compactness theorem - first order logic 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,
 A: 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.

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 $\mathbb{Q}$.)

