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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?

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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
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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

1 Answer 1

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.

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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! –  Arthur Fischer Feb 10 '13 at 19:40

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