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Using the fact that every countable ordinal is isomorphic to a closed subset of $\mathbb Q$, I find out that any countable limit ordinal is the union of a sequence of increasing ordinal. Now I'm trying to generalize the result to what follows:

Let $\alpha\in ON$ be a limit ordinal, and $\kappa:=\vert\alpha\vert$ the corresponding cardinal. Then there always exists a order-preserving function $f:\kappa\to ON$, such that $\bigcup\{f(\xi)\mid\xi\in\kappa\}=\alpha$.

But I have no ideal how to prove or disapprove my conjecture. Can anyone help me? Thanks in advance.

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    $\begingroup$ For any ordinal $\alpha$ we have $\alpha=\{\beta s.t. \beta<\alpha\}$ and, since for ordinals $\beta<\alpha$ iff $\beta\subset\alpha$ you have $\alpha=\cup_{\beta<\alpha}\beta$. The sum is therefore indexed by the ordinal $\alpha$ which has cardinality $\kappa$ $\endgroup$
    – user126154
    Apr 18, 2015 at 10:07
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    $\begingroup$ by the way, $\kappa$ is an cardinal ansd not an ordinal, so the sentece "order-preserving function $f:\kappa\to ON$" makes no sense. $\endgroup$
    – user126154
    Apr 18, 2015 at 10:08
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    $\begingroup$ @user126154: In set theory every cardinal is an ordinal (with the property that there is no bijection with any strictly smaller ordinal). $\endgroup$
    – user642796
    Apr 18, 2015 at 10:29
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    $\begingroup$ If you want limit ordinals consider the smallest uncountable cardinal $\alpha$. then $\alpha$ and $\alpha+\omega$ have both cardinality $\alpha$ but $\alpha$ has no cofinal embedding in $\omega$ because it is uncountable and it is the smallest such. $\endgroup$
    – user126154
    Apr 18, 2015 at 10:57
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    $\begingroup$ By the way, the fact that every countable limit ordinal is the limit of an $\omega$ sequence can be done directly by using the fact that the ordinal is countable. Just enumerate it, and construct the sequence by recursion over the enumeration. (And unlike the use of $\Bbb Q$, this method is easily generalizable to higher cardinalities.) $\endgroup$
    – Asaf Karagila ♦
    Apr 18, 2015 at 12:30

1 Answer 1

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This is not quite possible, but we can slightly fix this.

First let me point out some defining characteristic of cardinals. These are well-orders that every proper initial segment has a strictly smaller cardinality. On the other hand, $\omega_2+\omega_1$ has an initial segment equipollent to the entire order.

So now we run into a problem. If there is an order preserving function from $\omega_2$ into $\omega_2+\omega_1$, then at some point it had to enter the tail segment of order type $\omega_1$. But the domain of the function still has an end segment of type $\omega_2$, which means there is no order preserving function like that.

But at the same time it is obvious what such function might be for $\omega_5+\omega_5$, or $\omega_{53}^{\omega_3}+\omega_{53}$, or whatever.

The solution is to understand that every ordinal has a cofinality, which is exactly the smallest ordinal which has a cofinal embedding into the ordinal. Namely, $$\operatorname{cf}(\alpha)=\min\{\delta\mid\exists f\colon\delta\to\alpha\text{ order preserving and unbounded}\}.$$

You can easily show that $\operatorname{cf}(\alpha)\leq\alpha$, and for limit ordinals $\omega\leq\operatorname{cf}(\alpha)$ (I'm not counting $0$ as a limit ordinal). We can also show that $\operatorname{cf}(\alpha)=\operatorname{cf}(\operatorname{cf}(\alpha))$.

And if we are careful enough, we might be able to prove that $\operatorname{cf}(\alpha)$ is always a cardinal.

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  • $\begingroup$ Thanks for your elaborate explanation. Only a small puzzle remained. At the end you said that $\operatorname{cf}(\alpha)$ is always a cardinal (I suppose you meant $\alpha$ being limit ordinal). But you have already show that $\operatorname{cf}(\omega_2+\omega_1)>\omega_2$ and so $\operatorname{cf}(\omega_2+\omega_1)$ cannot be a cardinal. Am I misunderstanding what you meant? $\endgroup$
    – Censi LI
    Apr 18, 2015 at 12:29
  • $\begingroup$ No, I only showed that $\operatorname{cf}(\omega_2+\omega_1)\neq\omega_2$. There's nowhere in the rules that says that cofinality is a strictly increasing function. It's not. As for it being a cardinal, well, for successor ordinals the cofinality is $1$, and the cofinality of $0$ is just $0$. So it's always a cardinal. $\endgroup$
    – Asaf Karagila ♦
    Apr 18, 2015 at 12:32
  • $\begingroup$ Ah, I was misunderstanding the definition of cofinality... Thank you very much! $\endgroup$
    – Censi LI
    Apr 18, 2015 at 12:37

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