Natural numbers in set theory is $\{0,1,2,...\}?$ The set of natural numbers $\mathbb{N}$ in set theory is defined with the axiom of infinity as the smallest inductive set and then it is usually proven that $\mathbb{N}$ satisfies the Peano axioms and well-ordering. However, I have yet to see a treatment where it is proven that $\mathbb{N}$ is equal to $\{0,1,2,...\}$, i.e. $\mathbb{N}$ contains the successors of 0 and nothing else.
I guess to answer my question, I would need to construct a model of set theory in which $\mathbb{N}$ contains $0$, $1$, $2$, etc. plus some extra junk and show that this model indeed works. Maybe someone has done this already? Or maybe it can be proven directly?
Edit: There seems to be some confusion in the comments and answers below. After some thought, I believe that what I want to show is that $\mathbb{N}$ does not contain an element apart from 0 that is not the successor of anything else in $\mathbb{N}$. I think I can prove this by contradiction with the axiom of specification to get an inductive set smaller than $\mathbb{N}$.

The Peano Axioms contain the Axiom of Induction, which says, basically, that the natural numbers contain only $0$ and its successors. At least, what I understand you to mean when you say "the successors of $0$", because, and Trevor pointed out, $0$ has only one successor. One formal way of stating this axiom is that if there is a set $S$ where $0\in S$, and for any natural number $n$, if $n$ is in $S$, then the successor of $n$ is in S, then the set $S$ contains every natural number. In other words, the statements "$0$ is a natural number" and "if $n$ is a natural number, then the successor of $n$ is a natural number," when combined with the other axioms about what the successor is and why $0$ is special, define all of the natural numbers.
So, if you show that $\Bbb{N}$ satisfies the Peano Axioms, then "$\Bbb{N}$ contains the successors of $0$ and nothing else." The Axiom of Induction is in the Peano Axioms to exlude that "extra junk".
See this question.
