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I want to show that $(\alpha+\beta)+\gamma= \alpha+(\beta+\gamma)$.

I'm carrying out the proof using transfinite induction, I've been successful for the first two steps i.e- For the 0 case and the successor case. I'm struggling to prove associativity for the limit ordinal case.

Any help will be much appreciated, Thanks

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  • $\begingroup$ For a start, how would you $ define\;$ $a+b$ for ordinals $a,b$? $\endgroup$ – DanielWainfleet Apr 29 '18 at 8:10
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Hint: Proceed by induction on $\gamma$. Prove that for limit ordinals $\gamma$, $\beta + \gamma$ is always a limit ordinal. Now observe that, by our induction hypothesis,

$$ \begin{align*} (\alpha + \beta) + \gamma & = \sup \{ (\alpha + \beta) + \delta \mid \delta < \gamma \} \\ &= \sup \{ \alpha + (\beta + \delta) \mid \delta < \gamma \} \\ &= \alpha + \sup \{ \beta + \delta \mid \delta < \gamma \} \\ &= \alpha + (\beta + \gamma). \end{align*} $$

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  • $\begingroup$ Thank you very much, I have a question however. How do i go about proving that $\beta + \gamma$ is a limit ordinal? $\endgroup$ – Maths Apr 20 '18 at 14:10
  • $\begingroup$ @Maths Let $\epsilon < \beta+ \gamma = \sup \{ \beta + \delta \mid \delta < \gamma \}$. Then there is some $\delta < \gamma$ such that $\epsilon \le \beta + \delta$. But now $\epsilon +1 < \beta + (\delta + 2) \le \beta + \gamma$. So $\beta + \gamma$ is not a successor ordinal. $\endgroup$ – Stefan Mesken Apr 20 '18 at 14:13
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    $\begingroup$ @Maths If you look closely you'll find that I'm using associativity in my proof that $\beta + \gamma$ is a limit ordinal. Fortunately, I'm using this only strictly below $\gamma$ -- and we've verified that much by induction already. $\endgroup$ – Stefan Mesken Apr 20 '18 at 14:17

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