Let $(\mathbb{P},\le)$ be a partial order and consider the following definitions
- $G\subseteq\mathbb{P}$ is filter iff $\forall p,q\in G\exists r\in G(r\le p\wedge r\le q)$ and $\forall p\in\mathbb{P}\forall q\in G(q\le p\rightarrow p\in G)$.
- $D\subseteq \mathbb{P}$ is dense iff $\forall p\in\mathbb{P}\exists q\in D(q\leq p)$.
- $A\subseteq\mathbb{P}$ is antichain iff $\forall p,q\in A\neg\exists r\in \mathbb{P}(r\leq p\wedge r\leq q)$.
I want to proof the following two statement are equivalent
- $G\cap D\neq\emptyset$ for all dense $D\subseteq\mathbb{P}$.
- $G\cap A\neq\emptyset$ for all maximal antichain $A\subseteq\mathbb{P}$.
Proof: ($\rightarrow$). Let $A\subseteq\mathbb{P}$ be an antichain. Let $A'=\cup_{p\in A}p\downarrow$ where $p\downarrow:=\{q\in\mathbb{P}:q\leq p\}$. It is easy to see that $A'$ is dense (by maximality of $A$) so $G\cap A'\neq\emptyset$. Then exists $p\in A$ such that $G\cap p\downarrow\neq\emptyset$. As $G$ is filter, we can get $p\in G$, so $G\cap A\neq\emptyset$.
I have troubles with the proof of the another direction. Can someone give me a hint?
(please, don't give me the answer.)