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I am trying to get a better understanding of the Löwenheim-Skolem Theorem, and so I'm working through the Tarski/Vaught paper "Arithmetical Extensions of Relational Systems."

The paper can be found here: https://eudml.org/doc/88848 .

The problem: Let a system $\langle A,R\rangle$ consist of objects $A$ together with relations $R$. For corollary 1.7, they state: If system $\mathfrak{G}=\langle B,S\rangle$ is an elementary extension of system $\mathfrak{R}=\langle A,R\rangle$, then $\mathfrak{G}$ is elementary equivalent to $\mathfrak{R}$.

To me, the above is stated as a general case. However, according to Wikipedia, because $\mathfrak{G}$ is an elementary extension it is only elementary equivalent to $\mathfrak{R}$ if we restrict ourselves to the objects of $\mathfrak{R}$, namely $A$. Therefore the general statement above of Tarski/Vaught would not be true.

Here is the quote from Wikipedia: "If $N$ is a substructure of $M$, then both $N$ and $M$ can be interpreted as structures in the signature $\sigma_N$ consisting of $\sigma$ together with a new constant symbol for every element of $N$. $N$ is an elementary substructure of $M$ if and only if $N$ is a substructure of $M$ and $N$ and $M$ are elementarily equivalent as $\sigma_N$-structures."

There is a similar statement about the non-generality of the above corollary 1.7 on model-theory wiki here: http://modeltheory.wikia.com/wiki/Elementary_extension .

Am I missing something... is this statement of the great Tarski... wrong?!?!?!?!

2 Answers 2

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The condition of being an elementary substructure is a strengthening of the condition of being elementarily equivalent, not a restriction as you claim. We are not "restricted" to elements of $N$; rather, we are expanding the requirement of elementarily equivalence to allow ourselves to refer to elements of $N$.

To be more precise, to say that $M$ and $N$ are elementarily equivalent means that if $\varphi$ is any sentence, then $M\models\varphi$ iff $N\models\varphi$. Note that here $\varphi$ is restricted to be a sentence, so it cannot refer to individual elements of $M$ or $N$: all its variables must be quantified. On the other hand, to say that $N$ is an elementary submodel of $M$ means that if $\varphi(x_1,\dots,x_n)$ is any formula whose free variables are $x_1,\dots,x_n$ and $a_1,\dots,a_n\in N$, then $M\models\varphi(a_1,\dots,a_n)$ iff $N\models\varphi(a_1,\dots,a_n)$. This includes the case that $\varphi$ is a sentence (when $n=0$), but also allows $\varphi$ to have free variables that we interpret as specific elements of $N$. So this condition is stronger than elementary equivalence.

(Instead of allowing free variables for which you substitute elements of $N$, you can instead add constant symbols to your language that refer to each element of $N$, and use those constant symbols in place of the free variables. This is what the quote from Wikipedia is talking about.)

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First: I want to point out that the statement by Tarski and Vaught should be uncontroversial. It follows immediately from the definitions of elementary substructure and elementary equivalence (in fact, the proof given in their paper is a one-line reference to the definitions).

Let's say $N$ is an elementary substructure of $M$, written $N\preceq M$ (forgive my use of more modern notation). This means that $N$ and $M$ are $L$-structures, $N\subseteq M$, and for any $L$-formula $\varphi(\overline{x})$ with free variables from $\overline{x}$ and any tuple $\overline{a}$ from $N$ of the same length as $\overline{x}$, we have $N\models\varphi(\overline{a})$ if and only if $M\models\varphi(\overline{a})$.

When we restrict attention to $L$-sentences $\varphi$ ($L$-formulas with no free variables), we have $N\models\varphi$ if and only if $M\models\varphi$. And this is the definition of elementary equivalence.


Second: The quote from Wikipedia does not in any way contradict the Tarski-Vaught statement. Indeed, it strengthens it!

In the context of the Wikipedia quote, we have $L$-structures $N\subseteq M$. We expand the language $L$ to a new language $L_N$ by adding one constant symbol naming each element of $N$. Then $N$ and $M$ are $L_N$-structures in an obvious way.

Now Wikipedia asserts that $N\preceq M$ if and only if $N\subseteq M$ and $M$ and $N$ are elementarily equivalent as $L_N$-structures. Taking just one direction of the biconditional, $N\preceq M$ implies that $M$ and $N$ are elementarily equivalent as $L_N$-structures. But this implies that $M$ and $N$ are elementarily equivalent as $L$-structures, since every $L$-sentence is an $L_N$-sentence.

So the assertion on Wikipedia easily implies the Tarski-Vaught assertion that if $N$ is an elementary substructure of $M$, then $N$ and $M$ are elementarily equivalent.

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    Okay, I think I'm beginning to see where I was wrong. My question, does $M$ have elements not in $N$? If so, why is there no conceivable sentence of the form $\exists v\varphi (v) $ such that an element of $M$ not in $N$ satisfies this sentence, thereby making $M$ and $N$ not elementary equivalent?2017-02-11
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    @PhilipWhite $M$ could have additional elements, but - if $M$ is an elementary extension of $N$ - they won't be *definable*. For example, consider the structures $N=(\mathbb{Q}, <)$ and $M=(\mathbb{R}, <)$. Clearly $M$ has more elements than $N$; but in the language of linear order alone, can you *describe* a single one of them?2017-02-11
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    @PhilipWhite: Because such a sentence is itself a formula whose free variables have been substituted for elements of $N$ (vacuously, since there are no free variables). So since $N\preceq M$, it is true in $M$ iff it is true in $N$.2017-02-11
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    Okay, so the truth value of sentences cannot change from one model to its elementary extension because we can't pick out any of the new members in the elementary extension? So an elementary extension is like an elementary equivalence for sentences, but when we are talking about free variables we are talking only about objects from the submodel? So the elementary equivalence of elementary extensions is almost trivial then. Is that right?2017-02-11
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    "Almost trivial"? I would say it's completely trivial. Again, all you have to do is read the definitions.2017-02-11
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    Okay, I think I'm so close to being confident with this information. I'm sorry to be a drag, but I have one last question. Does this mean that all extensions (not elementary extensions) are elementary equivalent? (Because they have the same relations, which are already satisfied in the subsystem).2017-02-11
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    Of course not. I think you're confusing interpreting relation symbols with satisfying sentences.2017-02-11
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    Reading original sources can be a great experience, but since you're still confused about the very basic definitions, I would recommend starting with the first few chapters of a modern textbook (e.g. A Shorter Model Theory by Hodges) before diving into the writings of Tarski.2017-02-11
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    Thanks, I was trying to read "Set Theory and the Continuum Hypothesis" by cohen but I read online that some of his proofs are just informal sketches. My idea was to supplement the cohen book with proofs elsewhere. I read through Godel's original paper on the completeness of the functional calculus just fine, and I read through skolem's original paper just fine -- but I thought to get the full lowenheim-skolem theorem i would just read the tarski paper. I'll check out your recommendation, thanks.2017-02-11
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    @PhilipWhite Some examples of / facts about extensions you might find instructive. (1) It's easy to show that if $\varphi(y_1, . . . , y_k)$ is an existential formula - of the form $\exists x_1, . . . , x_n\psi(x_1, . . . , x_n, y_1, . . . , y_k)$ for $\psi$ quantifier-free - then *any* extension is $\varphi$-elementary: if $M\subseteq N$, $m_1, . . . , m_k\in M$, and $M\models\varphi(m)$, then $N\models\varphi(m_1, . . . , m_k)$. But (2): generally *arbitrary* sentences aren't preserved: consider $N=(\mathbb{Z}, <)$ and $M$ the substructure of even numbers. Then $M\models\neg\exists x(22017-02-11
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    (cont'd) but clearly $N\models\exists x(2$x=3$, which is a *new element*. The point is that in the extension $M\subset N$, there *are* new elements which are definable; not every extension is elementary. – 2017-02-11