# Properties of Jech's hierarchy of stationary sets (Exercise 8.13, 8.14 of Jech)

I must first preface that while this is indeed a question on an exercise, I believe this is advanced enough for MathOverflow.

Let $$\kappa$$ be a regular uncountable cardinal. Recall that the notion of a stationary subset makes sense for subsets of limit ordinals of uncountable cofinality. In Jech's Set Theory (Third Millenium Edition), he defined an ordering $$<$$ on stationary subsets $$S,T \subseteq \kappa$$ by (Definition 8.18, or (2.6) of the linked paper below): $$S < T \iff S \cap \alpha \text{ is stationary for almost all } \alpha \in T$$ Note that we implicitly assume that almost all $$\alpha \in T$$ are limit ordinals of uncountable cofinality. The standard examples of such stationary sets are of the form: $$E_\lambda^\kappa := \{\alpha < \kappa : \operatorname{cf}(\alpha) = \lambda\}$$ This was first introduced in his paper Stationary subsets of inaccessible cardinals. He proved in Lemma 8.19 (Theorem 2.4 of the paper) that $$<$$ is a well-founded relation, so it makes sense to define a rank function on this relation, $$o(S)$$, which he calls the order of the set.

He then proceeds to give two exercises on this matter:

1. (Exercise 8.13) If $$\lambda < \kappa$$ is the $$\alpha^\text{th}$$ regular cardinal, then $$o(E_\lambda^\kappa) = \alpha$$.

2. (Exercise 8.14) $$o(\kappa) \geq \kappa$$ if and only if $$\kappa$$ is weakly inaccessible; $$o(\kappa) \geq \kappa + 1$$ if and only if $$\kappa$$ is weakly Mahlo.

I have no idea how to solve either exercise. It may be helpful to note that in his paper, he defined the notion of a canonical stationary set (of order $$\nu$$), and mentioned without proof that the set $$E_\lambda^\kappa$$ is the canonical stationary set of order $$\lambda$$. However, I do not see why this is true.

• Just a small remark, you don't need to "implicitly assume almost all ordinals in $T$ are limit ordinals", since it is true that almost all ordinals are limit ordinals. Apr 26, 2022 at 8:28
• @AsafKaragila The implicit assumption includes "of uncountable cofinality", and that's definitely a constraint on $T$, since the ordinals of cofinality $\omega$ form a stationary set. Apr 26, 2022 at 14:03
• In the first of the two exercises, it would be better not to use $\alpha$ for two different things. Apr 26, 2022 at 14:04

Here were my solutions from a few years ago, when I worked through Jech's 1984 paper on this topic.

Exercise 8.13: If $$\lambda<\kappa$$ is the $$\alpha$$th regular cardinal, then $$o(E_{\lambda}^{\kappa})=\alpha$$.

Proof. Now, $$o(E_{\lambda}^{\kappa})\geq \alpha$$ follows from the fact that $$E_{\nu}^{\kappa} when $$\nu<\mu<\kappa$$ are regular cardinals. (This later fact follows from $$E_{\mu}^{\kappa}\subset {\rm Tr}(E_{\nu}^{\kappa})$$, by the previous exercise.)

The other inequality is by induction, and we follow the proof given in Jech [1984: Stationary Subsets of Inaccessible Cardinals]. (We will show that any stationary set of order $$\alpha$$ must have an element of cofinality the $$\alpha$$th regular cardinal.) All of our stationary sets will be restricted to limit ordinals.

For each ordinal $$\alpha<\kappa^{+}$$, we say that a stationary set $$E$$ is canonical of order $$\alpha$$ if (1) $$o(X)=\alpha$$ for every stationary set $$X\subset E$$, and (2) $$E$$ meets every stationary set of order $$\alpha$$. Such a canonical set, if it exists, is clearly unique (modulo nonsingular sets). If it exists, we write $$E_{\alpha}$$ for such a set. (We will eventually show $$E_{\alpha}=E_{\lambda}^{\kappa}$$.)

Lemma 3.1 from Jech's paper: For a stationary set $$S$$, if $$o(S)>\alpha$$ and $$E_{\alpha}$$ exists, then $$S\cap {\rm Tr}(E_{\alpha})\neq \emptyset$$.

Proof of Lemma 3.1. The proof is by contradiction, letting $$S$$ be a minimal counterexample. Let $$T be stationary of order $$\nu$$. Write $$T=A\cup B\cup C$$ where $$A=T\cap E_{\alpha}$$, $$B=T\cap {\rm Tr}(E_{\alpha})$$, and $$C=T-(A\cup B)$$. Since $$A\subset E_{\alpha}$$ we have $${\rm Tr}(A)\subset {\rm Tr}(E_{\alpha})$$, so $${\rm Tr}(A)$$ is disjoint from $$S$$. Next, $$B\subset {\rm Tr}(E_{\alpha})$$, hence $${\rm Tr}(B)\subset {\rm Tr}({\rm Tr}(E_{\alpha}))\subset {\rm Tr}(E_{\alpha})$$ is also disjoint from $$S$$. From $$T we have $$S\subset {\rm Tr}(C)\mod I_{\rm NS}$$. In particular, $$C$$ is stationary and $$C. Since taking the order reverses inclusion, we have $$o(C)\geq o(T)=\alpha$$. From heredity, we cannot have $$o(C)=\alpha$$, so the inequality is strict. But then we have a contradiction on the minimality of $$S$$. $$\square_{\rm Lemma\ 3.1}$$

Corollary 3.2 from Jech's paper: $${\rm Tr}(E_{\alpha})$$ is the largest set (modulo nonstationary sets) of order $$>\alpha$$, if it is non-empty modulo $$I_{\rm NS}$$. In other words, for any stationary set $$S$$, if $$o(S)>\alpha$$ then $$S>E_{\alpha}$$.

Proof of Corollary 3.2. Assume not, so there is some stationary set $$S$$ with $$o(S)>\alpha$$ but $$S$$ is not contained in $${\rm Tr}(E_{\alpha})$$ (modulo $$I_{\rm NS}$$). Let $$T:=S-{\rm Tr}(E_{\alpha})$$, which is a stationary set by the assumption. Further, $$T$$ is disjoint from $${\rm Tr}(E_{\alpha})$$ and $$o(T)\geq o(S)$$ (from order reversal from $$T\subset S$$), a contradiction to the previous lemma. $$\square_{\rm Corollary\ 3.2}$$

Lemma 3.3 from Jech's paper: The set $$E_{\alpha}$$ exists if and only if there exists a largest stationary set $$M_{\alpha}$$ of order $$\alpha$$. Then $$\begin{equation}\tag{A}\label{Eq:3.3fromJech} M_{\alpha}=E_{\alpha}\cup {\rm Tr}(E_{\alpha}) \end{equation}$$ and $$\begin{equation}\tag{B}\label{Eq:3.4fromJech} E_{\alpha}=M_{\alpha}-{\rm Tr}(M_{\alpha}). \end{equation}$$ The union in (\ref{Eq:3.3fromJech}) is disjoint, and $${\rm Tr}(M_{\alpha})\subset M_{\alpha}$$. (Both statements are taken modulo $$I_{\rm NS}$$.) Moreover, $${\rm Tr}(E_{\alpha})={\rm Tr}(M_{\alpha})$$.

Proof of Lemma 3.3. $$(\Rightarrow)$$: Assume $$E_{\alpha}$$ exists. Note that $$E_{\alpha}$$ and $${\rm Tr}(E_{\alpha})$$ are disjoint modulo $$I_{\rm NS}$$ since $$o({\rm Tr}(E_{\alpha}))>\alpha$$ and the heredity property on $$E_{\alpha}$$. Putting $$M:=E_{\alpha}\cup {\rm Tr}(E_{\alpha})$$, the order of $$M$$ is $$\alpha$$. [Prove by induction, using minimal counterexample, that $$o(S\cup T)=\min\{o(S),o(T)\}$$ for stationary sets. We call this the union property.] If $$o(S)=\alpha$$ write $$S=A\cup B$$ where $$A=S\cap E_{\alpha}$$ and $$B=S-A$$. If $$B$$ is nonstationary, then $$S=A\subset M$$ modulo $$I_{\rm NS}$$. If $$B$$ is stationary, by the second defining property on $$E_{\alpha}$$ and the union property we have $$o(B)>\alpha$$, and so $$B\subset {\rm Tr}(E_{\alpha})\mod I_{\rm NS}$$ by the previous corollary. Hence in this case we also have $$S\subset M\mod I_{\rm NS}$$; proving that $$M$$ is the largest set of order $$\alpha$$.

$$(\Leftarrow)$$: Assume $$M_{\alpha}$$ exists. By the union property [or by trivial considerations when $${\rm Tr}(M_{\alpha})$$ is nonstationary] it follows that $$M_{\alpha}\cup {\rm Tr}(M_{\alpha})$$ has order $$\alpha$$, so $${\rm Tr}(M_{\alpha})\subset M_{\alpha}$$ (modulo $$I_{\rm NS}$$) by maximality. Let $$E=M_{\alpha}-{\rm Tr}(M_{\alpha})$$. The order of $$E$$ is $$\alpha$$ since $$M_{\alpha}=E\cup {\rm Tr}(M_{\alpha})$$, by the union property, etc.

We first show that $$E$$ has order $$\alpha$$ hereditarily. Let $$S\subset E$$ be stationary. If $$o(S)>\alpha$$ then there is a stationary set $$T of order $$\alpha$$. By maximality of $$M_{\alpha}$$ we have $$T\subset M_{\alpha}\mod I_{\rm NS}$$, so $$S\subset {\rm Tr}(T)\subset {\rm Tr}(M)\mod I_{\rm NS}$$ which contradicts the fact that $$S$$ does not intersect $${\rm Tr}(M_{\alpha})$$.

Finally we show that $$E$$ meets every stationary set of order $$\alpha$$. Suppose, by way of contradiction, that $$S$$ is stationary of order $$\alpha$$ and $$S\cap E=\emptyset$$. We have $$S\subset M_{\alpha}\mod I_{\rm NS}$$, and since $$S\cap E=\emptyset$$ we have $$S\subset {\rm Tr}(M_{\alpha}\mod I_{\rm NS}$$, hence $$M, contradicting the order. This proves that $$E$$ is the canonical set of order $$\alpha$$.

Finally (working modulo $$I_{\rm NS}$$ everywhere), since $$E\subset M$$ we have $${\rm Tr}(E)\subset {\rm Tr}(M)$$, while the reverse containment follows from the previous corollary.$$\square_{\rm Lemma\ 3.3}$$

Corollary 3.4 from Jech's paper: Assuming $$M_{\alpha}$$ exists, for any stationary set $$S$$, then $$o(S)\geq \alpha$$ if and only if $$S\subset M_{\alpha}\mod I_{\rm NS}$$.

Proof of Corollary 3.4. This is just a direct application of the definition of $$M_{\alpha}$$ and the union property (taking the union with $$M_{\alpha}$$). [I don't see how it is a corollary.] $$\square_{\rm Corollary\ 3.4}$$

Lemma 3.5 from Jech's paper: [Note, it is misnumbered there.] Let $$\alpha$$ be a limit ordinal and assume that $$M_{\beta}$$ exists for all $$\beta<\alpha$$. Then $$M_{\alpha}$$ exists if and only if $${\rm inf}_{\beta<\alpha}M_{\beta}$$ exists, and is nonzero, in the Boolean algebra $$P(\kappa)/I_{\rm NS}$$.

Proof of Lemma 3.5. $$(\Rightarrow)$$: Assume $$M_{\alpha}$$ exists. We have $$M_{\beta}\supset M_{\alpha}\mod I_{\rm NS}$$ when $$\beta<\alpha$$ by the previous corollary, so $$M_{\alpha}$$ is a lower bound on the $$M_{\beta}$$. If $$S$$ is any stationary set which is a lower bound on each $$M_{\beta}$$ modulo $$I_{\rm NS}$$, then again by the previous corollary $$o(S)\geq \beta$$ for each $$\beta<\alpha$$, hence $$o(S)\geq \alpha$$, so $$S\subset M_{\alpha}\mod I_{\rm NS}$$ as desired.

$$(\Leftarrow)$$: If the inf exists and is nonzero, call it $$M$$. As it is contained in each $$M_{\beta}$$ (for $$\beta<\alpha$$) it has order at least $$\alpha$$. Given any stationary subset $$S$$ with $$o(S)=\alpha$$ then by the previous corollary $$S\subset M_{\beta}\mod I_{\rm NS}$$ for each $$\beta<\alpha$$, hence it lies inside the infimum. In particular, the infimum has exactly order $$\alpha$$ and is the largest set containing all such sets.$$\square_{\rm Lemma\ 3.5}$$

Lemma 3.6 from Jech's paper: If $$M_{\alpha}$$ exists and the height of $$\kappa$$ is at least $$\alpha+2$$ (i.e.\ there is a set of order $$\alpha+1$$), then $$M_{\alpha+1}={\rm Tr}(M_{\alpha})$$.

Proof of Lemma 3.6. We know $${\rm Tr}(S)\geq o(S)+1$$ (when $${\rm Tr}(S)$$ is stationary). Thus $$o({\rm Tr}(M_{\alpha}))\geq \alpha+1$$. By Corollary 3.4, $${\rm Tr}(M_{\alpha})\subset M_{\alpha+1}\mod I_{\rm NS}$$. Conversely, $$o(M_{\alpha+1})>\alpha$$ so by Corollary 3.2 $$M_{\alpha+1}\subset {\rm Tr}(E_{\alpha})={\rm Tr}(M_{\alpha})\mod I_{\rm NS}$$. $$\square_{\rm Lemma\ 3.6}$$

As diagonal intersection gives inf for sets of cardinality $$<\kappa^{+}$$, the previous two lemmas complete an induction, so we see that we can find the maximal sets $$M$$ inductively for $$\alpha<\kappa^{+}$$ (when the height of $$\kappa$$ is bigger than $$\alpha$$) by: $$M_{0}$$ is the limit points below $$\kappa$$, $$M_{\alpha+1}={\rm Tr}(M_{\alpha})$$, and for limits $$\alpha$$ we have $$M_{\alpha}=\triangle_{\beta<\alpha}M_{\beta}$$.

In the first paragraph we showed that the height of $$\kappa$$ is at least the ordinality of the sequence of regular infinite cardinals $$<\kappa$$. It is easy to check (via induction) using these definitions that $$E_{\alpha}=E_{\lambda}^{\kappa}$$ as claimed.$$\square_{\rm Exercise\ 8.13}$$

Exercise 8.14: The height of $$\kappa$$ is at least $$\kappa$$ if and only if $$\kappa$$ is weakly inaccessibly; the height of $$\kappa$$ is at least $$\kappa+1$$ if and only if $$\kappa$$ is weakly Mahlo.

Proof. First suppose $$\kappa$$ is an uncountable, regular, successor cardinal. Say $$\kappa$$ is the $$\alpha$$th regular cardinal. We have $$\alpha<\kappa$$. If $$\alpha$$ is a successor ordinal, say $$\alpha=\beta+1$$, let $$\lambda$$ be the $$\beta$$th regular cardinal. Then $$E_{\beta}=E_{\lambda}^{\kappa}$$ and $${\rm Tr}(E_{\beta})=\emptyset$$ by Exercise 8.12, so the height of $$\kappa$$ is $$\alpha$$ by Lemma 3.6 in the solution to the previous problem. If $$\alpha$$ is a limit ordinal, then since $$\triangle_{\beta<\alpha}E_{\alpha}=\triangle_{\lambda<\kappa:\lambda\text{ regular}}E_{\lambda}^{\kappa}=\emptyset$$, by Lemma 3.5 in the solution of the previous problem we have the height of $$\kappa$$ is $$\alpha$$.

Second suppose $$\kappa$$ is an uncountable, regular, limit cardinal. This means the sequence of limit cardinals $$<\kappa$$ is a cofinal sequence, so the height of $$\kappa$$ is at least $$\kappa$$, again by the previous problem.

Putting these together, we see $$h(\kappa)\geq \kappa$$ iff $$\kappa$$ is weakly inaccessible.

Now, fix $$\kappa$$ to be weakly inaccessible. The diagonal intersection of the sequence $$E_{\lambda}^{\kappa}$$ consists of regular cardinals $$<\kappa$$, so this is stationary if and only if $$\kappa$$ is weakly Mahlo. Hence by Lemma 3.5 of the previous solution, $$h(\kappa)\geq \kappa+1$$.

Conversely, if $$h(\kappa)\geq \kappa+1$$, the diagonal intersection of $$E_{\lambda}^{\kappa}$$ is stationary, hence there is a stationary sequence of regular cardinals $$<\kappa$$, so $$\kappa$$ is weakly Mahlo.$$\square_{\rm Exercise\ 8.14}$$

• Pace, thank you very much for taking the time to write out the detailed answer. However, I've spent the past few days digesting the proof in Monk's notes (finished shortly after I saw you posted the answer), so I'll accept my own answer instead. May 3, 2022 at 6:42

After some googling I found the notes by J. D. Monk, which have answered the questions in the span of Theorem 2.68 to 2.91.