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Specifically, I need to show there exists a structure $\mathcal{Z}'$ that is elementarily equivalent to $ \langle \mathbb{N} ; <; \cdot ; 0,1 \rangle $ (the standard model of arithmetic) but is not isomorphic to it.

(side question: is it 'elementary' or 'elementarily' equivalent, I've seen them both written, or do they mean different things?)

I've read that it's $\mathbb{N}$ with infinitely many densely ordered copies of $\mathbb{Z}$ after it, but I don't understand why. I only have very basic knowledge of Model Theory, so would appreciate as simple an argument as possible.

And then after, I need to show there exists a countable non-standard model of arithmetic, so any hints/tips would be great!

Eric Wofsey
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Naomi
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    It's "elementarily equivalent", and the relation is "elementary equivalence". – BrianO Nov 07 '15 at 18:11
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    You know the compactness theorem for first order logic, yes? – BrianO Nov 07 '15 at 18:12
  • Ah, that makes sense- thanks @BrianO – Naomi Nov 07 '15 at 18:13
  • Yes, I do @BrianO – Naomi Nov 07 '15 at 18:13
  • Extend the first-order language of arithmetic by adding an extra constant symbol $c$ (the idea being that this will denote a non-standard element of your non-standard model). Try to construct an extension of the theory of the standard model in your extended language with the property that every finite subset of your set of sentences is satisfiable, but any model of the complete set of sentences would have to be a nonstandard model. – Rupert Nov 07 '15 at 18:17
  • Figured you did ;) And do you know the (upward) Lowenheim-Skolem theorem? That will give you a large model of the theory of $(\mathbb{N},<,+,\cdot,0,1)$ (let's call this $\mathbb{N}$ too), which can't be isomorphic to it (too big). To get a nonstandard model, there's a standard trick of adding countably many sentences involving a new constant $c$, any finite subset of which is satisfiable in $\mathbb{N}$, but which can't all be satisfiable in $\mathbb{N}$ because they say that $c$ is bigger than every standard integer. By compactness, they have a model; downward L-S gives you a countable one. – BrianO Nov 07 '15 at 18:19
  • Once you have a countable nonstandard model, it's not hard to show that its order type has to be as you describe. – BrianO Nov 07 '15 at 18:26
  • Ah okay, not very familiar with Lowenheim-Skolem, but will brush up on it and then I think that all makes sense. Thanks! – Naomi Nov 07 '15 at 18:53
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    Possible duplicate of Non-standard models for Peano Axioms – Ross Millikan May 14 '17 at 16:02
  • @RossMillikan: To be elementarily equivalent to $\mathbb{N}$ is more than to satisfy the Peano axioms, so I'm happy with not treating this as a duplicate even though the title here mentions just "non-standard models". – hardmath May 15 '17 at 00:44

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