Let $\alpha ,\alpha^{'},\beta$ Be ordinals then prove that if $\alpha<\alpha^{'}\implies \alpha +\beta \le \alpha^{'}+\beta$

My attempt is to use induction on $\beta$, when $\beta=0$ then its trivial that $\alpha < \alpha^{'}$ by the problem we are trying to solve, my problem is how to proceed from here, for when its a successor and a limit ordinal, any help would be gratefully appreciated.

My definition of ordinal addition is as follows $\alpha +\beta= \alpha$ if $\beta=0$ it is equal to $S(\alpha+\gamma)$ if $\beta=S(\gamma)$ here S is referring to the successor, and is equal to $\sup_{\gamma<\beta}(\alpha+\gamma)$ if $\beta$ is a limit ordinal.

  • $\begingroup$ What are the definition of those inequalities? $\endgroup$ – Filburt Nov 6 '17 at 2:46
  • $\begingroup$ in You're counter example there both equal to $\omega$ note that the inequality sign is $\le$ ? which is satisfied as both sides are equal? & coincidently there is another question on this sheet that is the same as the one you proposed $\endgroup$ – user395952 Nov 6 '17 at 3:09
  • $\begingroup$ No problem, would you be able to assist me in solving the problem? $\endgroup$ – user395952 Nov 6 '17 at 3:21
  • $\begingroup$ @Gibberish: wrote out several proofs. Cheers. $\endgroup$ – Matematleta Nov 6 '17 at 4:33

I don't think you need induction for this. Note that $\alpha +\beta$ is, by definition, the unique ordinal $\gamma$ that is order isomorphic to $\alpha \sqcup \beta.$ So, if $\alpha<\alpha'$, then there is the obvious order-preserving embedding $\alpha \sqcup \beta\to \alpha' \sqcup \beta$. Now, since no ordinal can be order isomorphic to itself, it must be the case that $\alpha \sqcup \beta\le \alpha' \sqcup \beta.$

If you want to do it by induction, I think we can argue as follows:

If $\beta $ is a limit ordinal, we have, by definition, $\alpha + \beta = \text{sup}(\{\alpha + \delta \; |\; \delta < \beta\})$, and similarly for $\alpha' +\beta$. Then, with $\alpha<\alpha',$ the induction hypothesis says $\alpha + \delta \leq \alpha' + \delta$ for all $\delta < \beta$ and the claim follows on purely set-theoretic grounds.

Remark: It's easy to show that $\alpha\le \alpha'\Rightarrow \alpha+1\le \alpha'+1$

Now, suppose $\beta= \delta + 1.$ Then, using the the associativity law , the inductive hypothesis, and the remark, we compute $\alpha +\beta = \alpha +(\delta+1)=(\alpha +\delta)+1\le(\alpha'+\delta)+1= \alpha'+(\delta+1)=\alpha'+\beta.$

  • $\begingroup$ Is $\sqcup$ the union ? $\endgroup$ – user395952 Nov 6 '17 at 18:06
  • $\begingroup$ It's the disjoint union. More precisely, if we interpret ordinals $\alpha$ and $\beta$ as well-ordered sets, then $\alpha +\beta$ is the disjoint union with the lexicographical order $(r,s)<(x,y)⇔r<x$ or $r=x$ and $s<y$. i.e. we paste the two sets together so that all the elements of $\alpha$ are less than all the elements of $\beta$. $\endgroup$ – Matematleta Nov 6 '17 at 20:00

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