An infinite cardinal $\kappa$ is regular if $\mathrm{cf}(\kappa) = \kappa$.

It is known that if $\kappa$ is regular, then for any family $(\kappa_i)_{i \in I}$ of cardinals $\kappa_i < \kappa$ with $|I| < \kappa$ $\kappa \neq \sum_{i \in I} \kappa_i$.

I want to prove the converse: let $\kappa$ be an infinite cardinal; if for any family $(\kappa_i)_{i\in I}$ of cardinals $\kappa_i < \kappa$ with $|I| < \kappa$ we have $\kappa \neq \sum_{i \in I} \kappa_i$, then $\kappa$ is regular.

One should possbily proceed by contradiction. Assume that $\mathrm{cf}(\kappa) < \kappa$. Then there is an ordinal $\alpha < \kappa$ for which there is a cofinal map $f\colon\alpha\to\kappa$, that is, a map $f\colon \alpha\to\kappa$ so that $(\forall \beta < \kappa)(\exists \gamma < \alpha)(\beta \leq f(\gamma))$. From this we should somehow derive a contradicting by showing that $\kappa$ is equal to $\sum_{i \in I} \kappa_i$ for some family $(\kappa_i)_{i \in I}$ of cardinals.


1 Answer 1


There really is only one option that sticks out: $$\sum_{\gamma\in\alpha}f(\gamma)$$

  • $\begingroup$ Can you elaborate on why $\bigcup_{\gamma < \alpha} f(\gamma) = \kappa$? Indeed, $(\forall \beta < \kappa)(\exists \gamma < \alpha)(\beta \leq f(\gamma))$. Note that inequality $\leq$ is not strict (in was in my original question, an unfortunate typo) since it is what the definition of the a cofinal map requires. $\endgroup$
    – Jxt921
    Sep 24, 2018 at 16:44
  • 1
    $\begingroup$ @Jxt921 The strict inequality is a slight inconvenience but doesn't change anything really. If $\beta<\kappa$ then $\beta+1$ is still smaller than $\kappa$, and you can find a $\gamma<\alpha$ such that $\beta+1\leq f(\gamma)$ and therefore $\beta<f(\gamma)$. Now note that $\sum f(\gamma)$ cannot be smaller than $\kappa$ because any smaller ordinal is surpassed by some term in the sum. And it cannot be larger because $\sum f(\gamma)\leq\sum \kappa=\kappa$. $\endgroup$
    – Arthur
    Sep 24, 2018 at 19:38

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