For $A,B\subseteq \omega$ we write $A \subseteq^* B$ if $A\setminus B$ is finite. We call ${\cal T}\subseteq {\cal P}(\omega)$ a tower if it is linearly quasiordered with respect to $\subseteq^*$.
Using Zorn's Lemma, it is easy to see that every tower is contained in a maximal tower.
Does every maximal tower have the same cardinality? And if not, is there a cardinal $\kappa\leq 2^{\aleph_0}$ such that
The answer is yes, for I claim that every maximal chain has size continuum.
Suppose that $C$ is a chain of subsets of $\mathbb{N}$ which is maximal with respect to almost inclusion. Let's work in the quotient, so we consider only one member from each equivalence class.
First, notice that $C$ must be dense as a linear order, since otherwise, it will make a discrete step from some $A$ up to $B$, with $A\subset^* B$ and $B-A$ infinite, but no member from $C$ is strictly between $A$ and $B$ modulo finite. In this case, we can make a larger chain by adding half of $B-A$ to $A$ and creating a new member, violating the maximality of $C$. So it must be dense.
Second, I claim that if $$A_0\subset^*A_1\subset^*\cdots\quad\cdots\subset^*B_1\subset^*B_0$$ are sets from $C$, with the $A_n$'s increasing from below and the $B_m$'s decreasing from above, then there must be an element of $C$ above all the $A_n$'s and below all the $B_m$'s. This is just like filling an $(\omega,\omega)$ Hausdorff gap. By maximality, $C$ must fill such a gap.
But now, the point is that this property already implies that $C$ must have size continuum. You can find a subset of $C$ ordered like $\mathbb{Q}$, and then construct such gaps for every real number, which must be filled by a member of $C$.
