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


1. every tower has cardinality $\leq \kappa$, and
2. there is a tower with cardinality $\kappa$?



Cardinalities of maximal towers in ${\cal P}(\omega)$
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 $C$. 


