7
$\begingroup$

This MO thread has several examples of complete Boolean algebras that are not isomorphic to $\sigma$-algebra of sets. But what is a non-trivial example of a complete Boolean algebra that is isomorphic to a $\sigma$-algebra of set as Boolean algebras? Equivalently, is there a $\sigma$-algebra of set that is complete as a Boolean algebra?

A trivial example is the power set algebra $\mathcal{P}(X)$, or any algebra of set that is atomic and complete as Boolean algebra. Also, from the above post it seems a non-trivial example is probably not ccc. I am wondering if the tree algebra on some tree such as $2^{<\aleph_1}$ can help, but the tree algebra itself doesn't seem complete, or even $\sigma$. Maybe we can consider its completion?

Clarification: By a complete Boolean algebra I mean a Boolean algebra whose any subset has supremum. An algebra of set (on $X$) is a nonempty subset of $\mathcal{P}(X)$ closed under union and complementation. An algebra of set can be viewed as a Boolean algebra in an obvious way.

Edit: Now I have the feeling that the Boolean completion of the poset $(2^{<\aleph_1},\supseteq)$ should does it. This is the poset of partial map from $\aleph_1$ to $\{0,1\}$ with countable domain, under inverse inclusion (the largest element being the empty map). Every branch in $2^{<\aleph_1}$, equivalently every element in $2^{\aleph_1}$, determines an $\sigma$-complete ultrafilter in the Boolean completion because the branch has uncountable cofinality. The Boolean completion seems to be the regular-open algebra on $2^{\aleph_1}$ with countable support product topology. I'm not sure if it is extremally disconnected; probably not.

$\endgroup$
7
  • $\begingroup$ Please clarify: are you asking about $\sigma$-algebras of sets? And what do you mean by "complete"? That it is closed under arbitrary unions and intersection? Or complete as an abstract lattice? $\endgroup$
    – tomasz
    Commented Nov 26, 2021 at 20:15
  • $\begingroup$ After clarification, every example given in the MO thread you linked qualifies. Every Boolean algebra is isomorphic (as a Boolean algebra) to a Boolean algebra of sets (namely, the algebra of clopen subsets of its Stone space). If the Boolean algebra of sets is the power set, then the countable meet has to be the countable intersection, so the original algebra is isomorphic to a $\sigma$-algebra of sets. $\endgroup$
    – tomasz
    Commented Nov 26, 2021 at 20:35
  • $\begingroup$ @tomasz But I thought countable meet may not correspond to countable intersection under Stone representation, e.g., math.stackexchange.com/questions/3147558/… $\endgroup$
    – Lxm
    Commented Nov 26, 2021 at 20:40
  • $\begingroup$ It may not, but in this case (or any case when the Stone algebra is a $\sigma$-algebra of sets) it clearly does, which is my point. $\endgroup$
    – tomasz
    Commented Nov 26, 2021 at 21:02
  • 2
    $\begingroup$ @tomasz Simon's answer in that thread showed that the Boolean algebra of Lebesgue measurable sets modulo null sets is complete, and not isomorphic to any $\sigma$-algebra on any set $X$ (of course it is not isomorphic to $\mathcal{P}(X)$ since it is atomless while $\mathcal{P}(X)$ is atomic). $\endgroup$
    – Lxm
    Commented Nov 26, 2021 at 22:31

2 Answers 2

4
$\begingroup$

Here is a partial answer. In particular, I will show it is consistent with ZFC that no nontrivial examples exist, i.e. that any $\sigma$-algebra of sets which is complete as a Boolean algebra is isomorphic to a power set.

First, some general remarks on representing Boolean algebras as algebras of sets. Suppose $B$ is a Boolean algebra. Then to give a homomorphism $f:B\to\mathcal{P}(X)$ is the same thing as giving, for each $x\in X$, an ultrafilter $U_x$ on $B$ (and then $f$ is defined by $f(A)=\{x\in X:A\in U_x\}$). The homomorphism $f$ is injective iff these ultrafilters separate points of $B$, i.e. iff every nonzero element of $B$ is contained in some $U_x$. If $\kappa$ is a cardinal and $B$ is $\kappa$-complete, then $f$ is a $\kappa$-complete homomorphism iff each $U_x$ is a $\kappa$-complete ultrafilter. In particular, in the case $\kappa=\aleph_1$, this says that a Boolean $\sigma$-algebra $B$ is isomorphic to a $\sigma$-algebra of sets iff every nonzero element of $B$ is contained in some countably complete ultrafilter.

Now here's the key fact which makes it difficult for this to happen in a complete Boolean algebra.

Theorem: Let $B$ be a complete Boolean algebra and let $U$ be a nonprincipal countably complete ultrafilter on $B$. Then there exists a measurable cardinal $\kappa$ such that $U$ is $\kappa$-complete.

In the special case where $B$ is a power set, this is a well-known fact in the basic theory of measurable cardinals. The proof for a general complete Boolean algebra is similar.

Proof of Theorem: Let $\kappa$ be the least cardinal such that $U$ is not $\kappa^+$-complete (such a cardinal exists since $U$ is not principal; in particular $U$ is not $|U|^+$-complete since if $\bigwedge U$ were in $U$ then it would generate all of $U$). Let $(x_\alpha)_{\alpha<\kappa}$ be a sequence of elements of $U$ such that $\bigwedge_{\alpha<\kappa} x_\alpha\not\in U$. Adding $\neg\bigwedge x_\alpha$ as one more term at the start of the sequence, we may assume that in fact $\bigwedge_{\alpha<\kappa} x_\alpha=0$. By minimality of $\kappa$, for each $\alpha<\kappa$, $\bigwedge_{\beta<\alpha} x_\beta\in U$. So, replacing each $x_\alpha$ with $\bigwedge_{\beta<\alpha} x_\beta$, we may assume the sequence $(x_\alpha)$ is decreasing and continuous.

Now let $y_\alpha=x_{\alpha+1}\wedge \neg x_\alpha$. Then the $y_\alpha$ are pairwise disjoint, each $y_\alpha\not\in U$, and $\bigvee_{\beta<\alpha} y_\beta=\neg x_\alpha$. In particular, since $\bigwedge_{\alpha<\kappa}x_\alpha=0$, this means $\bigvee_{\alpha<\kappa}y_\alpha=1$. There is then a is a complete homomorphism $f:\mathcal{P}(\kappa)\to B$ defined by $f(S)=\bigvee_{\alpha\in S} y_\alpha$. Now consider the pullback ultrafilter $f^{-1}(U)$ on $\mathcal{P}(\kappa)$. Since $f$ is a complete homomorphism and $U$ is $\kappa$-complete, $f^{-1}(U)$ is also $\kappa$-complete. Since $y_\alpha\not\in U$ for each $\alpha$, $f^{-1}(U)$ is nonprincipal. Thus $\kappa$ is measurable. $\blacksquare$

Now suppose $B$ is a $\sigma$-algebra of sets that is complete as a Boolean algebra and is not isomorphic to a power set. In particular, then, $B$ is not atomic, so there exists some nonzero $a\in B$ which contains no atom. Since $B$ is a $\sigma$-algebra of sets, $a$ is contained in some countably complete ultrafilter $U$, which must be nonprincipal since there is no atom below $a$. By the Theorem, this implies the existence of a measurable cardinal.

So no such $B$ can exist if measurable cardinals do not exist. I don't know whether it is possible for such a $B$ to exist if measurable cardinals do exist.

$\endgroup$
-1
$\begingroup$

There are no non-trivial examples. An algebra of sets is a complete boolean algebra (cBa) iff it is isomorphic to a power set algebra.

For the interesting direction, fix an algebra $\mathcal{A}$ on a set $X$ such that $\mathcal{A}$ is a cBa. Define an equivalence relation $E$ on $X$ by $a E b$ iff $(\forall A \in \mathcal{A})(a \in A \iff b \in A)$. Let $\mathcal{E}$ be the set of $E$-equivalence classes. Now use the fact that $\mathcal{A}$ is a cBa to show that $\mathcal{A} = \{\bigcup \mathcal{W}: \mathcal{W} \subseteq \mathcal{E}\}$. To show this, first check that if $T \in \mathcal{E}$ and $a \in T$, then the $\subseteq$-infimum of the set of those members of $\mathcal{A}$ that contain $a$ is $T$. So $\mathcal{E} \subseteq \mathcal{A}$. Next show that if $\mathcal{W} \subseteq \mathcal{E}$, then $\bigcup \mathcal{W}$ is the $\subseteq$-supremum of $\mathcal{W}$. It follows that $(\mathcal{A}, \subseteq)$ is isomorphic to $(\mathcal{P}(\mathcal{E}), \subseteq)$.

$\endgroup$
2
  • 1
    $\begingroup$ 1. Why is the infimum in $\mathcal{A}$ equal to $T$, or why does $T$ belong to $\mathcal{A}$ in the first place? 2. The answer by Joseph in the MO thread says a $\sigma$-complete Boolean algebra is isomorphic to an algebra of sets iff each nonzero element is contained in some $\sigma$-complete ultrafilter. I'm not sure if he's really talking about $\sigma$-algebra or just algebra, but at least there should be nontrivial example of algebra, which is also why I thought about $2^{<\aleph_1}$. $\endgroup$
    – Lxm
    Commented Dec 2, 2021 at 0:29
  • 1
    $\begingroup$ This answer is false. A Boolean algebra is isomorphic to a powerset algebra if and only if it is complete and atomic. There are complete Boolean algebras which are not isomorphic to a powerset algebra. (For example, the complete Boolean algebra of regular opens of $\mathbb{R}$.) $\endgroup$
    – Pilcrow
    Commented Dec 23, 2021 at 21:15

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .