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I'm trying to solve the exercise in the title and I think it makes no sense. Here's what it says:

An onto map $f: X \to Y$ is called an elementary partial isomorphism between $\mathcal{A}$ and $\mathcal{B}$ if $X \subseteq A$, $Y \subseteq B$ and $$(\mathcal{A},\{x\}_{x\in X}) \equiv (\mathcal{B},\{f(x)\}_{x\in X})$$ $f$ is immediately extensible if for each $a\in A$ (respectively $b\in B$) there is a $b\in B$ (respectively $a\in A$) such that $$(\mathcal{A},\{x\}_{x\in X},a) \equiv (\mathcal{B},\{f(x)\}_{x\in X},b)$$ Suppose every finite, elementary partial isomorphism between $\mathcal{A}$ and $\mathcal{B}$ is immediately extensible. Show $\mathcal{A} \equiv\mathcal{B}$.

From what I can see there are two cases:

  • either there exists a finite, elementary partial isomorphism between $\mathcal{A}$ and $\mathcal{B}$, in which case $\mathcal{A} \equiv\mathcal{B}$ simply by removing the additional constants;
  • or there isn't, so by taking $X=\emptyset$, we obtain $\mathcal{A} \not\equiv\mathcal{B}$.

In either case, the extensibility condition plays no role. Am I getting it wrong?

Also, from what I've looked up, this seems related to Ehrenfeucht–Fraïssé games, but I'm not experienced enough to translate the statement back.

Any help? Thanks.

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For the first question, yes, the existence of an elementary (partial) function implies elementary equivalence.

For the second question, yes, this is related to Ehrenfeucht-Fraisse games: the Ehrenfeucht's theorem says that if $\mathcal A,\mathcal B$ are relational structures (of the same signature), then $\mathcal A\equiv \mathcal B$ if and only if every partial isomorphism (not elementary!) is immediately extensible in the analogous sense.

It seems to me like you might have misread the question: if the two structures are countable, then the condition you cited implies not just that $\mathcal A \equiv \mathcal B$, but that actually $\mathcal A\cong \mathcal B$, which is a much stronger condition. Are you sure this isn't what you're supposed to prove here?

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  • $\begingroup$ Thanks! Probably, Sacks mixed up his partial isomorphisms. (I have copied the question verbatim.) $\endgroup$ – Andrei Sipoș Nov 6 '14 at 20:27
  • $\begingroup$ @AndreiSipoș: It might also be some funny convention. Bear in mind that the book is quite old. $\endgroup$ – tomasz Nov 6 '14 at 20:43

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