**Preliminaries:**

Let $[\omega]^{\omega}$ be the set of all infinite subsets of $\omega$, the first countable ordinal (the set of the natural numbers).

- We say that $\mathcal A\subset [\omega]^{\omega}$ is open if for all $A, B \in [\omega]^{\omega}$ then if $A\subset^*B\in \mathcal A$ then $A\in \mathcal A$.
- We say that $\mathcal D\subset [\omega]^{\omega}$ is dense if for all $A \in [\omega]^\omega$ there exists $B \in \mathcal D$ such that $B\subset^*A$.

It's possible to show that these open sets define a topology on $[\omega]^\omega$ and that in this topology, the dense subsets are what we defined above.

Let $\mathfrak h$ be the first cardinal that there exists a colection $\mathscr A$ of dense open subsets of $[\omega]^\omega$ of cardinality $\mathfrak h$ such that $\bigcap \mathscr A=\emptyset$. It's possible to show that $\omega<\mathfrak h\leq \mathfrak c$, that $\mathfrak h$ is regular and it's consistent that $\mathfrak h<\mathfrak c$.

We may prove the following theorem, that is called the Base Matrix Tree Lemma:

There exists $\mathcal T\subset [\omega]^\omega$ such that:

- $\supset^*$ is a tree order on $\mathcal T$. The first level of $\mathcal T$ is $\{\omega\}$. The height of $\mathcal T$ is $\mathfrak h$.
- Every node of $\mathcal T$ has exactly $\mathfrak c$ successors.
- For every $A\in [\omega]^\omega$ there exists $B \in \mathcal T$ such that $B\subset A$.
- If $0<\alpha<\mathfrak h$, then the $\alpha$-th level of $\mathcal T$ is a mad family over $\omega$.

Reminder: Let $N$ be an infinite countable set. An almost disjoint family over $N$ is an infinite collection $\mathcal A$ of infinite subsets of $N$ such that for every $A, B \in \mathcal A$, $A\cap B$ is finite. A maximal almost disjoint family (mad family) over $N$ is an almost disjoint family that isn't properly contained in any almost disjoint family.

**Context:**

I am trying to understand the proof of theorem 4.2 of this article. I have understood every detail but the one I'm gonna ask for help.

**My Question:**

Suppose $\mathfrak h<\mathfrak c$ and fix $\mathcal T$ as above. For every $A \subset 2^{<\omega}$, let $\pi_A=\{n \in \omega: A\cap 2^n\neq \emptyset\}\subset \omega$. So there exists $\mathcal A\subset[2^{<\omega}]^\omega$ such that:

- $\mathcal A$ is a mad family (over $2^{<\omega}$)
- Every $A \in \mathcal A$ is either a chain or an antichain in $2^{<\omega}$.
- $\pi_A \in \mathcal T$ for all $A \in \mathcal A$.
- If $A, B \in \mathcal A$ and $A\neq B$, then $\pi_A\neq \pi_B$.

Question: Why does $\mathcal A$ exists?

**My Progress:**

The authors claim that the existence of $\mathcal A$ follows from a simple application of the Zorn's Lemma. So I defined: $$\mathcal U=\{\mathcal A \subset [2^{<\omega}]^\omega: |\mathcal A|\geq\omega, \mathcal A \text{ is an adf}, \mathcal A \text{ satisfies (2), (3), (4)}\}.$$

I managed to show that $\mathcal U$ is not empty and it's easy to see that the union of a chain in $\mathcal U$ is again an element of $\mathcal U$, therefore there exists a maximal element $\mathcal M$. It remains to show that $\mathcal M$ is mad, and this is where I'm having trouble. If $\mathcal M$ is not mad, then there exists $X\in [2^{<\omega}]^\omega$ such that for every $A \in \mathcal M$, $A\cap X$ is finite. Then it's not hard to show that there exists an infinite $X' \subset X$ such that $X'$ is either a chain or an antichain in $2^{<\omega}$ and such that $\pi_{X'}\in \mathcal T$. If we can shrink $X'$ so that for every $A \in \mathcal M$, $\pi_A\neq \pi_{X'}$ then we are done but this is where I'm stuck.