Are the Boolean algebras ${\cal P}(\omega)/(\text{fin})$ and ${\cal P}(\omega)/(\text{thin})$ isomorphic?

Question

A set $A\subseteq \omega$ is said to be thin if $$\lim\sup_{n\to\infty}\frac{|A\cap \{0,\ldots, n\}|}{n+1} = 0.$$

We say for $A, B\subseteq \omega$ that $A\simeq_\text{fin} B$ if the symmetric difference $A \, \triangle \, B$ is finite. It is easy to see that this an equivalence relation, and a routine verification shows that the resulting quotient is an atomless Boolean algebra. It is customarily denoted by ${\cal P}(\omega)/(\text{fin})$.

We can do exactly the same thing for thin sets by saying $A\simeq_{\text{thin}} B$ if $A \, \triangle \, B$ is thin. This gives rise to a Boolean algebra ${\cal P}(\omega)/(\text{thin})$.

Question. Are the Boolean algebras ${\cal P}(\omega)/(\text{fin})$ and ${\cal P}(\omega)/(\text{thin})$ isomorphic?

Answer 1

The answer is no.

In the Boolean algebra $P(\omega)/\text{Fin}$ every strictly descending sequence $A_0>A_1>A_2>\cdots$ has a nonzero lower bound, by the famous construction of Hausdorff. Namely, one can find an infinite set $A$ that is almost contained in every $A_n$. For example, let $A$ consist of the $n$th element of $A_0\cap A_1\cap\cdots\cap A_n$ for every $n$. This is an infinite set, which after the $n$th point is contained in $A_n$.

But the Boolean algebra $P(\omega)/\text{Thin}$, in contrast, does not exhibit this feature, since we can arrange that the density of the sets converges to $0$, and in this case there can be no set with nonzero density that is below them all.

Here is another way to argue:

Theorem. The Boolean algebras $P(\omega)/\text{Fin}$ and $P(\omega)/\text{Thin}$ are not isomorphic, because:

1. There is no countably infinite maximal antichain in $P(\omega)/\text{Fin}$, but
2. There is a countably infinite maximal antichain in $P(\omega)/\text{Thin}$.

Proof. For 1, suppose that $A_0, A_1, A_2,\ldots$ is a countably infinite antichain. So these form an almost-disjoint family of infinite sets. Let $A$ choose one element from $A_n-\bigcup_{k<n}A_k$. So $A$ is an infinite set almost disjoint from each of them, showing it wasn't maximal.

For 2, let $A_n$ be all the numbers of the form $2^nr$, where $r$ is odd. So $A_0$ has density 1/2 and $A_1$ has density 1/4 and so forth. This is a partion of $\omega$, but every set with positive density will have positive density intersection with one of them, since if the density of $A$ exceeds that of the tail beyond $n$, then it must overlap nontrivially with some $A_k$ for $k<n$. $\Box$

Ultimately, the two argument methods can be seen as the same. For example, with the antichain $A_n$ in (1), we may consider the descending sequence $A_0^c>(A_0\cup A_1)^c>(A_0\cup A_1\cup A_2)^c>\cdots$ and then apply the argument from before about descending sequences. And similarly in the density case.