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Prove that $0+\beta=\beta$ where $\beta$ is an ordinal.

My attempt was using transfinite induction, and is as follows

We do the induction on $\beta$ so when $\beta=0$ we have that $0+0=0=\beta$ so the case when $\beta=0$ holds.

Now when $\beta$ is a successor ordinal we have that $\beta=S(\gamma)$ for some ordinal $\gamma$ so $0+\beta=S(0+\gamma)=S(\gamma) =\beta$ so the case when $\beta$ is a successor also holds by the definition.

Now when $\beta$ is a limit ordinal specifically $\beta>1$ so we have that $0+\beta=\sup_{\gamma<\beta}(0+\gamma)$ =$\sup_{\gamma<\beta}\gamma$ =$\beta$ hence the case when $\beta$ is a limit ordinal holds also thus by the principle of transfinite induction it holds that $0+\beta=\beta$. $\forall \beta $ ordinals .

I'm unsure if i've understood the principle of transfinite induction correctly, so if someone could clarify if my proof is correct that would be great.

Note: My definition of ordinal addition is that of
$\alpha+\beta = \begin{cases} \text{$\alpha$,} &\quad\text{if $\beta=0$}, \\ \text{$S(\alpha+\gamma)$,} &\quad\text{if $\beta=S(\gamma)$} \\ \text{$\sup_{\gamma<\beta}(\alpha+\gamma)$} &\quad\text{if $\beta$ is a limit ordinal}\\ \end{cases}$

Where $S(\gamma)$ is the successor of $\gamma$

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  • $\begingroup$ $S(\alpha + \beta)$ should read $S(\alpha + \gamma)$ in the second case of your definition of ordinal addition. In your proof, you should make the case when $\beta$ is a successor ordinal match the definition: i.e., you should write "when $\beta$ is a successor ordinal, we have that $\beta=S(\gamma)$ for some ordinal $\gamma$ so $0 + \beta = S(0 + \gamma) = \ldots$". $\endgroup$ – Rob Arthan Nov 6 '17 at 21:23
  • $\begingroup$ Thanks for that i will update my post now. $\endgroup$ – user395952 Nov 6 '17 at 21:28
  • $\begingroup$ Would you be able to check if there are any more errors in my post? $\endgroup$ – user395952 Nov 6 '17 at 21:40
  • $\begingroup$ Your proof is fine now apart from the minor detail that it doesn't add anything useful when you say $\beta > 1$ in the limit ordinal case. $\endgroup$ – Rob Arthan Nov 6 '17 at 21:42
  • $\begingroup$ But if it's equal to 0 is that not also a limit ordinal and we've had the case when it's equal to 0? $\endgroup$ – user395952 Nov 6 '17 at 22:38

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