Question about the proof of this lemma: If $\alpha$, $\beta$ are ordinals, then either $\alpha \subset \beta$ or $\beta \subset \alpha.$

Question

Proof: Clearly $\alpha \cap \beta$ is an ordinal, $\alpha \cap \beta = \gamma.$ Then $\gamma = \alpha$ or $\gamma = \beta$. For, if not, then $\gamma \not= \alpha$ and $\gamma \not= \beta$. Then $\gamma \in \alpha$ and $\gamma \in \beta$ implying $\gamma \in \gamma$ which contradicts the definition of an ordinal (namely that set inclusion is a strict ordering of $\alpha$).

My question is that why does $\gamma \in \alpha$ and $\gamma \in \beta$ implies $\gamma \in \gamma$?

The lemma in question is #2.11 in Thomas Jech's Set Theory.

Answer 1

Note that $\gamma = \alpha \cap \beta$ by definition. Hence $$ \forall x \qquad x \in \gamma \iff x\in \alpha \land x \in \beta $$ Now apply this to $x = \gamma$.