Let $\alpha_1 \geq \ldots \geq \alpha_n$ be ordinal numbers. I am interested in necessary and sufficient conditions for the ordinal sum $\alpha_1 + \ldots + \alpha_n$ to be equal to the Hessenberg sum $\alpha_1 \oplus \ldots \alpha_n$, most quickly defined by collecting all the terms $\omega^{\gamma_i}$ of the Cantor normal forms of the $\alpha_i$'s and adding them in decreasing order.

Unless I am very much mistaken the answer is the following: for all $1 \leq i \leq n-1$, the smallest exponent $\gamma$ of a term $\omega^{\gamma}$ appearing in the Cantor normal form of $\alpha_i$ must be at least as large as the greatest exponent $\gamma'$ of a term $\omega^{\gamma'}$ appearing in the Cantor normal form of $\alpha_{i+1}$. And this holds just because if $\gamma' < \gamma$,

$\omega^{\gamma'} + \omega^{\gamma} = \omega^{\gamma} < \omega^{\gamma} + \omega^{\gamma'} = \omega^{\gamma'} \oplus \omega^{\gamma}$.

Nevertheless I ask the question because:

1) I want reassurance of this: I have essentially no experience with ordinal arithmetic.
2) Ideally I'd like to be able to cite a standard paper or text in which this result appears.

Bonus points if there happens to be a standard name for sequences of ordinals with this property: if I had to name it I would choose something like unlaced or nonoverlapping.

P.S.: The condition certainly holds if each $\alpha_i$ is of the form $\omega^{\gamma} + \ldots + \omega^{\gamma}$. Is there a name for such ordinals?


1 Answer 1


You are exactly right about the "asymmetrically absorptive" nature of standard ordinal addition (specifically with regard to Cantor normal form). Your condition is necessary and sufficient (sufficiency is easy, and you've shown necessity). I don't know of any standard name for such sequences, though. As for your $\omega^\gamma+\cdots+\omega^\gamma$ bit, that isn't appropriate for Cantor normal form. We require the exponents to be listed in strictly decreasing order.

  • $\begingroup$ $@$Cameron: thanks. Two things: (i) I'm still hoping for something to cite. (ii) In the literature I've encountered there are two versions of Cantor normal form, a "long" and a "short" one. In the long one you don't multiply the $\omega^{\gamma}$'s by integers: you write them out allowing finite repetitions. Actually though I wrote it out that way only to avoid the headache of deciding on whether to write $\omega^{\gamma} n$ or $n \omega^{\gamma}$, i.e., which convention to use on the definition of ordinal multiplication. $\endgroup$ May 9, 2013 at 4:44
  • $\begingroup$ In fact, I didn't say that I was giving Cantor normal form: I just said "of the form...". But no biggie... $\endgroup$ May 9, 2013 at 4:50
  • $\begingroup$ @Pete: (i) I'll see if I can track something down tomorrow. (ii) Ah! I see where you're coming from. Can't say I blame you, either. The lack of commutativity (and in particular the asymmetric absorptions) take some getting used to. Still, the way that the Cantor normal form is written (even in long form) is suggestive of how we need to do things. Except with finite ordinals, terms in a reduced sum (i.e. $\omega$ instead of $1+\omega$) must be written from biggest to smallest because of the aforementioned absorption. Likewise for products, so we need $\omega^\gamma n$. $\endgroup$ May 9, 2013 at 4:59
  • 1
    $\begingroup$ Ultimately, it has to do with the nature of lexicographic (or dictionary) order. For example, $\omega 2$ is the order type of $$\bigl\{\langle0,0\rangle,\langle0,1\rangle,\langle0,2\rangle,\dots, \langle1,0\rangle,\langle1,1\rangle,\langle1,2\rangle\dots\bigr\},$$ which is $\omega+\omega$. But $2\omega$ is the order type of $$\bigl\{\langle0,0\rangle,\langle0,1\rangle,\langle1,0\rangle,\langle1,1\rangle,\langle2,0\rangle,\langle2,1\rangle,...\bigr\},$$ which is $\omega$. Still, I can't blame you. It's a convention that takes some serious getting used to. I'll try to track down a reference. $\endgroup$ May 9, 2013 at 11:37
  • 2
    $\begingroup$ $@$Cameron: my point is that the standard convention on the order type of $\alpha \times \beta$ is that it is the reverse lexicographic order. I dug it up a little more, and it seems that Cantor himself changed his mind about the definition: see mathforum.org/kb/message.jspa?messageID=508086. In particular the last post in the thread give a good reason to take the standard convention: maybe that convinces me once and for all. $\endgroup$ May 9, 2013 at 16:18

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.