Are there $2^{\aleph_0}$ pairwise non-isomorphic Boolean algebra structures on $\omega$?

Question

Is there a collection of $2^{\aleph_0}$ pairwise non-isomorphic countable Boolean algebras?

Equivalently, are there $2^{\aleph_0}$ pairwise non-homeomorphic closed subsets in the Cantor space?

Answer 1

The answer is yes: there are $2^{\aleph_0}$ countable Boolean algebras up to isomorphism, or equivalently $2^{\aleph_0}$ homeomorphism class of metrizable totally disconnected compact Hausdorff spaces. This is the main result of:

Reichbach, M. The power of topological types of some classes of 0-dimensional sets. Proc. Amer. Math. Soc. 13 1962 17-23 (Open link).

It precisely consists of showing that there are $c=2^{\aleph_0}$ closed subsets in a Cantor space, modulo global homeomorphism. Adding a discrete countable subset accumulating onto the given closed subset yields the desired family of continuum many non-homeomorphic metrizable Stone [=totally disconnected compact Hausdorff] spaces.

(Note that it also directly implies that there are $\ge c$ isomorphism types of Boolean subalgebras in $2^\omega$. At the topological level, classifying Boolean algebras embedding into $2^{\aleph_0}$ is the same as classifying [nonempty] separable Stone spaces. For Stone spaces, the class of metrizable spaces is properly contained in the class of separable ones. By Reichbach's 1962 result the former (modulo homeomorphism) has cardinal $c$ while by Freniche's 1984 result given by Juan, the latter has cardinal $2^c$.)

Answer 2

Another reference for "Answer = Yes":

Chapter 12 of the Handbook of Boolean algebras has the title "The number of Boolean Algebras".

Don Monk, the author of this chapter, writes: "For almost all classes K of BAs which have been an object of intensive study, there are exactly $2^\kappa$ isomorphism types of members of K of each infinite power $\kappa$."

In particular, this is true for K=interval algebras. There are $2^\kappa$ many linear orders $L$ of cardinality $\kappa$ such that the corresponding interval algebras $Int(L)$ are pairwise non-isomorphic. (The elements of $Int(L)$ are the finite unions of intervals of $L$.)

Here is an explicit sketch of Monk's proof (which he calls "of folklore nature"):

• Call a BA $B$ "atomic" if you can find an atom below every positive element, and call $x\in B$ "atomless" if there is no atom below $x$.
• For any BA $B$, let $I(B)$ be the (possibly improper) ideal generated by the atoms together with the atomless elements. (An ideal is improper if it is equal to the whole BA)
• Write $D(B)$ (the "derivative" of $B$) for the Boolean algebra $B/I(B)$.
• Let $A_0=\omega$ be the linear order of natural numbers, and let $A_1=1+\eta+\omega$ be the linear order of the nonnegative rationals followed by a copy of the natural numbers. Then $Int(A_0)$ is atomic, and $Int(A_1)$ is not, so you have two nonisomorphic algebras.
• (In both these linear orders we call the smalles element $0$.)
• Check that both $I(Int(A_0))$ and $I(Int(A_1))$ are improper, i.e. the derivatives of $Int(A_0)$, $Int(A_1)$ are singletons. This fact is responsible for the "crucial point" below.
• Now for any $i,j\in \{0,1\}$ consider $A_{i}\times A_{j}$ with the "inverse" lexicographic order (first compare two second components in $A_{j}$, and only when they are equal compare the first components)
• $A_0\times A_0$ and $A_0\times A_1$ lead to atomic interval algebras, but $A_1\times A_0$ and $A_1\times A_1$ do not.
• A crucial point is that the first factor "disappears" when taking the derivative: $D(Int(A_i\times A_j))$ is naturally isomorphic to $Int(A_j)$, and therefore atomic iff $j=1$. So the four interval algebras derived from $A_i\times A_j$ are pairwise nonisomorphic.
• This proof idea can be generalized to infinite (weak) products. For notational simplicity I restrict the proof sketch to countably many factors.
• For any sequence $x=(x(0),x(1),\dots)\in 2^\omega$ let $A_x$ be the weak product $\prod_i^{\rm wk} A_{x(i)}$, i.e., the set of all functions $f$ defined on $\omega$ which satisfy $f(i)\in A_{x(i)}$ for all $i$, and $f(i)=0$ for almost all $i$.
• The set $A_x$ is countable, and linearly ordered as an "inverse" lexicographic product.
• The main work to do is to check that the $n$-th derivative of $Int(A_x)$ is naturally isomorphic to $Int(\prod_{i\ge n}^{\rm wk} A_{x(i)})$, and hence is atomic iff $x(i)=0$.
• This show that for different $x,y\in 2^\omega$ the Boolean algebras $Int(A_x)$ and $Int(A_y)$ are not isomorphic.

Essentially the same proof can be used for uncountable $\kappa$, where the result is probably more interesting. (I seem to recall that all countable BAs are interval algebras.)

Answer 3

This was a problem in the Scottish Book posed by Ulam. It was solved by

F. J. Freniche, The Number of Nonisomorphic Boolean subalgebras of a Power Set, Proc. Amer. Math. Soc., 91 (1984) 199-201.

the number of non isomorphic sub algebras is $2^{2^k}$ for any infinite cardinal $k$

Answer 4

I gave, in the comments of this answer, a construction of closed nowhere dense subsets $A_S$ of $[0,1]$ for each set $S\subset\mathbb N^*$ of positive integers, no two of which being homeomorphic. These properties imply that they are second countable Stone spaces, hence their algebra of clopen subsets is countable; moreover, according to Stone's representation theorem, no two of these algebras are isomorphic. YCor made the connection to this question in the aforementioned comments, and I, perhaps pretentiously, understood it as an invitation to write an answer here.

The construction is as follows. Let, inductively, $\Omega_1$ be a the countable compact subspace of $[0,1]$ with a single limit point at $0$ (say the closure of $\{2^{-n},n\in\mathbb N\}$), and $\Omega_{k+1}$ be a subspace of $[0,1]$ consisting of a sequence of copies of $\Omega_k$ accumulating at zero, together with zero itself. Below is a representation of (a countable dense subset of) $\Omega_2$, where each black circle is a single point (blue circles for illustration purposes).

If this helps, the usual order of $[0,1]$, restricted to $\Omega_k$, is the reverse order of $\omega^k+1$, where the +1 of $\omega^k+1$ corresponds to the zero of $\Omega_k\subset[0,1]$.

Let $K_k$ be the compact set consisting of the gluing of $\Omega_k$ and the usual Cantor set $C$, respectively at zero and at 1. Of course it is homeomorphic to a subset of $[0,1]$. Now construct $A_S\subset[0,1]$ by fitting a copy of $K_k$ in $(2^{-k},2^{-k+1})$ if and only if $k\in S$, and adding 0 (to make it compact if $S$ is infinite).

It is clear (at least to me) that $A_S$ is closed nowhere dense. It remains to show that two such subsets are not homeomorphic. Let us fix $S$ and see that we can recover it from the topology of $A:=A_S$ alone.

Write $\lim B$ for the set of limit points of $B$. I define $n$-limit points of $B$ as the elements of $\lim^{n}B\setminus\lim^{n-1}B$. By convention, 0-limit points are isolated points and $\infty$-limit points are elements of $\bigcap_{n\geq0}\lim^nB$. Let $\ell_n(A)$ be the set of $n$-limit points of $A$. Then $k\in S$ if and only if the set $$ \ell_\infty(A)\cap\overline{\ell_{k-1}(A)}\cap\left(\overline{\ell_k(A)}\right)^\complement $$ is non empty (in which case it is the gluing point of $K_k$).