Does $\mathbb{N}\cup\{\omega\}$ satisfy true arithmetic?

Question

Does $\mathbb{N}\cup\{\omega\}$ satisfy true arithmetic?(whereas $\omega$ is greater than any natural number) Even if not the standard model of natural number? I believe it does, but I am not quite sure because it leads me to that the successor of omega is the same as omega.

Answer 1

I assume that by true arithmetic, hence TA, you mean the set of sentences which are satisfied by the standard model $(\mathbb N,+,\cdot,<,0,1)$. That being said, your set is not even a model of PA:

Indeed, the sentence $\forall x (1+x\neq x)$ is provable in PA, and can't hold in your model, assuming addition involving $\omega$ is the same as ordinal addition.

Alternatively, if $M\models $PA extends $\mathbb N$ and $c\in M\setminus \mathbb N$, then $c+1\in M\setminus\mathbb N$. Then $\{c+n:n\in\mathbb N\}$is an infinite set contained in $M\setminus \mathbb N$, so there must always be infinitely many non-standard elements.