I am new to set theory and have been working through the proof that every measurable cardinal is Mahlo on page 135 of Jech's text. With the help of Asaf's comments (Measurable $\rightarrow$ Mahlo), I have been able to make sense of the first half of the proof.

However, I found the second half (that argues by contradiction that $\{\alpha < \kappa : \alpha \text{ is regular} \} \in D$) quite terse, and cannot quite follow what is going on.

Could someone please provide a detailed version of Jech's proof or perhaps a detailed alternative proof (that mimics the proof that every measurable cardinal is inaccessible, which I believe I understand better).

Thank you in advance for your help.

  • $\begingroup$ I don't have Jech available to me right now. But the simplest proof, in my opinion, is via ultrapowers. Are you familiar with them? $\endgroup$ Commented Mar 22, 2019 at 16:39

2 Answers 2


Here is a detailed explanation of Jech's proof.

Let $D$ be a normal measure on $\kappa$. Suppose, towards a contradiction, that $\kappa$ is not Mahlo. Then there is some club $C \subseteq \kappa$ such that $$ C \cap \{ \alpha < \kappa \mid \mathrm{cof}(\alpha) = \alpha \} = \emptyset. $$

Since $D$ is normal, it contains all clubs. In particular $C \in D$. Since $D$ is closed under intersections, we therefore must have that $\{ \alpha < \kappa \mid \mathrm{cof}(\alpha) = \alpha \} \not \in D$ and hence that $$ \{ \alpha < \kappa \mid \mathrm{cof}(\alpha) < \alpha \} = \kappa \setminus \{ \alpha < \kappa \mid \mathrm{cof}(\alpha) = \alpha \} \in D. $$

By normality there is some $\lambda < \kappa$ such that

$$ E_\lambda = \{ \alpha < \kappa \mid \mathrm{cof}(\alpha) = \lambda \} \in D. $$

By replacing $E_\lambda$ with $E_\lambda \setminus \lambda$ we may and shall assume that $E_\lambda \cap \lambda = \emptyset$

For each $\alpha \in E_\lambda$ fix a strictly increasing, cofinal function $$ f_\alpha \colon \lambda \to \alpha $$ Now, for each $\xi < \lambda$, the function $$ g_\xi \colon E_\lambda \to \kappa, \ \alpha \mapsto f_\alpha(\xi) $$ is decreasing. Hence there is some $A_\xi \in D$ and some $y_\xi < \kappa$ such that $f_\alpha(\xi) = y_\xi$ for all $\alpha \in A_\xi$.


$$ A = \bigcap_{\xi < \lambda} A_\xi. $$

Since $\lambda < \kappa$ we have that $A \in D$.

Now let $\alpha \in A$. For all $\xi < \lambda$ we have $f_\alpha(\xi) = y_\xi$ is independent of $\alpha$ (by the construction of $A_\xi$). But $$ \alpha = \sup_{\xi < \lambda} f_\alpha(\xi) = \sup_{\xi < \lambda} y_\xi $$ is completely determined by the sequence $(y_\xi \mid \xi < \lambda)$.

Hence $A$ contains at most one element. This is a contradiction, since $D$ is non-principal.

It follows that $\kappa$ is Mahlo after all!

  • $\begingroup$ Your two proofs here are very good and quite interesting too, Stefan. I'm a bit surprised that there isn't some simpler, and more intuitive, way way to prove the same thing, however, since neither Measurable nor Mahlo cardinals are difficult to define. I certainly have no idea what such a simpler proof might be, though, or whether it is even possible at all. $\endgroup$
    – Wd Fusroy
    Commented Oct 26, 2019 at 17:03

Here is a proof leveraging ultrapowers. It generalizes well to all sorts of situations which is why I recommend learning it at some point:

In the following let $U$ be a normal ultrafilter on a measurable cardinal $\kappa$ and let $$ \pi \colon V \to \mathrm{Ult}(V;U) $$ be the canonical ultrapower embedding (we regard $\mathrm{Ult}(V;U)$ as transitive).

Claim. Let $X \subseteq \kappa$. Then $\pi(X) \cap \kappa = X$.

Proof. For $\xi < \kappa$ we have $$ \xi \in X \iff \pi(\xi) = \xi \in \pi(X). $$ Q.E.D.

Claim. Let $C \subseteq \kappa$ be a club. Then $\kappa \in \pi(C)$.

Proof. By elementarity $$ \mathrm{Ult}(V;U) \models \pi(C) \text{ is a club in } \pi(\kappa) $$ and $\pi(C) \cap \kappa = C$ is unbounded below $\kappa < \pi(\kappa)$.

Thus $$ \mathrm{Ult}(V;U) \models \kappa \in \pi(C) $$ and (by $\Sigma_0$-absoluteness) hence $\kappa \in \pi(C)$. Q.E.D.

Claim. $\mathrm{Ult}(V;U) \models \kappa \text{ is regular}$.

Proof. $\kappa$ is regular in $V$, $\mathrm{Ult}(V;U) \subseteq V$ and regularity is downward-absolute (a short cofinal sequence in $\mathrm{Ult}(V;U)$ would also witness in $V$ that $\kappa$ is singular). Q.E.D.

Now combine all of this:

Let $C \subseteq \kappa$ be a club. Then $$ \mathrm{Ult}(V;U) \models \kappa \in \pi(C) \text{ and } \kappa \text{ is regular }. $$ In particular $$ \mathrm{Ult}(V;U) \models \pi(C) \text{ contains a regular cardinal}. $$ By the elementarity of $\pi$ we obtain that $$ V \models C \text{ contains a regular cardinal}. $$ Since $C$ was an arbitrary club in $\kappa$, it follows that $\kappa$ is Mahlo.

  • $\begingroup$ Thank you so much for this. I am slowing working my way though this proof. Could you explain what you mean by elementarity (of πœ‹) and how to we show this? To prove the claim that if πΆβŠ†πœ… is club then πœ…βˆˆπœ‹(𝐢), how does appealing to the elementarity of πœ‹ help? I am not so familiar with ultrapowers (but I have some knowledge of ultraproducts and ultrafilters). $\endgroup$
    – E. Green
    Commented Mar 23, 2019 at 1:31
  • $\begingroup$ @E.Green4321 The basics of elementary embedding and ultrapowers are covered in Jech's book. It's not something you can summarize in a few sentences. $\endgroup$ Commented Mar 23, 2019 at 13:52
  • $\begingroup$ Thank you so much for pointing me in the right direction and for your explanations. $\endgroup$
    – E. Green
    Commented Mar 23, 2019 at 20:13
  • $\begingroup$ @E.Green4321 You're welcome! $\endgroup$ Commented Mar 23, 2019 at 21:16

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