Axiom of regularity and Axiom of infinity Axiom of Infinity states the existence of the set:
$N=\{$
$\emptyset ,$
$\{ \emptyset \},$
$\{\{ \emptyset \}, \{\{ \emptyset \}\}\},$
$\{\{\{ \emptyset \},\{\{ \emptyset \}\}\}, \{\{\{ \emptyset \}, \{\{ \emptyset \}\}\}\}\},$
$ ... \}$
indeed they are the natural numbers.
Also, Axiom of regularity states that for every non-empty set x, x contains a member y that is disjoint from x.
My question is, what is the y for $N$?

The relevant $y$ is $\emptyset$.
We have $\emptyset\in \mathbb{N}$ - it's explicitly included - but trivially $\emptyset\cap\mathbb{N}=\emptyset$. Indeed, more generally we have:

$(*)\quad$ If $\emptyset \in A$, then $\emptyset$ is an element of $A$ disjoint from $A$ as required by Regularity.

Something like $\{\emptyset\}$ would not work: we have $$\{\emptyset\}\cap\mathbb{N}=\{\emptyset\}\not=\emptyset$$ since $\emptyset\in\mathbb{N}$. Indeed, $\emptyset$ is the only choice of $y$ which works.

Incidentally, it's worth meditating on $(*)$ for a bit.
Suppose $A$ were a counterexample to regularity. Well, then by $(*)$ we must have $\emptyset\not\in A$ (otherwise setting $y=\emptyset$ would show that $A$ isn't a counterexample to regularity).
OK, but then we have to have $\{\emptyset\}\not\in A$ either: if $\{\emptyset\}\in A$, then we would have $\{\emptyset\}\cap A=\emptyset$ since by the previous sentence we know that $\emptyset\not\in A$.
Similarly, this then tells us that $\{\emptyset,\{\emptyset\}\}\not\in A$. And so on and so forth: we can keep "bootstrapping up" to show that $A$ can't have any elements which "can be built up from $\emptyset$."
And that's exactly what regularity is about: in a precise sense, regularity says "every set can be built up from the emptyset."
(That precise sense is the following. By Replacement, we can iterate Powerset through the ordinals: we get for each ordinal $\alpha$ a set $V_\alpha$ such that $V_0=\emptyset$, $V_\lambda=\bigcup_{\alpha<\lambda}V_\alpha$ for $\lambda$ limit, and $V_{\alpha+1}=\mathcal{P}(V_\alpha)$. Over ZF-Regularity, the axiom of regularity is equivalent to "Every set is in some $V_\alpha$.")
