[Revised on September 29, 2022] [Remarks on atomless lattices added October 6, 2022]
It is incumbent on any treatment of epistemic modals to show what is wrong with such a statement as “It is raining but it might not be”. To prove the relevant theorem H&M introduce a special condition, Knowability.
This term, as well as intuitions about what the condition implies, immediately recalls Fitch’s Paradox of Knowability. Fitch argued that if every truth is knowable then every truth is known. That conclusion is startling because we are sure there are many propositions which are true but not known to be true, and we are inclined to think that what ever is the case could be known to be the case. But the argument is straightforward: for any proposition A consider A* = (A and it is not known that A). There is no possible world in which it is true that it is known that A*. So our ostensible certainty is refuted.
If we look for an analogue in epistemic modals, replacing “it is known that” by “it must be that” we can see immediately that Fitch is evaded in a way that he could not be evaded in classical theories of modality. For such a statement as “It is raining but it is not the case that it must be raining”, equivalent to ““It is raining but it might not be raining” is never true, not true in any possibility. So Fitch’s argument does not get off the ground.
But we also see in Holliday and Mandelkern’s theory that the possibility that every truth is known can be realized, and that this can actually play a role in illuminating epistemic modals.
Before making this precise, my thought quickly stated is that in the classical reading there are indeed many examples of true propositions to the effect that A and that it is not known that A, but not in the reading where “not” is the orthocomplement.
Propositions and the i-function
Recall here that we are dealing with the complete orthocomplemented lattice of propositions, which is formed by a closure operation on a set of possibilities. The zero and unit element are, respectively, the empty set and the set of all possibilities. For each possibility x there is a possibility i(x) such that “It must be that A” is true exactly if i(x) is in A. The first condition on the i-function is
Facticity. x is a refinement of i(x). Symbolically: x ⊑ i(x)
The second condition is
Knowability. for every possibility x there is y such that i(y) ⊑ x
Given Facticity that means that y ⊑ i(y) ⊑ x: all propositions true at x are true at i(y) and all propositions true there are true at y. What does it mean for the lattice of propositions for this condition to hold?
The simpler case is that of a complete atomistic lattice: A is an atom iff only the 0 element and A itself imply A. Recall here the earlier notation for the ‘span’ or ‘support’ of a possibility [x] = {y: y ⊑ x}. Let’s call a possibility x atomic exactly if z ⊑ x implies that [x] = [z]. So A is an atom iff A = [z] for some atomic possibility z.
If x is atomic and [x] has more than one member, those members are for all purposes in this theory the same, indiscernible, a harmless redundancy. In the example of a Euclidean space, where x is a vector, [x] = {y: y = kx for some number k}, and vectors which are multiples of each other belong to all the same subspaces and in physics do not represent different states. But the redundancy is easily removed too, so without loss of generality, I’ll add here:
Atom-uniqueness. If x is atomic then [x] has only one member.
Atomistic and atomless lattices
A lattice is atomistic (or atomic)exactly if each element is the join of a set of atoms. Specifically, in that case, then each possibility has a refinement which is atomic.
Lemma. If the lattice of propositions is atomistic then Knowability holds if and only if w = i(w) for each atomic possibility w.
Clearly if each possibility x has an atomic refinement w such that w ⊑ x and w = i(w) then Knowability holds. Conversely, if Knowability holds, and w is atomic then if y ⊑ i(y) ⊑ w then [y] = [w] = [i(y)], and so by Atom-uniqueness, w = i(w).
What if the lattice of propositions is not atomistic? Then any element x may have an infinite chain of refinements, and the condition has to be that for at least one element y in that chain, i(y) is also in that chain. But the same would apply to this y, and so we see an infinitely descending subchain of the form … y(j) ⊑ i(y(j) ⊑ y(k) ⊑ …. x.
If the lattice is atomless it is certainly possible for some element y to be such that y = i(y). In that case Knowability holds for all the elements x such that y refines x. But then, nevertheless, there is an element z that refines y, and a further element w such that w ⊑ i(w) ⊑ y, and hence also w ⊑ i(w) ⊑ x, So, if the lattice is atomless we can conclude that for each element y such that y ⊑ i(y) ⊑ x there is a distinct element w that refines y and w ⊑ i(w) ⊑ x. There is no bottom to it ….
So now what happens to Fitch’s paradox?
Suppose the lattice is atomistic, Knowability holds, x is in A, but also in ~□A. Then there is an atomic refinement w such that w ⊑ x, and all of the following are true at w: A, ~□A, □A, □~ □A. That is impossible. So there is no possibility x in which (A ∩ ~□A) is true. (Similarly, even if less transparently, if the lattice is not atomistic and Knowability holds.)
And yet of course it is the case that everything that is true in x is known at some other possibility, namely at the atomic possibility w which refines x, since i(w) ⊑ x.
Notice that argument I just gave does not go through if we just suppose that x is in A and x is not in □A. For our “not” in the metatheory is not just an orthocomplement, it is classical. In the case in which x is in A and i(x) is neither in A nor in ~A, which is not ruled out a priori. As pointed out in the previous post, the condition of i-regularity is required even to establish that {x: i(x) is in A} is a proposition. (And we must note that in H&M’s proof of 4.21 both Knowability and i-regularity are invoked).
Note. In view of the lemma it would seem that the i-function is not easily identifiable. In an atomistic lattice each element is the join of the atoms which refine it. Supposing that z is such that [z] = [x] ⊕ [y], where x and y are atomic so that x = i(x) and y = i(y), there cannot be in general a simple relation between i(z) and the pair i(x) and i(y). For the value of the i-function must in general be a ‘less informative’ possibility of which its argument is a non-trivial refinement.
NOTE. Reference is to Holliday and Mandelkern article, at https://arxiv.org/abs/2203.02872v3
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