Is this a correct definition of the natural numbers in ZF?

Question

Set $s$ is a natural number if $s$ is transitive and for every $x$, $y$ and $z$

• $y\in{s}\rightarrow(y$ is transitive$)$, and if
• $x\in{P}s\wedge(x$ is transitive$)\wedge{z}\in{P}x\wedge(z$ is transitive$)$ $\rightarrow{z}\in{x}\cup\{x\}$, and if
• $x\in{P}s\wedge(x$ is transitive$)\wedge\bigcup{x}\neq{x}$ $\rightarrow x=\bigcup{x}\cup\{\bigcup{x}\}$.

Then $\emptyset$ is the first natural number, the successor of natural number $s$ is defined by $s\cup\{s\}$ and the predecessor of number $s$ is defined by $\bigcup{s}$. All Peano axioms are straightforward proved, including the principle of mathematical induction. Also the natural number arithmetic is proved, for example if $a$ and $b$ are natural numbers then with induction is proved $a+b$ and $a\cdot{b}$ are natural numbers. The set $\omega$ of natural numbers is then postulated as $s\in\omega\leftrightarrow(s$ is a natural number).

Answer 1

This definition is not for a natural number, but rather for an ordinal. Let's see that $s=\omega$ satisfies this definition.

1. $\omega$ is transitive.
2. If $y\in\omega$ then $y$ is transitive.
3. If $x\subseteq\omega$ and transitive, and $z\subseteq x$ and transitive (in particular, both $x$ and $z$ are natural numbers), then $z\subseteq x\cup\{x\}$.
4. If $x$ is a transitive subset of $\omega$ and $\bigcup x\neq x$ then $x=\bigcup x\cup\{\bigcup x\}$. This is true for every ordinal, because if $x$ is not a successor then $\bigcup x=x$. So the implication holds vacuously.

In order to limit the definition to finite ordinals you need to require that every $x\subseteq s$ which is transitive and non-empty satisfies, $x\neq\bigcup x$. This means that every ordinal $\leq s$ is zero or a successor.