Ordinal isomorphic to $n\times\omega,n\in\omega$

Question

What ordinal is order isomorphic to $n\times\omega$ with the lexicographic order? I was thinking it is $\omega$ but I can't make an order preserving bijection between them. If it isn't $\omega$, which ordinal is it?

Answer 1

$ n \times \omega$ with the lexicographical order is just $\omega \times n$.

Indeed, for ordinals $\alpha, \beta$, the order "$<$" on $\alpha \times \beta$ is like the lexicographical order except that one puts the second position first : $$(a, b) < (a', b') :\Longleftrightarrow (b, a) <_\textrm{Lex} (b', a') $$

Answer 2

Is this what you mean by lexicographic order? For $3 \times \omega$, $$ 0\;0\\ 1\;0\\ 2\;0\\ 0\;1\\ 1\;1\\ 2\;1\\ 0\;2\\ 1\;2\\ 2\;2\\ 0\;3\\ 1\;3\\ 2\;3\\ \vdots $$ and from that we see that the order-type is $\omega$.

But if you mean the other lexicographic order, then the order-type is $\omega+\omega+\omega = \omega \;3$.