Ordinal arithmetic $(\omega+1) \cdot \omega$ and $\omega \cdot (\omega +1)$

Question

Here is where I am so far: $(\omega+1) \cdot \omega = \sup\{(\omega +1) \cdot n, n \in \omega\} = \omega^2$

and $\omega \cdot (\omega +1) = \omega \cdot \omega + \omega = \omega^2 + \omega$

Hence $((\omega+1) \cdot \omega) \in (\omega \cdot (\omega +1))$

Is this true? (I am least sure about the first line..)

Answer 1

If you're uncertain about the first line, why not expand it further? What is $(\omega+1)\cdot n$?

Quick calculation will show you that it is $\omega\cdot n+1$, and intuitively it's stacking $n$ copies of $\omega+1$ one after another, so the first $n-1$ copies "swallow" the $+1$, and you are left with $\omega\cdot n+1$.

So now what is $\sup\{\omega\cdot n+1\mid n\in\omega\}$? It is $\omega^2$, as you wrote.