How to identify limit ordinals?

Question

Definition: An ordinal number $\alpha$ is called a limit ordinal number if there is no ordinal number immediately preceding $\alpha$.

Now my lecture notes say that $\omega, 2\omega, \omega^2, \omega^\omega$ are limit ordinal numbers whereas $\omega+3,2^\omega+5$ are not which is intuitively clear. But is there a characterization of limit ordinals that may work when proving the case with some rigor? Or is it "just look and identify" sort of a thing using the ordering of the family of ordinal numbers?

Answer 1

You already have a characterization of limit ordinals. In any specific case you just have to verify that this characterization is fulfilled. Consider for example $\omega^2$:

$$ \omega^2 = \omega \cdot \omega = \sup \{ \omega \cdot n \mid n < \omega \}. $$

Hence, if $\alpha < \omega^2$, there is some $n < \omega$ such that $\alpha < \omega \cdot n$. But then $$ \alpha + 1 < \omega \cdot n + 1 < \omega \cdot (n+1) \le \omega^2. $$

Thus $\omega^2$ is a limit ordinal.

Answer 2

There can be several answers, depending on what you mean by "identify".

The simplest answer would be to look at the Cantor normal form of $\alpha$, and see if it has any finite ordinal there. If the answer is no, then $\alpha$ is a limit ordinal (or $0$, which may or may not be a limit ordinal depending on you convention) and otherwise it is a successor ordinal.

Another answer would be that $\alpha$ is a limit ordinal if and only if for every $\beta<\alpha$, $\beta+1<\alpha$ (with the same caveat about $0$ as before). Although it seems not to be exactly what you are looking for.