# Measurable Cardinals are Mahlo Cardinals

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).

• 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? – Stefan Mesken Mar 22 at 16:39

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.

• 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) and how to we show this? To prove the claim that if 𝐶⊆𝜅 is club then 𝜅∈𝜋(𝐶), how does appealing to the elementarity of 𝜋 help? I am not so familiar with ultrapowers (but I have some knowledge of ultraproducts and ultrafilters). – E.Green4321 Mar 23 at 1:31
• @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. – Stefan Mesken Mar 23 at 13:52
• Thank you so much for pointing me in the right direction and for your explanations. – E.Green4321 Mar 23 at 20:13
• @E.Green4321 You're welcome! – Stefan Mesken Mar 23 at 21:16

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 $$\alpha \in A_\xi$$.

Let

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

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 $$\alpha = \sup_{\xi < \lambda} f_\alpha(\xi) = y_\xi$$ 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!