I'm studying "Introduction to Set Theory" by Karel Hrbacek and Thomas Jech. In the proof that $\aleph_\alpha\cdot \aleph_\alpha=\aleph_\alpha$ for any ordinal $\alpha$, we want to prove $\aleph_\alpha\cdot \aleph_\alpha\leq \aleph_\alpha$ by transfinite induction.
We say that $(a,b)<(c,d)$ if
- $\max\{a,b\}<\max\{c,d\}$, or
- $\max\{a,b\}=\max\{c,d\}$ and $a<c$, or
- $\max\{a,b\}=\max\{c,d\}$ and $a=c$ and $b<d$.
We first argue that it suffices to show that for arbitrary $\alpha_1,\alpha_2\in\omega_\alpha,$ the set $$X=\{(\zeta_1,\zeta_2)\in\omega_\alpha\times \omega_\alpha : (\zeta_1,\zeta_2)<(\alpha_1,\alpha_2 )\}$$ has cardinality smaller than $\aleph_\alpha$.
To show this, we first set $\beta=\max\{\alpha_1,\alpha_2\}+1$. We can now observe that $\beta\in\omega_\alpha$ and $X\subseteq\beta\times \beta$.
Next, we choose $\gamma<\alpha$ for which $|\beta|\leq \aleph_\gamma$. This implies $|X|\leq |\beta|\cdot |\beta|\leq \aleph_\gamma\cdot \aleph_\gamma\leq \aleph_\gamma<\aleph_\alpha$, where the third inequality uses the induction hypothesis. This should finish the proof.
However, what guarantees that such $\gamma<\alpha$ exists at all?
$\omega_\alpha$ is an initial ordinal, so by definition $|\beta|<\aleph_\alpha$ for all $\beta<\omega_\alpha$. The cardinality of any well-ordered set is an $\aleph$, so $|\beta|=\aleph_\gamma$ for some ordinal $\gamma$, and since $|\beta|<\aleph_\alpha$, it must be the case that $\gamma<\alpha$.