# Is the successor of a limit ordinal undefined? [closed]

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.

• The successor of an even integer is an odd integer. Do you find that... odd? Sep 4, 2017 at 10:40
• Look at $\omega+1$ in the second image: madore.org/~david/weblog/… Sep 4, 2017 at 19:10
• Horribly confused. You appear to ask if the successor of a limit ordinal is well defined, then go on to say you see nothing wrong it. If you see nothing wrong, then what's the issue?! Sep 4, 2017 at 21:13
• @SimplyBeautifulArt: While the questioner has not explained what was worrying him, it seems to me that they may have felt that a limit ordinal ought not to have had a successor because there should be something final about it, perhaps in that when you reach a limit in other contexts you can go no further. Sep 4, 2017 at 22:22
• @PJTraill Perhaps so, but it is no way clear that this is the case or not, hence my vote to close as unclear. Sep 4, 2017 at 22:27

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