1
$\begingroup$

It is known that:

Theorem (Kelley, 1959). There exists a finite, strictly positive, finitely additive measure on a Boolean algebra $A$ if and only if $A^+$ is the union of a countable number of classes, each of which has a positive intersection number.

Let $A$ be a Boolean algebra and $C\subseteq A^+$ a nonempty set. For $c_0,\ldots,c_n\in C$ (not necessarily distinct), write

$\alpha^\ast(c_0,\ldots,c_n)=(n+1)^{-1} \text{max} \{|I|:I\subseteq n+1,\prod_{i\in I}c_i\not=0\}$.

The intersection number of $C$ is defined as

$\alpha(C)=\text{inf}\{\alpha^\ast(c_0,\ldots,c_n):n\in\omega,c_0,\ldots,c_n\in C\}$.

I am having a hard time understanding what is this intersection number concretely. For example, consider the power set of $4$, and let $C=\{\{0\},\{1,2\},\{1,3\},\{2,3\}\}$. What is the intersection nunber of $C$? (Fremlin ($\ast$), where I borrowed the definition of $\alpha^*$ and $\alpha$ from, asserts that it is $2/5$, but does not give any explanation.)

($\ast$) Fremlin, Measure Algebras, Handbook of Boolean Algebras, volume 3, 1989, pp. 877-980 (See, in particular, p. 949 for the definition of the intersection number.)

$\endgroup$
2
  • $\begingroup$ This old Prikry paper on intersection numbers may or may not be illuminating, long long time since I looked at it: ams.org/journals/proc/2001-129-02/S0002-9939-00-05561-1/… Kelley's intersection number is related somehow to the "fractional chromatic number of a hypergraph" whatever that may be. $\endgroup$
    – bof
    Commented Aug 31, 2021 at 21:04
  • $\begingroup$ In view of @bof's comment I have taken the liberty of adding the co.combinatorics tag - I hope that is OK. $\endgroup$
    – Yemon Choi
    Commented Sep 1, 2021 at 17:50

1 Answer 1

1
$\begingroup$

Years ago (2004) I was writing some notes for myself which included a section on some parts of Kelley's paper; looking at them now, I don't immediately see why things I omitted as "obvious" are obvious, but I reproduce what I wrote in case it is helpful. (I may try to come back and fill in more details later if required.)


The following definitions are made for subsets of $2^\Omega$, I trust that they would be equivalent to the definitions you gave above if one represents the given Boolean algebra $A$ as a Boolean sub-algebra of a suitable $2^\Omega$.

If $U_1,\dots, U_m$ are subsets of $\Omega$ (not necessarily distinct), and $\eta \in [0,1]$, we say the $m$-tuple ${\bf U}:=(U_1,\dots, U_m)$ is $\eta$-intersecting if $$\sum_{i=1}^m 1_{U_i} \leq \eta\ m 1_\Omega $$ and denote the least such $\eta$ by $\iota({\bf U})$. Note that if $k\cdot {\bf U}$ denotes ${\bf U}$ repeated $k$ times then $\iota(k\cdot {\bf U})=\iota({\bf U})$.

For ${\mathcal C}\subseteq 2^{\Omega}$ we define $\iota({\mathcal C})$ to be $\inf_{\bf U} \iota({\bf U})$ where the infimum is over all finite tuples ${\bf U}=(U_1,\dots, U_m)$ where $U_i\in {\mathcal C}$ for all $i$.

The following is what I have written in my old notes, without further justification/calculation. It might be explained/motivated by things in Kelley's paper but I do not have a copy to hand right now.

Think of $2^{\Omega}$ as the Hamming cube sitting inside $\ell_\infty^{|\Omega|}$, and ${\mathcal C}$ as a set of vertices (=extreme points of $2^{\Omega}$). Then $\iota({\mathcal C})$ is the "inradius", with respect to the $\ell_\infty$-norm, of the convex hull of ${\mathcal C}\cup\{{\bf 0}\}$. More precisely, $\iota({\mathcal C}) = \inf\{ \Vert x\Vert_\infty \colon x\in {\rm conv}({\mathcal C})\}$.

Example. Let $\Omega=\{1,2,3,4\}$ and let ${\mathcal C}$ be the following subset of $\ell_\infty^4$: $$ {\mathcal C}=\{ (1,0,0,0), (0,1,1,0), (0,1,0,1), (0,0,1,1)\}. $$ An element of the convex hull is $$ {\bf x} = (\lambda_1, \lambda_2+\lambda_3, \lambda_2+\lambda_4, \lambda_3+\lambda_4) $$ where $\lambda_i\in [0,1]$ for all $i$ and $\sum_i \lambda_i= 1$. Then $x_2+x_3+x_4=2(1-\lambda_1)$, so $$ \Vert {\bf x}\Vert_\infty \geq \max(|x_1|, \tfrac{1}{3} |x_2+x_3+x_4| ) =\max\{ \lambda_1, \tfrac{2}{3}(1-\lambda_1)\ \colon 0\leq\lambda_1\leq 1\} \qquad(*)$$ By drawing the lines $y=t$, $y=\frac{2}{3}(1-t)$ one sees that the RHS of $(*)$ is always greater than the y-coordinate of the point where these two lines meet; this intersection point is $(2/5, 2/5)$. This argument shows that $\Vert{\bf x}\vert_\infty \geq 2/5$, and by taking $\lambda_1=2/5$ and $\lambda_2=\lambda_3=\lambda_4=1/5$ we see that this lower bound can be attained at some ${\bf x}\in {\rm conv}({\mathcal C})$.

$\endgroup$
3
  • $\begingroup$ Thank you for having taken the trouble to answer, but I have a feeling that your argument is overcomplicated. There must be a simpler approach. $\endgroup$
    – Hugh
    Commented Sep 1, 2021 at 8:00
  • 1
    $\begingroup$ @Hugh if you are referring to my proof that a particular set system has intersection 2/5, then I was not aiming to find the shortest proof, but rather to show how the geometric perspective with convex hulls could be applied. $\endgroup$
    – Yemon Choi
    Commented Sep 1, 2021 at 17:46
  • 1
    $\begingroup$ @Hugh Your original question had a ttle which asked for "understanding" of the intersection number, and so my answer was meant to suggest that it can be viewed geometrically; indeed, the measure $\mu$ that is provided by the theorem you quote comes from a separating hyperplane argument IIRC. You even say "I am having a hard time understanding what is this intersection number concretely" and so I tried to offer a geometric formulation. If all you wanted was a walk-through of a simple solution to the exercise/example that you provided, then perhaps you could reword your question to that effect. $\endgroup$
    – Yemon Choi
    Commented Sep 1, 2021 at 17:47

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .