Why do we define the set of natural numbers as follows? 
Could someone explain why we define the set of natural numbers as follows:
Let $X=\{J\in \mathcal{P}(I) : (\emptyset\in J)\wedge(\forall x\in J,x \;\cup\{x\}\in J)\},$ where $I$ is the set guaranteed by the axiom of infinity. Then $X$ is a set by the power set axiom, axiom of infinity, and axiom of separation. Also, $\mathbb{N}\cup\{0\}=\bigcap_{x\in X} x$ is a set by separation (or by pairing, union, and separation) and so $(\mathbb{N}\cup \{0\})\backslash\{0\}$ is a set by separation.

Is the following incorrect:
Let $X=\{J\in I : (\emptyset\in J)\wedge(\forall x\in J,x \;\cup\{x\}\in J\},$ where $I$ is the set defined by the axiom of infinity. Then $X$ is a set by the axiom of infinity, and axiom of separation. Also, $\mathbb{N}\cup\{0\}=\bigcap_{x\in X} x$ is a set by separation (or by pairing, union, and separation) and so $(\mathbb{N}\cup \{0\})\backslash\{0\}$ is a set by separation.

Do we have to use $\mathbb{N}\cup \{0\}$ as we do not yet know that $\mathbb{N}$ is a set? Also, why do we have to use the power set of the infinite set $I$?
Thanks for your insight!

 A: We want $\Bbb N_0$ to be inductive, i.e., contain the number $0$ (i.e., the set $\emptyset$) and be closed under the successor operation, i.e., if $n\in\Bbb N_0$ then also $S(n)\in\Bbb N$, i.e., if $x\in \Bbb N_0$ then also $x\cup\{x\}\in \Bbb N_0$.
The set $I$ that is guaranteed to exist by Inf (but is not unique) has this properties. What disturbs us is that it may be too big - we want the smallest inductive set.
What follows is a standard construction: If $\Phi$ is a property such that the intersection of arbitrarily many (but at least one) sets with property $\Phi$ again has property $\Phi$, then we can consider the intersection of all subsets of some $U$ that have property $\Phi$
$$\tag1X_0:=\bigcap\{\,X\subseteq U\mid \Phi(X)\,\}.$$
(You can rewrite this in terms of power set and separation etc. so that the axioms used in the construction become clear).
The construction in $(1)$ is defined only if there exists at least one $X\subseteq U$ with $\Phi(X)$, of course. But in that case, the $X_0$ defined this way is clearly the smallest subset if $U$ with property $\Phi$: By the assumption that $\Phi$ is inherited by intersections, we see that $\Phi(X_0)$ holds. And by the definition of $\bigcap$, we have $X_0\subseteq X$ for all $X\subseteq U$ with property $\Phi$.
This is what has done in your textbook (though for some reason they seem to want to expell $0$ from $\Bbb N$) with $\Phi$ standing for the property of being inductive, i.e., 
$$\Phi(x)\equiv \emptyset\in X\land \forall y\colon y\in x\to y\cup\{y\}\in x .$$
What is needed in the end is not only that $\Bbb N$ is a set by virtue of us only applying suitable axioms - instead, we want $\Bbb N$ to be inductive. This is clear from the above construction as soon as you realize that the arbitrary intersection of inductive sets is indeed inductive: If in $(1)$, $\emptyset\in X$ for all $X$ on the right hand side, then also $\emptyset\in X_0$. And if $x\in X_0$, then $x\in X$ for all $X$ on the right, hence $x\cup\{x\}\in X$ for all $X$ on the right, hence $x\cup\{x\}\in X_0$.

You idea does produce a set - but not the set we want and in particular, the result depends on the $I$ we start from. In particular if $I$ is already the smallest inductive set, then none of its elements is inductive, hence your $X$ would be the empty set (and therey, $\bigcap_{x\in X}x$ isn't even defined). With different $I$ your mileage may vary: If $I$ is a larger infinite ordinal, then your $X$ is the set of all infinite elements of $I$ and $\bigcap X$ happens to be what we want. However, with yet other $I$ (think something like $\omega\cup\{a,a\cup\{a\},a\cup\{a\}\cup\{a\cup\{a\}\},\ldots\}$ with some arbitrary $a$ with $\emptyset\in a$), the result is again something totally different.
A: The point is that the set of natural numbers is the smallest inductive set (plus/minus a $0$, depending on your preference).
The axiom of infinity provides us with some inductive set, but not necessarily with the smallest. The idea is that intersections of inductive sets is again inductive, so once we have an inductive set, here $I$, if $J$ is any other inductive set, then $I\cap J$ is an inductive subset of $I$.
Therefore it is enough to intersect all the inductive subsets of $I$ itself in order to have the smallest one. And this is why we use $\mathcal P(I)$, and not $I$ itself.
