Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The following is a lemma in Just/Weese on page 179:

Lemma 17: Let $\kappa$ be an infinite cardinal. Then $\kappa$ is singular iff there exist an $\alpha < \kappa$ and a set of cardinals $\{\kappa_\xi : \xi < \alpha \}$ such that $\kappa_\xi < \kappa$ for every $\xi < \alpha$ and $\kappa = \sum_{\xi < \alpha}\kappa_\xi$.

I think the following constitutes a proof of the $\implies$ direction, can you please tell me where it's wrong? Thank you!

Let $\alpha = \mathrm{cf}(\kappa) < \kappa$. Then by a previous exercise (23 (f)) there exists a strictly increasing function $f: \alpha \to \kappa$ such that the range of $f$ is cofinal in $\kappa$. Now for $\xi < \alpha$ define $\kappa_\xi = |f(\xi)|$. Now we want to show that $\kappa = \sum_{\xi < \alpha}\kappa_\xi$. To see this observe that "$\ge$" immediately follows from theorem 14 on the same page.

The part I need you to check starts here:

To show $\le$, assume that we have strict inequality $ \sum_{\xi < \alpha}\kappa_\xi < \kappa $. We distinguish two cases: $\kappa$ is either an infinite successor cardinal or a limit cardinal. If $\kappa = \left ( \sum_{\xi < \alpha}\kappa_\xi \right )^+$ is an infinite successor cardinal then by corollary 15 (on the same page) $\kappa$ cannot be the union of fewer than $\kappa$ sets each of which has cardinality less than $\kappa$. But $\bigcup_{\xi < \alpha}f(\xi) = f[\alpha] = \kappa $ where $|f(\xi)| = \kappa_\xi < \kappa$ and $\alpha < \kappa$ which would be a contradiction. Hence $\kappa$ must be a limit cardinal so that there exists a cardinal $\eta$ with $\sum_{\xi < \alpha}\kappa_\xi < \eta < \kappa$. But $f$ is cofinal in $\kappa$ hence there is $\xi$ such that $f(\xi) \ge \eta$. Then $\kappa_\xi = |f(\xi)| \ge \eta$. Which is a contradiction.

We have:

enter image description here

enter image description here

And also, the reason why I am asking this question: in the proof in the book, $\kappa_\xi$ are defined as $\kappa_\xi = \left | f(\xi) \setminus \sum_{\eta < \xi} f(\eta) \right |$.

share|cite|improve this question
$\bigcup_{\xi<\alpha}f(\xi)=f(\alpha)=\kappa$ doesn’t make sense, since $\alpha\notin\operatorname{dom}f$. And how do you get from $\kappa$ being a limit cardinal to the existence of an $\eta$ such that $\sum_{\xi<\alpha}\kappa_\xi<\eta<\kappa$? You know that the sum is $\kappa$. – Brian M. Scott Dec 15 '12 at 11:44
@BrianM.Scott But if $f: X \to Y$ is a function why can't I write $f(X)$ to mean the image of $f$? In any case: I think this points out one idiotic mistake: $f$ isn't surjective so that $f(\alpha)$ is not necessarily $\kappa$. – Rudy the Reindeer Dec 15 '12 at 11:48
The definition $\kappa_\xi =\left | f(\xi)\setminus\sum_{\eta <\xi}f(\eta)\right |$ has the virtue of chopping $\kappa$ into disjoint pieces and taking their cardinalities. – Brian M. Scott Dec 15 '12 at 11:50
If you want $\{f(\xi):\xi\in\alpha\}$, write $f[\alpha]$ to avoid ambiguity. – Brian M. Scott Dec 15 '12 at 11:51
You’re right: I was temporarily confused when I wrote that about what you’d assumed at that point. – Brian M. Scott Dec 15 '12 at 13:30
up vote 2 down vote accepted

There is nothing wrong with your proof, except that "$\bigcup_{\xi < \alpha}f(\xi) = f[\alpha] = \kappa$" should be "$\bigcup_{\xi < \alpha}f(\xi) = \sup(f[\alpha]) = \kappa$."

However, a proof using $\kappa_\xi = \left | f(\xi) \setminus \sum_{\eta < \xi} f(\eta) \right |$ may have the advantages of not splitting into cases, and of not requiring the Axiom of Choice in successor cases. (Without AC, successor cardinals may be singular.)

share|cite|improve this answer

Your Answer


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.