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On page 4 of http://www.llf.cnrs.fr/sites/llf.cnrs.fr/files/biblio//the-same-ter.pdf

Richard Zuber writes:

"Given a fixed universe $E$, (where $|E| ≥ 2$), a type $n$ quantifier is a function from $n$-ary relations to truth values. A type $\langle 1 \rangle$ quantifier is a function from sets (sub-sets of $E$) to truth values, and thus it is a set of sub-sets of $E$. A type $\langle 1, 1 \rangle$ quantifier is a function from sets to type $\langle 1 \rangle$ quantifiers. In natural language semantics type $\langle 1 \rangle$ quantifiers are denotations of NPs and a type $\langle 1, 1 \rangle$ quantifiers are denotations of (unary nominal) determiners, that is expressions like $\textit{every, no, most, five}$, etc. Since both types of quantifiers form Boolean algebras they have Boolean complements (negations)."

In what sense do type $\langle 1 \rangle$ and $\langle 1, 1 \rangle$ quantifiers form Boolean algebras, as Zuber says? In what sense, for example, does the denotation of $\textit{every}$ form a boolean algebra?

Could it be that the Boolean algebra is formed by taking the boolean operations ON the set of subsets that is the denotation of a type $\langle 1, 1 \rangle$ quantifier? But then the boolean algebra would not be identical to the denotation of $every$ but a boolean algebra formed by operations performed on its denotation.

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It's not that a single quantifier forms a Boolean algebra; rather, the set of all quantifiers of a given type is a Boolean algebra (with the Boolean operations being performed pointwise, when you consider quantifiers as functions in the manner described). So for instance, the Boolean complement of the quantifier "every" is the quantifier "not every" (i.e., the function that takes a unary relation to true iff there is some input that makes the unary relation become false).

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    Ah yes, I see. Then Zuber's comment is actually ambiguous. It could mean that, say a type $\langle 1, 1 \rangle$ quantifier forms a boolean algebra alone. Or it could mean, as you point out, that a set of type $\langle 1, 1 \rangle$ quantifiers form a Boolean algebra. Still, in what sense is it that the quantifiers form the Boolean algebra, as opposed to the operations performed on the set of them? – 2017-02-20
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    I suppose by "form" Zuber did not mean "form alone" but "form (together with operations) a boolean algebra" – 2017-02-20
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    Yes, exactly. The set of quantifiers is the underlying set of a Boolean algebra, when you define Boolean operations as pointwise operations on the functions. – 2017-02-20
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    I realize that my description of "every" is treating it as a type $\langle 1\rangle$ quantifier, not as a type $\langle 1,1\rangle$ quantifier. It is not clear to me what natural way there is to consider "every" as a type $\langle 1,1\rangle$ quantifier (maybe it is the function that takes a set and returns the type $\langle 1\rangle$ quantifier that returns true on every superset of the original set?). In any case, the important point is that the Boolean operations are defined pointwise. – 2017-02-20
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    What do you mean $\textit{specifically in terms of quantifiers}$ by saying that the operations of are performed pointwise? The author also says at one point that "quantifiers are... functions forming (pointwise) a Boolean algebra and mapping a Boolean algebra into a Boolean algebra." What is meant by this? – 2017-03-01
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    If $B$ is a boolean algebra and $S$ is a set, then the set $B^S$ of all functions $S\to B$ becomes a Boolean algebra by defining $(f\wedge g)(x)=f(x)\wedge g(x)$ and $(f\vee g)(x)=f(x)\vee g(x)$ and so on for each $x\in S$. In the case of type $\langle 1\rangle$ quantifiers, for instance, $S$ would be the set of unary relations and $B$ would be $\{T,F\}$. – 2017-03-01