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 '20 at 0:40
  • 1
    $\begingroup$ 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 '20 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 $\gamma + \alpha < \gamma + \beta$.

  • $\begingroup$ Aren't you tacitly using $\alpha\le\beta\implies \gamma+\alpha\le\gamma+\beta$ in the last inequality? $\endgroup$
    – Reveillark
    Aug 9 '20 at 0:49
  • $\begingroup$ @Reveillark That follows almost immediately from the definition of ordinal addition. $\endgroup$ Aug 9 '20 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 '20 at 13:48
  • 1
    $\begingroup$ 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 '20 at 14:06
  • 1
    $\begingroup$ And the induction proof is so simple, that I would almost say it follows from the definitions. $\endgroup$ Aug 9 '20 at 14:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.