# Questions on: $G$ a generic ultrafilter on $B$ if and only if $G$ is a generic filter on $(B+,<)$?

Here are the definitions, and then I shall explain which parts of the implication I understand, and which parts I don't, which are the questions. The definitions are from Jech, as well as the question which is exercise 14.10.

Definitions:

A. $G$ is a generic ultrafilter on $B$ (complete Boolean algebra in ground model $M$) if and only if

1. $0 \notin G$, $1 \in G$,
2. if $a,b \in G$ then $a*b \in G$,
3. if $a < b$ and $a \in G$ then $b \in G$,
4. $\Pi X \in G$ whenever $X \in M$ and $X \subset G$, and
5. $\forall a \in B$, $a \in G$ or $\neg a \in G$.

B. $G$ is a generic filter on $(B^+,<)$ ($B^+$ is $B/\{0\}$, and $u\le v \iff u*v = u \iff u + v = v \iff u-v = 0$) if and only if

1. $G$ is nonempty,
2. if $p,q \in G$, then $\exists r \in G$; $r \le p$ and $r \le q$,
3. if $p \le q$ and $p \in G$ then $q \in G$, and
4. if $D \subset B^+$ is dense and $D \in M$, then $D \cap G \neq \emptyset$.

Proof and Questions:

$B \rightarrow A$

1. Since $G \subset B^+$, $0 \notin G$ since $G$ is nonempty; also and therefore since $G$ is upwards closed, $1 \in G$.
2. Let $a,b \in G$, then there is an $r \in G$ such that $r \le a$ and $r \le b$, but then $r \le a*b$ implies $a*b \in G$ since $G$ is upwards closed.
3. same-same
4. is it a density argument?
5. proof by contradiction?

$A \rightarrow B$

1. $G$ is nonempty since $1 \in G$.
2. If $a,b \in G$, then $a*b \in G$, and $a*b \le a$ and $a*b \le b$.
3. same-same
4. Let $D$ be dense in $B^+$, how do I show $G \cap D \ne \emptyset$?

To summarize, I'd like to know, but have not yet been able to prove, why a generic filter on $(B^+,<)$ is closed under boolean products, and why it is an ultrafilter on $B$. And, why a generic ultrafilter on a complete Boolean algebra $B$ meets the dense sets in the ground model.

-
Any relationship between generic ultrafilter and non-principal ultrafilter? – alancalvitti Jan 25 '13 at 2:43
I fiddled around with your latex a bit; I hope you don't mind. – Miha Habič Jan 25 '13 at 8:07

First, the closure under products. Let $G$ be generic on the poset $B^+$ and take $X\subseteq G,X\in M$. Let $$D=\{u\in B^+;\forall x\in X\colon u\leq x\}\cup\{u\in B^+;\exists x\in X\colon u\leq\lnot x\}$$ $D$ is dense; to see this, take $v\in B^+$. Now, either $v\leq x$ for all $x\in X$ or $v\cdot \lnot x\neq 0$ for some $x\in X$. In the former case we get $v\in D$ and in the latter we get $v\cdot\lnot x\in D$ with $v\cdot\neg x\leq v$. Now, because of genericity we find $u\in G\cap D$. But $u$ cannot come from the second piece of $D$, since it must be compatible with all elements of $X$. Therefore $u\leq x$ for all $x\in X$ and upward closure gives $\prod X\in G$.
Next, the maximality condition. Take $a\in B$ and let $D=\{u\in B^+; u\leq a\}\cup\{u\in B^+;u\leq\lnot a\}$. The same argument as before shows that $D$ is dense, so either $a\in G$ or $\lnot a\in G$.
Finally, let's see that a generic ultrafilter $G\subseteq B$ meets all dense sets in the ground model. Let $D$ be dense and assume $G\cap D=\emptyset$. Since $G$ is an ultrafilter, we must have $\{\lnot u;u\in D\}\subseteq G$ and, because of genericity, $\prod_{u\in D}\lnot u\in G$. If this product equals 0, we have found our contradiction. But even if $\prod_{u\in D}\lnot u>0$ we get into trouble. Since $D$ is dense, we find a $v\in D$ such that $0<v<\prod_{u\in D}\lnot u$, which is clearly absurd. All of this shows that $G\cap D$ is nonempty.
As a side note, it is very common to construct dense sets in the way shown: one piece contains the elements you want to capture and the other one makes the whole thing dense. A very similar argument shows that, if you have a condition in $G$ forcing a statement, you can assume that 1 already forces it.