TITLE: Are there $2^{\aleph_0}$ pairwise non-isomorphic Boolean algebra structures on $\omega$?
QUESTION [7 upvotes]: 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?

REPLY [2 votes]: 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 $K_k$).