Distribution

The notion of distributivity comes from the Latin word distribut-, meaning “divided up”, and has since evolved into how mathematics deals with things that are divided up. This starts back in school when we learn that $a\cdot (b+c)=ab+ac$. This property can be generalised in the language of Boolean algebras, still maintaining the intent of dealing with divided stuff, leading to the axiom of choice being a notion of distributivity as well!

As mentioned above, the distributive property for numbers is the well-known property $a\cdot (b+c)=ab+ac$, which holds (by definition) in any ring. From this property we also get that $(a+b)(c+d)=ac+ad+bc+bd$, which we can write as

$\Pi_{k<2}\Sigma_{l<2}a_{k,l} = \Sigma_{f:2\to 2}\Pi_{k<2} a_{k,f(k)}$.

This can then be generalised from 2 terms consisting of 2 numbers to m terms consisting of n numbers, yielding the following property:

$\Pi_{k

Alright, so far so good. Let’s move away from the numbers now and work in any boolean algebra $\mathbb B$ which we further require to be complete, meaning that all joins (sums) and meets (products) exist. We can then look at the following property of $\mathbb B$, where $X,Y$ are arbitrary sets:

$\Pi_{x\in X}\Sigma_{y\in Y}a_{x,y} = \Sigma_{f:X\to Y}\Pi_{x\in X} a_{x,f(x)},\qquad (2)$

i.e. exactly the same as before in $(1)$, but we’ve just replaced a few letters. We call this property $(X,Y)$-distributivity. If the numbers admitted a boolean algebra structure with addition and multiplication then $(1)$ would show that the numbers are $(m,n)$-distributive for all $m,n<\omega$.

But is this property just.. true? Working in the boolean algebra of all sets, $(2)$ is saying that

$\bigcap_{i\in I}\bigcup_{j\in J}X_{ij}=\bigcup_{f\in J^I}\bigcap_{i\in I}X_{if(i)}$,

which can be seen to be equivalent to the axiom of choice!  So this generalised kind of distributivity at least requires choice. But alas, we can find complete boolean algebras that aren’t distributive. Let’s have a look at the boolean algebra $\mathcal B/I_\mu$ of Borel sets modulo Lebesgue null sets with inclusion, which can be shown to be complete.

Define $A_0$ as the collection of all intervals of the form $[n,n+1]$, then define $A_1$ as the collection of all intervals of the form $[n,n+1/2]$ or $[n+1/2,n+1]$, and so on. Then the intersection $A:=\bigcap_n A_n$ only consists of singletons. Fix some measure one set $B$ and define $a_{n,k}$ such that

$\{a_{n,k}\mid k<\omega\}=\{B\cap a\mid a\in A_n\}\qquad (3)$.

Note then that $\sum_i a_{n,k}=B$ for every $n<\omega$, so that the left-hand side of $(2)$ is $B$ (here $X=Y=\omega$). Let’s look at the right-hand side. For every $f:\omega\to\omega$ let $B_f:=\prod_n a_{n,f(n)}$, which by $(3)$ is a singleton, which has measure zero. This means that $\sum_f B_f=0$ as well, but $0\neq B$. Thus $\mathcal B/I_\mu$ is not $(\omega,\omega)$-distributive.

Okay, so that was a non-example at least. Now, why do we care about distributivity?

Fact. A complete boolean algebra is $(\kappa,\lambda)$-distributive if and only if forcing with it doesn’t add any new functions $f:\kappa\to\lambda$.

Oh, right. That sure does look useful.

This for instance shows that $\mathcal B/I_\mu$ must add a new function $f:\omega\to\omega$, i.e. a new real number; these new real numbers are called random reals. The above also shows that forcing with a complete boolean algebra of sets will never add any new sequences if and only if the axiom of choice holds.