I am trying answer a question:
Let $\mathbb{P} = \{p \mid p \text{ is a finite function } \omega_2 \times \omega \to 2\}$ ordered by revesre inclusion, i.e. $p<q$ iff $p \supseteq q$. Does $\mathbb{P}$ have an uncountable antichain?
I know, that I'm supposed to use $\Delta$-system Lemma to deduce a negative answer:
If $A$ is an uncountable system of finite subsets of $\omega_1$, then there exists an uncountable $\Delta$-subsystem $B \subseteq A$.
I understand that I need to work with domains of functions but I don't quite see how to continue. Hints are appreciated. Thank you.
If $D$ is an uncountable subset of $\Bbb P$, then $\{\operatorname{dom}p\mid p\in D\}$ is an uncountable family of finite subsets of $\omega_2\times\omega$. So it contains an uncountable $\Delta$-system $A$ with a root $a$.
Now, given any two conditions in $D$ with domain in $A$, they are compatible if and only if they agree on their restriction to $a$. But since $a$ is finite, there can only be finitely many "types" and $\{p\in D\mid\operatorname{dom}p\in A\}$ is uncountable, so there is a "type" which is realised uncountably many times. In particular any two such conditions will be compatible.