Is it possible to translate theorems between isomorphic groups?

[Preno]

So why is isomorphism the appropriate measure of "equivalent for all practical purposes"?

Because if we are structuralists, we are only concerned with the roles the elements of a certain structure play, not their absolute identity outside of their interrelations defined by the structure. The integers with addition and the even integers with addition are the same group, because you can do the same groupy things with them. Whether an object is labelled 1 or 2 is irrelevant from the pov of the group - the 2 in the group of even integers has the same role as the 1 in the group of integers. An uncountable group elementarily equivalent with the even integers, however, is not the same group as the even integers, because only one of them has an injective homomorphism to Z.

Is it that X and Y are isomorphic if and only if they satisfy the same theorems of nth-order logic, for all n?

I don't think so. The models of second-order ZF are levels of the cumulative set hierarchy corresponding to some (strongly) inaccessible cardinal (I don't think all of them are models of second-order ZF, though), i.e. are of the form V_\alpha, \alpha strongly inaccessible (see the definition). That is, for every two models of second-order ZF, one is an initial segment of the other, so the two are obviously not isomorphic.

[Edit]Of course, if one holds that there are no strongly inaccessible cardinals (or only one), then this doesn't help much.

(Also, as has already been mentioned third- and higher-order logic is reducible to second-order logic.)

 

[Fredrik]

because you can do the same groupy things with them.

Yes, but is there perhaps a mathematical way to say "you can do the same groupy things with them"? My idea that two structures are isomorphic iff they "satisfy the same theorems" didn't work, because I didn't know about "elementarily equivalent structures". But isn't there a modified version of that idea that does work? Your counterexample involves a countable and an uncountable set. What if I change my suggestion to this instead? Two structures of the same type/signature are isomorphic iff their underlying sets have the same cardinality, and they satisfy the same sentences in the first-order language defined by the signature.

A sentence is a formula with no free variables, right? Maybe we should allow free variables too. Then we have to say something about functions that assign values to variables. Let X and Y be structures and |X| and |Y| their underlying sets. Suppose that they have the same cardinality. What if (X,f) and (Y,g_°f) and satisfy the same formulas for each value function f:V→|X| and each bijection g:|X|→|Y|. Isn't that enough to guarantee that X and Y are isomorphic?

 

[Preno]

Yes, but is there perhaps a mathematical way to say "you can do the same groupy things with them"?

Well, yes, namely that they're isomorphic. Why are you seeking another definition to justify that one?

My idea that two structures are isomorphic iff they "satisfy the same theorems" didn't work, because I didn't know about "elementarily equivalent structures". But isn't there a modified version of that idea that does work? Your counterexample involves a countable and an uncountable set. What if I change my suggestion to this instead? Two structures of the same type/signature are isomorphic iff their underlying sets have the same cardinality, and they satisfy the same sentences in the first-order language defined by the signature.

No, that doesn't work. Consider the theory of the successor function (with a single function S and constant 0) with the axioms:

\forall x \forall y (S(x)=S(y) \rightarrow x=y)
\forall x (x \neq 0 \rightarrow \exists y S(y)=x)
\forall x (S(x) \neq 0)
\forall x (S^m(x) \neq x) (this is an axiom schema containing one axiom for each natural m)

This theory has infinitely many mutually non-isomorphic countable models, and these are all elementarily equivalent, because by uncountable categoricity and the Vaught test (or by quantifier elimination), this is a complete theory.

A sentence is a formula with no free variables, right? Maybe we should allow free variables too. Then we have to say something about functions that assign values to variables. Let X and Y be structures and |X| and |Y| their underlying sets. Suppose that they have the same cardinality. What if (X,f) and (Y,g_°f) and satisfy the same formulas for each value function f:V→|X| and each bijection g:|X|→|Y|. Isn't that enough to guarantee that X and Y are isomorphic?

You're being a bit elliptic here, because a valuation is not a function V→|X|, but rather it assigns relations of the appropriate arity to each symbol in the language, but it seems to me that you're basically trying to define Y as the isomorphic image of X, so yes, trivially an isomorphic image of X is isomorphic to X.

 

[Fredrik]

Well, yes, namely that they're isomorphic. Why are you seeking another definition to justify that one?

I told you already: I want to be able to explain why two isomorphic structures can be considered "equivalent for all practical purposes".

No, that doesn't work. Consider the theory of the successor function (with a single function S and constant 0) with the axioms:

\forall x \forall y (S(x)=S(y) \rightarrow x=y)
\forall x (x \neq 0 \rightarrow \exists y S(y)=x)
\forall x (S(x) \neq 0)
\forall x (S^m(x) \neq x) (this is an axiom schema containing one axiom for each natural m)

This theory has infinitely many mutually non-isomorphic countable models, and these are all elementarily equivalent, because by uncountable categoricity and the Vaught test (or by quantifier elimination), this is a complete theory.

OK, thank you for that. I'm not familiar with...any of the things you mentioned after the word "because", but I understand that your point is that my idea doesn't work.

You're being a bit elliptic here, because a valuation is not a function V→|X|, but rather it assigns relations of the appropriate arity to each symbol in the language,

That's not the sort of function I'm talking about. I meant the same kind of function as the w in the definition here.

 

[Preno]

I told you already: I want to be able to explain why two isomorphic structures can be considered "equivalent for all practical purposes".

The definition of isomorphism says that the two structures are the same up to relabelling. Tbqh I don't see what other explanation you're seeking for the fact that they can be "considered equivalent for all practical purposes". Surely if we have a two-element group consisting of elements a,b and a two-element group consisting of elements c,d, then we don't need any additional explanation for why they're the same.

OK, thank you for that. I'm not familiar with...any of the things you mentioned after the word "because", but I understand that your point is that my idea doesn't work.

For all non-negative n, the theory has a model of the form N + n copies of Z, and these models are clearly not isomorphic. But I was saying it can be proven that the theory is complete (it decides all sentences), so any two models satisfy the same sentences.

That's not the sort of function I'm talking about. I meant the same kind of function as the w in the definition here.

Sorry, stupid mistake on my part. Yes, it's just a restatement of the definition of isomorphism, if you unpack the definitions, you get (for f such that f(x)=a, ...):

(a,b,\ldots) \in \varphi^X \Leftrightarrow (f(x),f(y),\ldots) \in \varphi^X \Leftrightarrow X \vDash_f \varphi(x,y,\ldots) \Leftrightarrow Y \vDash_{gf} \varphi(x,y,\ldots) \Leftrightarrow (g(f(x)), g(f(y)), \ldots ) \in \varphi^Y \Leftrightarrow (g(a),g(b),\ldots ) \in \varphi^Y

where \varphi is a n-ary relation symbol and \varphi^X,\varphi^Y its realizations in X, Y. Similarly for function symbols.

 

[Fredrik]

The definition of isomorphism says that the two structures are the same up to relabelling. Tbqh I don't see what other explanation you're seeking for the fact that they can be "considered equivalent for all practical purposes". Surely if we have a two-element group consisting of elements a,b and a two-element group consisting of elements c,d, then we don't need any additional explanation for why they're the same.

I understand what you're thinking, but I'm not so sure I agree with it. You're thinking that this is like when we use a Riemann integral to make the idea of "area under the graph" precise. It's not like the area is something else, and the integral just a technique to calculate it. The integral defines what we mean by "area" in this context. Similarly, the concept of "isomorphism" makes another idea precise. The thing is, I would say that the idea it makes precise is the idea of "relabeling", not the idea of "equivalent". It seems to me that there should be a way to make that idea precise and then prove that it's equivalent to "isomorphic". If the idea is "equivalent for all practical purposes", the mathematical concept that makes that idea precise should specify exactly what the "practical purposes" are.

 

[mathwonk]

Here is a statement true of G but not always of H: "The elements of the group are elements of G".

so I think you should be a little careful by the range of the word "theorem".