I have read the Kunen's set theory. But I get a struck to understand the proof of a theorem.

Let $\mathcal{A}\subset\mathcal{P}(\omega)$ be a family of pairwise almost disjoint sets, then we define $(\Bbb{P}_\mathcal{A},\le)$ as $\Bbb{P}_\mathcal{A}=\{\langle s,F\rangle : s\in [\omega]^{<\omega},\,F\in[\mathcal{A}]^{<\omega}\}$ and $\langle s,F\rangle \le \langle s',F'\rangle$ iff $s\subset s'$, $F\subset F'$ and $x\cap s'\subset s$ for each $x\in F$.

Lemma 2.8. In $\Bbb{P}_\mathcal{A}$, $\langle s,F\rangle$ and $\langle s',F'\rangle$ are compatible iff $x\cap s'\subset s$ for all $x\in F$ and $x\cap s\subset s'$ for all $x\in F'$.

Lemma 2.10. If $G$ is a filter in $\Bbb{P}_\mathcal{A}$ and $\langle s,F\rangle \in G$ then $x\cap d_G\subset s$ for each $x\in F$ (Where $d_G:=\bigcup\{s\mid\exists F:\langle s,F\rangle \in G\}$.)

Proof. If $\langle s',F'\rangle\in G$ then $\langle s',F'\rangle$ and $\langle s,F\rangle$ are compatible, so by Lemma 2.8, $x\cap s'\subset s$ for all $x\in F$.

I don't understand the proof of Lemma 2.10. Especially, I don't understand $\langle s',F'\rangle\in G$ implies the compatibility of $\langle s,F\rangle$ and $\langle s',F'\rangle$. In this book there is no proofs of this statement (as I know) and I don't know how to prove this statement. Thanks for any help.


Given $x \in F$, note that for any $\langle s^\prime , F^\prime \rangle \in G$ we have that $\langle s^\prime , F^\prime \rangle , \langle s , F \rangle$ are compatible -- since $G$ is a filter and $\langle s , F \rangle$ also belongs to $G$ -- so by Lemma 2.8 it follows that $x \cap s^\prime \subseteq s$.

It then follows that $x \cap d_G = x \cap \bigcup_{\langle s^\prime , F^\prime \rangle \in G} s^\prime = \bigcup_{\langle s^\prime , F^\prime \rangle \in G} ( x \cap s^\prime ) \subseteq s$.

Recall that given a poset $\langle \mathcal{P} , \preceq \rangle$, a (nonempty) subset $F \subseteq \mathcal{P}$ is a filter if

  • whenever $p,q \in \mathcal{P}$ and $p \preceq q$, then $p \in F$ implies $q \in F$;
  • whenever $p,q \in F$, then $p,q$ there is an $r \in F$ with $r \preceq p , q$.

This second condition implies that every two elements of a filter are compatible as elements of the poset.


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