(A further reflection on the paper by Holliday and Mandelkern (H&M), without reference to quantum mechanics.)
[Minor update about terminology on October 6, 2022.]
A. The law of double negation as clue to orthologic page 1
B. About compatibility-regularity page 1
C. About “must” and “might” page 3
D. Reflections on i-regularity page 4. (Note: this section corrected and updated on Sept. 8, 2022.)
A. The law of double negation as clue to orthologic
Let H be any set, and ⊥ an orthogonality ( i.e. symmetric, irreflexive) relation on H.
Define for the ortho-complement for subsets X of H: ~X = {x: for all y, if y is in X then x ⊥y}
Then ~~ is a closure operator on subsets of H, and a set X is closed iff X = ~~X.
The closed sets form an complete orthocomplemented lattice (Birkhoff references in preceding note). It is equally true of course that the elements of any complete orthocomplemented lattice satisfy the ‘law of double negation’: X = ~~X.
(Since there are many closure operations, let’s use “ortho-closed” for this one, when we want to be explicit.)
Note: for any set X, X is part of ~~X. The converse is often not true. However, there is also a ‘law of triple negation’ which holds for arbitrary subsets of H. That is, if Y = ~X then Y = ~~Y. I will discuss this below, where the reason will be easier to make clear.
B. About compatibility-regularity
I’ll use “compatible” for the complement of “orthogonal”, this to be understood in our classical meta-language: x and y are compatible exactly if they are not orthogonal. (Note on terminology: in lattice theory y is a ‘complement’ of x iff their meet is the zero element and their join the unit. In that sense, in a 2-dim Euclidean space, the straight lines through the origin are all each other’s complement, but of course only the ones at right angles to each other are each other’s ortho complement, and the rest are, in our present terminology, mutually compatible.)
The compatibility-regular sets, by H&M’s definition, are precisely the ortho-closed sets. The two ways of identifying propositions, as compatibility-regular sets of possibilities and as ortho-closed sets of possibilities, are the same.
Under what conditions is a set closed? To begin
- for any subset X of H, X ⊆ ~~X.
(Note: it would be standard to write “X⊥” rather than “~X”, but this symbolism shows the relation to the logical law of double negation, so I will use it here.)
Reason: If x is in X then x ⊥ y, if y is orthogonal to every member of X. But ⊥ is symmetric, so x is then orthogonal to any element y which is orthogonal to every member of X.
For the converse we need that if x is not in X then it is not orthogonal to all of ~X,
i.e. there is some element y such that y is in ~X but x is compatible with y. Hence, in view of 1.:
2. X = ~~X if and only if for each element x of H, if x is not in X then there is an element y such that y is compatible with x but y is orthogonal to X.
Lemma. y is orthogonal to X if and only if, for all z, if z is compatible with y then z is not in X.
From left to right is obvious: if y ⊥ X and z is in X then z ⊥ y. Starting from the right suppose the contrapositive: For all z, if z is in X then z ⊥ y. It follows that y is orthogonal to X.
Hence 2. is the same as:
3. X = ~~X if and only if for each element x of H, if x is not in X then there is an element y such that y is compatible with x and for all z, if z is compatible with y then z is not in X.
The right hand side of 3. is H&M’s definition of compatibility regularity. For their definition is:
Set X of possibilities is compatibility regular if and only if for any x, if x is not in X then there is a possibility y compatible with x such that, for any z, if z is compatible with y then z is not in X.
It may be more convenient then, if we tend to think in terms of orthogonality rather than its complement, write H&M’s definition as:
Set X of possibilities is compatibility regular if and only if for any x, if x is not in X then there is a possibility y compatible with x which is orthogonal to X.
So, to conclude this part:
4. A subset X of H is compatibility-regular if and only if it is ortho-closed.
Since H&M identify the propositions as the compatibility-regular sets of possibilities, this brings in train, of course, that the propositions form a complete orthocomplemented lattice.
By the way, given 2. and 3., it is now easier to see why the ‘law of triple negation’ holds.
Lemma. If X, Y are any sets of possibilities and Y = ~X then Y is ortho-closed.
(We might call this the law of triple negation, which extends beyond propositions to arbitrary sets: ~X = ~~~X.)
To show this we rely on 2. above.
Suppose that Y = ~X and that y is not in Y. Then y is not orthogonal to all of X, hence there is an element z of X which is compatible with y. But z, being a member of X, is orthogonal to all members of Y because orthogonality is symmetric. So if y is not in Y then there is a possibility z which is compatible with y and is orthogonal to Y. Hence Y = ~~Y.
This may help to identify propositions when new notions are introduced.
C. About must and might
H&M introduce a function i, I’ll call it the information function or i-function, which maps possibilities into possibilities. The first condition on this function is
Facticity. x refines i(x), that is, for any closed set A, if i(x) is in A, so is x.
Intuitively we think of i(x) as having less information in it, we can perhaps even think of it as selecting the propositions that are not only true but known in x.
H&M introduce “must” and “might” as operators on sets of possibilities by:
5. □A = {x: i(x) is in A}
6. ◊A = ~□~A
What is specially important is to show that the purported modal propositions are indeed propositions, that is, ortho-closed sets.
H&M give an equivalent to the definition of the “might” operator as:
7. ◊A = {x: (for all y)[if y is compatible with x, then (there is z)(z is compatible with i(y) and z is in A.}
Looking at 6., however, we can rephrase this as the equivalent:
8. ◊A = the orthocomplement of the set {x: i(x) is in ~A}
Recall the Lemma above. If X, Y are sets of possibilities and Y = ~X then Y is ortho-closed.
We conclude that ◊A is a proposition.
In modal logic it is customary to preserve the relationships: □A = ~◊~A and ◊A = ~□~. Since ~◊~A is the orthocomplement of the orthocomplement of the set {x: i(x) is in ~~A} = ~~ □~~A, we could take this equivalence as the definition of □A. In that case it follows that □A is also a proposition.
However, as Wes Holliday has pointed out, if we take ◊A as basic and define □A as ~ ◊~A, then □A = ~~{x: i(x) is in A}. That is, □A is then the closure of {x: i(x) is in A}, which is not guaranteed to be the same as {x: i(x) is in A}.
The alternative choice, that □A = {x: i(x) is in A}, follows another tradition in the semantics of modal logic, namely that necessity in a world is equated to truth in certain (other) worlds, selected by a ‘relative possibility’ relation. (Think also of the account of counterfactual conditionals through the selection of a ‘nearest’ possible world.)
So in their paper, H&M introduce a condition on the i-function, namely i-regularity, which ensures that {x: i(x) is in A} is a proposition.
D. Reflections on i-regularity
I will not comment on i-regularity directly here, but will look into just where such a special condition is needed, in order to ensure that □A, defined as the set {x: i(x) is in A}, as H&M do, is a proposition whenever A is a proposition.
What are the conditions for □A to be closed, if A is closed?
We already know that □A is part of ~~ □A, by 1. above. So, looking for inspiration to 2. above, what we need is
9. for all x, if x is not in □A then there is an element y which is compatible with x but orthogonal to □A.
If x is not in □A, i. e. i(x) is not in A, then there are various options for where x, i(x) can be. Some of these options we can deal with at once, and then we should end up with the option(s) that cannot be dealt with without imposing special new conditions on the i-function.
Lemma. If x is not in □A, but x is in ~ A, then (*) holds
For in that case, given that □A is part of A, x itself is compatible with x and orthogonal to □A.
Lemma. If x is not in □A, and x is not in A either, and also not in ~A, then (*) holds.
To show that it helps to think of x as more or less a certain proposition, namely, the set of all refinements of x.
Definition. [x] = ∩{B: x is in B}
(I am using A, B for propositions, that is, compatibility-regular sets, X, Y for arbitrary sets.)
I will call [x] the span of x; we could also, with reference to a related notion, call it the support of x.
Since the propositions form a complete lattice, [x] is a proposition.
Sub-Lemma. x is in A iff [x] ⊆ A. In addition, [x] = {y: y is a refinement of x}
Both are so because [x] is the smallest proposition to which x belongs, in the appropriate sense of “small”.
I f x is not a member of A and also not a member of ~A, then there must be a member x’ of [x] and a possibility y in A such that x’ is compatible with y. For if no member of ~A is compatible with any member of [x] then all members of ~A are orthogonal to all members of [x], and in that case [x] is orthogonal to all members of ~A, that is to say, x is in ~~A, that is, x is in A.
But since there is then an element y of ~A which is compatible with a member x’ of [x], note that x’ is a refinement of x. So y is also compatible with x. (I do not mean that compatibility is transitive in general. Rather, if any element z is orthogonal to x, then x is in ~[z], and thus all the refinements of x are in ~[z], so z is then orthogonal to every refinement of x. By contraposition, if z is compatible with a refinement of x then z is compatible with x.). So there is then an element y compatible with x which is orthogonal to A, and therefore orthogonal to □A, which is part of A.
So what remains? The case in which x is in A and i(x) is not in A.
It is clear that in this case i(x) is not in ~A either, since i(x) is compatible with x which is in A.
So the remaining case is only: x is in A and i(x) is neither in A or in ~A.
We need a condition to ensure that in this case, x is compatible with some element which is orthogonal to every element z such that i(z) is in A.
Therefore:
The set □A defined as {x: i(x) is in A} is ortho-closed if the following is the case: IF x is in A, and i(x) is neither in A or in ~A, THEN x is compatible with some y which is orthogonal to all of {z: i(z) is in A}.
Intuitively what is required then has to do with this case: A is true in x and it is not known in x that A, and also not known in x that ~A. Then, still speaking intuitively, x has a brother in a different but compatible possibility, where knowledge of A is absolutely excluded.
I imagine the following speech:
“I said: “It might not rain, and then again it might rain.” In fact it is raining. But what I said was true, because there was a possibility compatible with my situation in which it was definitely not the case that it must rain, a possibility in which knowledge that it would rain was absolutely precluded.”
All right, so now we know where there is a clear need for a condition on the i-function. That condition cannot make reference to any particular proposition, it has to be general.
To end then, here is the condition H&M chose:
Definition. i-regularity: For all y: if y is compatible with i(x) then there is u, compatible with x, such that (for all z)(if z is compatible with u then y is compatible with i(z))
The proof of proposition 4.21, that {x: i(x) is in A} is a proposition, combines appeal to the compatibility-regularity of A and the i-regularity of the i-function.
Note: reference is to Holliday and Mandelkern paper, at https://arxiv.org/abs/2203.02872v3
Bas van Fraassen's philosophy blog 