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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.

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    $\begingroup$ The successor of an even integer is an odd integer. Do you find that... odd? $\endgroup$
    – Asaf Karagila ♦
    Sep 4, 2017 at 10:40
  • $\begingroup$ Look at $\omega+1$ in the second image: madore.org/~david/weblog/… $\endgroup$
    – Carsten S
    Sep 4, 2017 at 19:10
  • $\begingroup$ 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?! $\endgroup$
    – Simply Beautiful Art
    Sep 4, 2017 at 21:13
  • $\begingroup$ @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. $\endgroup$
    – PJTraill
    Sep 4, 2017 at 22:22
  • $\begingroup$ @PJTraill Perhaps so, but it is no way clear that this is the case or not, hence my vote to close as unclear. $\endgroup$
    – Simply Beautiful Art
    Sep 4, 2017 at 22:27

2 Answers 2

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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.

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  • $\begingroup$ And clearly, the successor of a limit ordinal, is not a limit ordinal? $\endgroup$
    – Tristan Batchler
    Sep 4, 2017 at 10:32
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    $\begingroup$ That is correct. $\endgroup$
    – Kenny Lau
    Sep 4, 2017 at 10:33
  • $\begingroup$ Thanks! I'll accept your answer when I can. Gotta wait 10 more minutes apparently. :) $\endgroup$
    – Tristan Batchler
    Sep 4, 2017 at 10:34
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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.

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