# (p,q) versus Glivenko-Cantelli

Say a collection of sets $\mathcal{F}$ satisfies the (p,q) property if whenever $\mathcal{G}\subseteq\mathcal{F}$ with $|\mathcal{G}|\geq p$, there is a $\mathcal{H}\subseteq\mathcal{G}$ with $|\mathcal{H}|\geq q$ and $\bigcap\mathcal{H}\neq\emptyset$. A (p,q) theorem for some class of sets $\mathcal{C}$ gives a bound $T_{p,q}$ so that whenever $\mathcal{F}\subseteq\mathcal{C}$ has the (p,q) property, there is a set $S$ such that $|S|\leq T_{p,q}$ and whenever $F\in\mathcal{F}$, $F\cap S\neq\emptyset$.

The main classes I know of for which a (p,q) theorem has been proven are the convex sets in $\mathbb{R}^d$ and the sets of bounded VC dimension. But both of these are Glivenko-Cantelli classes, which means that if we select a sufficiently large finite set $S$ at random then, with probability $1$, every set in the class intersects with $S$ with fraction close to its measure. In particular, by choosing $S$ large enough, this means every set in the class of measure $\geq\epsilon$ contains a fraction of $S$ of size, say $(\epsilon/2)|S|$. (Technically $S$ is a multiset, but when the base set is large enough, this doesn't make a difference.)

Unless I'm misunderstanding something, the conclusion of Glivenko-Cantelli is much stronger than the conclusion of the (p,q) theorem: the claim holds for almost every choice of $S$, for all large sets in the class simultaneously, and each set has "large" intersection with $S$ instead of just one point. Also, Glivenko-Cantelli-type theorems seem to be easier to prove.

The only catch is that Glivenko-Cantelli only covers the large sets in the class, while the (p,q) theorem covers every single set in the class satisfying the (p,q) property.

This brings me to the problem: the main example of a collection $\mathcal{F}$ with the (p,q) property seems to be a collection of sets with measure $\geq\epsilon$ for some fixed $\epsilon$. In other words, exactly the collections already covered by Glivenko-Cantelli.

My question is, roughly, what more the (p,q) theorem does for us. Am I missing something about the statements, so that the (p,q) theorem provides more? Or are there interesting collections with the (p,q) property which aren't already covered by being a Glivenko-Cantelli class?

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I think the (p,q) theorem is completely different from Glivenko-Cantelli. Why do you say that the convex sets form a GL class? Consider the following family in R: Intervals containing 0 or 1. This is a (3,2) family but it is not GL, as it contains the {0} point. (Of course if you change the measure and concentrate it on {0,1}, then this does become a GL class, which is not surprising knowing the proof of the (p,q) theorem.)

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Whether or not the convex sets form a GC class depends on the measure (I was basing the statement on "Glivenko-Cantelli Theorems for Classes of Convex Sets", which shows that they are for various measures). I guess one way to apply the (p,q) theorem would be to choose a measure for which the convex sets fail to satisfy Glivenko-Cantelli, and use that measure to construct a (p,q) family, but if that's the only way to generate (p,q) families, it doesn't address the VC dimension case, since VC sets will satisfy Glivenko-Cantelli for any measure. –  Henry Towsner Jun 19 '12 at 21:53