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

If a cardinal $\kappa$ is regular then it cannot be written as a union of fewer than $\kappa$ sets, each of size less than $\kappa$.

This seems to be a very useful characterization. I have seen a proof or two, but can't grasp all the details... I am horrible with ordinal and cardinal arithmetic. Could someone please give an elementary (as much as possible) proof of this theorem for me?

share|cite|improve this question
What is your definition of regular cardinals? – Chris Eagle Feb 10 '13 at 17:58
$\kappa$ is a limit ordinal and $cf(\kappa)=\kappa$ – Forever Mozart Feb 10 '13 at 17:59
It might be wise to strengthen your cardinal arithmetics first, then. One should not attempt to eat a steak with a silly straw. – Asaf Karagila Feb 10 '13 at 18:01
up vote 5 down vote accepted

The simplest proof I can think of is as follows.

Suppose that $\kappa$ is a regular cardinal, let $\lambda < \kappa$ and let $\{ A_\xi : \xi < \lambda \}$ be a family of subsets of $\kappa$ each of cardinality $< \kappa$. As $\kappa$ is regular, then no $A_\xi$ is cofinal in $\kappa$ (since $|A_\xi| < \kappa = \mathrm{cf} ( \kappa)$), meaning that for each $\xi < \lambda$ there is an $\alpha_\xi < \kappa$ such that $A_\xi \subseteq \alpha_\xi$ — what I really mean here is $\beta < \alpha_\xi$ for all $\beta \in A_\xi$. Again by the regularity of $\kappa$ the family $\{ \alpha _\xi : \xi < \lambda \}$ cannot be cofinal in $\kappa$, and so there is an $\alpha < \kappa$ such that $\alpha_\xi < \alpha$ for all $\xi < \lambda$. But then it easily follows that $\bigcup_{\xi < \lambda} A_\xi \subseteq \alpha \subsetneq \kappa$.

There is no cardinal/ordinal arithmetic needed here; all we have used is the definition of cofinality. If there are particular points that are troubling you, please point them out and I can (try to) elucidate further.

share|cite|improve this answer
Ok this is exactly what I was looking for. I actually tried something very similar, applying the definition twice, but confused myself at the end. In your first step, I take it you are just assuming the sets A are subsets of kappa, using the cardinality bijection. If only you could see the other proof I was trying to understand! – Forever Mozart Feb 10 '13 at 18:59
@DavidL. In my first step I am just taking the usual von Neumann definition of an ordinal (the set of all ordinals strictly less than it) and noting that cardinals are just ordinals of a particular type. I'm quite glad I don't have to see the other proof you have been pouring over! – arjafi Feb 10 '13 at 19:40

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.