Ultrafilters and quantifier scope In https://www.jstor.org/stable/30226510, p.546, Zimmermann writes:
"It is easily seen (and well-known) that ultrafilters are exactly those quantifiers $F$ that lack scope with respect to arbitrary truth-functional connectives ©":
[($Fz) \hspace{0.3cm} z \in X \hspace{0.3cm} © \hspace{0.3cm} (Fz) z \in Y$ iff $(Fz) \thinspace [z \in X \hspace{0.3cm} © \hspace{0.3cm} z \in Y]$  
If this is so well-known, where is it proven? Why should this be the case? Zimmermann continues:
"Hence, due to the functional completeness of $\thinspace \textit{neither-nor}$, we find that $F$ is an ultrafilter over $A$ iff all subsets $X$ and $Y$ of $A$ satisfy: $(X \cup Y)^\complement \in F \thinspace$ iff $[X \notin F$ and ($Y \notin F$]" 
It is not clear to me what the functional completeness of $neither-nor$ has to do with the property of ultrafilters alluded to.
If anyone can explain these remarks of Zimmermann, I would be very happy.
 A: This is just a fast answer because I'm on my phone, but you can see that closure under intersection implies that the quantifier lacks scope with respect to $\land,$ and the ultrafilter property (every set or its complement is in the ultrafilter) implies that the quantifier lacks scope with respect to $\lnot.$ Since every truth-functional connective can be built up from $\land$ and $\lnot,$ it follows that every ultrafilter lacks scope with respect to every connective. 
Conversely, if a quantifier $Q$ on a set $A$ lacks scope with respect to all connectives, then it's closed under intersection and it satisfies the ultrafilter property (using lack of scope under  $\land$ and $\lnot).$ The ultrafilter property and lack of scope under $\lor$ implies that the entire set $A$ is in the set, so $\emptyset$ is not. The set must be a filter since if $X\in Q$ and $Y\not\in Q,$ then $X\cap Y^c\in Q,$ so $X$ can't be a subset of $Y.$
Finally, if a quantifier lacks scope with respect to NOR, then it lacks scope with respect to all connectives, since every connective can be built up from NOR. 
