Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I know this is a newbie question, so please bare with me :)

I'd like to prove within ZF axiomatic set theory that the addition of two ordinals is not commutative. In particular, I'd like to prove this counter example:

$\omega + 1 \neq 1 + \omega$

I have the following definition for addition on ordinal numbers (defined from transfinite induction):

(i) $\alpha + 0 = \alpha$

(ii) $\alpha + \beta' = (\alpha + \beta)'$

(iii) if $\beta$ is a limit ordinal then $\alpha + \beta = \bigcup_{\gamma \in \beta}(\alpha + \gamma)$

So my attempt was to start from the right side, which, intuitively would be something like this:

$1 + \omega = \bigcup_{\gamma \in \omega}(1 + \gamma) = \{2, 3, 4, ..\}$

My attempt at the left side started like this:

$\omega + 1 = (\omega + 0)' = \omega'$

And then I'm stuck. I'd like to think that the successor of $\omega$ is $\omega$ but with this definition how can I prove that? Also, if that's the case then there's a $1-1$ function that can map $\{1, 2, 3, ...\}$ to $\{2, 3, 4, ...\}$ and still preserve order, so shouldn't both sides of the addition be the same?

share|cite|improve this question
Show that $\omega+1$ has an element without an immediate predecessor, while $1+\omega$ doesn't. – Mariano Suárez-Alvarez Nov 19 '12 at 17:03
Or just a maximum. – Michael Greinecker Nov 19 '12 at 17:03
Yeah, that too :-) – Mariano Suárez-Alvarez Nov 19 '12 at 17:04

Recall the definition $\alpha+\beta=\sup\{\alpha+\gamma\mid\gamma<\beta\}$ when $\beta$ is a limit ordinal.

That means that $1+\omega=\sup\{1+n\mid n<\omega\}=\sup\{n\mid n<\omega\}=\omega$.

On the other hand, $\omega+1=(\omega+0)'=\omega+1\neq\omega$ because $\alpha\neq\alpha'$ for all $\alpha$. This is because $\alpha\in\alpha'$ therefore these sets are distinct. Therefore the ordinals are not order isomorphic either (because every well-ordered set is isomorphic to a unique ordinal).

share|cite|improve this answer
Wait, are you saying that I can't find a 1-1 function from $\langle \{2, 3, 4, ...\}, \in \rangle$ to $\langle N, \in \rangle$ with preserved order? How about $f(x) = x + 1$ ? What am I missing? – Nicolas Nov 19 '12 at 17:34
@Nicolas: I am saying that $\{2,3,4,\ldots\}$ is not an ordinal. In contrast $\omega\cup\{\omega\}$ is an ordinal. – Asaf Karagila Nov 19 '12 at 17:41
Thanks for your answer. I think I messed up the union definition. What this really means is that $1 + \omega = \cup_{(\gamma \in \omega)}(1 + \gamma) = (1 + 0) \cup (1 + 1) \cup (1 + 2) \cup ... = 1 \cup 2 \cup 3, ... = \{0\} \cup \{0, 1\} \cup \{0, 1, 2\} \cup ... = \omega$. So the real question is, $\omega \neq \omega \cup \{\omega\}$, how can I prove that? – Nicolas Nov 19 '12 at 18:00
@Nicolas: But that much is obvious. $\omega\notin\omega$ but $\omega\in\omega'$. – Asaf Karagila Nov 19 '12 at 18:00
Oh yes, I finally understood it, thanks for your help! – Nicolas Nov 19 '12 at 18:04

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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