But he gives no proof.

If we assume that $\kappa$ is Ramsey, then it isn't too difficult to show that every $\kappa$-block contains a $\kappa$-barrier.

So, by Shelah, $\kappa$ is Ramsey iff the $\kappa$-Galvin-Prikry theorem holds. But I haven't managed to prove what Shelah omitted.

So why does the $\kappa$-Galvin-Prikry theorem imply that $\kappa$ is Ramsey?

Or the same for any of the other equivalent versions I mentioned.

  • 19
    $\begingroup$ "But I haven't managed to prove what Shelah omitted" should be printed on t-shirts sold at ASL meetings because we've all been there... $\endgroup$ Jun 5, 2013 at 2:18

2 Answers 2


The Galvin-Prikry theorem for $\kappa > \omega$ was given by E.M. Kleinberg and R.A. Shore in their paper "On Large Cardinals and Partition Relations".

Let $\kappa > \omega$. If we denote by $\kappa \overset{\circ}{\rightarrow} (\kappa)^{\omega}_{\lambda}$ the assertion of $\kappa \rightarrow (\kappa)^{\omega}_{\lambda}$ restricted to $\kappa-$Borel functions, we have:

Theorem [Kleinberg-Shore, 1971]. Let $\kappa > \omega$. Then, $\kappa$ is Ramsey iff $\kappa \overset{\circ}{\rightarrow} (\kappa)^{\omega}_{\lambda}$ for each $\lambda < \kappa$.


I think it is the case, but anyway, it is at least possible that Shelah did not mean that the $\kappa$-Galvin-Prikry theorem implies that $\kappa$, what he may have meant is "in order to carry over the proof of the GP theorem to the $\kappa$ case one has to assume that $\kappa$ is Ramsey". I know that this is not an answer, but it seems to fit into the text of the paper.

  • $\begingroup$ Yes I suppose this may well be the case. Indeed, it seems that to carry over the proof you need $\kappa$ to be Ramsey. I guess the problem is less trivial than I imagined. (Since Shelah used barely a sentence, I thought it may be an easy proof for someone who knows more about Ramsey cardinals than me!) $\endgroup$
    – user33625
    Jun 24, 2013 at 12:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.