Borceux - Handbook of categorical algebra p.268


An infinite cardinal $\alpha$ is called regular if for every family of sets $\{X_i\}_{i\in I}$, $|I|<\alpha$ and $|X_i|<\alpha$ implies that $|\bigcup_{i\in I} X_i|<\alpha$.

It is stated in the text that for every family of cardinals $\{\alpha_i\}_{i\in I}$, there exists a regular cardinal $\alpha$ such that $\alpha_i< \alpha$ for all $i\in I$.

Is it possible to prove this under ZFC?

(This is equivalent to prove that given a cardinal $\beta$, there exists a regular cardinal $\alpha$ such that $\beta<\alpha$.)

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    $\begingroup$ Every successor cardinal $\kappa^+$ is regular. $\endgroup$ Jan 22 '17 at 21:51

Assuming the axiom of choice, if $\kappa$ is a cardinal, then $\kappa^+$ is regular:

Suppose that $f\colon\alpha\to\kappa^+$ for some $\alpha<\kappa^+$, then for all $\eta<\alpha$, $f(\eta)<\kappa^+$. Therefore $$|\sup\operatorname{rng}(f)|=\left|\bigcup\{f(\eta)\mid\eta<\alpha\}\right|=|\alpha|\sup\{|f(\eta)|\mid\eta<\alpha\}\leq\kappa\cdot\kappa=\kappa.$$

Therefore if $\alpha<\kappa^+$, and $f\colon\alpha\to\kappa^+$, then $\operatorname{rng}(f)$ is bounded in $\kappa^+$. In other words, $\kappa^+$ is regular.

Now suppose that $A$ is any set of cardinals, then $(\sup A)^+$ is a regular cardinal, strictly larger than all the members of $A$.

The axiom of choice is used here to prove the correctness of the infinitary sums in the cardinal arithmetic part of the show. Namely, for every $\eta<\alpha$, we need to choose an injection from $f(\eta)$ into $\kappa$. And indeed, without the axiom of choice it is consistent that $\omega_1$ is not regular. And in fact, assuming the consistency of large cardinals, it is possible that all cardinals have countable cofinality.

  • $\begingroup$ Thank you. It is off the topic, but is it unprovable that "an infinite cardinal greater than $\alpha_0$ is a successor of some cardinal" under ZFC? (If an inaccessible cardinal exists, this is false. right?) $\endgroup$
    – Rubertos
    Jan 22 '17 at 22:05
  • $\begingroup$ It is outright disprovable. Look at the $\omega$th successor. $\endgroup$
    – Asaf Karagila
    Jan 22 '17 at 22:09
  • $\begingroup$ @Rubertos "an infinite regular cardinal must be the successor of some cardinal" is the negation of the existence of weakly inaccessible cardinals. $\endgroup$ Jan 23 '17 at 13:44

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