Uncountable family of maximal antichains whose union is countable

Question

Given a ccc preorder $\mathbb{P}$ and a family $\mathcal{A}=\{A_\alpha:\alpha<\omega_1\}$ of pairwise distinct maximal antichains of $\mathbb{P}$, is it possible that $|\bigcup\mathcal{A}|\leq \aleph_0$?

I guess the answer is "no", but I did not found a way to prove it.

Thanks!

Answer 1

Let $\Bbb P={^{<\omega}\{0,1\}}$, ordered by $\subseteq$; certainly $\langle\Bbb P,\subseteq\rangle$ is ccc. For $\varphi\in{^\omega\{0,1\}}$ let

$$A_\varphi=\left\{(\varphi\upharpoonright n)^\frown\big(1-\varphi(n)\big):n\in\omega\right\}\;;$$

$A_\varphi$ is clearly an antichain in $\Bbb P$.

To see that $A_\varphi$ is maximal, let $s\in{^n\{0,1\}}$. If $s\ne\varphi\upharpoonright n$, there is a least $k<n$ such that $s(k)\ne\varphi(k)$, and

$$s\supseteq(\varphi\upharpoonright k)^\frown\big(1-\varphi(k)\big)\in A_\varphi\;,$$

so either $s\in A_\varphi$, or $A_\varphi\cup\{s\}$ is not an antichain.

If $s=\varphi\upharpoonright n$, then

$$s\subsetneqq(\varphi\upharpoonright n)^\frown\big(1-\varphi(n)\big)\in A_\varphi\;,$$

and $A_\varphi\cup\{s\}$ is not an antichain.

Thus, $\left\{A_\varphi:\varphi\in{^\omega\{0,1\}}\right\}$ is an uncountable family of maximal antichains in $\langle\Bbb P,\subseteq\rangle$.