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We define $ω$ to be the set of natural numbers, i.e. $ω=$ $\cap$ {$x$ | $0\in x \wedge \forall u\in x$ $ u+1\in x $}

Accordingly, I have managed to show that $ω \subset Ord$, where $Ord$ is the class of all ordinal numbers.

Since I was asked to prove that $\omega \in Ord$, it is only left to prove that $\omega$ itself is transitive (due to definition which states:

$x \in Ord$ iff x is transitive and every element of $x$ is transitive too.)

Does anyone have any idea of how to prove transitivity of $\omega$, where by transitivity I mean, if $x \in \omega$, then $x \subset \omega$...

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Under the von Neumann construction of the natural numbers, which is needed for this to be true with $\omega$ as defined, you take $0 = \varnothing$ and $n+1 = n \cup \{ n \}$ for all $n < \omega$. You can prove by induction on $n$ that, for all $x$, you have $x \in n$ if and only if $x \in \omega$ and $x < n$; the result then follows fairly easily.

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    Yes I have proven that $m\in n$ iff $m – 2017-02-17
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    I updated my answer to clarify something. The point is that the natural numbers are themselves sets of natural numbers by definition, and so $x \in \omega$ implies $x \subseteq \omega$ immediately. – 2017-02-17
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    I get it. Thanks a lot! – 2017-02-17