Prove that $\omega$ is not successor ordinal

Question

Prove that $\omega$ is not successor ordinal ?

I assumed that $\omega$ is successor ordinal, meaning there is ordinal $\beta$ such that $S(\beta) = \omega$ meaning $\beta \cup \{ \beta \} = \omega$

Resulting that $\beta \in \omega $ so $ \beta$ is finite, but how to proceed ?

Answer 1

Show $\beta$ is a maximum element of $\omega.$
Thus a contradiction.

Answer 2

First prove that $n\notin n$ for every $n\in\omega$.

For that define $b:=\{n\in\omega\mid n\notin n\}\subseteq\omega$ and prove by induction that $b=\omega$.

That comes to proving that $\varnothing\in b$ and $n\in b\implies n\cup\{n\}\in b$, allowing the conclusion that next to $b\subseteq\omega$ also $\omega\subseteq b$.

Now if $\omega=\beta\cup\{\beta\}$ then $\beta\in\omega$ and then also $\omega=\beta\cup\{\beta\}\in\omega$.

This because $\omega$ is closed under the successor operation.

But from $\omega\in\omega$ you can conclude now that $\omega\notin\omega$ and a contradiction is found.