Why isn't the inductive set _the_ set of natural numbers? ZFC's axiom of infinity states:
$$\exists x (\varnothing \in x \wedge \forall y \in x (y\cup \left \{y \right \} \in x)) $$
Isn't this set $ x $ really $\mathbb{N}$? It wouldn't be $\mathbb{N}$ if x would contain some set $z $ that is different from $\varnothing $, $ S (\varnothing)$, $ S (S (\varnothing)) $, etc. But we can't really prove that such a set $ z $ is an element of $ x $, or can we? I also know that this inductive set $ x $ is not unique, but that, in the light of my question, doesn't make any sense to me either. I feel like I'm missing a very peculiar technical point, and I don't know what it is. Any help is appreciated. 
 A: Another set which satisfies the axiom is $N''=\mathbb N \cup N'$ where $N'$ is the set containing $x_0 = \mathbb N$ (a set which is not in $\mathbb N$) and all the sets $x_{n+1} := x_n \cup \{ x_n\}$. If you define $n+1 := n \cup \{n\}$, $0 := \{\}$ and $\omega := \mathbb N$ you have
$$
N'' = \{0, 1, 2, \dots, \omega, \omega+1, \omega+2,\dots\}.
$$
A: Write 
$$\phi(x):\Leftrightarrow (\emptyset \in x\land \forall y\in x\,(y\cup\{y\}\in x)). $$
Then the Axiom of Infinity states 
$$\tag1 \exists I\,\phi(I).$$
From such a set $I$ you can derive (using the other axioms) the existence of
$$\tag2 \omega := \bigcap\{\,x\in\mathcal P(I):\phi(x)\,\}$$
and then show that this $\omega$ is what you want (and does not depend on the choice of $I$). However, trying to pack the minimality construction of $(2)$ into the axiom makes it look much more complicated than $(1)$ - and unnecessarily so, as we can prove just what we need. 
A: Axiom INF tells you that some set $x$ exists with property
$\phi\left(x\right)$ as defined in the answer of Hagen von Eitzen.
Applying axiom SEP you find that  $\omega:=\left\{ y\in x\mid\forall I\left[\phi\left(I\right)\Rightarrow y\in I\right]\right\} $
is a set. 
Here $\omega=\left\{ 0,1,2,\dots\right\} $
