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If $\alpha$ is any ordinal number does $\alpha + \omega$ have to be limit ordina, where $\omega$ is ordinal number of $\mathbb{N}$?

I know that $\omega+\alpha$ is not always limit ordinal since $\omega + 1$ is succesor of $\omega$.

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Normally, $\alpha+\omega$ is defined as $\sup\{\alpha+\beta\colon\,\beta<\omega\}$, from this you can see that it is a limit ordinal. Another definition defines $\alpha+\omega$ as the order type of $\{0\}\times\alpha\cup\{1\}\times\omega$ ordered lexicographically, but it is easily shown that these are equivalent. –  martin.koeberl Jun 19 '13 at 19:33

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up vote 6 down vote accepted

Yes.

This is more-or-less by definition of ordinal addition:

$$\alpha+\omega = \sup_{n<\omega} \alpha+n$$

Suppose that $\alpha+\omega$ were a successor, say $\beta+1$. Then since it is the supremum, we must have that there is an $n$ such that $\alpha+n = \beta+1$ (otherwise, $\beta$ would be an upper bound as well, contradicting that $\beta+1$ is the supremum).

But then clearly $\alpha+(n+1) =\beta+2 > \beta+1$, so $\beta+1$ can't be the supremum.

It follows that $\alpha+\omega$ is a limit ordinal.

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$\alpha + \beta$ is always a limit ordinal if $\beta$ is such.

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