Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Thank everybody for answering my previous questions: first, and second.

Here I would like to ask about some important thing which I do not understand clearly.

  1. Is it necessary for theory to have given interpretation in some universe by definition or not, it is not necessary for set of axiom to be a theory?

The meaning of question above is: is this a correct way of defining theory if we do not give domain of interpretation for it? What obstacles may arise from that? For example: As far as I know standard category theory do not qualify clearly which is its domain of interpretation, then, may we say that it is "defined theory" from a formal point of view? I have heard that there are some efforts to define category theory by means of metacategories which probably has as one of its aims to give clear definition of domain of interpretation.

So there another question arises for which I will try to give some introduction by example.

I may think about theory which when interpreted in set universe defined by some formulas, relations whatever may be consistent and have models, whilst in other universes do not. Is this a case? For example in answer about possible definition of domain of discourse for first order theories here Joel David Hamkins wrote: "If V is the universe of all sets, we can define certain classes in V, such as { x | φ(x) }, where φ is any property."

What if $\phi(x)$ is not property from first order logic?

For example we may consider sentence $S= \{x | \phi (x) \}$ as meaning:"for every theory x when schema of axioms of induction is present in axiom set". It is clearly not first order sequence, so maybe model which is interpreted in such universe is not first order model even if axiom of theory forms first order set? The only difference here is in a way we introduce a domain of interpretation so maybe this level of freedom is too much?

Usual definition of domain of interpretation which I have found in wikipedia here uses set "structure" but not state clearly that domain of discourse has to be first order set from the beginning! It is something which is not clear for me here. Should I believe that some "interpretations" of theory may be inconsistent whilst other may be consistent when we drop interpretation part as a definition and we leave it free?

  1. If we construct theory in first order logic ( for example as in my second question), and then we will try to use it on domain of objects defined by means of higher order logic, do we end with higher order theory, or rather first order theory which is saying something about second order logic objects? Or maybe the answer is" "it depends" and there should be stated additional requirements?

I have hope it is no obviously wrong question...


Note 1. As I do not want to be found very speculative, I will point to blog of Terence Tao here where You may find some information about nonfirstorderisability and specially this sentence:

which is part of the fundamental theorem of linear algebra, does not seem to be expressible as stated in first order set theory

So there are pretty useful practical statement, not very abstract, in normal mathematics which cannot be expressed in first order theory language.


Note 2. In wikipedia here I found the following remark:

"MK can be confused with second-order ZFC, ZFC with second-order logic (representing second-order objects in set rather than predicate language) as its background logic. The language of second-order ZFC is similar to that of MK (although a set and a class having the same extension can no longer be identified), and their syntactical resources for practical proof are almost identical (and are identical if MK includes the strong form of Limitation of Size). But the semantics of second-order ZFC are quite different from those of MK. For example, if MK is consistent then it has a countable first-order model, while second-order ZFC has no countable models."

share|improve this question
add comment

1 Answer 1

To answer your first question: No, a theory can be well-defined formally without having any specified domain of interpretation. But if it is a first-order theory, a technique developed by Godel (in his proof of the Completeness Theorem) allows a domain to be constructed for the theory. So we don't, at least, have to worry about theories with no domain at all (in the first order case).

For the second question: Consider a second-order theory with two domains, the reals and the countable ordinals. We add categorical axioms for these structures, and a function symbol $f$, with additional axioms stating that it is a bijection between the two domains. Then this theory has a model if and only if the Continuum Hypothesis holds. So we see that, in general, existence of models is universe-dependent, as you suggested.

As for your suggestion that a model of a first-order theory might not be first-order, if I am understanding you correctly, this is also true: the first-order theory of the natural numbers, for example, has nonstandard models as well as the standard one (this is the Louwenheim-Skolem Theorem), so it follows that the standard model cannot be characterized by first-order means.

But if by "first-order set" you just mean a set of the sort that appears in first-order set theories, then the Reflection Theorem ensures that any theory described by a "first-order set" which has a "second-order set" model must also have a "first-order set" model. So there is no loss of generality in allowing only "first-order set" models. [Now that I think about it, this probably is what you meant, especially given that you called out category theory as being particularly problematic.]

For your second question 1, you would generally get a first-order theory (possibly multi-sorted) about second-order objects (like Morse-Kelly set theory, in the wikipedia quote). Unless you specifically tell people to read it as a higher-order theory, of course.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.