Proof verification: For any ordinals $\gamma,\beta$, $\gamma\le \gamma+\beta$.

Question

I need to prove that for any ordinals $\gamma,\beta$, $\gamma\le \gamma+\beta$.

Proof by induction:

Base case: when $\beta=0, \gamma+\beta=\gamma+0=\alpha$ Thus $\gamma=\gamma+ \beta\le \gamma+\beta$.

Step case: Assume that for an ordinal $\beta$ we have $\gamma\le \gamma+\beta$, then for $\beta^+$, we have:

$\gamma\le \gamma+\beta$

$< \gamma+\beta\cup \lbrace \gamma+\beta\rbrace$

$=(\gamma+\beta)^+$

$=\gamma+\beta^+ $

Hence by induction, for any ordinals $\gamma,\beta, \gamma\le \gamma+\beta.$

Could someone please tell me if I am correct? Thanks!

Answer 1

You don't want to do ordinary induction, you want to do transfinite induction. In that case, your induction step requires checking on successor ordinals (which you've done) and for limit ordinals (those $\beta$ which do not have any $\beta' < \beta$ such that $(\beta')^+ = \beta$ (in your notation).