Prove that if b is a limit ordinal and $a<b$ then $a+b=b$ ($a$ is a successor ordinal).

This is what I tried:

If $a<b$ then there exists an ordinal $c$ so that $a+c=b$. It's clear that $c$ can't be a successor of an ordinal because then $b$ would be a successor of an ordinal, so it has to be a limit ordinal. Now I have to prove that $c=b$. I tried by contradiction, by assuming $c<b$ because $b<c$ can't happen, but it doesn't seem to lead anywhere. I also know that if $b$ is a limit ordinal and $a<b$ then $a+1<b$, so I guess I should use that?

Edit: I changed $a$ to a successor ordinal, thanks for the comment, if $a$ is a limit ordinal the statement is false.

  • $\begingroup$ It completely cut my question I don't understand. Could someone please fix that for me? $\endgroup$ – user15269 Jun 27 '18 at 12:48
  • $\begingroup$ You need put a space around the < symbol. Not doing that is what caused the "cut". $\endgroup$ – quasi Jun 27 '18 at 12:49
  • 6
    $\begingroup$ Your statement isn't true (for the standard interpretation of ordinal addition). Take for example $a = \omega$ and $b = \omega + \omega$. $\endgroup$ – Stefan Mesken Jun 27 '18 at 12:52
  • 3
    $\begingroup$ @Math $a=\omega+1$, $b=\omega+\omega $. $\endgroup$ – Andrés E. Caicedo Jun 27 '18 at 15:49

The result you are trying to prove is false. For example, if $a=\omega+1$ and $b=\omega+\omega$, then $a+b=\omega\cdot 3>b$.

Here is what is true: first, the key result you should establish (by induction) is that

An ordinal $\alpha>0$ has the property that for all $\beta<\alpha$, $\beta+\alpha=\alpha$ if and only if $\alpha$ is a power of $\omega$.

(For example, if $\beta<\omega^2$, then it has the form $\omega\cdot n+m$ for some $n,m<\omega$, and one quickly checks that $\beta+\omega^2=\omega^2$. On the other hand, for instance, $(\omega+1)+(\omega+\omega)>\omega+\omega$, as indicated above.)

These ordinals are called (additively) indecomposable, because you cannot write any such $\alpha$ in the form $\beta+\gamma$ where $\beta,\gamma<\alpha$.

Recall that every nonzero ordinal $\alpha$ admits a (unique) representation in what is called Cantor normal form: There are $k<\omega$, $\delta_0,\dots,\delta_k$ with $\alpha\ge\delta_k>\dots>\delta_0$, and nonzero $n_0,\dots,n_k<\omega$ such that $$\alpha=\omega^{\delta_k}\cdot n_k+\dots+\omega^{\delta_0}\cdot n_0.$$ Given such a representation of $\alpha$, call $\omega^{\delta_k}$ its principal part.

The characterization you want is this:

Given ordinals $\alpha>0$ and $\beta<\alpha$, we have that $\beta+\alpha=\alpha$ if and only if $\beta$ is strictly smaller than the principal part of $\alpha$.

(For example, $(\omega^2\cdot 3+\omega\cdot7+5)+(\omega^3\cdot 2+11)=\omega^3\cdot 2+11$ and $\omega^2\cdot 3+\omega\cdot7+5$ is stricty below the principal part $\omega^3$ of $\omega^3\cdot 2+11$. On the other hand, for instance, $\omega+1$ and $\omega+\omega$ both have the same principal part, namely $\omega$, and as mentioned several times already, $(\omega+1)+(\omega+\omega)>\omega+\omega$.)

This characterization is easy to verify using the result about indecomposable ordinals highlighted above. In turn, there are several answers on this site providing details of the result on indecomposable ordinals.


Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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