Suppose $G \subset \mathbb{P}$ is a $\mathbb{P}$-generic filter over a countable transitive model $M \ni \mathbb{P}$ (satisfying enough of ZFC). Then $G$ is an ultrafilter.

The idea of the proof, outlined in this answer is to look at $$ D = \{ q \in \mathbb{P} : q \leq p \} \cup \{ q \in \mathbb{P} : q \leq \neg p \} $$ and see that it is dense, but there's a problem with this argument when looking at a general poset $\mathbb{P}$, namely that in a general poset, we don't have negations/complements!

We can complete an arbitrary poset to a complete boolean algebra $\mathbb{B}$, but does this transformation equally turn $G \subset \mathbb{P}$ into a generic filter $G^\prime \subset \mathbb{B}$?

Furthermore, will showing that $G^\prime$ is an ultrafilter imply that $G$ itself is an ultrafilter?

Finally, if $\mathbb{P}$ is a boolean algebra in the original statement, I have an idea of how to proceed:

Take arbitrary $r \in \mathbb{P}$. Suppose $r \cdot p = 0$. Then $\neg (p \cdot r) = 1$, so $\neg p + \neg r = 1$. Hence, $\neg p \cdot r = r$, so $\neg p \leq r$. Likewise, if $r \cdot \neg p = 0$, then $r \leq p$. Therefore, either $r \leq p$ or $r \leq \neg p$. This shows that $D$ is dense.

But this argument feels wrong to me. I never actually use $D$ itself, and instead look only at $p$ and $\neg p$.

Since the argument via the boolean completion appears unnecessarily complicated, maybe it's better to show that $G \subset \mathbb{P}$ is maximal. In that case I think I should proceed by contradiction to obtain $G^\prime \supset G$ and build a dense set out of $G^\prime \setminus G$ to contradict that $G$ is a generic filter. However, I don't see how to build this dense set.

  • $\begingroup$ What does "ultrafilter" mean if you don't have a boolean algebra? A maximal proper filter? $\endgroup$ Apr 10 '17 at 1:11
  • $\begingroup$ Yes, a maximal proper filter. $\endgroup$ Apr 10 '17 at 1:22
  • $\begingroup$ In your argument, you showed $rp=0$ implies $r\leq\neg p$ and $r(\neg p)=0$ implies $r\leq p$. But typically neither $rp$ nor $r(\neg p)$ will be $0$. The trick is that if $rp\neq 0$, then $rp$ itself is an extension of $p$ in $D$. $\endgroup$ Apr 10 '17 at 1:26

For any $p\in\mathbb{P}$, let $A(p)=\{q\in\mathbb{P}:q\leq p\}$ and let $B(p)$ be the set of $q\in\mathbb{P}$ that are incompatible with $p$. Then I claim that $D(p)=A(p)\cup B(p)$ is dense (if $\mathbb{P}$ is Boolean then $B(p)=\{q:q\leq\neg p\}$, so this is your $D$). Indeed, if $r\in\mathbb{P}$ has no extension in $A(p)$, that means exactly that $r$ is incompatible with $p$, so $r\in B(p)$.

Thus if $G$ is generic, $G\cap D(p)\neq 0$ for all $p$. If $p\not\in G$, then $G\cap A(p)=\emptyset$. We thus must have $G\cap B(p)\neq\emptyset$, which means no filter containing $G$ can contain $p$. Since $p\not\in G$ was arbitrary, $G$ is maximal.


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