Patrick Grim has a well-known argument that there is no set of all true propositions. Suppose that argument is sound. For every set S of true propositions, there is a larger set S’ of true propositions. Since there is no set of all true propositions, it cannot be the case that anything knows every proposition in the set of all truths. Is this a problem for classical versions of omniscience? Here’s Alan Rhoda (see Enigman, too)
This creates a problem for omniscience if that notion is defined set-theoretically, e.g., believing every member of the set of all truths. One can argue along Cantorian lines that there can be no such thing as the set of all truths. That is, for every set of truths, one can construct new truths that are not already members of the set. As Enigman points out, if this is right, then it follows that an essentially epistemically static (EES) God - i.e., a God who cannot either acquire or lose beliefs - cannot be omniscient (in a set-theoretical sense). Such a God cannot know all truths. Moreover, there would have to exist truths that are forever outside the ken of an EES God.
I confess to not quite seeing the problem described here, unless it turns on that specific set-theoretic definition of omniscience. Suppose we constuct a series of sets such that every set S contains only true propositions and is a subset of the succeeding set S’. We begin with what we would ordinarily call a saturated set of propostions (what Plantinga would call a maximal or complete and consistent set of propositions) and let each subsequent and larger set be constructed in the way Grim descibes. There will be infinitely many larger and larger sets and no largest set of true propositions.
Each set S in the sequence is consistent and contains propositions that hold at the actual world. If we define omniscience set-theoretically, could there be a classically omnisicient being in the actual world? Let an omniscient being be the following:
O. N(x)(x is omniscient iff. ~(ES)(S is a set of true propositions & x does not know every member of S).
(O) states that necessarily, for all x, x is omniscient just in case there is no set of true propositions such that x does not know every member of that set. As you move up in the sequence of sets of propositions, it is never the case that you reach some set S such that an omniscient being does not know every member of (that is, every proposition in) S. Since there is no set S such that an omniscient being does not know every member of S, an omniscient being knows every true proposition. That is perfectly possible even if the sequence is infinite. So there is no problem for classical omnisicence defined set theoretically, even if there is no set of all true propositions. So, we arrive at two conclusions.
There is no set of all true propositions.
God knows every true proposition.
Addendum
O is not equivalent to O1, since the union of sets in S = the set of all sets of true propositions is not in S. We have conceded that the union of all sets in S is not a set, given Grim’s proof. But O1 entails that there is a set of all true propositions.
O1. N(x)(x is omniscient iff (Vp)(p is a true proposition only if x knows p)
This puts a somewhat interesting restriction on the logic of omniscience. There is one counterexample to the principle,
P. (Kp & Kq) only if K(p & q)
God will know each true proposition (and most conjunctions of true propostions), but he will not know the conjunction of all true propositions, since the conjunction of all true propositions (we should argue) is not a proposition at all.
Hi, you may have already read this, but here is a very interesting exchange between Grim and Plantinga on this very subject:
http://www.sunysb.edu/philosophy/faculty/pgrim/exchange.html
Very interesting. I'm convinced. Question. If this is right, doesn't it follow that God knows an infinite number of propositions to be true? I thought that William Lane Craig had argued persuasively that there are no actual infinites. Some of those arguments are supposed to be that you cannot construct an infinite collection by successive addition, but this isn't a feature of all of his arguments. So, can we just say once and for all that he can't say both that there are no actual infinites and that God is omniscient?
Jason,
Yes, I'm familiar with that exchange. Thanks, though, for the link.
Clayton,
Yes, definitely it involves God knowing infinitely many propositions. I don't find any of Craig's arguments persuasive on this score, since I think numbers are abstract objects and I'm pretty sure that there is a countably infinite set of natural numbers. Secondly, though, (and as you know) Craig's argument is a bit beside the point that Grim is makiing, and which I'm addressing, since Grim insists that God would know infinitely many propositions. But, lastly, let's suppose there are no actual infinites. In that case, certainly, the number of true propostions cannot exceed some large finite number. So, all there is to know is summed up in some finite number of actually true propositions. In that case, to know all there is to know, to be omniscient, you have to know no more than finitely many true propositions.
Hey Mike,
"Craig's argument is a bit beside the point", I know, I couldn't resist. It just strikes me as an exceptionally odd position for him to hold (omniscient being but no actual infinites) given that it seems so independently plausible to think that there's an infinite number of things to know and he's not willing to modify the concept of omniscience to amount to something less than knowing every true proposition.
. . . it seems so independently plausible to think that there's an infinite number of things to know and he's not willing to modify the concept of omniscience to amount to something less than knowing every true proposition.
It's almost undeniable, I think, that there are infinitely many things to know (e.g. to know that 2 comes after 1 and 3 after 2 and ....). But Craig has a response, I think. Ok, suppose it turns out that his argument is right. Then (strangely, I agree) there could not be infinitely many true propositions to know. You could still have an omniscient being; it's just that he would know only finitely many things. Not the most impressive omniscient being certainly, but one nonetheless.
It looks to me like the no-set objection is exactly analogous to another kind of (obviously wrong) argument. Consider "omniapplicability": a property is omniapplicable iff it applies to all objects. Someone might try to reason that there are no omniapplicable properties—so, e.g., self-identity couldn't be one:
The reply is obvious: omniapplicability doesn't mean applying to every member of the set of all objects: it just means applying to all objects—not the same thing if they don't form a set. As for omniapplicability, so for omniscience.
Hi Jeff,
Thanks. That's a very interesting analogy. The problem is actually a little trickier, I think. Grim's argument goes from omniscience + standard set-theoretical views of quantification to the existence of a set of all truths. The argument is basically that a set-theoretical semantics for any quantification over all propositions demands a set of all propositions. The trick is to specify omniscience in a way that avoids quantification over all propositions, since that gets us into Grim's problem. I suggest in (O) that quantification is over sets of true proposition, but I do not assume (I hope) that there is a set of all true propositions or a set of all sets of true propositions. (Actually, there would be no problem in quantifying over the set of all sets of true propositions. No paradox forthcoming there as far as I can see. Certianly, Grim's paradox does not arise.) I assume that there is no set of true propositions some of whose members are not known by God.
There might be another way to go here. You could go from a definition of non-omniscience, such as in (O'.)
O'. N(x)(x is non-omniscient iff. (ES)(S is a set of true propositions & x does not know every member of S).
(O') clear does not entail that there is a set of all sets of true propositions. We can then define omniscience as not non-omniscient.
Is it the case that God knows all true propositions, or is it the case that God knows all knowable propositions? Is there a difference? If so, does the latter disjunct avoid Grim's paradox?
Hi Raymond,
I think omniscience requires that God knows all propositions, not just all knowable propositions. This is what generates the Knower Paradox for ideal knowers. But I'm not addressing that difficult problem here.
Mike:
I. Parenthetically, I don't think the Open Theist should embrace this argument. For the argument, if successful, would equally show that it is impossible for God to know all the truths about present (assuming set theoretic truths are truths about the present).
II. I am confused. (O) is just equivalent to:
(O*) N(x)(x is omniscient iff ~(Ep)(p is a true proposition and x does not know p)
Both directions are very easy. If there is a set of truths (a truth being a true proposition) not all of which are known by x, there is a truth not known by x. And if there is a truth p not known by x, then {p} is a set not all of whose members are known by x.
III. Grim's argument is harder than this to get out of. It can be formulated without set theory. I myself suspect that to get out of it one may need to be an irrealist about propositions, or deny compositionality, or do some other move that I am willing to make but that few others are. It is, however, worth noting that if Grim's argument succeeds in showing that omniscience is impossible if it succeeds in showing that there is no property had by all and only the true propositions (e.g., truth). The latter is a really radical claim.
II. I am confused. (O) is just equivalent to:
(O*) N(x)(x is omniscient iff ~(Ep)(p is a true proposition and x does not know p)
I'm not sure how (O*) is equivalent to (O), if you accept, as I do, that there is no set of all true propositions. Since I accept that there is no set of all true propositions, I cannot intelligibly quantify over them as you do in (O*). But I do not quantify over the set of all true propositions in (O). I quantify over the set of all sets of true propositions. Grim has no argument that I know of to show that there is no set of all sets of true propositions. No set in the set of all sets of true propostions is the set of all true propositions, since there is no such set. All I have to deny is that the union of all of the sets I am quantifying over is itself a set. But that is no problem: take the union of the sets {p}, {p'}, {p''}, for every true proposition. Each of those is clearly a set, but their union is not. So the problem does not rearise for the set I am quantifying over.
Just to be clear, I don't deny that (O) entails that God knows all true propositions. In fact, that's the idea. I do deny that it entails that there is a set of all true propositions. Since the universal quantifier '(Vp)' standardly has a set-theoretical interpretation, I'm rejecting the interpretation of the English quantifier in "God knows every true proposition" as it appears in your (O*).
I frankly don't think Grim's paradox is all that deep. As Russell points out above (and you point out just above) it has the absurd consequence that it's false that all true propositions are proposition. That is not a "radical consequence" of Grim's view, as I see it, it's a patent reductio.
"I quantify over the set of all sets of true propositions. Grim has no argument that I know of to show that there is no set of all sets of true propositions."
Well, we can easily produce an argument on his behalf that there is no set of all sets of true propositions. (I have no idea why I got so many steps below when I wrote it out. Probably because most of the steps are obvious.)
1. Let T be the set of all sets of true propositions. (Premise for reductio)
2. Let k = Card(T). (Set theory guarantees this is well defined)
3. Let S be any set such that k 4. For any member s of S, let p(s) be the proposition that s is a member of S. (Definition)
5. p(s) is true for every s in S. (Obvious)
6. Let S* = { {p(s)} : s in S }. (Definition. {p(s)} is the singleton whose only member is p(s).)
7. Card(S*) = Card(S). (Obvious: the correspondence s --> {p(s)} is a one-to-one function from S onto S*.)
8. Every member of S* is a set of true propositions. (Obvious: every member of S* is a singleton {p(s)}, and every p(s) is true)
9. Therefore, S* is a set of sets of true propositions. (Restatement of 8)
10. Therefore, S* is a subset of T. (By 9 and the definition of T)
11. Therefore, Card(S*) is less than or equal to Card(T). (Standard set theory)
12. But Card(T) = k 13. Thus, Card(T)
I agree that Grim's argument has absurd consequences. But that's just because it's a paradox. (Grim's book talks more about truth than about omniscience. So, yes, he's presumably aware of the weird consequence that truth is not a property of propositions and all that.) While absurd consequences are enough to show us that an argument is unsound, they aren't enough to show us why it's unsound. In the case of the most interesting paradoxes, this is a non-trivial task.
Formally, Grim's argument involves a dressing-up of the Russell-like paradox of the property of not being self-instantiated (P is self-instantiated iff P is a property and P has P). But the latter paradox is not at all trivial.
I forgot that I was writing HTML-ish code, and so my < signs got eaten up.
Step 3 should read:
3. Let S be any set such that k < Card(T).
Step 12 should read something like:
12. But Card(T) < k = Card(S) = Card(S*).
Ooops. I am on a really bad roll--corrigenda to corrigenda (partly because I've got an appointment to rush to).
Step 3 should read:
3. Let S be any set such that k < Card(S)
Step 12 should read:
12. But Card(T) = k < Card(S) = Card(S*).
I think this is right.
Does it even require quantification over the set of all sets of true propositions? Suppose we index sets of true propositions to numbers, with sets of any single arbitrary truth n = 1, and then larger sets of two truths n = 2, etc. Then necessarily, for all x, x is omniscient just in case there is no index number n such that n indexes any sets the truths of which are not known by x. We are not quantifying over the set of all true propositions, nor are we quantifying over the set of all sets of true propositions; and N (when we include the trivial sets of no true propositions) seems a pretty benign set to be quantifying over.
It's a bit confusing. I'm not sure what this shows. I don't deny that there is a set of sets of true propsitions (premise 9) whose cardinality is less than the set of all sets of true propositions (premise 1). Obviously there is. But so what?
But, then, in your last post you say,
12. But Card(T) = k is less than Card(S) = Card(S*).
How could that square with (11) in the proof?
11. Therefore, Card(S*) is less than or equal to Card(T). (Standard set theory).
Mike:
This argument shows that there is no set of all sets of true propositions.
Yes, (11) contradicts (12). Since it's a proof by reductio, it's complete as soon as one arrives at a contradiction.
Let me rewrite the proof with the corrections to the typographic problems:
1. Let T be the set of all sets of true propositions. (Premise for reductio)
2. Let k = Card(T). (Set theory guarantees this is well defined)
3. Let S be any set such that k < Card(S). (S exists by diagonal argument)
4. For any member s of S, let p(s) be the proposition that s is a member of S. (Definition)
5. p(s) is true for every s in S. (Obvious)
6. Let S* = { {p(s)} : s in S }. (Definition. {p(s)} is the singleton whose only member is p(s).)
7. Card(S*) = Card(S). (Obvious: the correspondence s --> {p(s)} is a one-to-one function from S onto S*.)
8. Every member of S* is a set of true propositions. (Obvious: every member of S* is a singleton {p(s)}, and every p(s) is true)
9. Therefore, S* is a set of sets of true propositions. (Restatement of 8)
10. Therefore, S* is a subset of T. (By 9 and the definition of T)
11. Therefore, Card(S*) is less than or equal to Card(T). (Standard set theory)
12. But Card(T) = k < Card(S) = Card(S*). (By 2, 3 and 7)
Thus, 11 and 12 are mutually contradictory, and hence 1 must be rejected.
The basic idea of the proof is really simple. The reason why there can't be a set of all true propositions is that there are way too many propositions (at least one per set, and there is no set of sets). For exactly the same reason there can't be a set of all sets of true propositions, since there are at least as many sets of true propositions as there are true propositions (since for any true proposition p, the singleton {p} is set of true propositions).
Mike:
I am puzzled by this statement in your addendum: "the union of sets in S = the set of all sets of true propositions is not in S." If U is any set of sets, then the union of all the members of U is itself a set. (That's just ZF's Sum Axiom.) Thus, the union of the sets in your S is a set, call it T. Moreover, every member of T is a member of a member of S and every member of S is a set of true propositions. Thus, every member of T is a true proposition. Hence, T is a set of true propositions. Thus, T is a member of S. Moreover, T is the set of all sets of true propositions, if S is a set. Hence, the set of all sets of true propositions is in S.
Does it even require quantification over the set of all sets of true propositions? Suppose we index sets of true propositions to numbers, with sets of any single arbitrary truth n = 1, and then larger sets of two truths n = 2, etc. Then necessarily, for all x, x is omniscient just in case there is no index number n such that n indexes any sets the truths of which are not known by x.
Branden,
No, it need not be a set. But then (O) is mistaken, since the quantifier is interpreted (standardly) set-theoretically. We might as well talk instead about the class of all true propositions. Grim's view is that he can probably run the argument again for classes of propositions, or any other collection.
I am puzzled by this statement in your addendum: "the union of sets in S = the set of all sets of true propositions is not in S." If U is any set of sets, then the union of all the members of U is itself a set. (That's just ZF's Sum Axiom.)
If it were true that S contained every set of true propositions, then their union would be the set of all true propositions, which is not a set. So, perhaps the ZF axiom itself shows that there is no such S.
My suggestion is quantification over a set (the set of natural numbers in particular) -- although it is a bit different from O, of course. My point, though, was that even if there is a problem with the set of all sets of true propositions the general approach might be salvageable.
Hi Brandon,
Suppose we do quantify over the naturals and index the naturals to the sets of true propositions. Call that set S. It would then follow that S has some cardinality k, right? But then Alex's argument shows that there is a set S* of sets of true propositions whose cardinality is larger than k. That juat means that your set will fail to index some sets of true propositions.
Hi, Mike,
That really wasn't what I meant. There would be no set S; such a set is ex hypothesi impossible and doesn't enter into the claim at all. All we're doing is noting that if a set has 1 and only 1 true proposition, it has another, i.e., the proposition it contains is known by God; then that if a set has 2 and only 2 propositions, both are known by God; and so forth. (We could stick with n=1, but I thought it was clearer not to restrict it to that.) Every set dealt with is indeed such that it doesn't include some true propositions; in fact, none of the sets dealt with, except the set of natural numbers, is an infinite set. Unless we're going to argue that set theory itself is incoherent, we have to admit that it's possible for a claim of the form
if a set of true propositions has property x, it has property y
to apply to each set of true propositions that might come up. (Substitute 'is a set' for x and 'is empty or has at least one element' for y.) So there's nothing formally problematic for an approach that says that if a set of true propositions has true propositions, its true propositions are known, even though there is no set of all such true propositions.
But in any case, nothing hangs on my particular suggestion; my point, again, was just to consider whether it's possible (while still quantifying only over sets) that you wouldn't have to quantify over the set of all sets of true propositions in order to obtain the two desiderata of there being no set of all true propositions known by God and true propositions not known by God being an empty set.
Brandon,
I guess this made it sound like you were quantifying over the set of naturals.
Suppose we do quantify over the naturals and index the naturals to the sets of true propositions.
But if you want the things known not to form a set, then there's no need to talk about sets of propositions at all. We can just talk about propositions directly and say of the collection/class of true propositions that they are believed/known by God. Similarly, the sets that you are adding indices to are, taken together, if not a set, then some sort of collection or class of objects. I don't think Grim has a problem with there being a class of such things, though he offers a promisory note that he can run his set-theoretic argument again.
The things known can form sets; this is different from the things known forming a set, and both are different from the thing known being a set. The real question, I take it, is whether we can characterize omniscience, in a way that in some reasonable sense can be called "knowing all truths," without taking all truths to form a set (and, if we take Grim's broader argument, without holding that there are true propositions about all true propositions, etc.).
Similarly, the sets that you are adding indices to are, taken together, if not a set, then some sort of collection or class of objects. I don't think Grim has a problem with there being a class of such things, though he offers a promisory note that he can run his set-theoretic argument again.
I'm not sure what you mean by 'taken together' here; what's being taken together -- the sets indexed with any particular n? But we really aren't taking them together -- we're only recognizing that sets of true propositions exist (which no one denies) and that if they are of a certain sort, they have a certain property (which is not formally problematic and so, if problematic, would have to be shown tob e such). Or are we taking together all sets of true propositions? But why? In the claim we aren't ever talking about all sets of true propositions, just sets that have a particular characteristic. All we're doing is talking quantification over the set of natural numbers and relations between the natural numbers and such sets. The claim that they would form a totality or collection is either stipulation or something that follows by necessity from the claims; if we have (as we do ex hypothesi) solid reason to think that there is no such totality, we have no good reason for stipulating it, and that leaves only demonstrating that it follows from the claim itself. Since it doesn't appeal directly to any such totality, it isn't obvious that it does so. (I agree, though, that it might turn out in fact to do so; one of the reasons it was only a suggestion.)
I do think Grim would have a problem with there being a class of all sets of true propositions; there are a number of places where he makes very strong claims that would seem to suggest this. The basic problem he raises us is not really about omniscience (objections to omniscience never are, it seems), but about how we can talk about truths and knowledge given that truths about all truths, sets of all and only true propositions, totality of things that can be known, etc., are all confused and incoherent; and it's hard to see how it could do this if we could just have classes of all sets of true propositions and the like.
But I'm left a bit confused; I thought I had understood what you were doing in the post, namely, arguing that it is possible to have an account of 'knowing all truths' that didn't lead to the paradoxes Grim claims it does because it didn't make use of a set of all truths or anything sufficiently analogous to allow the paradoxes. But now I'm not quite sure I did understand, since the point you raise looks like it would apply directly to your argument as well. What am I missing?
But I'm left a bit confused; I thought I had understood what you were doing in the post, namely, arguing that it is possible to have an account of 'knowing all truths' that didn't lead to the paradoxes Grim claims it does because it didn't make use of a set of all truths or anything sufficiently analogous to allow the paradoxes. But now I'm not quite sure I did understand, since the point you raise looks like it would apply directly to your argument as well. What am I missing?
Right, that was the idea. Alex offered a nifty proof that the direction I was going would not solve the problem. In general, we can't solve this problem using standard quantification theory whose domains are sets of objects. Probably the way to solve it is to assume an unrestricted domain over which to quantify. I'm not sure whether the resulting domain would be a class of objects, though I suspect it would be. Grim has made claims to the effect that he could generate the same problems there, but I (for one) have nott seen him do it. That aside, I'd have to look closer at issues surrounding talk about everything or absolute generality (not that we need a domain that large, but it would help to see the problems with such a domain) before I could say anything interestingly related to everything that an omnisicent being would know. There's a lot of interesting work on absolute generality these days, and I'm anxious to have a closer look at it.
Thanks, Mike, that clarifies things (it's been a rough week of grading and Christmas chores, so sorry about the delay in responding). I'm skeptical of set-theoretical accounts of omniscience (or, indeed, of quantification, for reasons too long to go into here); but I'm also not (yet) convinced that such accounts are as limited in this regard as often thought, partly because (as suggested above) I see no reason at this point to think that, even if we have to have a set-theoretical account, we have to quantify over all truths in order to claim that no truth is unknown by omniscience; any more than we have to quantify over a set of all possible geometrical shapes in order to say that a square circle is impossible (or over a set of all sets to say that no set is both finite and isomorphic to a proper subset of itself). It's certainly an area where there are lots of tangled issues.
Brandon:
That's an interesting suggestion. So one says that a truth unknown by God is impossible, just as a circle that is square is impossible. The natural take on the latter is quantificational: L~(Ex)(c is a circle and x is a square). But one might, instead of taking a quantification view, simply hold that circularity and squareness are incompatible. And similarly, truth and unknownness-by-God are incompatible.
I think that may be right. I think a similar thing is to be said about claims such that a proposition that is neither true nor false is impossible, etc.