# Question on a proof of elementarily equivalent but not isomorphic

Recall that the theory of $\mathfrak{A},$ written as $Th\, \mathfrak{A},$ is the set of all sentences true in $\mathfrak{A}.$

In Enderton's A Mathematical Introduction to Logic, page $152,$ the author provided the following example.

Example: Consider the structure $$\mathfrak{R} = (\mathbb{N};0,S,<,+,\cdot).$$ We claim that there is a countable language $\mathfrak{R}_0,$ elementarily equivalent to $\mathfrak{R}$ (so that $\mathfrak{R}_0$ and $\mathfrak{R}$ satisfy exactly the same sentences) but not isomorphic to $\mathfrak{R}.$

Proof: We will construct $\mathfrak{R}_0$ by using the compactness theorem. Expand the language by adding a new constant symbol $c.$ Let $$\Sigma=\{0<c,S0<c,SS0<c,...\}.$$ We claim that $\Sigma\cup Th \,\mathfrak{R}$ has a model. For consider a finite subset. That finite subset is true in $$\mathfrak{R}_k=\{\mathbb{N};0,S,<,+,\cdot,k\}$$ where $k=c^{\mathfrak{R}_k}$ for some large $k.$ So by the compactness theorem $\Sigma\cup Th\, \mathfrak{R}$ has a model.

By the Lowenheim-Skolem Theorem, $\Sigma\cup Th\,\mathfrak{R}$ has a coutable model $$\mathfrak{M} = (|\mathfrak{M}|; 0^{\mathfrak{M}},S^{\mathfrak{M}}, <^{\mathfrak{M}}, +^{\mathfrak{M}}, \cdot^{\mathfrak{M}}, c^{\mathfrak{M}}).$$ Let $\mathfrak{R}_0$ be the restriction of $\mathfrak{M}$ to the original language: $$\mathfrak{R}_0 = (|\mathfrak{M}|;0^{\mathfrak{M}},S^{\mathfrak{M}},<^{\mathfrak{M}},+^{\mathfrak{M}},\cdot^{\mathfrak{M}}).$$ Since $\mathfrak{R}_0$ is a model of $Th\, \mathfrak{R},$ we have $\mathfrak{R}_0\equiv \mathfrak{R}.$ We leave it to the reader to verify that $\mathfrak{R}_0$ is not isomorphic to $\mathfrak{R}.$

I have two questions regarding the proof above.

Questions:

$(1):$ Given a finite subset of $\Sigma\cup Th\, \mathfrak{R},$ why is it true in $$\mathfrak{R}_k=\{\mathbb{N};0,S,<,+,\cdot,k\}?$$ $(2):$ How to show that $\mathfrak{R}_0$ is not isomorphic to $\mathfrak{R}?$

• For large enough $k$, $k> S...S0$ for all $S...S0$ appearing in the finite fragment in question. As for the isomorphism, they don't satisfy the same infinitary sentences; if you want a proof not using those, simply note that a morphism sends $0$to $0$, $S0$ to $S0$ and so on, so that $c$ is not in the image of said morphism – Max Mar 2 '18 at 12:33
• @Max: I get your first sentence on why the finite subset in question is true. But I do not quite get your second part. Perhaps you can write an answer? – Idonknow Mar 2 '18 at 12:49
• Here's an outline: Suppose there is an isomorphism $f$. Then $c = f(SS....S0)$ for [the interpretation of] some term $SS....SS0$. But the non-standard structure satisfies $SS...SS0<c$ and ($x<y$ implies $x<> y$) – Ned Mar 2 '18 at 14:12

(1) as explained in the comments, for a finite segment of this theory, there are only finitely many axioms of the form $c> S...S0$. Calling "$k$th axiom" the axiom $c> S...S0$ with $k$ $S$'s, taking a large enough $k$, bigger than all $j$'s such that the $j$th axiom appears in the finite segment, you can interpret $c$ as $k+1$, and then your structure will be a model of the given finite segment.
(2) Let $f : \mathfrak{R} \to \mathfrak{R}_0$ be any morphism. Then by definition, $f(S...S0) = S...S0$ (simple induction). But $\mathfrak{R}$ has an underlying set composed only of the $S...S0$'s for finitely many $S$'s. In particular, since $x<y \implies x\neq y$, for all $n\in \mathbb{N}$, $f(n) < c$ and so $f(n) \neq c$: $c$ is not in the image and so $f$ is not surjective, hence not an isomorphism
• Since $\mathfrak{R}_0$ is a restriction of $\mathfrak{M}$ to the original language, why is $c$ still in $\mathfrak{R}_0?$ Because in its universe, I do not see $c^{\mathfrak{M}}.$ – Idonknow Mar 3 '18 at 13:45
• It does not matter that the language doesn't know $c$ exists anymore, the element $c^{\mathfrak{M}}$ is a fixed element $|\mathfrak{M}| = |\mathfrak{R}_0|$ – Max Mar 3 '18 at 15:28
• I just used $c$ as an abbreviation, I should have written $c^{\mathfrak{M}}$ instead – Max Mar 3 '18 at 15:29