# Instance of non-isomorphic frameworks which are elementarily equal

I simply began finding out model theory on my very own, and also I was asking yourself if there are any kind of intriguing instances of 2 frameworks of a language L which are not isomorphic, yet are elementarily equal (this suggests that any kind of L-sentence pleased by among them is pleased by the 2nd).

I am making use of the notaion of David Marker's publication "Model theory: an intro".

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2019-05-07 00:55:23
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First, I'm grateful you read my publication! :)

Allow me make a number of talk about Pete's solution - - this is my very first time below and also I do not see just how to leave remarks.

2) Any 2 thick straight orders without endpoints are elementarily equal. Specifically $(Q,<)$ and also $(R,<)$ are elementarily equal. So there is no first order means of sharing the efficiency of the reals.

3) Any 2 algebraically shut areas of the very same feature are elementarily equal. So the algebraic numbers is elementarily equal to the intricate numbers. This suggests you can confirm first order features of the algebraic numbers making use of facility evaluation or the intricate numbers making use of Galois Theory or countability.

4) Similarly the reals area is elementarily equal to the actual algebraic numbers or to the area of actual Puiseux collection. One can as an example make use of the Puiseux collection to confirm asymptotic buildings of semialgebraic features.

Ultimately, Pete's comment 5) concerning boundless versions of the concept of limited areas being elementarily equal isn't fairly appropriate. This is just real if the loved one algebraic closure of the prime areas are isomorphic. As an example,

a) take an ultrapower of limited areas $F_p$ where the ultrafilter having 2,4,8, ... after that the resulting version has particular 2, while if the ultrafilter has the set of tops, after that the ultraproduct has particular 0.

b) if the ultrapower has the set of tops conforming to 1 mod 4, in the ultraproduct - 1 is a square, while if the ultraproduct has the set of tops conforming to 3 mod 4 after that in the ultraproduct - 1 is not a square.

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2019-05-09 04:37:33
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There are an actual numbers well worth of countable versions of Peano math, all elementarily equal to the common version, yet all pairwise nonisomorphic.

Every nonstandard instance (nonstandard methods not isomorphic to the common version) is identified by the reality that it has "definitely huge" numbers. That is, each version M has an approved duplicate of N = common naturals inside. Nonetheless, each nonstandard version M has a component p (actually, countably several such components) which the version points is larger than every little thing in its duplicate of N.

They are developed by typical density debates coupled with the descending Lowenheim - Skolem theory.

Allow represent the axioms for Peano Arithmetic and also set T = Th (N), the concept of N (i.e., the set of all first order declarations which hold true in the common analysis.

Ultimately, add a constant c to the language and also allow phi_n be the declaration c > n, or even more suitably, c > SSSS.S (0 ), where S is the follower function of and also there are n S's in the expression.

Currently, take into consideration the set of axioms union T union phi_n for every single n. By a typical density argument, this set of axioms corresponds, therefore by the Godel efficiency theory, there is a version M of all the axioms all at once. By descending Lowenheim - Skolem, we can think M is countable.

Currently, ask on your own, what can c be? Well, given that M pleases phi_n for every single n, we have to have c > n for each and every "common" all-natural number in M, i.e., c is boundless!

What's actually trendy is that by experimenting with various sort of phi_n, one can locate versions that have components which are divisible by NO "typical" prime number. One can additionally locate versions that have components which are all at once divisible by EVERY "typical" prime number!

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2019-05-08 21:33:59
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Take any kind of full concept $T$ that has 2 versions of of various cardinalities. After that the versions are primary matching (they both version $T$) yet they can not be isomorphic due to the fact that isomorphisms maintain cardinality.

By the Lowenheim - Skolem theory, every concept with boundless versions has versions of various cardinalities, so actually any kind of full concept with boundless versions will certainly benefit $T$.

Individual note : Whenever I educate pupils concerning cardinality, I constantly mention this type of application. Due to the fact that cardinality is maintained by bijections, revealing 2 things (teams, rings, versions, etc) have various cardinalities is a very easy means to show they are not isomorphic. I locate this a really engaging factor to respect cardinality beyond set concept.

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2019-05-08 20:44:35
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Laws, yes [1 ] : if there weren't such instances, there would not be a subject called model theory and also consequently Marker's publication would certainly not exist.

There are, nonetheless, not much specific instances which can be offered, with evidence, right after the definition of primary equivalence, an instructional trouble that I encounered lately when I educated a brief summer season training course on model theory. When I first offered the definition, all I had the ability to think of was the adhering to argument : for any kind of language $L$, the class of $L$ - frameworks is a correct class [any kind of nonempty set can be made right into the "cosmos", or underlying set of, an $L$ - framework ] whereas given that there go to the majority of xx_math_3 various concepts in the language $L$, the $L$ - frameworks approximately primary equivalence have to create a set.

Yet if you simply desire instances without evidence, sure, below goes :

1) In the vacant language, 2 collections are elementarily equal iff they are both boundless or both limited of the very same cardinality.

2) Any 2 thick straight orders without endpoints are elementarily equal. [The very same holds for DLOs with endpoints, yet both courses of frameworks are not elementarily equal per various other. ]

3) Any 2 algebraically shut areas of the very same feature are elementarily equal.

4) Any 2 actual - shut areas are elementarily equal.

5) Any 2 boundless versions of the concept of limited areas are elementarily equal. (Nope! See Dave Marker's solution.)

In each instance, such frameworks exist of every boundless cardinality, so the class of isomorphism courses of such frameworks is a correct class, not a set.

[1 ] : "M - O - O - N, that spells model theory!"

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2019-05-08 20:41:06
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Take an appearance at the Wikipedia write-up on Real closed fields. Quickly, they are areas that are first - order matching to the area of the actual numbers. An instance is the area of actual numbers that are origins of polynomials with sensible coefficients. That's virtually all I find out about them, however.

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2019-05-08 20:22:41
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