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I want to prove that $(\alpha+\beta)+\gamma=\alpha+(\beta+\gamma)$. now the method I first attempted this was by transfinite induction which is what i'm trying to get a better understanding of, The addition of ordinal numbers definition that I am using is $\alpha+\beta = \begin{cases} \text{$\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 $\gamma$

So my attempt at the proof is

Using induction on $\gamma$ we get that when $\gamma=0$ the result is trivial so suppose that $\gamma=\delta+1$ then we get that $(\alpha+\beta)+\gamma=(\alpha+\beta)+(\delta+1)=((\alpha+\beta)+\delta)+1$ $=(\alpha+(\beta+\delta))+1$ $=(\alpha+(\beta+\delta)+1)$ $=\alpha+(\beta+(\delta+1)$ $=\alpha+(\beta+\gamma)$ so the case when $\gamma$ is a successor is satisfied.

Now when $\gamma$ is a limit, specifically when $\gamma>1$. Then $\beta+\gamma$ is a limit and therefore so is $\alpha+(\beta+\gamma)$ and $(\alpha+\beta)+\gamma$ so we have that; $(\alpha+\beta)+\gamma=\sup_{\epsilon<\gamma}((\alpha+\beta)+\epsilon)$ $=\sup_{\epsilon+\beta<\beta+\gamma}((\alpha+\beta)+\epsilon)$ $=\sup_{\beta+\epsilon<\beta+\gamma}(\alpha+(\beta+\epsilon))$ $=\sup_{\delta<\beta+\gamma}(\alpha+\delta)$ $=\alpha+(\beta+\gamma)$ So it is satisfied when $\gamma$ is a limit ordinal, Thus $(\alpha+\beta)+\gamma=\alpha+(\beta+\gamma)$ for all ordinals $\alpha,\beta,\gamma$.

My question is how/why in the limiting case do we change the inequality $\sup_{\epsilon<\gamma}((\alpha+\beta)+\epsilon))$ to $\sup_{\epsilon+\beta<\beta+\gamma}((\alpha+\beta)+\epsilon))$. Any clarification would be great (I kind of guessed for the limiting step, and it turned out to be correct, but I don't understand it properly).

In response to the comment, If $\gamma$ is a limit then $\forall \alpha$ $\alpha+\gamma $ is a limit

proof $\gamma \ne 0$ so $\alpha+\gamma \geq \gamma >0$, i,e $\alpha+\gamma \ne 0$. so let $x\in \alpha+\gamma$. then show that $x+1<\alpha+\gamma$ , $x\in \alpha+\gamma =\bigcup_{\beta<\gamma}(\alpha+\beta)$, i.e there is $\beta < \gamma$ such that $x \in \alpha+\beta$. by a previous lemma, $x+1\leq \alpha+\beta$. if $x+1 \in \alpha+\beta, x+1<\alpha+\gamma$. so suppose $\alpha+\beta=x+1$. then since $\gamma$ is a limit , $\beta+1<\gamma$ and by definition $\alpha+(\beta+1)=(\alpha+\beta)+1$ and $x+1 \in (\alpha+\beta)+1$, hence $x+1 \in \alpha+\gamma$.

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    $\begingroup$ But you already did the successor step above (using $+1$ notation), didn't you? If $\gamma=S(\delta)$, then $(\alpha+\beta)+\gamma=S((\alpha+\beta)+\delta) = S(\alpha+(\beta+\delta))=\alpha+S(\beta+\delta)=\alpha+(\beta+S(\delta))=\alpha+(\beta+\gamma)$ $\endgroup$ – Hagen von Eitzen Nov 11 '17 at 20:50
  • $\begingroup$ i added my response to a previous comment. $\endgroup$ – user395952 Nov 11 '17 at 20:51
  • $\begingroup$ @HagenvonEitzen i did, but i was getting confused with the different notation. Thank you for explaining it using that specific notation. $\endgroup$ – user395952 Nov 11 '17 at 20:52
  • $\begingroup$ @HagenvonEitzen i edited my post to change the question i originally asked as that one was quite obvious looking back on it after your response, i was wondering if you could answer the new one? thanks!. $\endgroup$ – user395952 Nov 11 '17 at 21:08
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Perhaps a different perspective helps.

Given ordinals $\alpha,\beta$, we have that $\alpha+\beta$ is the minimal element of the class $$\mathcal S_{\alpha,\beta}:=\{\,\xi\in\operatorname{On}\mid \alpha\le \xi,\,\forall \eta<\beta\colon \alpha+\eta<\xi \,\}.$$ In particular, $\beta<\beta'$ implies $\alpha+\beta\notin \mathcal S_{\alpha,\beta'}$ and hence $\alpha+\beta<\alpha+\beta'$ (just in case this was not yet established), also $\alpha+\beta=\alpha+\beta'\iff \beta=\beta'$.

Now fix $\alpha,\beta$ and consider the class $$ \mathcal T=\{\,\xi\in\operatorname{On}\mid (\alpha+\beta)+\xi\ne \alpha+(\beta+\xi)\,\}.$$ Our goal is to show that $\mathcal T=\emptyset$. So assume otherwise and let $\gamma=\min\mathcal T$.

Clearly, $\beta+\gamma\ge \beta$ and hence $\alpha+(\beta+\gamma)\ge \alpha+\beta$. And for all $\eta<\gamma$, we have first $\beta+\eta<\beta+\gamma$ and then $(\alpha+\beta)+\eta=\alpha+(\beta+\eta)<\alpha+(\beta+\gamma)$. We conclude $\alpha+(\beta+\gamma)\in\mathcal S_{\alpha+\beta,\gamma}$ and hence $$ (\alpha+\beta)+\gamma\le \alpha+(\beta+\gamma).$$ Assume $ (\alpha+\beta)+\gamma< \alpha+(\beta+\gamma)$. As clearly $ (\alpha+\beta)+\gamma\ge\alpha$, this implies $$\exists\eta<\beta+\gamma\colon \alpha+\eta\ge (\alpha+\beta)+\gamma.$$ Note that $\alpha+\eta\ge (\alpha+\beta)+\gamma\ge\alpha+\beta$ makes $\eta\ge\beta$. Then $\eta<\beta+\gamma$ implies $\eta\notin \mathcal S_{\beta,\gamma}$, i.e., $$\exists \delta< \gamma\colon \beta+\delta\ge \eta.$$ But then $$(\alpha+\beta)+\gamma>(\alpha+\beta)+\delta=\alpha+(\beta+\delta)\ge\alpha+\eta\ge (\alpha+\beta)+\gamma, $$ contradiction.

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