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On wiki page on Fraïssé limit, it says that neither $⟨\Bbb{N}, < ⟩$ nor $⟨\Bbb{Z}, < ⟩$ are the Fraïssé limit of FCh (Fraïssé class) because although both of them are countable and have FCh as their age (the class of all finitely generated substructures), neither one is homogeneous.

Then it gives such examples as substructures $⟨ { 1 , 3 } , < ⟩$ and $⟨ { 5 , 6 } , < ⟩$, and the isomorphism $1 ↦ 5, 3 ↦ 6$ between them. It concludes that this cannot be extended to an automorphism of $⟨\Bbb{N}, < ⟩$ or $⟨\Bbb{Z}, < ⟩$, since there is no element to which we could map $2$, while still preserving the order.

I think the substructure $⟨ { 1 , 3 } , < ⟩$ and $⟨ { 5 , 6 } , < ⟩$ are sums of $a+3b$ and $5a+6b$. So the isomorphism is $\phi(a+3b)=5\phi(a)+6\phi(b)$. But if we take $a=-1, b=1$ and $a'=-2,b'=2$, then $\phi(2)=2$. So what does it mean really?

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    $\begingroup$ $\phi(a+3b) = 5\phi(a) + 6\phi(b)$ still doesn't give a well-defined function, since an integer can be written as $a + 3b$ in multiple ways... e.g. $5 = 2 + 3\cdot 1 = 5 + 3\cdot 0$. $\endgroup$ May 25, 2020 at 2:38

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The last paragraph makes no sense at all to me. How do you propose to read $\phi(a+3b) = 5a'+6b'$ as the definition of a function $\mathbb{Z}\to\mathbb{Z}$?

In any case, if $\phi\colon \mathbb{Z}\to \mathbb{Z}$ is a map with the property that $\phi(1) = 5$, $\phi(3) = 6$, and $\phi(2) = 2$, then $\phi$ is not an isomorphism, since $1 < 2$, but $\phi(1) = 5 > 2 = \phi(2)$, so $\phi$ does not preserve $<$.

In fact, for any integer $n$, if $\phi\colon \mathbb{Z}\to \mathbb{Z}$ is a map with the property that $\phi(1) = 5$, $\phi(3) = 6$, and $\phi(2) = n$, then $\phi$ is not an isomorphism. This is because $1 < 2 < 3$, so for $\phi$ to be an isomorphism, we'd have to have $5 = \phi(1) < \phi(2) < \phi(3) = 6$, but there is no integer strictly between $5$ and $6$.

Edit: Upon rereading, I've realized what you mean by "sums of $a + 3b$". In the structure $(\mathbb{Z};0,+)$, the substructure generated by $1$ and $3$ would be $\{a+3b\mid a,b\in \mathbb{N}\}$ (which is just $\mathbb{N}$). But note that $+$ is not in the language! The only symbol in the language is $<$, and the substructure generated by $1$ and $3$ is just $\{1,3\}$. The isomorphism $(\{1,3\},<)\to (\{5,6\},<)$ is exactly as stated: $1\mapsto 5$, $3\mapsto 6$.

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  • $\begingroup$ @hermes Correct. It has only two elements and it is isomorphic to $\{5,6\}$ and neither contains $2$. But the claim is that this isomorphism does not extend to an automorphism of $\mathbb{Z}$ (or $\mathbb{N}$) which does contain $2$! $\endgroup$ May 25, 2020 at 2:43
  • $\begingroup$ I gave you a proof in the 3rd paragraph of my answer... $\endgroup$ May 25, 2020 at 2:51
  • $\begingroup$ @hermes No, absolutely not. $\endgroup$ Aug 4, 2020 at 1:04
  • $\begingroup$ @hermes It's just the definition. "$\varphi$ is order preserving" means "if $x<y$, then $\varphi(x)<\varphi(y)$". $\endgroup$ Aug 4, 2020 at 3:41
  • $\begingroup$ Ok, order preserving is different from order type which causes this confusion. $\endgroup$
    – anon
    Aug 4, 2020 at 16:16

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