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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?

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    $\begingroup$ Generally, the sum, product, or power of two limit ordinals tends to be limit ordinals, which might be helpful to prove. Also, you probably meant $\omega2$ instead of $2\omega$ at the beginning of the second paragraph. $\endgroup$ Commented Jun 10, 2017 at 17:00
  • $\begingroup$ What I meant was $\omega+\omega$. $\endgroup$
    – Janitha357
    Commented Jun 10, 2017 at 17:06
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    $\begingroup$ Then I believe that should be $\omega2$. See here for more details. $\endgroup$ Commented Jun 10, 2017 at 17:07

2 Answers 2

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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.

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    $\begingroup$ The last paragraph seems to be what I was looking for. $\endgroup$
    – Janitha357
    Commented Jun 10, 2017 at 17:03
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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.

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