Question about Boolean algebra and ultrafilters In the following $B$ denotes a Boolean algebra and $\bar{x}$ is the complement of $x$.
In my notes there is the following theorem:
If $U \subset B$ is an ultrafilter on $B$ then for every $x \in B$ we either have $x \in U$ or $\bar{x} \in U$.
The proof they give starts as follows:
"Assume neither $x$ nor $\bar{x}$ are in $U$. Then there exist $y_1 , \dots , y_n, z_1, \dots , z_m \in U$ such that $x \land y_1 \land \dots \land y_n = 0$ and $\bar{x} \land z_1 \land \dots \land z_m = 0$."
Question: I would like to know why they want more than just one $y$ and one $z$ because I think I can prove this claim as follows (thanks for pointing out where I go wrong):
Theorem: If $U \subset B$ is an ultrafilter on $B$ then for every $x \in B$ we either have $x \in U$ or $\bar{x} \in U$.
Proof: Assume neither $x$ nor $\bar{x}$ are in $U$. Then there is a $y \in U$ such that $x \land y = 0$ and a $z \in U$ such that $\bar{x} \land z = 0$. Using the fact that $x \land y = 0 \iff y \leq \bar{x}$ we then get $y \leq \bar{x}$ and $z \leq \bar{\bar{x}} = x$ and hence $y \land z \leq \bar{x} \land x = 0$ so $y \land z = 0$ which would be a contradiction to $U$ being an ultrafilter.
Claim used in proof: If neither $x$ nor $\bar{x}$ are in $U$. Then there is a $y \in U$ such that $x \land y = 0$ and a $z \in U$ such that $\bar{x} \land z = 0$.
Proof: Assume neither $x$ nor $\bar{x}$ are in $U$.
Then we define a filter $\tilde{U}$ that properly contains $U$ as follows: $\tilde{U}:= U \cup \{x \} \cup \{ b \in B \mid b \geq x \} \cup \{ x \land b \mid b \in U \text{ or } b \geq x \}$. 
(i) $\tilde{U}$ is a proper subset of $B$ since $\emptyset \notin \tilde{U}$.
(ii) $\tilde{U}$ is a proper superset of $U$ since $x \notin U$.
(iii) $\tilde{U}$ is upwards closed by construction
(iv) $\tilde{U}$ is closed with respect to $\land$ by construction
This would be a filter properly containing $U$ which would be a contradiction to $U$ being an ultrafilter. 
Thanks for your help.
 A: The argument is overall right. But the definition of $\tilde{U}$ seems overly complicated. So I will write out a complete solution.
There are various equivalent definitions of filter on a Boolean Algebra. The one I prefer says (i) $0$ is not in, (ii) if $a$ and $b$ are in so is $a\land b$, and (iii) if $a$ is in, and $b\ge a$, then $b$ is in.  Instead of condition (iii), one can use (iii)' if $a$ and $b$  are in, so is $a\lor b$. But then one should immediately prove (iii)!
Let $U$ be an ultrafilter. We want to show that for any $x$, one of $x$ or $\bar{x}$ is in $U$.
Suppose neither $x$ nor $\bar{x}$ is in $U$. Let $V$ be the set all $v$ such that for some $u$ in $U$, $v\ge x\land u$.  Clearly $U$ is a proper subset of $V$, since $V$ contains $x$ but $U$ doesn't.  If we can prove that $V$ is a filter, we will have contradicted the fact that $U$ is maximal.
First we deal with Condition (i) of the definition of filter. Suppose that $0$ is in $V$. Then $0=x\land u$ for some $u$ in $U$, and therefore $\bar{x}\ge u$, so $\bar{x}$ is in $U$, contradicting our assumption that it isn't. 
Next, we verify Condition (ii). If $a\in V$ and $b\in V$, then $a\ge x\land c$, and $b\ge x\land d$ for some $c,d$ in $U$. But then $a\land b\ge x\land(c\land d)$,  so by definition $a\land b$ is in $V$. 
Finally, Condition (iii) of the definition of filter is built into the definition of $V$.
Remark: On your question about the use of $y_1\land\cdots \land y_n$ instead of a simple $y$, it makes no difference, since the meet of a finite number of elements in the filter is in the filter. I can only speculate that the argument you refer to starts from a filter base rather than a filter.  
