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.

  • $\begingroup$ 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. $\endgroup$ – zeh Feb 21 '14 at 9:04
  • $\begingroup$ 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. $\endgroup$ – Joseph Van Name Feb 21 '14 at 15:58
  • $\begingroup$ Other related link: terrytao.wordpress.com/2009/01/12/… $\endgroup$ – YCor Apr 28 '19 at 17:11
  • $\begingroup$ 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 $\endgroup$ – 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.

  1. Sikorski, Roman Remarks on some topological spaces of high power. Fund. Math. 37, (1950). 125–136.
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  • $\begingroup$ Thanks a lot, that is a useful answer and reference. Btw, what was the example of not every sigma-algebra is a booblean algebra?. $\endgroup$ – zeh Feb 22 '14 at 7:00
  • $\begingroup$ mathoverflow.net/questions/158271/… gives $\sigma$-complete Boolean algebras which are not $\sigma$-algebras. $\endgroup$ – Joseph Van Name Feb 22 '14 at 13:58
  • $\begingroup$ 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. $\endgroup$ – Dean Young May 5 '19 at 22:32

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