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I am wondering if the category of abstract $\sigma$-algebras is contravariantly equivalent to the pro-category of the category of countable sets.

Note: I have tweaked this question a little. Now it is asking something which is different, but which I think is correct.

To make this precise, I set these definitions:

Definition: An abstract $\sigma$-algebra is a boolean algebra with countable meets and joins. A map of abstract $\sigma$-algebras $f : A \rightarrow B$ is a map of boolean algebras which preserves countable meets and joins.

The class of abstract $\sigma$-algebras, along with the class of maps of abstract $\sigma$-algebras, form a category, which I call $\sigma$-alg.

Let $\sigma \text{-set}$ be the pro-category of the full subcategory of set whose objects are countable sets. My question is whether there an equivalence of categories between the pro category $Pro(\sigma \text{-set})$ and $\sigma \text{-alg}$.

Let $\text{Hom}_{\sigma \text{-set}}(-, \mathbb{F}_2) : \sigma \text{-set} \rightarrow \sigma \text{-alg}$ be the functor sending a countable set $X$ to the $\sigma$-algebra of $\text{Hom}_{\sigma \text{-set}}(X, \mathbb{F}_2)$ of set-maps from $X$ to $\mathbb{F}_2$. We put a partial order on this where, for $\phi, \psi \in \text{Hom}_{\sigma \text{-set}}(X, \mathbb{F}_2)$, $\phi \leq \psi$ when $\phi(x) \leq \psi(x)$ for each $x \in X$ (we put $0 < 1$ in $\mathbb{F}_2$). Let $C$ be full subcategory of $\sigma \text{-alg}$ whose objects are isomorphic to $\text{Hom}_{\sigma \text{-set}}(X, \mathbb{F}_2)$ for some countable set $X$.

We must then show three things:

A) $\text{Hom}_{\sigma \text{-set}}(-, \mathbb{F}_2) : \sigma \text{-set} \rightarrow C$ is a categorical equivalence.

B) $\sigma$-alg is the ind-category of $C$.

C) Additionally, I am interested to show directly that the pro-category of $\sigma$-set is the full subcategory of topological spaces whose objects are completely regular, such that every open cover has a countable subcover, and such that the intersection of countably many open sets is open.

See "Remarks on some topological spaces of high power" by Roman Sikorski for a proof of a similar equivalence. I think the results in this paper show that $\sigma$-algebras are equivalent to the full subcategory of topological spaces whose objects are

1) Completely regular

2) Every open cover of $X$ has a countable subcover.

3) The intersection of countably many open sets is open.

Is this the same as being an object in the pro-category of $\sigma$-set?

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  • $\begingroup$ Presumably you want a contravariant equivalence? Stone duality is contravariant. $\endgroup$
    – Qiaochu Yuan
    Commented May 6, 2019 at 9:56
  • $\begingroup$ @QiaochuYuan Yes. I think I have it; Write $C$ for the subcategory of Boolean algebras with countable joins and meets, where maps preserve countable joins and meets. $F = \text{Hom}_{\sigma \text{-set}} (-, \mathbb{F}_2) : \sigma \text{-set} \rightarrow C$ is fully faithful, let $D$ be the subcategory of $C$ whose objects are isomorphic to an object of the form $F(X)$ for some $X$. Then I want to show that $C$ is the ind-category of $D$, and that the pro-category of $\sigma$-set is the full subcategory of top whose objects satisfy (1) (2) and (3) above. $\endgroup$
    – user30211
    Commented May 6, 2019 at 18:07
  • $\begingroup$ Above I use $\sigma$-set to mean the full subcategory of set consisting of countable sets. $\endgroup$
    – user30211
    Commented May 6, 2019 at 18:31
  • $\begingroup$ @QiaochuYuan And by the way, what you were getting at is completely right- I had the question set up as a covariant correspondence, which couldn't work since the functors in question are contravariant. $\endgroup$
    – user30211
    Commented May 6, 2019 at 19:48

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