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 $\beta<\alpha$.)

  • 1
    $\begingroup$ Every successor cardinal $\kappa^+$ is regular. $\endgroup$ – Rene Schipperus 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. $\endgroup$ – Pedro Sánchez Terraf Jan 23 '17 at 13:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.