# Club Guessing At Inaccessibles

Disclaimer: So after having typed this up, I’m a little shaky on a few details here still (mainly with keeping track of indices). But, I think I now have a better idea of how this proof should look if I attempt to rewrite it at some point in the future.

In this post, I want to work through the construction of a club guessing sequence which appears as Claim 0.14 of Sh413. Recall:

Definition: Let $\lambda$ be a regular uncountable cardinal, and let $S\subseteq \lambda$ be stationary. We say that $\bar C=\langle C_\delta : \delta \in S\rangle$ is an $S$-club system if each $C_\delta$ is a club subset of $\delta$.

For the construction, we need a few definitions which have become common-place in the literature:

Definition: If $C$ is a club set of ordinals, and $\alpha\in nacc(C)$, set

$Gap(\alpha, C)=(\sup(C\cap\alpha), \alpha)$.

Definition: If $C$ and $E$ are sets of ordinals with $E\cap \sup(C)$ closed in $\sup(C)$, we define:

$Drop(C,E)=\{\sup(\alpha\cap E) : \alpha\in C\setminus\min(E)+1\}$

Roughly what we’re doing here is dropping ordinals from $C$ above $min(E)$ down to their sup in $E$ in order to create a closed subset of $\sup(C)$.

Definition: Let $C, E\subseteq\lambda$ for a regular uncountable, and let $\langle e_\alpha : \alpha<\lambda\rangle$ be a $\lambda$-club system. For each $\alpha\in nacc(C)\cap acc(E)$, define

$Fill(\alpha, C, E)=Drop(e_\alpha, E)\cap Gap(\alpha, C)$

We suppress mention of the $\lambda$-club system in our notation for $Fill$ because it doesn’t really matter. What’s important is that $Fill(\alpha, C, E)$ gives us a way of looking at the gap between $sup (C\cap\alpha)$ and $\sup (E\cap\alpha)=\alpha$, and filling it in with a club subset of $Gap(\alpha, C)$. So we will just think of $\langle e_\alpha : \alpha\in \lambda\rangle$ as a fixed collection of clubs which we will draw upon to fill in gaps in an appropriate way.

Now, these describe operations which Shelah refers to as $gl^1$ and $gl^0$ in Cardinal Arithmetic. We’re going be using something called $gl^2$ over the course of our construction, which is the result of filling in multiple gaps and gluing the result together.

Definition: Let $C, E\subseteq\lambda$ for a regular uncountable, and let $\langle e_\alpha : \alpha<\lambda\rangle$ be a $\lambda$-club system. Further, let $A\subseteq\lambda$ be a stationary set of limit ordinals. We define $gl_n(C,E,A)$ by induction on $n<\omega$.

• $gl^2_0(C,E,A)=Drop(C,E)$
• $gl^2_{n+1}(C,E,A)=gl_n(C,E,A)\cup\bigcup_{\beta\in nacc(gl^2_n(C,E,A))\setminus A}\big(Fill(\beta,E,gl^2_n(C,E,A))\cup\{\sup(\alpha\cap E) : \sup(\alpha\cap E)\geq\sup (\alpha\cap e_\beta)\}\big)$
• $gl^2(C,E,A)=\bigcup_{n<\omega}gl^2_n(C,E,A)$.

This looks horrendous, but the main idea is that at each stage we look at the gaps in the previous stage outside of $A$, and fill those gaps in. Then at the end we glue it all together (hence gl) and provided that $C$ is club in some $\delta\in acc(E)$, the end result is a club of the same $\delta$. With this definition in hand, we’re ready to state the theorem and carry out the construction.

Claim 0.14 of Sh413: Assume

1. $\lambda$ is inaccessible.
2. $A\subseteq \lambda$ is a stationary set of limit ordinals such that, if $\delta<\lambda$ and $A\cap\delta$ is stationary in $\delta$, then $\delta\in A$.
3. $J$ is a $\sigma$-indecomposable ideal on $\lambda$ extending the non-stationary ideal.
4. $S\notin J$ and $S\cap A=\emptyset$.
5. $\omega<\sigma=cf(\sigma)<\lambda$ and $\delta\in S\implies cf(\delta)\neq\sigma$.

Then for some $S$-club system $\bar C=\langle C_\delta : \delta\in S\rangle$ we have that, for any $E\subseteq\lambda$:

$\{\delta\in S : \delta=\sup(E\cap nacc(C_\delta)\cap A)\}\notin J$.

Proof: We begin by fixing a $\lambda$-club system $\bar e=\langle e_\alpha : \alpha<\lambda\rangle$ with the property that $ot(e_\alpha)=cf(\alpha)$ and such that $e_\delta\cap A=\emptyset$ for any limit $\delta\in \lambda \setminus A$ (recall that $A$ does not reflect outside of itself).

For any club $E\subseteq\lambda$, define $C_{\delta,E}:=gl^2(e_\delta,E,A)$ if $\delta\in acc(E)$, and $C_{\delta,E}=e_\delta$ otherwise. If there is some club $E$ of $\lambda$ such that $\bar C_E=\langle C_{\delta, E}: \delta\in S\rangle$ is as required, then we are done. So suppose otherwise. Then for very club $E\subseteq\lambda$, there is a club $D(E)\subseteq\lambda$ such that

$Y_E:=\{\delta\in S : \sup(D(E)\cap A\cap nacc (C_{\delta, E}))=\delta\in J$.

Note that shrinking $D(E)$ only makes $Y_E$ smaller, so we may replace $D(E)$ with $D(E)\cap E$ in order to assume that $D(E)\subseteq E$. We now inductively pick sets $E_\epsilon$ by induction on $\epsilon<\sigma$ such that:

1. $E_\epsilon\subseteq\lambda$ is club;
2. $\xi<\epsilon\implies E_\epsilon\subseteq\ E_\xi$;
3. If $\epsilon=\xi+1$, then $E_\epsilon\subseteq D(E_\epsilon)$.

This is simple. We first let $E_0=\lambda$, and at successor stages we take $E_\epsilon=D_\xi$. At limit stages we let $E_\epsilon=\bigcap_{\xi<\epsilon}E_\xi$.

Next, let $E=\bigcap_{\epsilon<\sigma}E_\epsilon$ which is itself club in $\lambda$, and so $E\cap A$ is stationary in $\lambda$. Thus, the set $E'=\{\delta\in E : \delta=\sup(E\cap A\cap \delta)\}$ is club in $\lambda$. The following claim will finish the proof.

Claim: For every $\delta\in S\cap E'$, there is some $\epsilon_\delta<\sigma$ such that for every $\epsilon_\delta\leq \epsilon<\sigma$, we have $\delta\in Y_{E_\epsilon}$.

To see that this claim suffices, we begin by letting $Y_{\epsilon}=\bigcap_{\epsilon\leq\xi<\sigma}Y_{E_\xi}$ for each $\epsilon<\sigma$. Of course, $Y_\epsilon\in J$ as $Y_\epsilon\subseteq Y_{E_\epsilon}\in J$ by assumption, and $\epsilon_1<\epsilon_2\implies Y_{\epsilon_1}\subseteq Y_{\epsilon_2}$. As $J$ is $\sigma$-indecomposable, it’s closed under increasing unions of length $\sigma$ and in particular $\bigcup_{\epsilon<\sigma} Y_{\epsilon}\in J$. On the other hand, the claim tells us that $\delta\in Y_{\epsilon_\delta}$ for every $\delta\in S\cap E'$, and so $S\cap E'\in J$. But by assumption we have that $S\notin J$, and as $J$ extends the non-stationary ideal we have a contradiction.

Proof of Claim: We begin by fixing, for each $\delta\in S\cap E'$, an increasing cofinal sequence $\langle \beta^i_\delta : i of elements of $A\cap E\cap S$. Fix $\delta\in S\cap E'$, and note that by assumption $\delta\notin A$, and so $e_\delta\cap A=\emptyset$. In particular, $\{\beta^i_\delta : i. Now if we look at the construction of $gl^2(e_\delta, E_\epsilon, A)$, we see that since $\beta^i_\delta\in A\setminus e_\delta$, we only add boundedly-many elements of $\beta^i_\delta$ to $gl^2_n(e_\delta, E_\epsilon, A)$ for each $\epsilon<\sigma$ and $i< cf(\delta)$.

For each $i, $\epsilon<\sigma$, and $n<\omega$, let $\beta^i_\delta(n,\epsilon)=\min(gl^2_n(e_\delta, E_\epsilon, A)\setminus\beta^i_\delta)$ and note that since $gl^2_n\subseteq gl^2_{n+1}$, it follows that the sequence $\langle \beta^i_\delta(n,\epsilon) : n<\omega\rangle$ is decreasing and hence eventually constant (say it stabilizes at $n(i,\delta,\epsilon$). Now, since $\beta^i_\delta\cap gl^2_n(c_\delta, E_\epsilon, A)$ is bounded in $\beta^i_\delta$, it follows that $\beta^i_\delta(n,\epsilon)\in nacc(C_{\delta, E_\epsilon})$ for each $n>n(\i,\delta,\epsilon)$. In fact, we also have that $\beta^i_\delta(n,\epsilon)$ is in $A$ for each $n\geq n(i, \delta, \epsilon)$. Otherwise, we would have that $\beta^i_\delta(n,\epsilon)\in nacc (gl^2_n(e_\delta, E_\epsilon, A)\setminus A$, and so we would have ended up adding a club subset of $\beta^i_\delta(n,\epsilon)$ in the next stage. So far we have shown: For each $i, and $\epsilon<\sigma$, there is some $n(i,\delta,\epsilon)<\omega$ such that, for every $\beta^i_\delta(n,\epsilon)$ we have

1. $\beta^i_\delta(\epsilon, n(i,\delta,\epsilon))=\beta^i_\delta(\epsilon,n)$ and
2. $\beta^i_\delta(\epsilon, n)\in nacc(C_{\delta, E_\epsilon})\cap A$.

Next, we look at this sequence from the other direction. That is, we consider $\langle \beta^i_\delta(\epsilon, n) : \epsilon<\delta\rangle$ for a fixed $n$. The best way to visualize this situation is that for each $\delta\in S\cap E'$ and $i, we have a $\sigma\times\omega$ matrix $M_{i,\delta}$ where the $\epsilon, n$ entry is $\beta^i_\delta(\epsilon, n)$. So far, we’ve shown that the values are decreasing and eventually constant along the rows, and at this point we’d like to show the same for the columns. We begin by noting that the sequence of the sets $E_\epsilon$is decreasing, and so we have that $\sup(\alpha\cap E_{\epsilon_1})<\sup (\alpha\cap E_{\epsilon_2})$ whenever $\epsilon_2>\epsilon_1$. Therefore, the sequence $\langle \beta^i_\delta(\epsilon, n) : \epsilon<\delta\rangle$ is decreasing and eventually stabilizes at some $\epsilon(i,\delta, n)<\sigma$. In particular, this tells us that

$(\forall \xi>\epsilon(i,\delta, n))(\min (C_{\delta, E_\xi}\setminus \beta^i_\delta)=\min (C_{\delta, E_{\epsilon(i,\delta, n)}}\setminus \beta^i_\delta)=\beta^i_\delta(\epsilon(i,\delta, n), n))$

Now, since $\delta\in S$, we know that $cf(\delta)\neq\sigma$ and so for each $\delta$, we can find some $\epsilon_\delta$ such that $\sup\{i: \epsilon(i,\delta, n)\leq\epsilon_\delta \}=cf(\delta)$. This is pretty simple, as we just let $\epsilon_\delta$ be the supremum of $\epsilon(i,\delta, n)$ when running over $i< cf(\delta)$. Then this supremum will be below $\sigma$ as the cofinality of $\delta$ is wrong.

At this point, we would like to show that for every $\delta\in S\cap E'$, and $\epsilon\geq\epsilon_\delta$, we get that $\delta\in Y_{E_\epsilon}$. In particular, we need to show that $\delta=\sup(D(E_\epsilon)\cap A\cap nacc(C_{\delta,E_\epsilon})$. Recall that our construction of $C_{\delta,E_\epsilon}$ actually gives us the elements of $C_{\delta,E_\epsilon}$ are of the form $\sup(\alpha\cap E_{\epsilon}$. As $D(E_\epsilon)\subseteq E_\epsilon$, it follows that we only need to show that $\delta=\sup(A\cap nacc(C_{\delta, E_\epsilon}))$. Recall that stabilization along columns tells us that if $\epsilon_\delta\leq\epsilon<\sigma$:

$\min (C_{\delta, E_\epsilon}\setminus \beta^i_\delta)=\min (C_{\delta, E_{\epsilon(i,\delta, n)}}\setminus \beta^i_\delta)=\beta^i_\delta(\epsilon(i,\delta, n), n)$.

Finally, since $\langle \beta^i_\delta : i is increasing and cofinal in $\delta$, and $\beta^i_\delta<\beta^i_\delta(\epsilon,n)<\delta$, it follows that:

$\delta=\sup\{min(C_{\delta, E_\epsilon}\setminus \beta^i_\delta: \epsilon(i,\delta, n)\leq\epsilon_\delta\}$.

Since we already showed that the supremum of the indices $i$ involved in the above supremum has supremum equal to $cf(\delta)$. Finally stabilization along the rows told us that eventually $\beta^i_\delta(\epsilon, n)$ ends up in $A\cap nacc C_{\delta, E_\epsilon}$ for all sufficiently large $n<\omega$. So, we can just pick the correct $\beta^i_\delta(\epsilon, n)$ to achieve a supremum of $\delta$. That completes the proof of the claim, and consequently the proof of the whole thing.