Suppose $\alpha$ is a finite ordinal. Prove $\alpha\in\omega$

Question

I am reading the proof of this statement:

Below are theorems and statement mentioned in the proof: (i): Suppose $\beta$ is an infinite ordinal, then $\omega\leq \beta$ and $|\beta^\dagger| = |\beta|$

What I don't understand is why $\omega\subseteq n$ if $m\not\in \omega$. If $m\not\in\omega$, we have $\omega\subseteq m$ by 3.4.6. But $m\not\in n$, how is $\omega \in n$? And I have similar question for $n\subseteq m$. And I wonder why there is an injective function from $m^\dagger$ to $n$ especially we already have $n\subset m^\dagger$? Thank you

Answer 1

1. If $\omega\subseteq m$, let $x\in \omega\subseteq m$. Since $m\in\alpha$ and $\alpha$ is transitive, $x\in \alpha$. Since $x\in m$, it follows by minimality of $m$ that $x\not\in\alpha\setminus n$. Therefore $x\in n$. Hence every member of $\omega$ is a member of $n$ so $\omega\subseteq n$.
2. Since $m\in\alpha\setminus n$, $m\not\in n$. By the trichotomy, $n\subseteq m$.
3. Since $m\in\alpha$ and $\alpha$ is transitive, $m\subseteq\alpha$, and both together imply that $m^\dagger = m\cup\{m\}\subseteq \alpha$. We're assuming there is an injective function $h:\alpha\to n$, the the restriction of $h$ to $m^\dagger$ is an injective function of $m^\dagger$ to $n$.