# Why does the generalised Galvin-Prikry Theorem only hold at Ramsey cardinals?

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.

• "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... – Todd Eisworth Jun 5 '13 at 2:18

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.
• 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!) – user33625 Jun 24 '13 at 12:35