Ordinal $10^\omega$

Question

$10^\omega$ = $10 \cdot 10 \cdot 10 \cdot ...= \lim_{\alpha \lt \omega} (10^\alpha) = \omega$. Are my thoughts correct? Is this sufficient explanation, given the ordinal arithemtic proved from ZFC?

Answer 1

I'd say it would be better if you referred to the appropriate definition of ordinal arithmetic.

For an ordinal $\alpha$ and a limit ordinal $\beta$ (like e.g. $\beta = \omega$) exponentiation is defined as

$\alpha^\beta = \bigcup_{\gamma < \beta} \alpha^\gamma$

Replacing the appropriate values in the definition we therefore have:

$10^{\omega} = \bigcup_{n < \omega} 10^n$

Since for $n \le m$ we have the set inclusion $10^n \subseteq 10^m \subseteq \omega$ it should now become apparent that $\bigcup_{n < \omega} 10^n = \bigcup_{n < \omega} n = \omega$.