I'm messing around with ultrafilters to try to understand them better. It seems as if ultrafilters work well as sets of designated truth values in matrix semanticses for classical logic.
This also seems like a pretty simple way to characterize what an ultrafilter is (if we're restricting attention to Boolean lattices).
So my question is twofold.
- Is this a correct characterization of what an ultrafilter is (in the special case of a Boolean lattice)?
- Assuming it's true, is this a trivial consequence of some more general result? (For example, is it a super obvious consequence of the definition of an ultrafilter?)
What follows is my attempt to prove this myself. It is additional context and a record of what I've tried to do so far, not a core part of the question itself.
Let $X$ be a set. Let $\Lambda$ be $2^X$ with $\le$ given by $\subset$. Let $\Xi$ be $2^{2^X}$.
As a Boolean lattice / powerset algebra, $\Lambda$ is equipped with natural candidates for $\land$, $\lor$, and $\lnot$ (namely $\cap, \cup, {}^c$ respectively). In order to make $\Lambda$ into a matrix semantics for classical propositional logic though, we need a $D \in \Xi$ that picks out the designated elements of $\Lambda$.
I think we get classical logic back if and only if $D$ is an ultrafilter.
First, I'll talk about what it means for $D$ to respect classical logic. Classical logic is what we are using in the background (i.e. it is Truth™), so our prospective matrix semantics is valid if and only if it respects Truth™.
First, it must hold for every $e$ in $\Lambda$ that $e$ is in $D$ if and only if $e^c$ is not in $D$.
If we squint a bit this is equivalent to insisting on consistency (at most one of $\{e, e^c\}$ is present) and completeness (at least one of $\{e, e^c\}$ is present).
We also want it to hold that $a \land b$ is in $D$ if and only if $a$ is in $D$ and $b$ is in $D$.
If $a$ is in $T$ and $b$ is in $T$, then $a \land b$ is in $T$ is equivalent to the following.
$$ \text{$T$ is closed under intersection.} $$
If $a \land b$ is in $T$, then $a$ is in $T$ and $b$ is in $T$ is equivalent, by contrapositive, De Morgan's laws, and substituting $x^c$ with $x$ for all variables $x$, to the following.
$$ \text{If $a$ is in $T$ or $b$ is in $T$, then $a \lor b$ is in $T$} $$
This condition is equivalent to upward closure. If we pick $a$ first, we can pick a $b$ to get any $a' \supset a$ that we want.
Let $U$ be an ultrafilter; it satisfies each of our four conditions.
completeness
Suppose $U$ is incomplete. Therefore, there exists an $e$ such that $\{e, e^c\} \cap U = \varnothing$. $\langle e, U \rangle$ is a filter and a proper extension of $U$. $\langle e, U \rangle$ is not $\Lambda$ because it does not contain $e^c$. This contradicts the definition of an ultrafilter. Thus $U$ is complete.
consistency
Suppose $U$ is inconsistent. There exists an $e \in U$ such that $e^c \in U$. There therefore exists a $z$ in $U$ such that $z \le e$ and $z \le e^c$. Thus $z \le \varnothing$. Thus $z = \varnothing$. By upward closure of $U$, $U$ is equal to $\Lambda$. This is a contradiction.
closure under intersection
Let $x$ and $y$ be elements of $U$. There exists a $z$ such that $z \le x$ and $z \le y$. If $z$ is $x \cap y$, we're done. If not, $z$ is a subset of $x \land y$ and thus $x \land y$ is in $U$ by upward closure of $U$.
upward closure
$U$ is upwardly closed by being an ultrafilter.
Suppose $A \in \Xi$ is complete, consistent, closed under intersection, and upwardly closed. I will now show that it is an ultrafilter.
filter
$A$ is a filter. For all ordered pairs of elements $(x, y)$, $x \cap y \le x$ and $x \cap y \le y$. We also have upward closure.
not empty
$A$ is not $\varnothing$. For an arbitrarily chose $e$, either $e$ is in $A$ or $e^c$ is in $A$.
not whole space
$A$ is not $\Lambda$. If $e$ is in $A$, then $e^c$ is not in $A$.
maximal among proper filters
Let $B$ be a proper filter properly extending $A$.
By completeness of $A$, there is an $e$ such that $\{e, e^c\} \subset B$.
Therefore $\varnothing$ is in $B$ and thus $B = \Lambda$. This is a contradiction because $B$ is a proper filter on $\Lambda$.
Therefore $A$ is an ultrafilter.
