# Proof that there are enough regular cardinals

Borceux - Handbook of categorical algebra p.268

Definition

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 $$\beta<\alpha$$.)

• Every successor cardinal $\kappa^+$ is regular. 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.

• 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?) Jan 22 '17 at 22:05
• It is outright disprovable. Look at the $\omega$th successor. Jan 22 '17 at 22:09
• @Rubertos "an infinite regular cardinal must be the successor of some cardinal" is the negation of the existence of weakly inaccessible cardinals. Jan 23 '17 at 13:44