Thursday, November 5, 2020

Is there a set of all set-theoretic truths?

Is there a set of all set-theoretic truths? This would be the set of sentences (in some encoding scheme, such as Goedel numbers) in the language of set theory that are true.

There is a serious epistemic possibility of a negative answer. If ZF is consistent, then there is a model M of ZFC such that every object in M is definable, i.e., for every object a of M, there is a defining formula ϕ(x) that is satisfied by a and by a alone in M (and if there is a transitive model of ZF, then M can be taken to be transitive). In such a model, it follows from Tarski’s Indefinability of Truth that there is no set of all set-theoretic truths. For if there were such a set, then that set would be definable, and we could use the definition of that set to define truth. So, if ZF is consistent, there is a model M of ZFC that does not contain a set of all the truths in M.

Interestingly, however, there is also a serious epistemic possibility of a positive answer. If ZF is consistent, then there is a model M of ZFC that does contain a set of all the truths in M. Here is a proof. If ZF is consistent, so is ZFC. Let ZFCT be a theory whose language is the language of set theory with an extra constant T, and whose axioms are the axioms of ZFC with the schemas of Separation and Replacement restricted to formulas of ZFC (i.e., formulas not using T), plus the axiom:

  1. x(x ∈ T → S(x))

where S(x) is a sentence saying that x is the code for a sentence (this is a syntactic matter, so it can be specified explicitly), and the axiom schema that has for every sentence ϕ with code n:

  1. ϕ ↔ n ∈ T.

Any finite collection of the axioms of ZFCT is consistent. For let M be a model of ZFC (if ZF is consistent, so is ZFC, so it has a model). Then all the axioms of ZFC will be satisfied in M. Furthermore, for any finite subset of the additional axioms of ZFCT, there is an interpretation of the constant T under which those axioms are true. To see this, suppose that our finite subset contains (1) (no harm throwing that in if it’s not there) and the instances ϕi ↔ ni ∈ T of (2) for i = 1, ..., m. It is provable from ZF and hence true in M that there is a set t such that x ∈ t if and only if x = n1 and ϕ1, or x = n2 and ϕ2, …, or x = nm and ϕm.

Moreover, any such set can be proved in ZF to satisfy:

  1. x(x ∈ t → S(t)).

Interpreting T to be that set t in M will make the finite subset of the additional axioms true.

So, by compactness, ZFCT has an interpretation I in some model M. In M there will be an object t such that t = I(T). That object t will be a set of all the truths in M that do not contain the constant T. Now consider the interpretation I of ZFC in M, which is I without any assignment of a value to the constant T (since T is not a constant of ZFC). Then ZFC will be true in M under I. Moreover, the object t in M will be a set of all the truths in M.

So, if ZF is consistent, then there is a model of ZFC with a set of all set-theoretic truths and a model of ZFC without a set of all set-theoretic truths.

The latter claim may seem to violate the Tarski Indefinability of Truth. But it doesn’t. For that set of all truths will not itself be definable. It will exist, but there won’t be a formula of set theory that picks it out. There is nothing mathematically new in what I said above, but it is an interesting illustration of how one can come close to violating Indefinability of Truth without actually violating it.

Now, what if we take a Platonic view of the truths of set theory? Should we then say that there really is a set of all set-theoretic truths? Intuitively, I think so. Otherwise, our class of all sets is intuitively “missing” a subset of the set of all sentences. I am inclined to think that the Axioms of Separation and Replacement should be extended to include formulas of English (and other human languages), not just the formulas expressible in set-theoretic language. And the existence of the set of all set-theoretic truths follows from an application of Separation to the sentence “n is the code for a sentence of set theory that is true”.

2 comments:

  1. What about "definable in the English language"? Applying Separation we get the set of all definable countable ordinals. But then consider the least countable ordinal not in that set... it seems I've just defined it.

    Might this be a good reason to deny the existence of a set of definable countable ordinals?



    ReplyDelete