Proof involving transfinite induction on ordinals

Question

Prove that if $\omega\le \alpha, 0\le n <\omega$ and $0<m<\omega$ then $(\alpha+n)*m=\alpha*m+n$

My attempt is as follows:

By induction on $n$, the base case when $n=0$ is trivial so, looking at the successor step we set $n=\delta+1$ so then $(\alpha+(\delta+1))*m=(\alpha*m)+(\delta+1)=(\alpha*m)+n$ so the step when $n$ is a successor holds.

Now when $n$ is a limit ordinal we have that $(\alpha+n)*m=(\alpha+\sup_{\epsilon<n}\{\epsilon\})*m=(\alpha*m)+\sup_{\epsilon<n}\{\epsilon\}=(\alpha*m)+n$ as required thus the statement is true and has been proved?

Was just wondering if anyone could check if the above is correct and if not would be able to provide where i have gone wrong and how to fix it

Answer 1

(1) In the theory of ordinals, $\omega$ is not a variable, but a constant. It is the least infinite ordinal. Therefore $n < \omega$ means that $n$ is finite. In particular, it is not a limit ordinal. So you don't even need to worry about what happens when $n$ is a limit ordinal. (And also, your proof is by ordinary induction, not transfinite induction).

(2) your proof for non-limit $n$ is wrong. You say

$(\alpha+(\delta+1)*m=(\alpha*m)+(\delta+1)=(\alpha*m)+n$

But showing that $(\alpha+(\delta+1))\cdot m$ is indeed $(\alpha\cdot m)+(\delta+1)$ is what you are supposed to do. What you already know (by supposition) is only that $(\alpha+\delta)\cdot m = \alpha\cdot m +\delta$. From there you have to prove that it holds for $\delta + 1$ as well, not just state it.

Edit (removed incorrect statements User1006 pointed out)

Answer 2

Since $\alpha >\omega>n$, it can be proved as: \begin{align} (\alpha +n)\cdot m&=\underbrace{(\alpha +n)+(\alpha +n)+\cdots+(\alpha +n)}_{m} \\ &=\underbrace{\alpha +(n+\alpha)+\cdots+(n+\alpha) }_{m}+n \\ &=\underbrace{\alpha +\alpha+\cdots+\alpha }_{m}+n \\ &=\alpha \cdot m+n \end{align}