Proof of MM implies non-stationary ideal on $\aleph_1$ is $\aleph_2$ saturated.

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I am trying to understand the proof of thm 37.16 of Jech on page 687. I don't understand the first 4 lines, why does that suffice to proof the theorem? I don't see how they are related. It sais the following:

Assume MM and let $\{A_i : i \in W \} $ be a maximal almost disjoint collection of stationary subsets of $\omega_1$. We shall find a set $Z\subset W$ of size $\leq \aleph_1$ such that $\sum_{i\in Z} A_i$ contains a closed unbounded set. This implies the theorem.

Here the sum is the diagonal union. In the comments it was suggested that almost disjoint in this case means two different $A_i$'s have a non-stationary intersection.

Thanks!

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Let's consider the Boolean algebra $\Bbb B=\mathcal P(\omega_1)/\rm NS$ (one can also just consider this as a notion of forcing). Saturation simply means that it has the $\aleph_2$-chain condition.

So take a maximal family of mutually stationary sets, which is exactly a maximal antichain in $\Bbb B$. If there is a family of size $\aleph_1$ whose union contains a club, that means that this family would already constitute a partition of $1_{\Bbb B}$ (in forcing terms: this family would induce a maxiaml antichain already), because $1_{\Bbb B}$ is exactly the club filter on $\omega_1$.

In other words, it means that every maximal antichain in $\Bbb B$ is of size $<\aleph_2$, which is exactly what we want to prove.