I'm trying to understand how to bring ordinals of the form $(\omega +n)\omega$ to normal form. Unfortunately, I am confused by what seem to be inconsistent answers to very similar questions:
- In a comment to the answer of this question, it is said $(\omega+2)\omega=\omega^2 +2$.
- On the other hand, in this question it is said $(\omega+1)\omega=\omega^2\neq \omega^2 +1$.
The answer to the first question says that the same method of calculating $(\omega+n)k$ by writing out $(\omega+n)+\cdots +(\omega +n)$ $k$ times and using associativity also works when $k$ is replaced by $\omega$. If I imagine doing this, then the answer does indeed turn out to be $\omega^2 +2$.
The detailed calculation in the answer to the second question I can't seem to follow, but it does not appear to be plagued by any handwavyness.
I tried to draw the ordinals $(\omega+2)\omega,\omega^2+2$ to "see" if they're the same, but it looks like $\omega^2+2$ has a maximal element while $(\omega+2)\omega$ does not:
To add to my confusion, the course instructor explicitly said $(\omega +2)\omega=\omega ^2+2$ without explaining much.
What's going on here? What is the correct, formal way to bring such ordinals to normal form?
$$(\omega + 2) \omega = \omega^2$$
By definition of ordinal multiplication,
$$(\omega + 2)\omega = \sup \{ (\omega + 2)n : n \in \omega \}$$
Now, $$(\omega + 2) 2 = (\omega + 2) 1 + (\omega + 2) = \omega + \underbrace{2 + \omega}_{\omega} + 2 = \omega + \omega + 2 = \omega2 + 2$$
$$(\omega + 2)3 = (\omega+2)2 + (\omega+2) = (\omega2+2) + (\omega+2) = \omega3+2$$
Inductively, you can show that $(\omega + 2)n = \omega n + 2$.
That is, $$(\omega + 2) \omega = \sup \{ \omega n + 2 : n \in \omega \}$$
This is $\omega^2$. Indeed, every $\omega n + 2 \leq \omega^2$, while for any $\alpha < \omega^2$ we may express $\alpha = \omega n + m$ (see bottom), and then $\omega (n+1) + 2$ is bigger.
Why is the first answer wrong?
The first answer is wrong because you can "lose limiting tails". For instance, $\sup \{ n + 1 : n \in \omega \}$ is $\omega$ which is not a successor, but each $n+1$ is a successor. We have lost the tail $1$ in taking the limit. In general, on taking the sup, you lose the ability to shuffle symbols around in the way that the first post said you could: $(\omega + n) + \dots + (\omega + n) + \dots$ can't necessarily be manipulated in the same way as $(\omega + n) + \dots + (\omega + n)$ can. Again in the $n+1$ example, in a way which matches more closely the first answer, $$\underbrace{1 + \dots + 1}_{\text{$n$ times}} = (n-1) + 1$$ but $$\underbrace{1+\dots + 1}_{\text{$n$ times}} + 1 + 1 + \dots = \omega$$
We have lost the "tail" 1. Naively, if you took the sup in the manner of the first answer, you'd get $\omega + 1$, which is wrong. The limits happening to the left of the "tail" 1 turn out to swamp the tail. In fact, you'd expect this to happen: you'd expect the result of this limiting process to give a limit ordinal. It seems implausible in your original question that multiplying by $\omega$ could give you a successor ordinal.
Appendix: justification that $\alpha < \omega^2$ may be written as $\omega n + m$
This is by induction. It's true for $0$, trivially. It's true for successors, trivially - set $m \mapsto m+1$ and $n \mapsto n$. It's true for limits $ \lambda < \omega^2$: $\lambda$ is the sup of $\{ \alpha : \alpha < \lambda \}$. Then it is the sup of $\{ \omega n_{\alpha} + m_{\alpha} : \alpha < \lambda \}$, where $m_{\alpha}, n_{\alpha} \in \omega$, and the $n_{\alpha}$ are $\leq$ some $n \in \omega$ because otherwise the sup $\lambda$ would be $\geq \omega^2$. We'll let $n$ be the least such bound on the $n_{\alpha}$. I claim that $\lambda$ is in fact $\omega (n+1)$. Indeed, it's certainly $> \omega n$ because $\omega n + m_{\alpha}$ is in the list, for some $\alpha$; and it's $\leq \omega (n+1)$ because every $\omega n_{\alpha} + m_{\alpha} < \omega (n+1)$. There's only one limit ordinal that satisfies those constraints, and that's $\omega (n+1)$.