# Regular cardinals and unions

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?

-
What is your definition of regular cardinals? –  Chris Eagle Feb 10 '13 at 17:58
$\kappa$ is a limit ordinal and $cf(\kappa)=\kappa$ –  Tom Cruise 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

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 $\bigcup_{\xi < \lambda} A_\xi \subseteq \alpha \subsetneq \kappa$.