Families of pairwise incomparable subsets of the integers Certain maximal objects whose existence follows from Zorn's Lemma have received some
set-theoretic attention.
Examples are maximal independent families and maximal almost disjoint families.
There is quite a bit of information out there about the two cardinal invariants $\mathfrak i$
and $\mathfrak a$, the minimal sizes of maximal independent families of subsets of $\omega$
and of maximal almost disjoint families of subsets of $\omega$.
However, I have not seen anything concerning maximal incomparable families:
Let $\mathcal A$ be a family of subsets of $\omega$ that are both infinite and coinfinite.
We call $\mathcal A$ incomparable if the elements of $\mathcal A$ are pairwise incomparable with respect to the relation $\subseteq^*$ where $A\subseteq^*B$
if $A\setminus B$ is finite.  Note that both independent families and almost disjoint families are incomparable.
Question:  What is the minimal size of a maximal incomparable family of subsets of $\omega$?
I am guessing that this cardinal invariant is equal to $2^{\aleph_0}$ or equal to $\mathfrak i$, but mostly for psychological/social reasons:  It should have been looked at before, unless it is equal to one of the known invariants.  
 A: Michael Hrusak has recently answered my question:  The cardinal invariant that I am looking for is indeed $2^{\aleph_0}$.  
We are looking for the minimal size of a maximal infinite family of pairwise incomparable elements of the Boolean algebra $\mathcal P(\omega)/fin$.  We can ask the same question for any infinite Boolean algebra $B$.  For a Boolean algebra $B$ and $b\in B$ let $B\upharpoonright b=\{a\in B:a\leq b\}$.  Let $d(B)$ be the minimal size of a dense subset of $B$.
The following is true for any infinite Boolean algebra $B$:  The minimal size of a maximal infinite family of pairwise incomparable elements of $B$ is at least $\min\{d(B\upharpoonright b):b\in B\setminus\{0\}\}$.  Note that in the case of $B=\mathcal P(\omega)/fin$ we have $d(B\upharpoonright b)=2^{\aleph_0}$ for all non-zero $b\in B$, hence this fact solves the original problem.
To see this fact, let $A$ be an infinite set of pairwise incomparable elements of $B$ 
of size $<\min\{d(B\upharpoonright b):b\in B\setminus\{0\}\}$ and let $a\in A$.  Since $A$ is infinite, both $a$ and $-a$ are different from $0$.
Let $\langle A\rangle$ denote the Boolean algebra generated by $A$.
Since $A$ is infinite, $\langle A\rangle$ is of the same size as $A$.  By the size of $A$, 
there are $x\in B\upharpoonright -a$ and $y\in B\upharpoonright a$, both non-zero,
such that no non-zero element of $\langle A\rangle$ is below $x$ or $y$.
Let $a'=x\vee(a-y)$.  An easy computation shows that $a'$ is incomparable with all elements of $A$, showing that $A$ was not maximal.
