# Model Theory: relation between two theories with the same models?

1. Apr 10, 2013

This is a question out of model theory. (This preamble is due to the fact that "model" and "theory" are used in different ways in different fields.) Is there any specific relationship between two theories which have the same models? A knee-jerk reaction would be to say that they are isomorphic, but upon reflection this does not seem likely. (This question is not to get mixed up with the fact that two models having the same theory are isomorphic.)
Thanks.

2. Apr 10, 2013

### Bacle2

But, isn't a model for a collection of axioms also a model for any subcollection?

You start with a collection of axioms, and find a structure so that all axioms are verified

using Tarski def. of Truth (my def. of a model) . If all axioms are satisfied, wouldn't every

subcollection of axioms also be satisfied?

3. Apr 10, 2013

Yes, of course, but this does not answer the question.
First, whereas a model satisfying a collection of axioms will, by definition, also satisfy any subcollection, the converse is not necessarily true: a model for one collection of axioms will not necessarily satisfy a supercollection of those axioms. So a collection of axioms and a subcollection usually will not have the same models.
Secondly, you could have two collections of axioms both being satisfied by the same models, yet neither collection is a subcollection of the other.
So, back to my original question.....

4. Apr 10, 2013

### Bacle2

Maybe I don't understand your use of isomorphic theories; I know what isomorphic models are, but not what isomorphic theories are. How do you define it?

5. Apr 10, 2013

First, I just said that the "isomorphic" response would be a knee-jerk reaction, which implies that the knee-jerker would not have a good definition in mind, and as I stated, it would not have been a good response. My question was precisely to avoid this, and I am looking for a proper term for two theories with the same models.

However, if I were backed into a corner and had to give a definition of "isomorphic theories":grumpy:, I suppose I would remember that theories are simply strings of symbols, so that there would be a bijection f that took constant symbols of one theory to constant symbols of the other, relation symbols of one to relation symbols of the other, etc. so that f(aRb) = f(a)f(R)f(b), and so on and so forth. It would be very restrictive, of course, and not at all appropriate as an answer for my question.

So, back to my original question, ignoring the part about what is not an answer....

6. Apr 17, 2013

### yossell

If theories are sets of sentences closed under logical implication, and if we restrict our attention to first order theories, then wouldn't two theories with the same models be identical? For if they have the same models, they will be logically equivalent and so - by soundness and completeness - sentence S will be a theorem of one if and only if it is a theorem of the other.

7. Apr 17, 2013

Yossell, thanks, you are right, that makes sense. I guess I was thinking of a slightly different question, concerning not the theories but the axioms chosen for the theories. That is, my question should have been stated: what is the relationship between different axiom systems for the same theory? I am thinking along the lines of the relationship between two different bases for a vector space: the bases are not equal, even though they generate the same vector space, but one can indicate a relationship of rotation, etc. between bases. Any thoughts on this?

8. Apr 18, 2013

### yossell

nomadreid -- that's an interesting idea, but I can't think of anything similar. Of course, the axioms of one theory will be provable from the axioms of the other, and vice versa, but I can't think of anything analogous to bases of a vector space. The trouble is one and the same theory can be axiomatised in so many different ways, different axiomatisations even differing in the number of axioms that appear within the axiomatisation, that I find it hard to imagine there's any such similar relationship here.

9. Apr 19, 2013