If $\alpha\ne\beta$ are ordinals and $\alpha\subset\beta$ show $\alpha\in\beta$

Solution 1:

There are two things here.

1. An ordinal $\xi$ is equal to the set $\{\zeta\mid \zeta<\xi\}$. Exactly because the ordering of ordinals is given by $\in$.

2. Since $\gamma$ is an element of $\beta$, and $\alpha=\gamma$ it follows that $\alpha$ is an element of $\beta$.

Solution 2:

By definition, the relation $<$ on $\beta$ is $\in$. That is, for $x,y\in\beta$, $x<y$ is defined to mean $x\in y$. So we have $$\alpha=\{\xi\in\beta:\xi<\gamma\}=\{\xi\in\beta:\xi\in\gamma\}=\gamma\cap\beta.$$ But since $\beta$ is transitive and $\gamma\in\beta$, $\gamma\subseteq\beta$, so $\gamma\cap\beta=\gamma$.

Since $\alpha=\gamma$ and by definition $\gamma$ is an element of $\beta$ (namely, the least element of $\beta-\alpha$), we conclude that $\alpha\in\beta$.