Simplify this expression of ordinal numbers: $(ω+1) ω²$

Question

Let $ω$ be the ordinal of the natural numbers. Simplify $(ω+1) ω²$

Answer 1

Recall the definition of ordinal multiplication ($\delta$ is a limit ordinal, $\beta$ is any ordinal):

$$\alpha\cdot0=0\qquad\alpha\cdot(\beta+1)=(\alpha\cdot\beta)+\alpha\qquad\alpha\cdot\delta=\sup_{\beta<\delta}\alpha\cdot\beta$$

By transfinite induction, let's prove that $(\omega+1)\cdot\alpha=\omega\cdot\alpha+1$ if $\alpha$ is a successor and $(\omega+1)\cdot\alpha=\omega\cdot\alpha$ otherwise.

$$(\omega+1)\cdot0=0=\omega\cdot0$$

\begin{align} (\omega+1)\cdot(\beta+1)&=(\omega+1)\cdot\beta+(\omega+1)\\ &=((\omega+1)\cdot\beta+\omega)+1\\ &=((\omega\cdot\beta+1)+\omega)+1\\ &=(\omega\cdot\beta+\omega)+1\\ &=\omega\cdot(\beta+1)+1\end{align}

If $\beta$ is not a successor, then skip the third step above. (Note that $(\alpha+1)+\omega=\alpha+\omega$ was used; this is because the supremum of $\alpha+1+n$ for $n<\omega$ is $\alpha+\omega$.) Finally, for limit ordinals, the supremum of $\omega\cdot\beta+1$ for $\beta<\delta$ is $\omega\cdot\delta$, so we are done. Plugging in $\alpha=\omega^2$ gives $(\omega+1)\cdot\omega^2=\omega\cdot\omega^2=\omega^3$.

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Alternatively, we can give an order isomorphism $f$ between the set $\omega^2\times(\omega+1)=\{(\omega a+b,c):a,b\in\omega,c\in\omega+1\}$ under dictionary order, with the first factor as the most significant position, and $\omega^3$.

If $c=\omega$, then let $f(\omega a+b,\omega)=\omega^2a+\omega (b+1)$. Else if $b=0$, then let $f(\omega a,c)=\omega^2a+c$. Otherwise let $f(\omega a+b,c)=\omega^2 a+\omega b+(c+1)$.