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According to Wikipedia, if $\alpha$ is a countable limit ordinal, then $\mathrm{cf}(\alpha)=\omega$. It is intuitively clear to me that it should be so. Certainly the cofinality of such an ordinal must be $\geq\omega$. We can now look at what the ordinal ends with. If it has a single $\omega$ at the end, then that is a cofinal subset and we are done. It may, however, not end in $\omega$. It could, for example, end in $\omega$ $\omega$s or $\omega^2$. Then we can pick one element of each $\omega$ constituting the $\omega^2$, and the set of these elements is cofinal in $\alpha$, and its type is $\omega.$ If $\alpha$ ends in $\omega^3$, we can take one element of each $\omega^2$ constituting the $\omega^2$, and get a cofinal subset of type $\omega.$

It seems to me that this reasoning should work. It should probably be done inductively. How should I do it?

Bartek
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    Cofinality is always a regular cardinal. – Eran May 12 '13 at 12:55
  • @Eran I can see why the regular part is true. But why must it be a cardinal? – Bartek May 12 '13 at 12:58
  • By definition, a map from the least ordinal. – Eran May 12 '13 at 13:00
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    @Eran I don't understand. Cofinality must be the least ordinal among those that are types of cofinal subsets. How do I know that this is implies that it is also the least ordinal of some cardinality? – Bartek May 12 '13 at 13:04
  • If there is a cofinal $\alpha$-sequence in $\beta$ and a cofinal $\beta$-sequence in $\gamma$, then there is a cofinal $\alpha$-sequence in $\gamma$. Therefore cofinalities are initial ordinals. – Zhen Lin May 12 '13 at 13:11
  • @Zhen I'm really sorry I'm so obtuse, but I still think there's something to prove here. Suppose $\mathrm{cf}(\alpha)$ is not initial, that is there is $\delta<\alpha$, $|\delta|=|\alpha|.$ According to what you say I should get a contradiction with the transitivity of "is cofinal in". But I don't see an obvious contradiction. – Bartek May 12 '13 at 13:22

2 Answers2

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There are two approaches here:

  1. $\omega\leq\operatorname{cf}(\alpha)\leq|\alpha|\leq\alpha$, for limit ordinals $\alpha$. If you show this inequality then you are done, since countable ordinals have cardinality $\omega$.

  2. You can actually construct a cofinal $\omega$-sequence. Let $f\colon\omega\to\alpha$ be a bijection. Let $\alpha_0=f(0)$, and let $\alpha_{n+1}=f(k)$ such that $k$ is the least for which $f(k)>\alpha_n$. Because ordinals are well-ordered there are no decreasing sequences, and therefore there exists such $k$. It is not hard to show that the sequence $\alpha_n$ is cofinal.

Asaf Karagila
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  • Thank you very much. In the first approach, why do I need the $|\alpha|\leq\alpha$ part? Certainly $\omega\leq\mathrm{cf}(\alpha)$, and $|\alpha|=\omega$ so if $\mathrm{cf}(\alpha)\leq|\alpha|,$ then we have $\omega\leq\mathrm{cf}(\alpha)\leq\omega$, so $\mathrm{cf}(\alpha)=\omega.$ Why do I need the last inequality? – Bartek May 12 '13 at 13:42
  • @Bartek: Hmm, you don't. It's just true, and I just wanted $\alpha$ to appear in the formula in a plain form as well. :-) – Asaf Karagila May 12 '13 at 13:47
  • OK, I see. I will have to think about $\mathrm{cf}(\alpha)\leq|\alpha|.$ This is what people people were saying in the comments too. I don't see it yet, but I will think about it. – Bartek May 12 '13 at 13:49
  • Bartek, the proof of that is exactly the generalized proof of the second point in my answer. – Asaf Karagila May 12 '13 at 14:00
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    We do not need the well-ordered thing here, it is really a distraction. You have that $\alpha$ is a countable linear order without a largest element, and $f:\omega\to\alpha$ is a bijection. This is enough to recursively find $g:\omega\to\alpha$ cofinal. – Andrés E. Caicedo May 12 '13 at 22:48
  • @Andres Isn't $f$ already cofinal? It is a surjection on a limit ordinal $\alpha$ and thus $f[\omega]$ is unbounded in $\alpha$, therefore cofinal in it. – Rafał Gruszczyński Nov 11 '16 at 22:40
  • @Mad Hatter: But it's not monotone. – Asaf Karagila Nov 11 '16 at 22:42
  • @Asaf Why does it have to be monotone? – Rafał Gruszczyński Nov 11 '16 at 22:45
  • @Mad Hatter: It's the definition of cofinality. – Asaf Karagila Nov 11 '16 at 22:47
  • @Asaf According to one of the characterizations of cofinality, the cofinality of $\kappa$ is the smallest ordinal $\alpha$ such that there is a function from it to $\kappa$ such that the image of $\alpha$ under $f$ is unbounded in $\kappa$. But it does not entail monotonicity of $f$, does it? – Rafał Gruszczyński Nov 11 '16 at 22:54
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    @Mad Hatter: Correct, it implies there is a monotone function. But as is the case with equivalent definitions, it depends which one you assume, and which one you prove to be equivalent. The standard definition is an unbounded monotone function. – Asaf Karagila Nov 11 '16 at 22:57
  • @Asaf OK, thanks for the clarification. – Rafał Gruszczyński Nov 11 '16 at 23:02
  • @Asaf How can I be sure that the choices made in the 2nd point are not bounded by some $f(m)$? I cannot see it. Any hint, please. – Rafał Gruszczyński Nov 17 '16 at 22:46
  • @Mad Hatter: They cannot be bounded by some $f(m)$, because by the $m$th step you had to choose an ordinal at least as large as $f(m)$ (prove this by induction), and $\alpha$ is a limit ordinal. – Asaf Karagila Nov 18 '16 at 04:41
  • @Asaf The inductive step is where I get stuck. I must show that $(\forall m\in\omega)(\exists k\in\omega)f(m)\leqslant\alpha_k$. For $0$, $f(0)<\alpha_1$. Assume $f(n)\leqslant\alpha_k$ and consider $f(n+1)$ such that $\alpha_k<f(n+1)$. But according to the definition $\alpha_{k+1}=f(\min{m\mid f(m)>\alpha_k})$, so $\alpha_{k+1}\leqslant f(n+1)$ and if strictly smaller I hit the same problem with $\alpha_{k+2}$. Where do I make a mistake? – Rafał Gruszczyński Nov 18 '16 at 09:46
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    @Mad Hatter: Let's see by induction on $k$ that if $m<k$, then $f(m)\leq a_k$, shall we? For $0$ this is true. Certainly. Now assume this holds for $k$, we took $a_{k+1}$ to be the least such that $a_{k+1}>a_k$. But this means that $a_{k+1}$ is not $f(m)$ for any $m\leq k$, by the induction hypothesis. So $f(m)<a_{k+1}$ for all $m\leq k$, since $a_k$ was greater than all the $f(m)$'s, it follows that $a_k$ itself is at least as large as $f(k)$ itself. Therefore $a_{k+1}$ is also greater than $f(k)$ and therefore at least as large as $f(k+1)$, by virtue of our choice. – Asaf Karagila Nov 18 '16 at 10:37
  • @Mad Hatter: There is a slightly easier variant, that will work with any countable ordering, where we take $a_k$ to be the least in the enumeration such that $a_k$ is greater than $a_{k-1}$ and all the $f(m)$ for $m<k$. This ensures that the sequence is strictly increasing, and the proof is slightly easier. But that is somewhat redundant if you want to streamline the argument. – Asaf Karagila Nov 18 '16 at 10:38
  • @Asaf Thanks a lot! One more question. As Andrés remarked above, you can actually skip the minimality requirement and just take $\alpha_{k+1}\in{f(m)\mid f(m)>\alpha_k}$ (it doesn't harm the argument, does it?). You then need (?) at least the dependent choices to construct the sequence. And your construction and proof do not call for any version of AC, do they? – Rafał Gruszczyński Nov 18 '16 at 10:59
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    @Mad Hatter: No, this doesn't harm the argument, and since there are definable choice functions in this case (as you're going through an ordinal), you don't even need the axiom of choice. But since we can specify something relatively concrete, why not? (Also, note that this choice depends on the enumeration, which is in fact somewhat arbitrary.) You will need AC if you want to find a cofinal well-ordered sequence in an arbitrary linearly ordered set. – Asaf Karagila Nov 18 '16 at 11:02
  • @Asaf Thanks one more time for the help :) – Rafał Gruszczyński Nov 18 '16 at 11:23
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Hint:Let $\gamma$ be a countable ordinal. Enumarate $\gamma=\{\alpha_n:n<\omega\}$, construct $\beta_n$ as $\beta_n=\alpha_{m}$, where $m$ is the least $m<\omega$ with $\alpha_m>\beta_{n-1}$, then what is $\lim_{n\rightarrow \omega}\beta_n$?

Camilo Arosemena-Serrato
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  • Thank you for the answer. What are $\alpha_n$ and $\gamma?$ – Bartek May 12 '13 at 13:25
  • OK, I understand. The supremum is $\gamma$, right? So $\beta_n$ is cofinal in $\gamma,$ and its type is cleary $\omega.$ Thanks! – Bartek May 12 '13 at 13:32