Question about the order of a Stationary subset of $ \kappa$

Question

Greets

I'm trying to prove one part of exercise 8.14 of Jech's "Set Theory", namely that if $o(k)\geq k$, then $k$ is weakly inaccessible, where $\kappa$ is regular; $o(\kappa)$ is defined as follows:

For any stationary subsets $S,T$ we define $S<T$ iff $S\cap \alpha$ is stationary in $\alpha$ for almost all $\alpha\in T$, Jech proves that $<$ is a well-founded relation, and thus he defines for any stationary $X$, $o(X)=\operatorname{sup}\{o(Y)+1: Y<X\}$, and $o(\kappa)=\operatorname{sup}\{o(Y)+1: Y$ is a stationary subset of $\kappa\}$.

Some preliminaries to this exercise are :

• The set of closed unbounded subsets fo $\kappa$ is closed under intersections of less than $\kappa$ elements, and it is also closed under diagonal intersections.
• If $\lambda<\kappa$ is the $\alpha$th regular cardinal, then $o(E_{\lambda}^{\kappa})=\alpha$.

To prove this part of the exercise I need only to show that if $X,Y,Z$ are stationary subsets of $\kappa$ such that $X<Y$ and $o(Y)=o(Z)$, then $X<Z$, I don't know whether this property is true or not, if it is true, I would like to see a proof of it, or a hint to solve Jech's exercise.

Thanks

Answer 1

I will give an answer to Jech's exercise.

Let $F:\mathrm{Ord}\to\mathrm{Card}$ be the enumeration of the regular cardinals.

First of all observe that if $A<B$ and $C\subseteq B$ (and $C$ is stationary) then $A<C$. This follows almost immediately: If $D$ is a club set such that for all $\alpha\in B\cap D$ we have that $A\cap\alpha$ is stationary, then the same club gives that for all $\alpha\in C\cap D$ that $A\cap\alpha$ is stationary. From this follows that if $C\subseteq B$ then $o(C)\geq o(B)$.

Claim 1: If $A$ is stationary on $\kappa$, then almost all elements of $A$ have cofinality less that $\kappa$.

Proof:

If $\kappa$ is regular this is trivial. If $\kappa$ is not regular then take $f:\mathrm{cf}(\kappa)\to\kappa$ a normal and unbounded function. $f_{-1}(A)$ is stationary on $\mathrm{cf}(\kappa)$. Notice that $\mathrm{cf}(\kappa)>\mathrm{cf}(\beta)\geq\mathrm{cf}(f(\beta))$ for all $\beta\in\mathrm{cf}(\kappa)$. Also notice that $f[\mathrm{cf}(\kappa)]$ is club in $\kappa$. Hence $A\setminus f[\mathrm{cf}(\kappa)]$ the set of potential elements whose cofinality may be greater than or equal to $\kappa$ is non-stationary on $\kappa$.

Claim 2: If $A$ is a stationary subset of $\kappa$ whose elements have cofinality $\lambda$, then $o(A)=F^{-1}(\lambda)$.

Proof:

The proof is by induction on $F^{-1}(\lambda)$. For $0$ we have that all elements of $A$ have cofinality $\omega$, and hence its order is $0$ be definition.

Assume it is true for all $\beta<\alpha$ and let $A$ be a stationary subset of $\kappa$ such that all the elements of $A$ have cofinality $F(\alpha)$. First of all $o(A)\geq \alpha$ since $E^\kappa_{F(\beta)}<A$ for all $\beta<\alpha$.

Take $B<A$. Let $C$ be a club such that for all $\xi\in A\cap C$, $B\cap\xi$ is stationary. I claim that $B'=B\cap\{\xi\in\kappa : \mathrm{cf}(\xi)<F(\alpha)\}$ is stationary. Indeed if not, there is a club $H$ such that $B'\cap H=\varnothing$. Then for $\xi\in A\cap C\cap H$, we have that $H\cap\xi$ is a club subset of $\xi$ and $B'\cap H\cap\xi=\varnothing$, contradicting Claim 1, that almost all elements of $B\cap\xi$ have cofinality less than $F(\alpha)$.

Define the function $g:B'\setminus(F(\alpha)+1)\to\kappa$ as $g(\gamma)=\mathrm{cf}(\gamma)$. Notice that $g$ is regressive and thus by Fodor's Theorem $g$ is constant on a stationary set $B''\subseteq B'$. Then all elements of $B''$ have cofinality $F(\beta)$, where $\beta<\alpha$, and by the induction hypothesis $o(B'')=\beta$. By the observation above $o(B'')\geq o(B')\geq o(B)$. Hence for all $B<A$ we have $o(B)<\alpha$ and thus $o(A)\leq\alpha$. This completes the proof of Claim 2.

Claim 3: If $\kappa$ is a regular cardinal and $F^{-1}(\kappa)<\kappa$ then $o(\kappa)<\kappa$.

Proof:

Take $A$ a stationary subset of $\kappa$. Since $F^{-1}(\kappa)<\kappa$, the set of regular cardinals below $\kappa$ is bounded in $\kappa$. Hence almost all elements of $A$ are singular cardinals, so let's assume that $A'\subseteq A$ is a stationary subset of $\kappa$ that contains only singular cardinals. Then as in Claim 2, define a map that sends all elements of $A'$ to their cofinality. By Fodor's Theorem there is a stationary set $A''\subseteq A'$ such that all its elements have cofinality $\lambda<\kappa$. Then $o(A'')=F^{-1}(\lambda)<F^{-1}(\kappa)<\kappa$. From the observation in the beginning we have that $o(A)\leq o(A'')$. Thus $o(A)<F^{-1}(\kappa)$, hence $o(\kappa)\leq F^{-1}(\kappa)$.

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Claim 3 is enough to show the contrapositive of what you want, since if $\kappa$ is a regular successor then $F^{-1}(\kappa)<\kappa$.