# 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.

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Filters are closed under finite intersection (meet), so there is no difference between $y_1\land \cdots \land y_n$ and $y$. Your condition on the extension seems more elaborate than necessary, but I am most familiar with ultrafilter construction on powerset. If not answered, more in a couple of hours. –  André Nicolas Jan 31 '12 at 14:21
@AndréNicolas Thanks! Then you're saying my proof is right? This would answer my question. You don't have to type up a full answer. –  Rudy the Reindeer Jan 31 '12 at 14:51

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.

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Thank you. "Filter base" was not mentioned in the lecture. –  Rudy the Reindeer Feb 1 '12 at 10:36