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On page 89 in A Friendly Introduction to Mathematical Logic, the author writes that the standard model $\mathfrak{N}$ for $\mathcal{L}_{NT}$ is elementarily equivalent to a model $\mathfrak{A}$ that has an element of the universe $c$ that is larger than all other numbers.

I'm new to mathematical logic, but I understand that elementarily equivalent means the two structures have the same set of true sentences. However, it seems to me that the following sentence is true in $\mathfrak{A}$ but not in $\mathfrak{N}$. What am I missing?

$\exists x\ \forall y\ (x=y \vee y<x)$

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    $\begingroup$ The key phrase is "all other numbers," which is hiding an implicit mistaken interpretation. As Carl's answer says, $\mathfrak{A}$ contains an element $c$ which is bigger than all standard numbers (that is, all "truly finite" elements; or if you prefer, all elements in the image of the unique homomorphism $\mathfrak{N}\rightarrow\mathfrak{A}$), but this does not mean that $c$ is bigger than all elements of $\mathfrak{A}$. $\endgroup$ – Noah Schweber Sep 27 '18 at 15:41
  • $\begingroup$ Must the universe of $\mathfrak{A}$ contain other nonstandard numbers? $\endgroup$ – Katie Johnson Sep 27 '18 at 19:26
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    $\begingroup$ Yup, lots. Any nonstandard element has a successor, after all, which must be nonstandard. And a predecessor, and a square, and a ... All the arithmetic structure of $\mathfrak{N}$ exists in $\mathfrak{A}$ as well, even for the nonstandard elements, since $\mathfrak{N}\equiv\mathfrak{A}$. $\endgroup$ – Noah Schweber Sep 27 '18 at 19:28
  • $\begingroup$ Yes, of course. Thank you!! :) $\endgroup$ – Katie Johnson Sep 27 '18 at 19:38
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In the notes, I don't see the claim that $c$ is larger than all other numbers of $\mathfrak{A}$. The number $c$ in $\mathfrak{A}$ is larger than $0$, $S(0)$, $S(S(0))$, etc., - so $c$ is greater than every element of $\mathfrak{N}$. But there will be other elements of $\mathfrak{A}$ that are larger than $c$. Not every element of $\mathfrak{A}$ is of the form $S^n(0)$ for some $n \in \mathfrak{N}$.

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  • $\begingroup$ Thank you! I'm starting to understand. But can't we define $\mathfrak{A}$ so that the universe is exactly $\mathbb{N} \cup \{c\}$? I was imagining $\mathfrak{A}$ as the naturals with one extra nonstandard element thrown in, but it sounds like that's not the case. $\endgroup$ – Katie Johnson Sep 27 '18 at 19:24
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    $\begingroup$ @KatieJohnson No, you cannot do this. Remember, it's not enough to just build some structure; we also have to argue somehow that it's elementarily equivalent to $\mathfrak{N}$. This is a very strong condition, and we can't just handwave it away. $\endgroup$ – Noah Schweber Sep 27 '18 at 19:29
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    $\begingroup$ @Katie Johnson: In particular, we would need to have elements $S(c)$, $S(S(c))$, etc. Also, the original model satisfies "every element that is not zero is the successor of another element", so $c$ has to be the successor of some element, and that element has to be the successor of another element, etc. $\endgroup$ – Carl Mummert Sep 27 '18 at 20:34

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