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.

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

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 $\bigwedge_{m \in \omega}\bigvee_{n \in \omega} a_{m,n}$.

  • $\begingroup$ .@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. $\endgroup$ – LJGC Jun 28 '19 at 8:10
  • $\begingroup$ 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 $\frak{A}$? $\endgroup$ – LJGC Jun 28 '19 at 8:10
  • $\begingroup$ @LJGC Are you asking me this because you want an answer to your previous question: mathoverflow.net/questions/321481/… ? $\endgroup$ – Robert Furber Jun 28 '19 at 10:15
  • $\begingroup$ @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 $\mathfrak{A}$. $\endgroup$ – Robert Furber Jun 28 '19 at 10:19
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    $\begingroup$ @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. $\endgroup$ – Robert Furber Jun 28 '19 at 10:38

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