Prove that for any ordinals $\gamma,\beta$, $\gamma\le \gamma+\beta$.

Question

I have tried using induction, but have been told that there are quite a lot of things which I have not considered. So now I am confused and asking for a formal proof. I am quite new to transfinite induction. Could someone please show me explicitly how should I deal with that? Thanks!

Answer 1

Transfinite induction on $\beta$.

For $\beta=0$, we trivially have $\gamma \leq \gamma = \gamma+0$.

For $\beta = \alpha+1$, we have $\gamma \leq \gamma + \alpha$ by our induction hypothesis, and then $\gamma \leq (\gamma+\alpha)+1 = \gamma+(\alpha+1) = \gamma+\beta$.

For $\beta$ a non-zero limit ordinal, we have $\gamma \leq \gamma+\alpha$ for all $\alpha<\beta$ by our induction hypothesis. Thus, $\gamma \leq \sup\{\gamma+\alpha\mid \alpha < \beta\}=\gamma+\beta$.