1

Definition of omniscience: X is omniscient = for the whole preposition P, X knows the value of P.

  1. Let T be the set of all true propositions. An omniscient being knows all these propositions (T).

  2. Let P(T) be the set of all subsets of T. For each of these subsets let Q(i) be the true proposition that states that a certain element of the set i belongs to i.

  3. Then there are more truths than elements of T (by the argument of Cantor (1)).

  4. Then, there are true propositions that an omniscient being who knows all true propositions does not know. Which is absurd.

Conclusion: God is not omniscient because the set of true propositions is greater than the set of true propositions known to God.

1: The set of the parts has more elements than the original set.

11
  • 3
    Can you unpack what you mean by "the argument of Cantor (1)" and how that justifies point 3 of the argument? (Not my downvote, by the way.)
    – Mark
    Jun 10, 2023 at 12:04
  • 9
    The currently accepted answer is incorrect, and omniscience is irrelevant here. Even without any mention of it, 3 contradicts 1. The problem with the argument is in the assumption that all true propositions form a set. The contradiction shows that they do not, they form instead a proper class. As the concept of cardinality, and hence Cantor's diagonal argument, do not apply to proper classes, the argument fails at step 3.
    – Conifold
    Jun 10, 2023 at 13:25
  • 3
    @eques Sure it can, when the argument's flaw does not depend on the (mis)perceived point, as is the case here.
    – Conifold
    Jun 10, 2023 at 14:14
  • 1
    @eques If X is mentioned, no matter how irrelevant, it is not "entirely", right? After all, it was mentioned. But the question is "How can one refute this argument?" And to answering this question, omniscience is irrelevant. Entirely. One can literally erase it from 1-4 and the argument will be refuted by the exact same flaw.
    – Conifold
    Jun 10, 2023 at 15:05
  • 1
    All mention of god can be removed from the question. It is equivalent to the question: "Can there be a set of all true propositions" it goes: Assume T is that set. Then we do some shenanigans with it to make Q. Q contains element that are true but not in T. Therefore T with such a definition cannot exist. Any Gods or omniscience are irrelevant to that question.
    – tkruse
    Jun 12, 2023 at 21:32

6 Answers 6

9

Going with T as the set of all truths, let t, t', t'', etc. be elements of that set. Let {t, t'} be some subset of T. In fact, let's specify {t, t'} as (tt'), i.e. as the conjunction of the two individual truths. Now, if this subset isn't in T, then it's not true, i.e. the conjunction of two individual truths isn't itself true! So much for conjunction, then, perhaps.

The way the original argument (as reported in the SEP) goes, the idea is that there is a separate truth-statement for every assertion (x ∈ ℘(T)), so that there are as many such statements, which are supposed to be true, as there are x. In fact, such an issue has often motivated intuitionists to reject the powerset axiom for infinite sets in the first place (predicativists have a different, but related, set of objections in play, here), see Storer[10], esp. §§3.3, 5.1, and 5.3. Anyway, if there is a separate, "It is true that x ∈ ℘(T)," for every x, then T would be both a subset and a superset of ℘(T), which if not contradictory, is at least exceedingly curious. But perhaps there are not separable true statements for every x as such, but a plurally quantified statement that at once maps all possible instances of the scheme, "x ∈ ℘(T)," into T, so that there is only one true statement for all the x as such.

Now again, if ℘(T) has elements that are subsets of T that aren't in T, then those subsets, despite being combinations of truths, don't represent truths themselves. An omniscient being can hardly be faulted for not knowing things that aren't true, I suppose. An easier way out would just be to deny either the powerset axiom for T itself (i.e. there is no set of all subsets of T because there aren't "all subsets" of T in the first place) or anything, e.g. separation/collection, that could be used to extract subsets from T and then theorematically compose them into a non-axiomatic ℘(T).

Or even more easily, just deny that there is a set of all truths, but there is a proper class (where this isn't a proper set) of them, and say that an omniscient being knows the proper class of all truths internally (even we might be said to know of the class externally, although we are not omniscient and so don't know what all its elements are individually).

ADDENDUM: Skolem's paradox

Whether a set of all truths about elements of some powerset must be greater than a set of all truths about a base set is something touched upon in connection with Skolem's paradox. Per first-order model theory:

The downward Löwenheim-Skolem theorem: Suppose L is a first-order language which has κ formulas, A is an L-structure and λ is a cardinal which is at least κ but less than the cardinality of A. Suppose also that X is a set of at most λ elements of A. Then A has an elementary substructure which has cardinality exactly λ and contains all the elements in X.

So we might think that a set of all truths about elements of a powerset must run through more truths than there are elements of a base set, but it turns out that we don't need to do this; it is possible to "collapse" (a word beloved by set theorists!) an uncountable model to a countable one (a model is countable when it has only countably many sentences to its name). So there should be a way to find a countable set of truths that model-theoretically "covers" uncountably many sentences about elements of powersets. Now I think there are things called "Löwenheim-Skolem numbers for logics" such that they replace the "collapsed to countable" description with "collapsed to λ" such that theories in these logics must have at least λ-many sentences (that's my attempt to understand the descriptions, anyway), but so again, if we take a theory of size λ for a set of all truths (per the theory), we should be able to go down from 2λ nevertheless.

So again, an omniscient being would perhaps be able to directly comprehend a set of all truths, for a set of size λ, without having to directly comprehend, in the same act, a set-of-subtruths of size 2λ.

2
  • 2
    This is a much better answer than the accepted one. Jun 10, 2023 at 14:54
  • What is a powerset? Answer: The powerset of a set S, call ir P, is the set of all subsets of S. If ... Jun 14, 2023 at 7:26
7

The definition of omniscient including the form given as "X is omniscient = for the whole preposition P, X knows the value of P" means that the truths Q(i) are contained in T because T is all true propositions.

In order for the argument to work, it would be necessary to show that the propositions defined in number 2 are excluded from the set defined in 1, but that's not the case. In so far as they describe something which is true, they are included in 1.

This suggests that T is ill-defined. Using set or proposition in the definition of T leads to a contradiction. @Conifoid suggested a proper class instead of a set and @causative noted that a side-effect of that would be that statements Q(i) would not be propositions and thus outside the definition of omniscience.

Further, if this argument held, it would mean that omniscience means knowing all true facts but not knowing a structural fact about those facts (namely that a fact belongs to an arbitrary subset of facts) which is absurd. Given that state, nothing would prevent defining a higher omniscience which subsumed that limitation -- this would be true omniscience then.

Hence, this argument amounts to a theological strawman of sorts. It sets up a definition of omniscience, attempts to show the argument as contradictory and then concludes that God cannot be omniscient. However, it doesn't establish that what is meant by "God is omniscient" in Christian theology is in fact covered by the given definition.

13
  • 3
    This is incorrect. The power set of a set is always larger than the set itself, which means it must always contain elements that the set itself does not contain. So, yes, some propositions defined in number 2 are excluded from the set defined in 1.
    – causative
    Jun 10, 2023 at 13:08
  • 2
    @causative the fact that a power set is always larger is an axiom, sure, but this necessarily leads to paradoxes in set theories. It is not possible for T to contain all true statements and then there be more true statements. There is no qualification intrinsic to the definition of T which excludes propositions of the type Q(i)
    – eques
    Jun 10, 2023 at 13:15
  • 2
    No, the fact that a power set is always larger is a theorem of Cantor, not an axiom. You can't just discard it. If you want to have the set axioms, Cantor's theorem is an inevitable consequence.
    – causative
    Jun 10, 2023 at 13:21
  • 2
    I didn't just discard it. I pointed out that the definition of T necessarily includes propositions of type Q(i). If Q(i) is a true proposition has to be in T -- T doesn't exclude propositions that define the existence of propositions in other subsets.
    – eques
    Jun 10, 2023 at 13:25
  • 4
    @DavidGudeman I am familiar enough with the diagonal argument. Yes, Cantor's argument says the power set must be larger and yes there cannot be a bijection between T and P(T). My point, which you appear to fail to grasp is that it ultimately is irrelevant. The argument does result in a contradiction, but the implication of that contradiction is that the premised definition of omniscience is flawed not that omniscience is not a divine attribute
    – eques
    Jun 10, 2023 at 16:59
7

This is a variation of Russell’s Paradox, which creates a paradoxical set from a natural-language specification. We do not seem to be talking about sets that are well-formed in the set theory Cantor proved diagonalization in. The conjunction of all true statements is represented by the set of all true statements. It is a true statement, so the set of true statements would contain itself. But no valid set in ZF set theory may contain itself. Therefore, it is not valid to apply theorems from ZF set theory to this so-called set. The premises of Cantor’s Power-Set Theorem are not sound for it.

Note that this paradox has nothing to do with “omniscience,” but (purportedly) shows set theory to be self-contradictory. We already know that any set theory contradicts itself if it has a “set of all sets that do not contain themselves.” This just shows that the same is true of a set theory with predicate logic and a “set of all true statements.” The proof claims that there are more true statements than there are true statements. Nothing about it depends on anyone knowing or not knowing anything, just that the true propositions that are not in “the set of all true propositions” supposedly exist.

We could, though, define a limited form of knowledge that we can use set theory to prove things about. For example, if we define that to mean knowing the truth value of all statements with finite length in some formal language—that is, all questions that could ever be asked—this paradox disappears. (Although that would have some strange implications of its own, such as that this type of knowledge cannot include even a single fact about any but an infinitesimal portion of the individual real numbers.) All propositions could then be numbered, and there is a countable number of finite sets of natural numbers, so the set of all finite conjunctions of propositions and the set of all propositions are both isomorphic to the set of natural numbers, so they are the same size (in the same sense that there are as many even numbers as there are natural numbers).

I think, though, that a better way out here is to say that “all true statements” does not have the correct structure to be a “set.” It’s implicitly a specification on statements about sets, of which there is at least one for every set. But we already know that we can’t declare sets that can contain sets as large as themselves, or we get Russell’s Paradox.

1

I'm going to focus on this part:

Let P(T) be the set of all subsets of T. For each of these subsets let Q(i) be the true proposition that states that a certain element of the set i belongs to i.

So, P(T) is indeed larger than T. But you might want to examine the elements of P(T) before thinking it is reasonable.

P(T) intersect with finite subsets has the same cardinality as T.

P(T) intersect with effectively described subsets has the same cardinality as T.

Ie, let Z be the set of all propositions. Suppose you have an algorithm A that, given a proposition, says to include it or not. This describes a subset of Z, call it S_A, that is effectively described.

Take P(T) intersected with the set of effectively described subsets of Z. This set has the same cardinality as T.

Or, in short: The elements of P(T) that make it larger than T are the ones whose elements cannot be effectively described.

This is because T is countable, and the set of effectively described subsets of Z are also countable. P(T) intersect a countable set is merely countable.

Now, there is no effectively described bijection between T and effectively describable P(T), as you can use Cantor on any such bijection and get an effectively described element of effectively describable P(T) that is missing; the Cantor diagonalization takes your effective description of the bijection and turns it into an effective description of an element of P(T).

So, while you can prove your Omniscent being does not know the truth value of some element of P(T), nobody can ever produce that element of P(T) to demonstrate that Omniscent being's lack of knowledge. They can prove its existence, but cannot describe its contents.

-2

Infinity plus one is still infinity. For T be the set of all true propositions, then it must include propositions that can be inferred from T.

--- edit to be more explicit about the bad maths.

"3. Then there are more truths than elements of T" therefore T could not have been the set of all truths to start with.

A set of any infinite property is not a closed set and therefore can not have a countable number of elements, otherwise given for example the set of all even numbers, you could always add two to the largest number to get an even number that was not in the set. Clearly this is ridiculous and would render set theory useless. A set of any infinite property does not and can not have a fixed number of elements, and a number that does not exist can not be compared to any other number.

Let T be the set of all true propositions. Therefore, any truth t must exist in T, and the truth t1, that t exists in T, must also exist in T, and the truth t2 that t1 exists in T must also exist in T, ad infinitum. There are no truths that can not exist in T, therefore set theory provides no contradiction to the possibility of omniscience.

3
  • "Infinity plus one is still infinity," is somewhat ambiguous. For example, ℵ0 + 1 = ℵ0, 1 + ω = ω, but ω + 1 ≠ ω. OTOH ω + 1 is still an infinite ordinal. Apparently ORD + 1 can retain the noncommutative difference from 1 + ORD, although I'm still bewildered by that possible fact. Jun 12, 2023 at 13:39
  • ω is a transfinite ordinal, not a natural numbers, so you can't use the definition of addition over natural numbers to reason about what either ω + 1 or 1 + ω means. Whether "infinity plus one is still infinity" depends on the actual definition of "infinity" being used.
    – chepner
    Jun 12, 2023 at 14:07
  • 1
    While true, the nature of such proud typically is not that a set Q is "greater in size" than set T, but that Q must contain at least one element not in T. The question does not describe that well.
    – tkruse
    Jun 12, 2023 at 21:28
-3

There are already some good answers and comments, but I will add that 1 probably begs the question against that which monotheists consider the premise of God's omnicience.

A serious discussion would differentiate between God knowing the truth and God being the truth. If God is the truth, everything he wills becomes true, and nothing becomes true against his will.

So, instead of true propositions existing ontologically without prerequisite of which God knows all...

 God
 / \
truth
 tru
  t

...God is the prerequisite of all true propositions

  t
 tru
truth
 / \
 God
3
  • As it’s currently written, your answer is unclear. Please edit to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center.
    – Community Bot
    Jun 13, 2023 at 16:25
  • 1
    "God being the truth" is not a very useful statement as it requires you to change the definitions of the words you use and neither quote offered relates to omniscience which is the topic of this particular conversation. My beer glass is empty. That is a truth independent of my belief in God or Gods belief in me. Suggesting that it is empty because God willed it so is not very useful because it implies that God could will it full again, something that no amount of prayer is going to achieve, but a short walk to the fridge can fix.
    – Paul Smith
    Jun 14, 2023 at 0:32
  • @PaulSmith It is a useful statement because the premises in the original question assume a different definition of "omniscience" than that proposed by most monotheists, i.e. premise 1 changed the definition of truth (with regard to divine omniscience), not me. Hence, it begs the question by assuming the classic understanding of omniscience is incorrect. Omniscience, understood by most monotheists, means the glass and the beer and all truth about them originate in the divine will: they are not discovered post-existence, but willed pre-existence.
    – qxn
    Feb 12 at 4:24

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