How to prove that a subset of an ordinal is an element of the ordinal by transitivity.

Question

I'm reading Jech's set theory, and in chapter two (page 19) in defining ordinals it states

a set is transitive if $\alpha \in \beta \implies \alpha \subset \beta $

it then defines an ordinal as a transitive set well ordered by $\alpha \in \beta \Leftrightarrow \alpha <\beta $

It then goes on to state that if $\alpha,\beta$ are ordinals and $\alpha \subset \beta$ such that $\alpha \not= \beta$ then $\alpha \in \beta$.

It proves this by saying take $\gamma$ to be the least element of $\beta - \alpha$ then $\alpha$ is the segment of $\beta$ given by $\gamma$, using transitivity.

I'm wondering how we can rule out that there's some $x \in \alpha$ such that $x > \gamma$, so that even though the set $s_\gamma=\{ y | y <\gamma \}$ is contained in $\alpha$ (by the fact that gamma is the least element of $\beta-\alpha$), perhaps $\alpha \not\subset s_\gamma$.

Answer 1

If we had that, we'd have γ ∈ x ∈ α. But α is an ordinal, hence transitive. So γ ∈ α, a contradiction.

Answer 2

Let $\gamma$ be the least element of the set $\beta - \alpha$. Then each $\delta ∈ \gamma$ is an element of the set $\alpha$. If not, $\delta$ is a element of the set $\beta - \alpha$ by transitivity of $\beta$, and is smaller than $\gamma$, a contradiction.

Conversely, each $\delta ∈ \alpha$ is an element of $\gamma$. If not, $\gamma$ is an element of an element of $\alpha$ by well-ordering, and $\gamma ∈ \alpha$ by transitivity of $\alpha$, a contradiction.

So, $\alpha = \gamma ∈ \beta$.