Is the successor of a limit ordinal undefined?

Question

Suppose $\beta$ is a limit ordinal. Then for all $\gamma < \beta$, $\mathbf{s}(\gamma) < \beta$.

Now I'm wondering if it makes sense to consider $\alpha = \mathbf{s}(\beta)$. If so, $\alpha$ is not a limit ordinal as it is a successor ordinal. It seems unintuitive, but I also don't see anything wrong with it, as $\alpha = \beta \cup \{ \beta \}$ appears to be well defined.

Am I missing something here? Cheers.

Answer 1

The successor of a limit ordinal is well-defined: $s(\beta) = \beta \cup \{\beta\}$ where $\beta$ is the limit ordinal.

However, a limit ordinal is not the successor of any ordinal.

Answer 2

In set theory, a function on ordinals is continuous if the value of the function at a limit ordinal is the limit of the values of that function on smaller ordinals. If you find the fact that limit ordinals have successors to be counterintuitive, then you seem to be thinking in terms of continuous functions. The successor function is instead a nice example of a discontinuous function on ordinals.