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

  • 1
    $\begingroup$ You could prove it by induction over $\beta$. $\endgroup$ Aug 9, 2020 at 0:40
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    $\begingroup$ 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 $\Gamma=\mathsf{ON}$. $\endgroup$ Aug 9, 2020 at 0:42

2 Answers 2


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 $\gamma + \alpha < \gamma + \beta$.

  • $\begingroup$ Aren't you tacitly using $\alpha\le\beta\implies \gamma+\alpha\le\gamma+\beta$ in the last inequality? $\endgroup$
    – Reveillark
    Aug 9, 2020 at 0:49
  • $\begingroup$ @Reveillark That follows almost immediately from the definition of ordinal addition. $\endgroup$ Aug 9, 2020 at 1:43
  • $\begingroup$ 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. $\endgroup$
    – Ali Dursun
    Aug 9, 2020 at 13:48
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    $\begingroup$ 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. $\endgroup$ Aug 9, 2020 at 14:06
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    $\begingroup$ And the induction proof is so simple, that I would almost say it follows from the definitions. $\endgroup$ Aug 9, 2020 at 14:07

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