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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.

2 Answers 2

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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 $\sum_{\xi < \delta} \kappa_\xi < \kappa$.

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    If $\lambda^+=\kappa$ then for all $\alpha<\kappa$ we haveh $|\alpha|\leq\lambda$, in particular $\sup\{|\alpha|:\alpha<\kappa\}=\lambda<\kappa$.2012-08-21
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    @Asaf: I think I wrote that above (well, except for the "$< \kappa$" part at the very end).2012-08-21
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    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)2012-08-21
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    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. :-)2012-08-21
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    @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.2012-08-21
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    @Arthur: What is the significance of continually saying "increasing sequence that is cofinal". Is it not sufficient to say "set that is cofinal"? The set has no sense or order between elements, so it's meaningless to say they are increasing, but doesn't "cofinal" capture everything that "increasing sequence" does - that for every element in parent set, there is an element in the subset that is greater?2012-08-21
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    @Travis: Let's look at the first instance of "(increasing) sequence" in my answer above. Note that while I may have used "set braces", since the family is indexed by the ordinals $< \delta$, what I actually have in mind a function $i : \delta \to \mathrm{On}$ given by $\xi \mapsto \alpha_\xi$, and this function/sequence may be considered increasing accordingly. You are right that their use is pretty non-essential, especially in the first case, but sometimes it's nice to have some additional structure. (cont...)2012-08-21
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    @Travis: (...inued) In the second place I used "(increasing) sequence" I was getting a sequence of cardinals cofinal in the limit cardinal $\kappa$. Later I said "Given $\delta < \mu$ note that $\kappa_\xi < \kappa_\delta$ for all $\xi < \delta$...." I could only do this if I knew the sequence was increasing. Of course, I could have got around this little snag by some circumlocution ("Given $\delta < \mu$ there is a $\varepsilon < \mu$ such that $\kappa_\xi < \kappa_\varepsilon$ for all $\xi < \delta$....") but, IMHO, it would not be as natural or clean.2012-08-21
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    Not necessary, but equivalent and useful. Thanks for all your help! Your posts are very clear and insightful.2012-08-21
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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.

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    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.2012-08-21
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    @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 $\geq\kappa$?2012-08-21
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    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.2012-08-21
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    @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.2012-08-21
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    @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.2012-08-21
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    @Travis: When I took my first serious course in set theory I recall sitting and trying to give a better proof of the equivalence between "ordinal cofinality" and "cardinal cofinality". I don't recall needing cases, and I think that my "trick" with ordinal summation of the cardinals was really useful. I'm too busy to find that argument again nowadays, though. But maybe you can with this argument.2012-08-21