compactness theorem - first order logic

Question

A basic question that keeps bugging me. Using compactness theorem, we can show that there exists a model of $Th(\mathbb N)$ [with the signature {<}] with an infinite element. But in $\mathbb N$ it is true that $\forall x \exists y (x<y)$, so am I to conclude that that model has more than one infinite element? Otherwise how can it be a model elementary equivalent to $\mathbb N$? The same question applys to infinitesimals elements in $\mathbb R$.

thanks,

Answer 1

You're quite right - every nonstandard model of $Th(\mathbb{N})$ has lots of infinite elements! For example, if $\mathcal{N}$ is a nonstandard model and $a\in\mathcal{N}$ is an infinite element, then:

• $a^2, a^3,...$ all exist in $\mathcal{N}$ and are infinite.

• A number $b$ such that either $2b=a$ or $2b+1=a$ exists (depending on the parity of $a$ in $\mathcal{N}$), and this $b$ is infinite.

• Similarly, a number $c$ such that $c^2<a$ but $(c+1)^2\ge a$ exists in $\mathcal{N}$, and is infinite.

And so on.

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In fact, the following is a good exercise:

Let $(\mathcal{N}, +, \times, <)$ be a nonstandard model of arithmetic. Then the underlying ordering of $\mathcal{N}$ - that is, $(\mathcal{N}, <)$ - looks like $\omega+\zeta\cdot \eta$; that is, we have an initial segment of the "correct" order type, and then infinitely many densely-packed parts of our model that are discrete. (Here "$\omega$" is the order-type of $\mathbb{N}$, "$\zeta$" is the order-type of $\mathbb{Z}$, and "$\eta$" is the order-type of $\mathbb{Q}$.)