In ZFC set theory, we define the set of the natural numbers as follows: By the axiom of infinity, an inductive set exists. Let I be an inductive set. Then, $\mathbb{N}$ is defined as $\{ x\in I |\forall J \:(J\; is \;inductive \rightarrow x\in J)\}$ or in some books it is even defined as $\{ x|\forall J \:(J\; is \;inductive \rightarrow x\in J)\}$. It feels to me that this definition is recursive and therefore it is kind of a logical fallacy. Here is my argument:
Let us accept the first one as our definition of $\mathbb{N}$. Then, I am asking myself that “Is $1$ in $\mathbb{N}$?”. Then, I should check whether the sentence $\forall J \:(J\; is \;inductive \rightarrow 1\in J)$ is true or not. According to this sentence we should check whether $1\in J$ for every inductive set $J$. But we know that the set $\mathbb{N}$ is inductive. So, we should also check that whether $1\in \mathbb{N}$ or not. As a result, we get that “In order to know that whether $1\in \mathbb{N}$, we should know first whether $1\in \mathbb{N}$ or not.” This is kind of a contradictive. Isn’t there a self-reference here? Can anyone please what is the problem here, if there is any?
In ZFC, inductive(x) is roughly "0 is a member of x, and for all members y of x, the successor of $y$ is also a member of x".
The axiom of infinity says "there exists an inductive set". Let's call it I.
The definition stated above for $\mathbb{N}$ says roughly "$\mathbb{N}$ is a subset of $I$ which is also the intersection of all inductive sets".
$\mathbb{N}$ so defined is separated out of $I$ by the axiom of separation, because ZFC does not allow for defining arbitrary sets via formulas to avoid the Russell paradox.
It is possible to prove from the definitions that
I think what confuses you is that 2. is Peano's induction axiom - but notice that is not part of the definitions. It is a theorem of ZFC that $\mathbb{N}$ satisfies Peano's induction axiom. There is no recursion in the definition of $\mathbb{N}$. Just application of ZFC axioms.