# Prove that if $\alpha<\beta$ then $\gamma + \alpha < \gamma + \beta$ for ordinals.

I know the statement "if $$\alpha<\beta$$, then $$\alpha + \gamma < \beta + \gamma$$ " is wrong, since $$0<1$$ but $$0+\omega = \omega = 1 + \omega$$. But what about "if $$\alpha<\beta$$, then $$\gamma + \alpha < \gamma + \beta$$ " ?

Here is definitions I used (Thomas Jech, Set Theory):

$$\alpha + 0 = \alpha$$ for all $$\alpha\in\mathrm{Ord}$$,

$$\alpha+(\beta + 1) = (\alpha + \beta) + 1$$ for all $$\alpha,\beta\in\mathrm{Ord}$$,

$$\alpha + \beta = \lim_{\xi\to\beta}(\alpha + \xi)$$ for $$\alpha,\beta\in\mathrm{Ord}$$ and $$\beta$$ is a limit ordinal.

and in general, limit is defined as $$\lim_{\xi\to\beta}\gamma_\xi=\sup\{\gamma_\xi: \xi<\beta\}$$ only if the sequence $$\gamma_\xi$$ is nondecreasing and $$\beta$$ is a limit ordinal.

I think that the statement is true, since I get there when I was trying to prove $$\alpha + \gamma = \lim_{\xi\to\gamma}(\alpha+\xi)$$ is defined well when $$\gamma$$ is a limit ordinal. Firstly, I tried to prove the sequence $$\langle\alpha + \xi: \xi<\gamma\rangle$$ is increasing since the limit is defined only if the sequence is nondecreasing, in Thomas Jech's book. But I have to show that "if $$\xi_1<\xi_2$$, then $$\alpha+\xi_1<\alpha+\xi_2$$" to show that the sequence is increasing (of course showing that $$\alpha+\xi_1\leq\alpha+\xi_2$$ is enough but I believe that $$\alpha+\xi_1<\alpha+\xi_2$$).

I tried to prove the statement, and I'm stuck.

Let $$\Gamma$$ be the class of all ordinals $$\gamma$$ satisfying the statement "$$\forall\alpha,\beta\in\mathrm{Ord}\left(\alpha<\beta \Longrightarrow\gamma + \alpha < \gamma + \beta\right)$$ ".

$$(i)$$ $$0\in\Gamma$$, since $$0 + \eta = \eta$$ for all $$\eta\in \mathrm{Ord}$$. (I previously proved this).

$$(ii)$$ Assume that $$\gamma\in\Gamma$$. Then $$(\gamma + 1) + \alpha = \gamma + (1 + \alpha) <^? \gamma + (1+\beta) = (\gamma + 1) + \beta$$ (But I couldn't show the inequality).

$$(iii)$$ Assume that for all ordinals $$\xi<\gamma$$, $$\xi\in\Gamma$$, i.e., if $$\alpha<\beta$$ then $$\xi+\alpha < \xi+\beta$$. Then, $$\gamma+\alpha$$... (and that is it, i couldn't continue, since definition doesn't say anything about addition when limit ordinal is at the left side)

Thanks a lot!

Induction on $$\beta$$ worked well!

And I love the limit case of the transfinite induction:

Let $$\beta$$ be a limit ordinal and "if $$\alpha<\xi$$ then $$\gamma + \alpha < \gamma + \xi$$" for all $$\xi<\beta$$.

This means the sequence $$\langle\gamma + \xi : \xi<\beta\rangle$$ is increasing, so we can use the definition of limit: $$\gamma + \beta = \lim_{\xi\to\beta}(\gamma + \xi) = \sup\{\gamma+\xi:\xi<\beta\}$$

Also, we know that there are ordinals $$\theta_1$$ and $$\theta_2$$ such that $$\alpha<\theta_1<\theta_2<\beta$$, if $$\alpha < \beta$$, since $$\beta$$ is a limit ordinal. Notice that $$\gamma + \theta_1,\gamma + \theta_2\in\{\gamma + \xi: \xi < \beta\}$$ so $$\gamma + \theta_1<\gamma + \theta_2\leq\sup\{\gamma + \xi: \xi<\beta\}$$.

so, if $$\alpha<\beta$$, then $$\gamma+\alpha<\gamma+\theta_1<\sup\{\gamma + \xi: \xi<\beta\} = \lim_{\xi\to\beta}(\gamma + \xi) = \gamma +\beta$$.

• You could prove it by induction over $\beta$. Aug 9, 2020 at 0:40
• Fix ordinals $\alpha$ and $\beta$ and let $\Gamma$ be the class of all ordinals $\gamma$ such that $\gamma\le\beta$ or $\alpha+\beta<\alpha+\gamma$. Show by induction that $\Gamma=\mathsf{ON}$. Aug 9, 2020 at 0:42

This is very easy to understand when using the order-theoretic definition of $$\alpha+\beta$$ as the linear order which is the initial segment $$\alpha$$, followed by the tail segment $$\beta$$.

Remember now that there is at most a single embedding of one ordinal into another whose image is an initial segment.

We note that $$\gamma$$ is a joint initial segment of both $$\gamma+\alpha$$ and $$\gamma+\beta$$, so the identity function is the only embedding. Now we may proceed by considering the embedding of $$\alpha$$ into $$\beta$$, which itself is also the identity, and using it to extend the embedding of $$\gamma\to\gamma$$.

This shows, easily, that $$\gamma+\alpha\leq\gamma+\beta$$. But now we remember that the embedding $$\alpha\to\beta$$ was not surjective, so the embedding we got is not surjective. Since that is the only embedding onto an initial segment, it must be that $$\gamma+\alpha<\gamma+\beta$$.

Of course, this doesn't help you if you're trying to prove the inequality from the recursive definition. But it gives a good picture as to what's going on.

Now fix $$\gamma$$, and prove by induction on $$\beta$$, that for all $$\alpha<\beta$$, $$\gamma+\alpha<\gamma+\beta$$; and conclude this is true for all $$\alpha,\beta,$$ and $$\gamma$$.

If $$\alpha < \beta$$, then $$\alpha + 1 \le \beta$$. Then $$(\gamma + \alpha) + 1 = \gamma + (\alpha + 1) \le \gamma + \beta$$, so $$\gamma + \alpha < \gamma + \beta$$.

• Aren't you tacitly using $\alpha\le\beta\implies \gamma+\alpha\le\gamma+\beta$ in the last inequality? Aug 9, 2020 at 0:49
• @Reveillark That follows almost immediately from the definition of ordinal addition. Aug 9, 2020 at 1:43
• Thanks for the answer! But, you are assuming "if $\alpha+1\leq\beta$ then $\gamma + (\alpha + 1) \leq \gamma +\beta$, which is a different variation of the problem, if I'm not wrong. I did it with induction on $\beta$ actually, but I want to know if there is another, algebraic proof like you have tried to do. Aug 9, 2020 at 13:48
• One could prove that if $\alpha \le \beta$, then there exists an ordinal $\gamma$ such that $\alpha + \gamma = \beta$. But the proof I have in mind for this ultimately uses that ordinal addition is monotone. Aug 9, 2020 at 14:06
• And the induction proof is so simple, that I would almost say it follows from the definitions. Aug 9, 2020 at 14:07