I have seen alternative methods of this proof, with one being: let $n$ be the set of all natural numbers. Then (1) $\omega$ is an ordinal, (2) If $\alpha$ is an ordinal and $\beta \in \alpha$, then $\beta$ is an ordinal. Therefore since $n \in \omega$, it follows that $n$ is an ordinal. (I'm not sure if this is definitely correct but this is what I worked out)
I wanted to know how to prove this by induction instead. I have so far, using standard induction format, that I need to prove (i) the empty set is an ordinal, and (ii) if $n$ is an ordinal then $n$ $\cup$ $\{n\}$ is an ordinal.
I am not sure how to explicitly prove (i) and (ii) so any help would be greatly appreciated. Thank you.
There are several equivalent formulations of what it means to be an ordinal. For example:
So now to prove that $\omega$ is an ordinal you need to pick the definition that you want to use, and show that $\omega$ satisfies this definition. How to do so depends on the definition itself.
If you pick the third definition, for example, then you can prove by induction that indeed every $n$ is an ordinal, then prove that $\omega$ is a transitive set.
If you pick the first definition, then you need to prove that every two integers are comparable by $\in$ and that $\omega$ is transitive. This can be done by induction.
If you pick the second definition, then you need to prove that every finite $n$ is a transitive, then show that $\omega$ is transitive. Again, you can do this by induction.
So the point is that in order to prove that $\varnothing$ is an ordinal, etc. you need to pick a definition of an ordinal, and then show that it holds for $\varnothing$ (all of them hold pretty much vacuously), and then by induction prove that if $n$ is an ordinal, so is $n\cup\{n\}$. And this depends on the definition of an ordinal.