# Upper bound on natural ordinal sum

I am wondering if there are any known bounds on how large $$a\oplus b$$ can get, where $$\oplus$$ is natural ordinal addition? I know that $$a+b \le a\oplus b$$, but is it true, for example, that $$a\oplus b \le a+b+b$$? What bounds have been put on $$a\oplus b$$?

The bound mentioned in the OP holds: that is, $$\alpha\oplus\beta\le\alpha+\beta+\beta,$$ even with strict inequality if $$\beta>0$$. Since $$\oplus$$ is commutative, we actually obtain $$\alpha\oplus\beta\le\min\{\alpha+\beta+\beta,\beta+\alpha+\alpha\},$$ with strict inequality if $$\alpha,\beta>0$$.
To see this, let $$\gamma$$ be the leading exponent in the Cantor normal form of $$\beta$$ so that $$\omega^\gamma m\le\beta<\omega^\gamma(m+1)$$ for some $$0, and let $$\alpha'$$ denote the initial part of the CNF of $$\alpha$$ consisting of terms with exponent $$>\gamma$$ so that $$\alpha'+\omega^\gamma n\le\alpha<\alpha'+\omega^\gamma(n+1)$$ for some $$n<\omega$$. Then $$\alpha\oplus\beta<\alpha'+\omega^\gamma(n+m+1)\le\alpha'+\omega^\gamma n+\omega^\gamma(2m)\le\alpha+\beta+\beta.$$
• I’ll refrain from further edits. But the bound can be shown optimal in that if $t(\alpha,\beta)$ is any expression using $\alpha$, $\beta$, $+$, $\min$, $\max$, and arbitrary ordinal constants, such that $t(\alpha,\beta)\ge\alpha\oplus\beta$ for all $\alpha,\beta$, then $t(\alpha,\beta)\ge\min\{\alpha+\beta+\beta,\beta+\alpha+\alpha\}$ for all $\alpha,\beta$. This may well be true even for expressions using $\cdot$, but this might require much more work to check. Commented Jul 16 at 12:50
• Hmm, with $\cdot$, there are exceptions if $\alpha$ or $\beta$ is $1$: we have $\alpha\oplus\beta\le\alpha+\beta^2$, which is smaller than the bound above if $\beta=1\le\alpha$. But I think the optimality should hold for $\alpha,\beta\ge2$. Commented Jul 16 at 13:04
• Would expressions like $a\oplus b < a+b+a$ fall under the same condition? My context is that I am writing a computer verified proof, and I need specific bounds on what $(a\oplus b)-a-b$ can be. Commented Jul 16 at 15:44
• As noted in comments to the other (now deleted) answer, $\alpha\oplus\beta\le\alpha+\beta+\alpha$ is false in general: it fails e.g. for $\alpha=\omega$, $\beta=\omega^2+1$. Commented Jul 16 at 15:53