Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Suppose we have a set of ordinals such that their supremum is a cardinal (not in the original set).

$\sup\{\alpha\}=\kappa$

I'm interested in the supremum of the cardinalities of those ordinals: $\sup\{|\alpha|\}$. I've seen many proofs that seem to assume that $\sup\{|\alpha|\}=\kappa$ also. I can see how this would be the case if the set $\{\alpha\}$ contained an infinite subset of cardinals whose supremum was also $\kappa$, but I can think of a simple counterexample that would seem to indicate that's not always the case.

Take our set to be $\{\alpha\mid\omega\le\alpha\lt\omega_1\}$. The supremum of this set is clearly $\omega_1$ (with cardinality $\aleph_1$ if you prefer). But each ordinal $\alpha$ is countably infinite, so $\forall\alpha,|\alpha|=\omega$ (or $\aleph_0$ if you insist). But then $$\sup\{|\alpha|\}=\sup\{\omega\}=\omega\ne\omega_1=\sup\{\alpha\}.$$

What am I missing?





Motivation: I'm trying to show that if $\kappa$ is an infinite cardinal, these definitions for cofinality are equivalent:

$$cof(\kappa)=\inf\{\beta\mid\{\alpha_\xi\}_{\xi\lt\beta}\text{ is cofinal in }\kappa\}\equiv\inf\{\delta\mid\sum_{\xi<\delta}\kappa_\xi=\kappa\}.$$

To establish the RHS, (after taking $\{\kappa_\xi\}=\{|\alpha_\xi|\}$) it's easy to show the summation is bounded above by $\kappa$. To show that it is also bounded below by $\kappa$, an example proof I found claims that since each $$\kappa_\xi\le\sum_{\xi<\delta}\kappa_\xi$$ then the summation, as an upper bound of all of the $\kappa_\xi$'s, must be larger than the smallest upper bound (supremum) of the $\kappa_\xi$'s, so $$\kappa\le\sup\{\alpha_\xi\}=\sup\{|\alpha_\xi|\}=\sup\{\kappa_\xi\}\le\sum_{\xi<\delta}\kappa_\xi$$ but I'm not so sure about the first equality. Any tips would be appreciated.

share|improve this question

2 Answers 2

up vote 2 down vote accepted

I'll first make a comment about the concern in the title of the question, and it seems to break down into successor and limit cases.

  • If $\kappa = \lambda^+$ is a successor cardinal, then $| \alpha | \leq \lambda$ for all ordinals $\alpha < \kappa$, and it easily follows that $\sup \{ | \alpha | : \alpha < \kappa \} = \lambda$. (This is what you noticed for $\kappa = \aleph_1 = \aleph_0^+$.)
  • If $\kappa$ is a limit cardinal, then $\mu^+ < \kappa$ for each $\mu < \kappa$. From here it easily follows that $\sup \{ | \alpha | : \alpha < \kappa \} = \kappa$.

The above dichotomy also holds whenever $\{ \alpha_\xi \}_{\xi < \nu}$ is an arbitrary family of ordinals cofinal in $\kappa$.

Of course, it is relatively easy to show that successor cardinals are regular: If $\delta < \kappa = \lambda^+$ and $\{ \alpha_\xi \}_{\xi < \delta}$ is an (increasing) sequence of ordinals $< \kappa$, then let $\beta = \sup_{\xi < \delta} \alpha_\xi = \bigcup_{\xi < \delta} \alpha_\xi$. Note that $$| \beta | = \left| \bigcup_{\xi < \delta} \alpha_\xi \right| \leq \sum_{\xi < \delta} | \alpha_\xi | \leq \sum_{\xi < \delta} \lambda = | \delta | \cdot \lambda = \lambda < \kappa.$$ Thus $\{ \alpha_\xi \}_{\xi < \delta}$ is not cofinal in $\kappa$. Note that this also contains the idea for proving the summation characterisation of cofinality in this case: if $\{ \kappa_\xi \}_{\xi < \delta}$ is a family of cardinals $< \kappa = \lambda^+$, then $\sum_{\xi < \delta} \kappa_\xi \leq \sum_{\xi < \delta} \lambda = | \delta | \cdot \lambda = \max \{ |\delta| , \lambda \}$, so if this sum equals $\kappa = \lambda^+$, it must be that $| \delta | = \kappa$.)

For a limit cardinal $\kappa$, letting $\mu = \mathrm{cof} ( \kappa )$, note that there is an (increasing) sequence $\{ \kappa_\xi \}_{\xi < \mu}$ of cardinals $< \kappa$ which is cofinal in $\kappa$. It is relatively easy to show that $\sum_{\xi < \mu} \kappa_\xi = \kappa$.

  • Clearly $\kappa_\eta \leq \sum_{\xi < \mu} \kappa_\xi$ for all $\eta < \mu$, and therefore $\kappa \leq \sum_{\xi < \mu} \kappa_\xi$.
  • Given $\delta < \mu$ note that $\kappa_\xi < \kappa_\delta$ for all $\xi < \delta$ and so $\sum_{\xi < \delta} \kappa_\xi \leq \sum_{\xi < \delta} \kappa_\delta = | \delta | \cdot \kappa_\delta = \max \{ | \delta | , \kappa_\delta \}$. As $\kappa_\delta < \kappa$, and $| \delta | < \kappa$, it follows that $\sum_{\xi < \delta} \kappa_\xi < \kappa$. Thus, $\sum_{\xi < \mu} \kappa_\xi \leq \kappa$.

But suppose $\delta < \mu$ and $\{ \kappa_\xi \}_{\xi < \delta}$ is any sequence of cardinals $< \kappa$. Then $\{ \kappa_\xi \}_{\xi < \delta}$ is not cofinal in $\kappa$, and so there is a cardinal $\nu < \kappa$ such that $\kappa_\xi \leq \nu$ for all $\xi < \delta$. Then $$\sum_{\xi < \delta} \kappa_\xi \leq \sum_{\xi < \delta} \nu = | \delta | \cdot \nu = \max \{ | \delta | , \nu \}.$$ As $\nu , | \delta | < \kappa$, we have that $\sum_{\xi < \delta} \kappa_\xi < \kappa$.

share|improve this answer
    
If $\lambda^+=\kappa$ then for all $\alpha<\kappa$ we haveh $|\alpha|\leq\lambda$, in particular $\sup\{|\alpha|:\alpha<\kappa\}=\lambda<\kappa$. –  Asaf Karagila Aug 21 '12 at 8:52
    
@Asaf: I think I wrote that above (well, except for the "$< \kappa$" part at the very end). –  Arthur Fischer Aug 21 '12 at 9:03
    
Arthur, it seems that I should work better on read. I confused $\kappa$ and $\lambda$ mid-sentence... :-) (I hope to see you when I visit in two weeks, btw) –  Asaf Karagila Aug 21 '12 at 9:17
    
So it does depend on whether $\kappa$ is a limit cardinal or not. It seems to me the proof was applying the concept indiscriminately to all infinite cardinals. Either I'm still missunderstanding the proof, or it was hastily done (for classroom display). I'm going to take some time and digest the rest of what you wrote on cofinality, but since you addressed the actual question, I'll mark it as answered. You two know each other IRL? Cool. :-) –  Travis Bemrose Aug 21 '12 at 15:14
1  
@Travis: The result (i.e., the characterisation of $\mathrm{cof}(\kappa)$ in terms of cardinal sums) doesn't depend on what $\kappa$ is. But it does seem (IMHO) that the proof needs to be split into the successor/limit cases, for the reason you noted. I don't think this is too surprising, however, since in the successor case so much more is known about $\kappa$ (in particular, we know what $\kappa$'s cofinality is). The limit case is the more elegant case, and I can see someone choosing to ignore some details in a classroom setting, even if this choice is pedagogically unsound. –  Arthur Fischer Aug 21 '12 at 15:32

Well, as you noted the supremum of the cardinalities is never larger than the supremum of ordinals.

To have a better understanding of this equivalence I suggest to divide into two cases:

  1. If $\kappa$ is regular then the least $\beta$ is $\kappa$ itself, and since we have: $$\inf\left\{\beta\mid\exists \{\alpha_\xi\}_{\xi<\beta}\text{ cofinal in }\kappa\right\}=\kappa\geq\inf\left\{\delta\mid\exists\{\kappa_\xi\}_{\xi<\delta}: \sum_{\xi<\delta}\kappa_\xi=\kappa\right\}$$ Note that $\kappa\geq\inf$ because we can always write $\kappa$ as sum of $\kappa$ singletons, so trivially we have the above.

    On the other hand if $\{\kappa_\xi\}_{\xi<\delta}$ is a sequence of cardinals of minimal order type then the ordinals $\alpha_\xi=\left(\sum_{\zeta<\xi}\kappa_\zeta\right)+\kappa_\xi$ (this sum is an ordinal sum!) form a cofinal sequence in $\kappa$, and therefore the equality follows.

  2. If $\kappa$ is singular then we know it can be written as a limit of cardinals, then we indeed have $\sup\{\alpha_\xi\mid\xi<\delta\}=\sup\{|\alpha_\xi|\mid\xi<\delta\}$ as wanted.

    It is therefore sufficient to show that there is a minimal cofinal sequence made of cardinals, but this is simple.

share|improve this answer
    
Everything is clear (including for regular $\kappa$, LHS=$\kappa$), except for the part you call trivial. Proofs gloss over and I'm missing something. Rather than using 2 cases, I can show for both cases that LHS$\le$RHS. Trying to show that LHS$\ge$RHS, I take the smallest $\beta$ from the LHS and use it in place of $\delta$ in the RHS. Showing $\beta\in\{RHS\}$ shows $\beta\ge\inf\{RHS\}$. Showing $\sum_{\xi\lt\beta}\kappa_\xi\le\kappa$ is easy. Trying to show $\kappa\le\sum_{\xi\lt\beta}\kappa_\xi$ is what led to my question above. –  Travis Bemrose Aug 21 '12 at 15:54
    
@Travis: I'm not sure which part is not clear to you. Is it the fact that the first $\inf$ is above the second, or that they are both $\geq\kappa$? –  Asaf Karagila Aug 21 '12 at 16:43
    
Well, both I guess; I suppose I've not fully grasped the implications of a sum of cardinals in the RHS. As I wrote, I've nearly proven everything just comparing the LHS and RHS for both regular and singular cardinals, not comparing either to $\kappa$. I see that sandwiching the RHS between the LHS and $\kappa$ gives equality (for regular cardinals), but I don't see why it is. I've almost been able to show the LHS$\ge$RHS relationship, but not quite. Doing so for both regular and singular cardinals would finish my proof, but if I need to rewrite using cases I will. –  Travis Bemrose Aug 21 '12 at 17:08
    
@Travis: Hm. Oh right. I have a misprint there. You're right, it's not trivial, it's wrong! :-D I'll remedy that in a moment. –  Asaf Karagila Aug 21 '12 at 17:15
    
@Asef: Ah, yes. Your large expression makes sense now. :-) I'm trying to wrap my head around your ordinal sum, but I need to take a break and rest my brain. After lunch I'll review this and read what Arthur wrote about cofinality and decide if I need to rewrite my proof with cases or not. Thanks for the help. –  Travis Bemrose Aug 21 '12 at 17:34

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.