Think of a unital, commutative, quantale as a kind of navigator.
A navigator understands ’getting from one place to another’
Different navigators understand different aspects (whether one can get from A to B, how much time it will take, ...)
What they share in common is that, given routes A to B and B to C, they understand how to get a route A to C.
Because of Proposition 2.96, a quantale has all meets and joins (even though we define it simply as having all joins).
A unital commutative quantale (or just quantale, in this book)
A symmetric monoidal closed preorder \(\mathcal{V}=(V,\leq,I,\otimes,\multimap)\) that has all joins.
\(\bigvee A\) exists for all \(A \subseteq V\)
Denote the empty join as \(0 := \bigvee \varnothing\)
We saw that Cost is monoidal closed in Example 2.83
To check if Cost is a quantale, we take an arbitrary set of elements and ask if it has a join.
Because \(\geq\) is a total order, we can take the infimum or greatest lower bound, as the join.
\(\bigvee\{2.5,2.05,2.005,...\} = 2\).
We need a \(0\), which is something which is related to everything (the first join condition is vacuous). Because the preorder relation is \(\geq\) in Cost we need something greater than everything, so \(0 = \infty\).
Thus Cost is a quantale.
Let \((P,\leq)\) be a preorder. It has all joins iff it has all meets.
Meets and joins are dual, so it suffices to prove one of the directions (the opposite category shows that having all meets having all joins, if we are able to prove that having all joins means having all meets in the original preorder).
Suppose there are all joins and \(A \subseteq P\) is a subset for which we want the meet.
Consider the set \(M_A := \{p \in P\ |\ \forall a \in A: p \leq a \}\) (everything below all of \(A\) - these are candidates for the meet of \(A\))
The first condition for the meet is satisfied by all. We get the actual meet by taking \(\bigvee M_A\) which is guaranteed to exist. Because this element is greater than or equal to all elements that are \(\leq A\), it satisfies the second condition for the meet.
Suppose \(\mathcal{V}=(V,\leq,I,\otimes)\) is a symmetric monoidal preorder that has all joins.
Then \(\mathcal{V}\) is closed, i.e. has a \(\multimap\) operation and hence is a quantale, iff \(\otimes\) distributes over joins
i.e. if Eq (2) from P2.87 holds: \((v \otimes \bigvee A)\cong \bigvee_{a \in A} v \otimes a\).
We proved one direction in P2.87
We need to show that \((v \otimes \bigvee A)\cong \bigvee_{a \in A} v \otimes a\) (and all joins existing) implies that there exists a \(\multimap\) operation that satisfies the closed property: \(\forall a,v,w \in V: (a \otimes v) \leq w\) iff \(a \leq (v \multimap w)\).
The adjoint functor theorem for preorders states that monotone maps preserve joins iff they’re left adjoint, so \(V \xrightarrow{-\otimes v} V\) must have a right adjoint g, which, being a Galois connection, will satisfy the property \((a \otimes v) \leq w \iff a \leq g(w)\) (this is the monoidal closed property).
Let’s rename \(g \equiv v \multimap -\). The adjoint functor theorem even gives us a construction for the right adjoint, namely: \(v \multimap w:=\bigvee\{a \in V\ |\ a \otimes v \leq w\}\).
What is \(\bigvee \varnothing\), called \(0\), in the case of:
\(\mathcal{V}=\mathbf{Bool}=\{\mathbb{B},\leq, true,\land\}\)
\(\mathcal{V}=\mathbf{Cost}=([0,\infty],\geq,0,+)\)
What is the join \(x \vee y\) for Bool and Cost?
\(False\) and \(\infty\) respectively
Logical or and \(min\) respectively
Recall the power set symmetric monoidal preorder \((P(S),\subseteq, S, \cap)\) Is this a quantale?
Yes, \(0=\varnothing\) (it is related to everything) and the join of any pair of subsets is well-defined as their union. By Proposition 2.98, this means it is a quantale.