I don't understand the following basic lemma (taken from p.15, Models and Ultraproducts, Bell and Slomson (1969)):
Lemma 3.1: If $F$ is a filter in a Boolean algebra $B$, $F$ is an ultrafilter iff for each $x \in B$ either $x \in F$ or $x^* \in F$, but not both
PROOF. Note first that if for some $x \in B$ both $x$ and $x^*$ are in $F$, then $0 = x \land x^* \in F$ and so $F$ is not proper. Now suppose that for each $x \in B$ either $x$ or $x^*$ is in $F$. Let $G$ be a filter properly containing $F$. Then there is some $x$ in $G-F$. Since $x \notin F$, $x^* \in F \subseteq G$
Why must it follow from $x \notin F$, that $x^* \in F \subseteq G$? I also don't understand this passage (the continuation of Lemma 3.1):
Conversely, suppose that $F$ is an ultrafilter and $x \notin F$. Let $G$ be the filter generated by $F \cup \{x\} $. Since $F$ is an ultrafilter $G$ is not proper. Therefore $F \cup \{x\} $ does not have the fip (finite intersection property) and so for some $X \in S_{\omega}(F)$, $inf(X) \land x = 0$. Hence $inf(X) \leq x^*$. $Inf(X)$ is in $F$ and therefore we can conclude that $x^* \in F$, thus completing the proof.
Why does it follow from the fact that $G$ is the filter generated by $F \cup \{x\} $ and the fact that $F$ is an ultrafilter, that $G$ is not proper? And then, does $F \cup \{x\} $ not have the fip because it is not a proper filter (and therefore not a filter?) ? I also don't understand the last part of the argument.
Totally confused!
The first part of the proof is proving the claim "suppose $F$ satisfies $\langle\forall x\in B$, either $x\in F$ or $x^*\in F\rangle$; then $F$ is an ultrafilter." In the proof, they choose an $x\in G\setminus F$; by the hypothesis, since $x\not\in F$, we must have $x^*\in F$. But then we have $x\in G$ (because $x\in G\setminus F$), but also $x^*\in G$ (because $x^*\in F\subset G$), so $\mathbf{0}\in G$ and $G$ is improper.
As for the second part — the statement that "if $F$ is an ultrafilter, then $\langle\forall x\in B$, either $x\in F$ or $x^*\in F\rangle$" — the definition of an ultrafilter is that it's maximal; i.e., for any filter $G$ with $F\subset G$, $G$ is improper. If $G$ is (the filter generated by) $F\cup\{x\}$, then we have $F\subset F\cup\{x\}\subset G$ and therefore $G$ must by definition be improper.