How do you best simplify $(\omega + 1) \cdot \omega^{2}$?

Question

I am currently reading "An Introduction to Set Theory" by Hrbacek and Jech. I have come to an exercise that I am having trouble with. The exercise asks the reader to simplify three ordinal arithmetic expressions. I am having trouble simplifying the last expression. The last expression is given below : \begin{equation} (\omega + 1) \cdot \omega^{2} \end{equation} Here $\omega$ represents the natural numbers (the smallest limit ordinal). I am not sure how to simplify this, my attempt so far is given below : \begin{align} (\omega + 1) \cdot \omega^{2} & = (\omega + 1) \cdot (\omega^{2}) \\ & = (\omega + 1) \cdot (\omega \cdot \omega)\\ & = \left( (\omega + 1) \cdot \omega \right) \cdot \omega \\ & = \left( \sup\left( \{(\omega + 1) \cdot n \; \mid \; n \in \boldsymbol{N} \} \right) \right) \cdot \omega \end{align} I'm not sure where to go from here (assuming what I've done so far is a reasonable path to the solution).

Can someone help with this ?

Answer 1

Following on from where you left off:

For finite $n$, we have \begin{align*} (\omega+1)\cdot n &= \underbrace{(\omega+1)+(\omega+1)+ \dots + (\omega+1)}_{n\text{ times}}\\ &= \omega + \underbrace{(1+\omega) + \dots + (1+\omega)}_{n-1\text{ times}}+1\\ &= \omega\cdot n + 1. \end{align*} So $(\omega+1)\cdot \omega = \sup_{n\in \omega} (\omega\cdot n + 1) = \sup_{n\in \omega} (\omega\cdot n) = \omega^2$. (The second equality follows from the observation that $\omega\cdot n \leq \omega\cdot n+1\leq \omega\cdot (n+1)$ for all $n$, so the sequences are mutually cofinal.)

Thus $(\omega+1)\cdot \omega^2 = ((\omega+1)\cdot \omega)\cdot \omega = \omega^2\cdot \omega = \omega^3$.

Answer 2

For any successor ordinal $\:\alpha<\omega^2$, there is \begin{align} (\omega+1)\cdot\alpha&=\underbrace{(\omega +1)+(\omega +1)+\cdots+(\omega +1)}_{\alpha} \\ &=\underbrace{\omega +(1+\omega)+\cdots+(1+\omega) }_{\alpha}+1 \\ &=\underbrace{\omega +\omega+\cdots+\omega}_{\alpha}+1 \\ &=\omega \cdot \alpha+1 \\ &<\omega^3 \end{align} Since for any successor ordinal $\:\gamma<\omega^3$, there is a successor ordinal $\:\alpha<\omega^2\:$ that $$ (\omega+1) \cdot \alpha=\omega \cdot \alpha+\alpha>\gamma\quad\text{and}\quad(\omega+1) \cdot \alpha<\omega^3 $$ So we have $$ (\omega +1)\cdot \omega^2=\sup\{(\omega+1)\alpha:\alpha<\omega^2\}=\omega^3 $$