# Is the standard model for the language of number theory elementarily equivalent to one with a nonstandard element?

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

• 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 $\mathfrak{A}$. – Noah Schweber Sep 27 '18 at 15:41
• Must the universe of $\mathfrak{A}$ contain other nonstandard numbers? – Katie Johnson Sep 27 '18 at 19:26
• 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 $\mathfrak{N}\equiv\mathfrak{A}$. – Noah Schweber Sep 27 '18 at 19:28
• Yes, of course. Thank you!! :) – Katie Johnson Sep 27 '18 at 19:38

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 $$n \in \mathfrak{N}$$.

• 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. – Katie Johnson Sep 27 '18 at 19:24
• @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. – Noah Schweber Sep 27 '18 at 19:29
• @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. – Carl Mummert Sep 27 '18 at 20:34