# For every cardinal $\kappa$, $\kappa^+$ is regular

Again I'm struggling with a proof from this introduction to cardinals.

Lemma 2.6. For every cardinal $$\kappa$$, $$\kappa^+$$ is regular.

Proof. If not, then there would be a cofinal map $$f:\lambda\to\kappa$$, where $$\lambda\leq \kappa$$. This would write $$\kappa^+$$ as a $$\kappa$$ union of sets, each of which size $$\leq \kappa$$. This contradicts theorem 1.8.

I think it should be $$f:\lambda\to\kappa^+$$, right?

But my main question is how the "this would write" conclusion works. How do we get from a cofinal map to a $$\kappa$$ union? I've read the entire paper and I still don't get it.

Also, on a side note, is this theorem equivalent to the statement that for any successor ordinal $$\alpha$$, we have $$\text{cf}(\omega_\alpha)=\omega_\alpha$$?

You can probably find the proof that every successor cardinal is regular in many places, for example, here: A Successor Cardinal is Regular. The discussion of the use of Axiom of Choice in the proof of this result can be found here: The regularity of successor cardinal.

I will only write about the part in this particular proof which you singled out as unclear to you.

Suppose that $f \colon \lambda \to \kappa^+$ is cofinal. This means that $\{f(\alpha); \alpha<\lambda\}$ is a cofinal set in $\kappa^+$, i.e., $$\sup_{\alpha<\lambda} f(\alpha) = \kappa^+.\tag{1}$$ This is just a different way of writing $$\bigcup_{\alpha<\lambda} f(\alpha) = \kappa^+.\tag{2}$$

Notice that we have used $|\lambda|$-many sets and $|\lambda|\le\kappa$.

And we also have $f(\alpha)<\kappa^+$. Since $\kappa^+$ is the smallest ordinal with cardinality $\kappa^+$, cardinality of each $f(\alpha)$ is at most $\kappa$.

So $\kappa^+$ is written $(2)$ as union of at most $\kappa$ sets, each of them having cardinality at most $\kappa$.

• Perfect! I really failed to see $(1)$. :S Commented Apr 25, 2016 at 18:26