If $\alpha$ and $\beta$ are ordinals then $\alpha \in \beta \Leftrightarrow \alpha \subsetneq \beta$

Question

Def 1. $x$ is $\underline{transitive}$ if $\forall y \forall z (z \in y \in x \Rightarrow z \in x)$.

Def 2. $x$ is $\underline{ordinal}$ if $x$ is transitive and all elements of $x$ are transitive.

Def 3. $S(x) = x \cup \{x\}$.

If $\alpha$ and $\beta$ are ordinals then $\alpha \in \beta \Leftrightarrow \alpha \subsetneq \beta$.

1. $\Rightarrow$. By transitivity.

2. $\Leftarrow$. It seems that we have to take $\alpha$ and prove that $S(\alpha) \subset \beta$. $S(\alpha)$ is also ordinal if $\alpha$ was ordinal. If we can do that it will imply that ${\alpha} \in \beta$. But how to make it precisely?

Answer 1

I found a theorem that was proved on lectures.

Th. If $\alpha$ and $\beta$ are ordinals then either $\alpha \in \beta$, or $\beta \in \alpha$, or $\alpha = \beta$.

Okay, we have that $\alpha \subsetneq \beta$. Obviously that $\alpha \neq \beta$. Suppose that $\beta \in \alpha$. Then $\alpha \subsetneq \beta \in \alpha$. But this implies that $\beta \in \beta$. Contradiction with axiom of regularity! Hence, last possible variant is $\alpha \in \beta$.

Answer 2

Consider the definition of $S(\alpha) = \alpha \cup \{\alpha\}$. It suffices to show $S(\alpha)\subset \beta$. Take any $x\in S(\alpha)$. Either $x\in \alpha$ or $x=\alpha$. In the first case, $x\in \beta$, because $\alpha \subset \beta$ as $\beta$ is an ordinal. In the second case $x=\alpha$, so $x\in\beta$. Thus $S(\alpha) \subset \beta$.