In my textbook, there is a theorem along with its proof as follows:
If $\alpha_1,\alpha_2,\beta$ are ordinals, then $\alpha_1<\alpha_2\iff \beta+\alpha_1<\beta+\alpha_2$.
Proof:
1. $\alpha_1<\alpha_2\implies \beta+\alpha_1<\beta+\alpha_2$
We prove this statement by induction on $\alpha_2$. Assume that $\forall \gamma<\alpha_2:\alpha_1<\gamma\implies\beta+\alpha_1<$ $\beta+\gamma$.
If $\alpha_2=\delta+1$, then $\alpha_1\le\delta$. By IH, we obtain $\beta+\alpha_1\le\beta+\delta<(\beta+\delta)+1=$ $\beta+(\delta+1)=\beta+\alpha_2$.
If $\alpha_2$ is a limit ordinal, then $\alpha_1+1<\alpha_2$ and $\beta+\alpha_1<(\beta+\alpha_1)+1=\beta+(\alpha_1+1)$ $\le \sup\{\beta+\delta\mid\delta<\alpha_2\}=\beta+\alpha_2$. Thus $\beta+\alpha_1<\beta+\alpha_2$.
2. $\beta+\alpha_1<\beta+\alpha_2\implies \alpha_1<\alpha_2$
If $\alpha_2<\alpha_1$, then $\beta+\alpha_2<\beta+\alpha_1$, which is a contradiction.
If $\alpha_2=\alpha_1$, then $\beta+\alpha_2=\beta+\alpha_1$, which is a contradiction.
Thus $\alpha_2>\alpha_1$
My problem:
I'm able to understand almost the whole proof, except for one point:
$\beta+(\alpha_1+1)\le \sup\{\beta+\delta\mid\delta<\alpha_2\}$
First, i don't understand how the authors derive that inequality.
Second, it seems weird to me that the authors even don't use the Inductive Hypothesis in this part, while the proof is by transfinite induction.
Any elaboration is greatly appreciated!