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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.

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IIRC, both $\mathfrak{a}<\mathfrak{i}$ and $\mathfrak{i}<\mathfrak{a}$ are consistent, so your cardinal can't be unconditionally equal to either one. – Nathan Jan 16 '14 at 15:03
I don't understand. This reasoning would apply if maximal independent families and maximal almost disjoint families were maximal incomparable. But I don't see why this should be the case. Independence and almost disjointness are both stronger than incomparability, but when we maximize, we maximize with respect to different properties. – Stefan Geschke Jan 16 '14 at 23:05
I think your invariant is $\ge \mathfrak{r}$, the reaping number. Suppose $\mathcal{A}$ is a maximal incomparable family, and let $\mathcal{B}$ consist of all members of $\mathcal{A}$ and, in addition, all complements of members of $\mathcal{A}$. If $x$ splits every member of $\mathcal{B}$, then $x\cap (\omega\setminus a)$ and $(\omega\setminus x)\cap a$ are infinite for every $a\in\mathcal{A}$, which means that $\mathcal{A}\cup\{x\}$ is incomparable. Hence $\mathcal{B}$ is unsplittable. – Paul McKenney Jan 17 '14 at 19:36
@Paul: Thank you for the comment. I was actually aware of this and should have mentioned it in the question. The first thing I observed along these lines was that MA implies that this invariant is large. Just use Cohen forcing. But the argument then boils down to the fact that Cohen reals are splitting. – Stefan Geschke Jan 17 '14 at 20:21

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.

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