# Infinite distributive laws in atomless free sigma-algebra

Let $$\frak{A}$$ be the free $$\sigma$$-algebra on $$\omega_1$$ free $$\sigma$$-generators. Then $$\frak{A}$$ is not completely distributive because it is atomless. However, is it $$\omega$$-distributive in the sense that

$$\bigwedge_{m\in\omega}\bigvee_{n\in\omega} A_{m,n}=\bigvee_{\alpha\in\omega^\omega}\bigwedge_{m\in\omega}A_{m,\alpha(m)}$$

where $$\omega^\omega$$ denotes the set of all mappings of $$\omega$$ into $$\omega$$?

Of course, since in every Boolean algebra, we always have

$$\bigwedge_{m\in\omega}\bigvee_{n\in\omega} A_{m,n}\geq\bigvee_{\alpha\in\omega^\omega}\bigwedge_{m\in\omega}A_{m,\alpha(m)}$$

the question reduces to ask whether

$$\bigwedge_{m\in\omega}\bigvee_{n\in\omega} A_{m,n}\leq\bigvee_{\alpha\in\omega^\omega}\bigwedge_{m\in\omega}A_{m,\alpha(m)}$$

holds in the free $$\sigma$$-algebra on $$\omega_1$$ free $$\sigma$$-generators.

• What do you mean by $\bigwedge_{n\in\omega}\bigvee_{n\in\omega} A_{n,n}$? Perhaps $\bigwedge_{n\in\omega}\bigvee_{k\in\omega} A_{n,k}$? – Goldstern Jun 27 '19 at 22:46

## 1 Answer

This holds because $$\mathfrak{A}$$ is a concrete $$\sigma$$-algebra, being the Baire $$\sigma$$-algebra of $$2^{\omega_1}$$. In fact, the cardinality of $$\omega_1$$ plays no role whatsoever and $$\omega_1$$ could be replaced by any set.

It is easy to prove using a couple of lemmas.

Lemma 1: Let $$\Sigma$$ be a concrete $$\sigma$$-algebra on a set $$X$$, and $$(S_i)_{i \in I}$$ a family of subsets (not necessarily countable). If $$\bigcup_{i \in I} S_i$$ is in $$\Sigma$$, then $$\bigcup_{i \in I}S_i = \bigvee_{i \in I}S_i$$ in $$\Sigma$$.

This is easily proved using the definition of supremum. The second lemma is:

Lemma 2: Let $$X$$ be a set, $$(S_{i,j})_{i \in I,j \in J}$$ a family of subsets of $$X$$. Then $$\bigcap_{i \in I}\bigcup_{j \in J} a_{i,j} = \bigcup_{f \in J^I} \bigcap_{i \in I} a_{i,f(i)}$$

This is proved by showing that an element of $$X$$ is in the left hand side iff it is in the right hand side (using the axiom of choice to construct a suitable function at the right moment).

The proof that $$\mathfrak{A}$$ is $$\omega$$-distributive then goes like this: $$\bigwedge_{m \in \omega} \bigvee_{n \in \omega} a_{m,n} = \bigcap_{m \in \omega} \bigcup_{n \in \omega} a_{m,n} = \bigcup_{f \in \omega^\omega}\bigcap_{m \in \omega} a_{m,f(m)} = \bigvee_{f \in \omega^\omega} \bigwedge_{m \in \omega} a_{m,f(m)}$$

One last thing - as $$\mathfrak{A}$$ is only $$\sigma$$-complete, the statement of $$\omega$$-distributivity is actually that if $$(a_{m,n})_{m,n \in \omega}$$ is a family of elements of $$\mathfrak{A}$$, then $$\bigvee_{f \in \omega^\omega}\bigwedge_{m \in \omega}a_{m,f(m)}$$ exists and is equal to $$\bigwedge_{m \in \omega}\bigvee_{n \in \omega} a_{m,n}$$.

• .@RobertFurber. You implicitly use Stone duality in your argument, which is always a good idea. However, the $\sigma$-field you have in mind is the $\sigma$-field of Baire subsets of the Stone space of the free Boolean algebra on $\omega_1$ free $\sigma$-generators (namely, the Cantor cube $2^{\omega_1}$). I could just as well use the Stone space of $\frak{A}$ instead, in which case $\frak{A}$ is not isomorphic to a $\sigma$-field. – LJGC Jun 28 '19 at 8:10
• I do not want to sound contentious, and I am ready to accept your answer (which arranges me BTW), but, in principle, why would I choose the Stone space of the free BA over the Stone space of $\frak{A}$? – LJGC Jun 28 '19 at 8:10
• @LJGC Are you asking me this because you want an answer to your previous question: mathoverflow.net/questions/321481/… ? – Robert Furber Jun 28 '19 at 10:15
• @LJGC To answer your question about why I use $2^{\omega_1}$ instead of the Stone space of $\mathfrak{A}$, it is simply because $2^{\omega_1}$ represents $\mathfrak{A}$ in a way that maps countable joins and meets to countable unions and intersections, whereas the Stone space does not. The fact that $2^{\omega_1}$, equipped with the product topology, is the Stone space of the free Boolean algebra on $\omega_1$ generators is a bit of a distraction -- what I am using is that it is what I call the Sikorski space of the $\sigma$-complete Boolean algebra $\mathfrak{A}$. – Robert Furber Jun 28 '19 at 10:19
• @LJGC One last thing. In your first comment, you say "in which case $\mathfrak{A}$ is not isomorphic to a $\sigma$-field". This is not right, it is only the case that the representation of $\mathfrak{A}$ as the clopens of its Stone space is not a $\sigma$-field. It is still isomorphic to a $\sigma$-field, namely the Baire $\sigma$-field of $2^{\omega_1}$. Facts about Boolean algebras expressible in the language of Boolean algebra, such as $\sigma$-distributivity, are invariant under isomorphism, so we can use any representation of $\mathfrak{A}$ at all to prove $\sigma$-distributivity. – Robert Furber Jun 28 '19 at 10:38