# The universal algebra of a $\sigma$-algebra

I am searching for the 'dual' algebraic structure of a $$\sigma$$-algebra. The notion of duality is like in the case of the Boolean algebra and set algebra.

If $$X$$ is a set, the complement and intersection on the power set of $$X$$ is called a set algebra and the series of equations that define a Boolean algebra is the dual of this structure.

I found this link that seems related to my question: Is there such a thing as the sigma-completion of a Boolean algebra?

but still it does not solve my problem.

• I thought about that, and I thought I just could add an axiom scheme like that. But not every sigma-algebra is a set algebra (I think), so that answer does not feel right. – zeh Feb 21 '14 at 9:04
• Are you asking for a type of duality for $\sigma$-complete Boolean algebras that is analogous to the duality between Boolean algebras and compact totally disconnected spaces? If so, then there are several ways to generalize Stone duality to $\sigma$-complete Boolean algebras. However, the situation is more complicated for $\sigma$-complete Boolean algebras than for Boolean algebras since $\sigma$-complete Boolean algebras do not necessarily have $\sigma$-complete ultrafilters. – Joseph Van Name Feb 21 '14 at 15:58
• Other related link: terrytao.wordpress.com/2009/01/12/… – YCor Apr 28 '19 at 17:11
• Here's a copy of a comment to an erased answer by Emil Jeřábek (Feb 21 '14), which I copy because it's of general interest: "There are $\sigma$-complete Boolean algebras that are not isomorphic to any $\sigma$-algebra. In particular, $\sigma$-algebras satisfy the distributive law $\bigwedge_{n=0}^\infty\bigvee_{m=0}^\infty a_{n,m}=\bigvee_{f:\omega\to\omega}\bigwedge_{n=0}^\infty a_{n,f(n)}$ (that is, the RHS exists even though the join is uncountable, and it equals the LHS), whereas it is easy to construct complete BA where this fails." See also mathoverflow.net/a/158316/14094 – YCor Apr 28 '19 at 17:17

In the paper [1], Sikorski constructs a duality that generalizes Stone duality to certain $\sigma$-complete Boolean algebras and more generally certain $\kappa$-complete Boolean algebras. I shall outline the duality mentioned in Sikorski's paper here.
Suppose that $\lambda$ is a cardinal. Then a Boolean algebra $B$ is said to be $\lambda$-complete if the least upper bound $\bigvee R$ exists whenever $|R|<\lambda$. A filter $Z$ on a $\lambda$-complete Boolean algebra $B$ is said to be a $\lambda$-complete filter if whenever $|R|<\lambda$ and $R\subseteq Z$, then $\bigwedge R\in Z$ as well. We shall call a $\lambda$-complete Boolean algebra strongly $\lambda$-representable if every $\lambda$-complete filter can be extended to a $\lambda$-complete ultrafilter. A $P_{\lambda}$-space is a completely regular space such that the intersection of less than $\lambda$ many open sets is open, and a topological space $X$ is said to be $\lambda$-compact if every open cover of $X$ has a subcover of cardinality less than $\lambda$. Sikorski gave a correspondence between all $\lambda$-compact $P_{\lambda}$-spaces and all $\lambda$-representable $\lambda$-complete Boolean algebras. The proof of this result is exactly the same as the proof of the duality between Boolean algebras and compact totally disconnected spaces.
• mathoverflow.net/questions/158271/… gives $\sigma$-complete Boolean algebras which are not $\sigma$-algebras. – Joseph Van Name Feb 22 '14 at 13:58
• Is the category of $\lambda$-compact $P_{\lambda}$-spaces the pro category of sets of cardinality less than $\lambda$? If this is directly analogous to stone space duality, I would suspect so, since the category of stone spaces is the pro-category of finite discrete spaces. – Dean Young May 5 '19 at 22:32