# 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.

# Another Update

Okay, so I’ve actually finished working through the proof that any inaccessible Jónsson cardinal must be Mahlo ($\omega$-Mahlo even), but I haven’t had much time to write stuff up. The main issue is that I’ve also been working my way through Sh413, which contains the result that any $\lambda$ which is an inaccessible Jónsson cardinal must be $\lambda\times\omega$-Mahlo.

What I’ll try to do is write up the proof that any inaccessible Jónsson cardinal must be Mahlo. That’s a bit more work than the thread I was following earlier, but not by much. One thing I will do is take the existence of the desired club guessing sequences for granted. I feel moderately comfortable in doing this because the construction of these sequences is incredibly similar to the construction in the case that $\lambda=\mu^+$ in EiSh819.

From there, I want to try and write up the result of working through the first section of Sh413. There’s a lot of material in that first section which ends up being tertiary to the main result, and I would like to write up a “straight shot” proof of the result. In particular, there is a really cool construction of a club guessing sequence provided that we have a stationary set $S$ which does not reflect outside of itself (Claim 0.14 from Sh413). Actually I might write that part up sooner rather than later.

# Club Guessing Ideals

I’m currently working my way through portions of Chapter III and IV of Cardinal Arithmetic to get a hold on some of the material regarding Jónsson algebras on inaccessibles. For now though, I want to make a short post which talks about the club guessing ideals which feature prominently in these chapters. In particular, I want to use the statement of Claim 1.9 from Chapter III of Cardinal Arithmetic to motivate the definitions of these ideals. Let’s start by recalling the statement:

Theorem (Shelah): Suppose that $\lambda$ is an inaccessible cardinal such that:

1. There is a stationary $S\subseteq \lambda$ such that $S$ does not reflect at inaccessible cardinals.
2. There is a sequence $\bar C=\langle C_\delta : \delta\in S\rangle$ such that each $C_\delta$ is a club subset of $\delta$, and for every club $E\subseteq \lambda$, there are stationarily-many $\delta\in S$ such that $E\cap nacc (C_\delta)$ is unbounded in $\delta$ and there is no regular $\gamma<|\delta|$ with $cf(\alpha)<\gamma$ for all $\alpha\in E\cap nacc(C_\delta)$.
3. For each $\delta\in S$, the set of regular Jónsson cardinals below $\delta$ has bounded intersection with $nacc (C_\delta)$.

Then, $\lambda$ carries a Jónsson algebra.

The statement of the theorem is a bit of a mess, but that’s partially my fault for trying to use as little notation as possible. So, let’s start with the notation.

Definition: 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 $\delta$.

The statement of the above theorem is about particular types of $S$-club systems. In particular, we’re asking that $\bar C$ anticipate clubs stationarily-often in the following sense:

given any club $E\subseteq \lambda$, there are stationarily-many $\delta\in S$ such that $nacc(C_\delta)\cap E$ is “not too small”.

This particular notion of not being too small leads us to associate to our $S$-club system $\bar C$, a system of ideals $\bar I=\langle I_\delta : \delta \in S\rangle$ as follows.

Each $I_\delta$ is an ideal over $C_\delta$ generated by the following sets:

1. $acc( C_\delta)$;
2. $\{\alpha\in C_\delta : \alpha <\beta\}$ for each $\beta<\delta$;
3. $\{\alpha\in C_\delta : cf(\alpha)<\gamma\}$ for each regular $\gamma<\delta$;

Then we see that the manner in which we want $\bar C$ to guess clubs is that we need $C_\delta\cap E\notin I_\delta$ stationarily-often for any club $E\subseteq\lambda$. So let’s say that the pair $(\bar C, \bar I)$ guesses clubs if this happens. Then the statement of the theorem becomes:

Theorem (Shelah): Suppose that $\lambda$ is an inaccessible cardinal such that:

1. There is a stationary $S\subseteq \lambda$ such that $S$ does not reflect at inaccessible cardinals.
2. There is a sequence $\bar C=\langle C_\delta : \delta\in S\rangle$ such that the pair $(\bar C, \bar I)$ guesses clubs.
3. For each $\delta\in S$, the set of regular Jónsson cardinals below $\delta$ is in $I_\delta$.

Then $\lambda$ is not Jónsson.

I want to point out that we could ask about guessing clubs relative to other sequences of ideals, and Shelah does precisely that. The ideals we’ve concocted however, are the ones most often considered as they seem to be the ones most directly relevant to the problem of producing Jónsson algebras. Now the next thing to note is that the crux of the proof was the fact that the elementary submodel $M$ inconsideration sees enough points of $nacc(C_\delta)\cap E$ where $E$ was a particular club subset of $\lambda$. So we didn’t need the full force of condition 3. If we let $A=\{ \alpha < \lambda : \alpha \text{ is regular and not J\'onsson}\}$  we only needed to produce a club guessing sequence $\bar C=\langle C_\delta : \delta \in S\rangle$ such that:

For every club $E\subseteq \lambda$, there are stationarily-many $\delta\in S$ such that $E\cap A\cap C_\delta\notin I_\delta$.

In other words, we are asking that $\langle C_\delta \cap A : \delta\in S\rangle$ guesses clubs. This leads us to yet another ideal that features prominently in the literature, and captures precisely when this doesn’t happen.

Definition: Let $S\subseteq\lambda$ be stationary, $\bar C=\langle C_\delta : \delta \in S\rangle$ be an $S$-club system, and let $\bar I=\langle I_\delta : \delta \in S\rangle$ be a sequence of ideals (with $I_\delta$ an ideal on $C_\delta$). We define the ideal $id_p(\bar C, \bar I)$ by saying that $A\in id_p(\bar C, \bar I)$ if and only if there is some club $E\subseteq\lambda$ such that for every $\delta\in S\cap E$, we have $E\cap A\cap C_\delta\in I_\delta$.

This allows us to state yet another version of the theorem (if we let $\bar I$ be defined as before):

Theorem (Shelah): Suppose that $\lambda$ is an inaccessible cardinal such that:

1. There is a stationary $S\subseteq \lambda$ such that $S$ does not reflect at inaccessible cardinals.
2. There is a sequence $\bar C=\langle C_\delta : \delta\in S\rangle$ such that the pair $(\bar C, \bar I)$ guesses clubs.
3. The set of regular cardinals which carry a Jónsson algebra is not in $id_p(\bar C, \bar I)$.

Then $\lambda$ is not Jónsson.

In the next entry, I want to (using the above theorem) give a proof that any inaccessible Jónsson cardinal must either be an inaccessible limit of Jónsson cardinals or a Mahlo cardinal. This is not near the state of the art, but it’s a good place to start, as the proof involves showing how to produce the sorts of club guessing sequences that we need.

# Jónsson Algebras on Inaccessibles

I want to use this post to work through a result of Shelah’s the uses club guessing to produce Jónsson cardinals on inaccessible cardinals. My main reason is that a few of the arguments that appear get used a number of times in Cardinal Arithmetic. First, some definitions and motivation (anyone interested more in the result itself can just skip down):

Lemma/Definition: Let $\lambda$ be an infinite cardinal. Then we say that $\lambda$ carries a Jónsson algebra if one of the following equivalent conditions hold:

1. There is an algebra (in the set-theoretic sense) $\mathfrak{A}=(A, (f_n)_{n<\omega})$ such that $|A|=\lambda$ with no proper subalgebras of the same cardinality (such an algebra is called a Jónsson algebra).
2. For any (equivalently for some) regular $\theta\geq\lambda^+$, and any countable expansion $\mathfrak{A}$ of $(H(\theta),\in,<_\theta)$, if $M\prec \mathfrak{A}$ is such that $\lambda\in M$ and $|M\cap\lambda|=\lambda$, then $\lambda\subseteq M$.
3. $\lambda\not\rightarrow[\lambda]^{<\omega}_\lambda$.

The proof that these three conditions are equivalent is pretty standard, and can be found in a number of places (Kanamori’s The Higher Infinite being one of them).

We say that a cardinal $\lambda$ is Jónsson if it carries no Jónsson algebras. It turns out that these things are interesting enough that there are three chapters in Cardinal Arithmetic with the purpose of clarifying the situation at Jónsson cardinals. Here are some quick facts:

1. All measurable cardinals are Jónsson, but the existence of a Jónsson cardinal implies that $0^\sharp$ exists;
2. It is equiconsistent with the existence of a measurable cardinal that there is a singular Jónsson cardinal;
3. If $\lambda$ is a regular Jónsson cardinal, then every stationary subset of $\lambda$ reflects;
4. If $\lambda=\mu^+$ is Jónsson for $\mu$ singular, then $\mu$ is the limit of regular Jónsson cardinals.

I’ve been particularly interested in the consistency of “There exists a singular $\mu$ such that $\mu^+$ is Jónsson”. In particular, I’ve recently been trying to get a good picture of what the first such $\mu$ should look like. For example, we know that it has to be countable, and a limit of inaccessible Jónsson cardinals. On the other hand, I have no idea whether or not there actually needs to be $\mu$-many Jónsson cardinals below $\mu$. So, getting a good idea of what’s going on here would require me to know what inaccessible Jónsson cardinals below such a $\mu$ look like. Luckily, Shelah’s devoted a lot of work in Cardinal Arithmetic to this kind of stuff.

I want to start with a result that appears relatively early on in chapter 3 of Cardinal Arithmetic. There is an analogous result for successors of singular cardinals, and in fact Shelah proves both results at the same time. Unfortunately, while the proofs do follow a similar pattern, they are not similar enough to warrant this treatment. So, I decided to tease out the part about inaccessible cardinals.

Theorem (Shelah): Suppose that $\lambda$ is an inaccessible cardinal such that:

1. There is a stationary $S\subseteq \lambda$ such that $S$ does not reflect at inaccessible cardinals.
2. There is a sequence $\bar C=\langle C_\delta : \delta\in S\rangle$ such that each $C_\delta$ is a club subset of $\delta$, and for every club $E\subseteq \lambda$, there are stationarily-many $\delta\in S$ such that $E\cap nacc (C_\delta)$ is unbounded in $\delta$ and there is no regular $\gamma<|\delta|$ with $cf(\alpha)<\gamma$ for all $\alpha\in E\cap nacc(C_\delta)$.
3. For each $\delta\in S$, the set of regular Jónsson cardinals below $\delta$ has bounded intersection with $nacc (C_\delta)$.

Then, $\lambda$ carries a Jónsson algebra.

What this is telling us is that, if we have a stationary subset of $\lambda$ which does not reflect at inaccessibles, and we can find a system of clubs on $S$ which guesses clubs in a sufficiently nice way (but off of Jónsson cardinals), then we can find a Jónsson algebra on $\lambda$. So the problem of finding Jónsson algebras actually comes down to building club guessing sequences along particular stationary sets.

Proof: Let $\theta\geq \lambda^+$ be large enough, and let $\mathfrak{A}=(H(\theta),\in,<_\theta, S,\bar C)$. Let $M\prec \mathfrak{A}$ be such that $|M\cap \lambda|=\lambda$ and $\lambda\in M$. In order to show that $\lambda\subseteq M$, we will show that there are arbitrarily large $\sigma<\lambda$ for which $\sigma\subseteq M$. Let $\sigma<\lambda$ be given and set

$E=\{\alpha<\lambda : \sup(\alpha\cap M)=\alpha\}$.

Note that $E$ is a club subset of $\lambda$ so we can find stationarily-many $\delta\in S\cap E$ such that $E\cap nacc(C_\delta)$ is unbounded in $\delta$. Next note that since $\lambda$ is a limit cardinal, we know that the set of cardinals below $\lambda$ is club in $\lambda$. So we can pick $\delta$ such that:

• $\delta\in S\cap E$;
• $E\cap nacc(C_\delta)$ is unbounded in $\delta$;
• $\delta$ is a cardinal;
• $\delta>\sigma$.

The following claim will finish the proof:

Claim: There is some regular $\kappa<\delta$ such that $M$ contains all points of $E\cap nacc(C_\delta)$ with cofinality greater than $\kappa$.

To see that the claim suffices, note that this allows us to pick some $\alpha\in nacc(C_\delta)\cap E)$ such that:

• $cf(\alpha)>\sigma$;
• $cf(\alpha)$ carries a Jónsson algebra;
• $\alpha\in M$.

By condition 2. in the hypotheses of the theorem, we can find elements of arbitrarily large cofinality below $\delta$ in there. Then, the claim and assumption 3. allow us to pick an appropriate such $\alpha$. But then, since $\alpha\in M$, we know that $cf(\alpha)$ is also in $M$ as it’s definable from parameters. Further, since $\sup M\cap \alpha=\alpha$, we also get that $\sup M\cap cf(\alpha)=cf(\alpha)$. This leaves us in the situation that $|M\cap cf(\alpha)|=cf(\alpha)$ and $cf(\alpha)\in M$, and since $cf(\alpha)$ carries a Jónsson algebra we immediately get that $\sigma\subseteq cf(\alpha)\subseteq M$.

Proof of Claim: We have two cases to deal with, of which we will take care of the easier one first.

Case 1 ($\delta\in M$): This case is relatively easy since the fact that $\delta, S, \bar C\in M$ then tells us that $C_\delta\in M$. So let $\beta\in nacc(C_\delta)\cap E$, and let $\beta^*\in M$ be such that $\beta^*>\sup (C_\delta\cap\beta)$, but $\beta^*<\beta$. This is possible by definition of $E$ combined with the fact that $\beta$ is a non-accumulation point of $C_\delta$. But then, $\beta=\min\{\alpha\in C_\delta : \alpha>\beta^*\}$, that is $\beta$ is the least element of $C_\delta$ that gets above $\beta^*$. So $\beta$ is definable from parameters in $M$ and is thus itself in $M$. In this case then, $E\cap nacc(C_\delta)\subseteq M$.

Case 2 ($\delta\notin M$): This one is a bit trickier since we don’t have $C_\delta$ lying around in our model to help us out. Our first job is to find a good candidate for $\kappa$ in the claim. With that said, define $\beta_\delta=\min(M\cap\lambda\setminus\delta)$, which is the least element of $M\cap \lambda$ which gets above $\delta$. Note that $\beta_\delta$ is a limit ordinal of uncountable cofinality, else we will be able to violate the minimality of $\beta$ (as $\delta$ is not in $M$). Further, since $\delta$ is a cardinal then $\beta_\delta$ must also be a cardinal since $|\beta_\delta|\geq\delta$ and $\delta\notin M$. Thus, since $|\beta_\delta|\in M$, if $\beta_\delta$ is not a cardinal then $\beta_\delta>|\beta_\delta|\geq \delta$ violates the minimality of $\beta_\delta$. So $\beta_\delta$ must be a cardinal of uncountable cofinality.

We now claim that $S$ reflects at $\beta_\delta$. That will give us that $\beta_\delta$ is a singular cardinal, and thus that $cf(\beta_\delta)<\delta$. To see that $S$ reflects at $\beta_\delta$, note that it suffices to show that this holds in $M$. Otherwise, if there is some club $d\subseteq \beta_\delta$ such that $S\cap d=\emptyset$, then we will be able to find such a club in $M$ by elementarity (as both $\beta_\delta$ and $S$ are in $M$). Along those lines, let $d\subseteq\beta_\delta$ be club in $\beta_\delta$ with $d\in M$. We will show that $\delta\in acc(d)$, so let $\alpha<\delta$ be given. As $\delta\in E$, we can find some $\beta\in M$ such that $\alpha<\beta<\delta$. Further as $M$ thinks $d$ is unbounded in $\beta_\delta$, we can find $\gamma\in d\cap M$ such that $\gamma>\beta$. On the other hand, the minimality of $\beta_\delta$ tells us that $\gamma<\delta$ and so we’ve found $\gamma\in d$ such that $\alpha<\gamma<\delta$. Thus, $\delta$ is an accumulation point of $d$ and so is itself in $d$.

At this point, we have shown that $\beta_\delta$ must be a singular cardinal. By the minimality of $\beta_\delta$, we then have that $cf(\beta_\delta)<\delta$. We will now show that every element of $E\cap nacc (C_\delta)$ with cofinality above $cf(\beta_\delta)$ is in $M$.

So let $\beta\in E\cap nacc(C_\delta)$ be such that $cf(\beta)>cf(\beta_\delta)$, and let $d\subseteq \beta_\delta$ be club in $\beta_\delta$ with $d\in M$ and such that $|d|=cf(\beta_\delta)$. Note that $\beta$ is not a limit point of $d$, as $cf(\beta)>|d|$, so we can find an ordinal $\beta_0$ satisfying:

• $\beta_0\in M$;
• $sup(C_\delta\cap\beta)<\beta_0<\beta$;
• $sup (d\cap\beta)<\beta_0$.

Given such an ordinal $\beta_0$, we can define the set

$A=\{\min (C_\epsilon\setminus\beta_0) : \epsilon\in d\cap S\}$.

Note that since all parameters used to define $A$ are in $M$, we have that $A$ is in $M$ as well. Further, we have that $|A|\leq |d| and so $A\cap\beta$ is bounded below $\beta$. Thus, we can find an ordinal $\beta_1$ satisfying:

• $\beta_1\in M$;
• $\beta_0<\beta_1<\beta$;
• $A\cap [\beta_1,\beta)=\emptyset$.

Now define

$d^*=\{\epsilon\in d\cap S\setminus\beta_1 : \min(C_\epsilon\setminus\beta_0)=\min(C_\epsilon\setminus\beta_1)\}$.

Again, we note that all parameters used to define $d^*$ are in $M$ and so $d^*$ is also in $M$. Further, recall that we have shown that $\delta\in S\cap d$ since $d\subseteq\beta_\delta$ is club and $d\in M$. Note that

$\min(C_\delta\setminus\beta_0)=\beta=\min (C_\delta\setminus\beta_1)$,

since both $\beta_0$ and $\beta_1$ sit above $\sup (C_\delta\cap\beta)$ but below $\beta$. Thus, $\delta\in d^*$. On the other hand, if $\min(C_\epsilon\setminus\beta_0)<\beta$, we know that $A\cap [\beta_1,\beta)=\emptyset$, and so in fact $\min(C_\epsilon\setminus\beta_0)<\beta_1$. Thus if $\epsilon\in d^*$, then $\min(C_\epsilon\setminus\beta_0)\geq\beta$. That is, $\beta$ is the minimum value $\min(C_\epsilon\setminus\beta_0)$ can take on for $\epsilon\in d^*$. But then $\beta$ is definable from parameters in $M$ and so is also in $M$. That completes the proof of the claim, as well as the theorem.

So one cool thing about the proof of this theorem is that most of the arguments here appear frequently when relating Jónsson algebras to club guessing. The primary vehicle this relation is actually the claim which ended up being the major crux of the proof. Basically, it allows us to conclude that, if $M$ is a candidate for witnessing that $\lambda$ is not Jónsson, then $M$ will be able to see enough of the ladder systems which guess clubs. So the ladder systems that are guessing clubs (as opposed to the clubs that are guessed themselves) end up being the vehicle for moving up Jónssonness. Interestingly enough, the sets along which scales appear also serve a similar purpose when considering successors os singular cardinals. Additionally, our argument showing that $\delta$ sits in every club $d\subseteq\beta_\delta$ such that $d\in M$ also appears when showing that every stationary subset of a regular Jónsson cardinal must reflect.

# Scales from I[\lambda]

I want to make good on the promise of linking up scales to square-like principles that I made a couple of posts ago. In particular I want to sketch how one produces scales using the machinery of $I[\lambda]$. If we believe that $I[\lambda]$ contains a lot of stationary sets, then most of the work was actually done by working through Claim 2.6A of Chapter 1 from Cardinal Arithmetic:

Theorem: Let $I$ be an ideal on a set $A$ of regular cardinals with $\kappa>|A|^+$ regular. Assume that:

1. $\lambda>\kappa^{++}$ is regular such that there is some stationary $S\subseteq S^\lambda_{\kappa^+}$ which has a continuity condition $\bar C$;
2. $\vec f=\langle f_\alpha : \alpha<\lambda\rangle$ is a sequence of functions from $A$ to the ordinals;
3. $\vec f$ obeys $\bar C$.

Then $\vec f$ has a $\leq_I$-exact upper bound.

We just have to show that it’s we can construct sequences which obey continuity conditions, and then there’s a relatively standard argument which allows us to move immediately from an exact upper bound for an appropriate sequence to a scale. Let’s briefly recall what continuity conditions are, and how they relate to $I[\lambda]$:

Theorem (Shelah): Let $\lambda$ be a regular cardinal. Then for $S\subseteq \lambda$, we have $S\in I[\lambda]$ if and only if there is a sequence $\bar C= \langle C_\alpha : \alpha<\lambda\rangle$ and a club $E\subseteq\lambda$ such that:

1. Each $C_\alpha$ is a closed (but not necessarily unbounded) subset of $\alpha$;
2. if $\beta\in nacc(C_\alpha)$ then $C_\beta=\alpha\cap C_\alpha$;
3. If $\delta\in S\cap E$, then $\delta$ is singular, and $C_\delta$ is a club subset of $\delta$ of order type $cf(\delta)$.

The sequence $\bar C$ is called a continuity condition for $S$, and functions somewhat like a square sequence over $S$. The major difference is that we only have coherency on the non-accumulation points which is a significant weakening, but still allows them to be useful enough. So the fact that $I[\lambda]$ contains a stationary subset of $S^\lambda_\kappa$ for every regular$\kappa$ such that $\kappa^{++}<\lambda$ can be regarded as a weak fragment of square which is true in ZFC.

Since we want to produce scales, our focus will be on $I[\mu^+]$ for $\mu$ singular. In particular, we have that for every regular $\kappa<\mu$, there is a stationary $S\subseteq S^{\mu^+}_{\kappa}$ such that $S\in I[\mu^+]$. Further, we also have that if $\mu$ is strong limit, then $S^{\mu^+}_{\leq cf(\mu)}\in I[\mu^+]$, though this won’t be particularly important to us (a proof of this can be found in Todd Eisworth’s Handbook chapter for the interested).

First, we show how to produce sequences that obey continuity conditions. So, fix a set of regular cardinals $A\subseteq \mu$ cofinal in $\mu$ with $ot(A)=cf(\mu)$ and such that $|A|^+<\min A$. Following standard notation, we will let $J^{bd}[A]$ denote the ideal of bounded subsets of $A$.

We first show that $\prod A/J^{bd}[A]$ is $\mu^+$ directed. To see this, note that is suffices to show that $\prod A/J^{bd}[A]$ is $\mu$-directed. For any set $F\subseteq \prod A/J^{bd}[A]$ such that $|F|=\mu$, then we can rewrite $F=\bigcup_{i where $|F_i|<\mu$. From there, we use $\mu$-directedness to produce bounds $f_i\in\prod A/J^{bd}[A]$ for each $F_i$, and then bound $\{f_i : i by $f\in \prod A/J^{bd}[A]$. For $\mu$-directedness, let $F\subseteq \prod A/J^{bd}[A]$ be such that $|F|<\mu$. Then let $f$ be defined by $f(a)=\sup \{g(a) : g\in F\}$, and note that $f$ is defined almost everywhere since each $a\in A$ is regular and $A$ is cofinal in $\mu$. Clearly then $f+1$ is a $<_{J^{bd}[A]}$-upper bound for $F$.

Now let $\kappa<\mu$ be regular, and let $S\subseteq S^{\mu^+}_\kappa$ be such that $S\in I[\lambda]$ with $\bar C=\langle C_\alpha : \alpha<\lambda\rangle$ a witnessing continuity condition. We first recall what it means for a sequence $\vec f=\langle f_\alpha : \alpha<\mu^+\rangle$ to obey $\bar C$:

Definition: We say $\vec f$ weakly obeys $\bar C$ if:

If $\alpha<\lambda$ is such that $ot(C_\alpha)\leq \kappa$, then for each $\beta\in nacc(C_\alpha)$, we have $f_\beta(i) for each $i<\kappa$.

This definition looks like a weakening of the one originally given, but it’s all that was required for the proof of Claim 2.6A to go through. Now we inductively define a sequence $\vec f$ which obeys $\bar C$ as follows. We first let $f_0$ be any function in $\prod A/J^{bd}[A]$. At stage $\alpha$, we suppose that $f_\beta$ has been defined for each $\beta<\alpha$. We let $f_\alpha'$ be a $<_{J^{bd}[A]}$-upper bound for $\{f_\beta : \beta<\alpha\}$ as guaranteed by $\mu^+$-directedness. If $C_\alpha$ is empty or $ot(C_\alpha)>\kappa$, then we just set $f_\alpha=f_\alpha'$. Otherwise, we let $f_\alpha$ be defined by setting $f_\alpha(a)=\max\{f_\alpha'(a), \sup_{\beta\in C_\alpha}f_\beta(a)\}+1$. Note that since $\kappa<\mu$, we know that $f_\alpha(a)$ is defined almost everywhere. It is also clear by construction that $\langle f_\alpha : \alpha<\mu^+\rangle$ is a $<_{J^{bd}[A]}$-increasing sequence which weakly obeys $\bar C$ and so we are done.

I also want to note that we could have started with a fixed sequence $\vec g=\langle g_\alpha : \alpha<\mu^+\rangle$, and asked that not only $\vec f$ weakly obey $\bar C$, but also that $g_\alpha for each $\alpha<\mu^+$. So for example if $\vec g$ weakly obeyed some other continuity condition $\bar D$, then the resulting $\vec f$ would weakly obey both $\bar D$ and $\bar C$. Further, if $\vec f$ obeys a continuity condition for a stationary subset of $S^{\mu^+}_\kappa$, then one can show that the exact upper bound $f$ produced by Claim 2.6A satisfies:

$\{\ a\in A : cf(f(a))<\kappa\}\in J^{bd}[A]$.

Okay, with all of this in hand, we can produce a $<_{J^{bd}[A]}$-increasing sequence of functions $\vec f=\langle f_\alpha : \alpha<\mu^+\rangle$ in $\prod A/J^{bd}[A]$ with the following properties:

1. $\vec f$ has an exact upper bound $f$;
2. For every regular $\kappa$ with $\min(A)\leq \kappa<\mu$, the set $\{a\in A : cf(f(a))<\kappa\}\in J^{bd}[A]$.

So, we then have that sequence $\vec f$ witnesses that $\prod_{a\in A} f(a)/J^{bd}[A]$ has true cofinality $\mu^+$. Now, by possibly altering $f$ on a null set we may assume $\min \{f(a):a\in A\}>|A|$. Let $B=\{cf(f(a)) : a\in A\}$, and note by condition 2, that $B$ is cofinal in $\mu$ and has order type $cf(\mu)$. A relatively standard argument then allows us to conclude that $tcf (\prod B/J^{bd}[B])=\mu^+$, and letting the witnessing sequence be $\vec h=\langle h_\alpha : \alpha<\lambda\rangle$, we get that $(B,\vec h)$ is a scale on $\mu$.

Honestly, parts of this sketch are pretty bare-bones, but the idea was to show that Claim 2.6A (once appropriately modified) is the only really difficult part behind producing scales. In fact, that claim plays the same role that the trichotomy theorem does for the theory of exact upper bounds. In particular, it shows us that, provided we can construct certain sorts of sequences, we can then get nice exact upper bounds. It just turns out that these sequences, once we have enough of the $I[\lambda]$ combinatorics in hand, are relatively easy to produce. From there, it’s just standard arguments showing that we really only need exact upper bounds to do a lot of the things we want. An alternative approach to exact upper bounds (outside of $I[\lambda]$ or trichotomy) is also furnished through what Abraham and Magidor call $(*)_\kappa$. It turns out that $(*)_\kappa$ is incredibly similar to having continuity conditions for a stationary subset of $S^\lambda_\kappa$ lying around.

Overall though, all three of these approaches are doing roughly the same thing.

# Continuity Conditions and I[\lambda]

In the previous post, I worked through a result of Shelah’s that allows us to produce exact upper bounds from continuity conditions. I want to use this post to briefly talk about where these things are coming from. As usual, for regular $\kappa<\lambda$, we denote:

$S^\lambda_\kappa=\{\alpha<\lambda : cf(\alpha)=\kappa\}$

Further, for a set $C$ of ordinals, we denote:

$acc(C)=\{\alpha\in C : \alpha=\sup (\alpha\cap C)\}$

$nacc(C)=C\setminus acc(C)$.

Definition: Let $\lambda$ be a regular cardinal, and let $\vec a=\langle a_\alpha : \alpha<\lambda\rangle$ be a sequence of elements of $[\lambda]^{<\lambda}$. Given a limit ordinal $\delta<\lambda$, we say that $\delta$ is approachable with respect to $\vec a$ if there is an unbounded $A\subseteq \delta$ of order type $cf(\delta)$ such that every initial segment of $A$ is enumerated prior to stage $\delta$. More precisely:

For every $\alpha<\delta$, there exists a $\beta<\delta$ such that $A\cap\alpha=a_\beta$.

Definition: Let $\lambda$ be a regular cardinal and define $I[\lambda]$ to be the collection of $S\subseteq\lambda$ such that there is a sequence $\vec a=\langle a_\alpha :\alpha<\lambda$ of elements of $[\lambda]^{<\lambda}$ and a club $E\subseteq\lambda$ such that every $\delta\in E\cap S$ is singular and approachable with respect to $\vec a$.

So the idea is that an ordinal is approachable with respect to some sequence above if there is some unbounded set whose initial segments get captured in a timely manner. A set of ordinals $S$ is in $I[\lambda]$ if almost every (modulo clubs) ordinal in $S$ is uniformly approachable, and this uniformity is captured by a single sequence.

Proposition: $I[\lambda]$ is a (possible improper) normal ideal over $\lambda$.

One thing to note is that, if $\lambda\in I[\lambda]$, then there is a club of singular ordinals which are all approachable by way of a single sequence $\vec a$. So one can imagine that if this is indeed possible, then $\lambda$ must have some nice combinatorial structure. It turns out that this is indeed possible, and this yields a square-like principle. The following alternative characterization of $I[\lambda]$ makes this more evident.

Theorem (Shelah): Let $\lambda$ be a regular cardinal. Then for $S\subseteq \lambda$, we have $S\in I[\lambda$ if and only if there is a sequence $\bar C= \langle C_\alpha : \alpha<\lambda\rangle$ and a club $E\subseteq\lambda$ such that:

1. Each $C_\alpha$ is a closed (but not necessarily unbounded) subset of $\alpha$;
2. if $\beta\in nacc(C_\alpha)$ then $C_\beta=\alpha\cap C_\alpha$;
3. If $\delta\in S\cap E$, then $\delta$ is singular, and $C_\delta$ is a club subset of $\delta$ of order type $cf(\delta)$.

With this in hand, we now note that $I[\lambda]$ is actually quite large.

Theorem (Shelah): Suppose that $\kappa^+<\sigma<\lambda$ for regular cardinals $\kappa,\sigma,\lambda$. Then there is a stationary $S\subseteq S^\lambda_\kappa$ in $I[\lambda]$ such that $S\cap \theta$ is stationary for stationarily-many $\theta\in S^\sigma_\kappa$.

Corollary: Suppose that $\kappa^{++}<\lambda$ for regular $\lambda,\kappa$. Then there is a stationary $S \subseteq S^\lambda_\kappa$ such that $S\in I[\lambda]$.

Thus, we see that a continuity condition is just a witness that these particular stationary sets live in $I[\lambda]$. Beyond giving us a link between squares and scales (which I want to fill out in the next post), $I[\lambda]$ is interesting in its own right. I won’t get into it much for now, but Todd Eisworth’s handbook chapter has a nice exposition on $I[\lambda]$ and its applications.

# Continuity Conditions and Exact Upper Bounds

I want to use this post to work through some material which solidifies the relationship between square-like principles and scales. We begin with a theorem due to Shelah.

Theorem: Let $\lambda, \kappa$ be regular cardinals with $\kappa^{++}<\lambda$. Then there is a stationary set $S\subseteq S^{\lambda}_{\kappa}$, and a sequence $\bar C=\langle C_\alpha : \alpha<\lambda\rangle$ such that:

1. Each $C_\alpha$ is a closed subset of $\alpha$;
2. If $\beta\in nacc(C_\alpha)$, then $C_\beta=C_\alpha\cap\beta$;
3. If $\delta\in S$, then $C_\delta$ is a club subset of $\delta$ with order-type $cf(\delta)=\kappa$.

The above sequence $\bar C$ is referred to as a continuity condition for $S$ by Shelah in Cardinal Arithmetic. We can think of these continuity conditions as weak fragments of square, which always hold in ZFC. The interesting thing is that continuity conditions allow us to produce exact upper bounds.

Definition: Let $\bar C=\langle C_\alpha : \alpha<\lambda$ be a continuity condition for some stationary set $S\subseteq S^\lambda_\kappa$ as above, and let $\vec f=\langle f_\alpha :\alpha<\lambda \rangle$ be a sequence of functions from $\kappa$ to the ordinals. We say $\vec f$ weakly obeys $\bar C$ if:

If $\alpha<\lambda$, then for each $\beta\in nacc(C_\alpha)$, we have $f_\beta(i) for each $i<\kappa$.

One thing to note is that despite having been published in 1994, Cardinal Arithmetic has no new material past 1989. In particular, the existence of continuity conditions over stationary sets was not known in the case that $\lambda=\mu^+$ for $\mu$ singular when cardinal arithmetic was sealed. However, Todd Eisworth’s chapter in the handbook has a nice exposition about the above theorem. On the other hand, the following is included in Cardinal Arithmetic (appearing as Claim 2.6A on page 16):

Theorem: Let $I$ be an ideal on a regular cardinal $\kappa$. Assume that:

1. $\lambda>\kappa^+$ is regular such that there is some stationary $S\subseteq S^\lambda_{\kappa^+}$ which has a continuity condition $\bar C$;
2. $\vec f=\langle f_\alpha : \alpha<\lambda\rangle$ is a sequence of functions from $\kappa$ to the ordinals;
3. $\vec f$ obeys $\bar C$.

Then $\vec f$ has a $\leq_I$-exact upper bound.

Proof: We first produce a $\leq_I$-least upper bound, and then show that this bound must be exact. In order to produce the desired lub, we inductively produce better and better upper bounds $g_\xi$ which are better and better approximations to a lub as follows:

Stage $\xi=0$: Let $g_0$ be defined by $g_0(i)=\sup_{\alpha<\lambda} f_\alpha(i) + 1$.

Stage $\xi=\eta+1$: Suppose that $g_\eta$ has been defined, and is a $\leq_I$-upper bound for $\vec f$. If $g_\eta$ is a least-upper bound, then we can terminate the induction. Otherwise, there is some $g_{\eta+1}\leq_I g_\eta$ which is an upper bound for $\vec f$ such that $g_{\eta+1}\neq_I g_\eta$.

Stage $\xi$ limit: Suppose that $g_\eta$ has been defined for each $\eta<\xi$. For each $i<\kappa$, define the set $S_\xi(i)=\{g_\eta(i) : \eta<\xi\}$, and for each $\alpha<\lambda$, define $f^\xi_\alpha$ ,the projection of $f_\alpha$ to $S_\xi=\langle S_\xi(i) : i<\kappa$, by setting $f^\xi_\alpha(i)=\min(S_\xi(i)\setminus f_\alpha(i))$.

Note that each $f^\xi_\alpha$ is defined $I$-almost everywhere, and that the sequence $\vec{f}^\xi=\langle f^\xi_\alpha: \alpha<\lambda\rangle$ is $\leq_I$-increasing. We claim that there exists some $\alpha_\xi<\lambda$ such that for each $\alpha\in[\alpha_\xi, \lambda)$, we have $f^\xi_\alpha=_I f^\xi_{\alpha_\xi}$, and we set $g_\xi=f^\xi_\alpha$. To see this, assume otherwise.

As the sequence $\vec f^\xi$ never stabilizes we can find a club set $E\subseteq \lambda$ such that

$\alpha,\beta\in E$ and $\alpha<\beta$ implies $f^\xi_\alpha\neq_I f^\xi_\beta$.

Let $\delta\in acc(E)\cap S$, then $C_\delta\subseteq \delta$ is club in $\delta$ of order-type $\kappa^+$. We then inductively define an increasing sequence $\{\beta_\epsilon : \epsilon<\kappa^+\}$ of ordinals in $nacc( C_\delta)$ as follows:

We let $\beta_0\in nacc(C_\delta)$. If $\beta_\epsilon$ has been defined, then we can find $\beta_{\epsilon+1}\in nacc(C_\delta)$ be such that $f_{\beta_\epsilon}\neq_I f_{\beta_{\epsilon+1}}$. Just pick $\beta<\delta$ such that $\beta\in E$ and $\beta>\beta_\epsilon$ (which we can do since $\delta$ is an accumulation point of $E$), and let $\beta_{\epsilon+1}>\beta$ be in $nacc(C_\delta)$. At limit stages, let $\beta_{\epsilon}'=\sup_{\gamma<\epsilon}\beta_\gamma\in C_\delta$, and let $\beta_\epsilon>\beta_\epsilon'$ be in $nacc(C_\delta)$. Then our collection has the following properties:

1. $\beta_\epsilon\cap C_\delta=C_{\beta_\epsilon}$;
2. $f^\xi_{\beta_\epsilon}\neq_I f^\xi_{\beta_{\epsilon+1}}$.

Now for each $\epsilon<\kappa^+$, define the set

$t_\epsilon=\{i<\kappa : f^\xi_{\beta_\epsilon}(i).

By condition 2, each $t_\epsilon\notin I$, and so for each $\epsilon<\kappa^+$ we can pick $i(\epsilon)\in t_\epsilon$. Since we have $\kappa^+$-many of these sets, and $i$ ranges through $\kappa$, there must be some unbounded subset of $\kappa^+$ for which $i(\epsilon)$ is constant. Call this constant value $i(*)$. Now suppose $\epsilon(1)<\epsilon(2)<\kappa^+$ are such that $i(\epsilon(1))=i(\epsilon(2))=i(*)$, then $\beta_{\epsilon(1)+1}\in C_\delta\cap\beta_{\epsilon(2)}=C_{\beta_(\epsilon(2))}$ (in fact, it’s in $nacc(C_{\beta(\epsilon(2))}$. Since $\vec f$ obeys $\bar C$, we have that $f_{\beta_{\epsilon(1)+1}}(i(*)), and so $f^\xi_{\beta_{\epsilon(1)+1}}(i(*))\leq f^\xi_{\beta_{\epsilon(2)}}(i(*))$. Finally, since $i(\epsilon(1))=i(\epsilon(2))$ are both in $t_{\epsilon(1))}$ and $t_{\epsilon(2))}$ respectively, we get the following string of inequalities:

$f^\xi_{\beta_{\epsilon(1)}}(i(*)).

But then, the sequence $\langle f^\xi_{\beta_{\epsilon +1}}: i(\epsilon)=i(*)\rangle$ is strictly increasing along $S_\xi(i)$, which has size $\kappa$. Since $i(\epsilon)=i(*)$ happens $\kappa^+$-many times, this is absurd. Thus, the induction can be carried through limit stages.

Next, we claim that this induction cannot have been carried through $\kappa^+$-many stages. Otherwise, assume that $g_\xi$ has been defined for each $\xi<\kappa^+$, and let $\alpha(*)=\sup_{\xi<\kappa^+} \alpha_{\xi}<\lambda$. Next note since the sequence $\langle S_\xi (i) : \xi<\kappa^+$ is $\subseteq$-increasing for each $i<\kappa$, that the sequence $\langle f^\xi_{\alpha(*)} : \xi<\kappa^+\rangle$ must be decreasing and hence stabilize. Thus, we see that $\langle f^\xi_{\alpha(*)}: \xi<\kappa^+\rangle$ must also be eventually constant. On the other hand, $f^\xi_{\alpha(*)}=_I g_\xi$ for each $\xi<\kappa^+$, contradicting the fact that $g_\xi\neq_I g_\eta$ for $\xi,\eta$ large enough.

Therefore, $\vec f$ has a $\leq_I$-upper bound $g$. If we can show that $g$ is exact, then we are done. So suppose otherwise and let $h$ witness this. That is, $h$ is a function from $\kappa$ to the ordinals such that $h<_I g$, but there is no $\alpha<\lambda$ such that $h<_I f_\alpha$. So for each $\lambda$ define $t_\alpha=\{ i<\kappa : h(i)\leq f_\alpha(i)\}$, which is $I$-positive by assumption. Further, note that $\langle t_\alpha : \alpha<\lambda\rangle$ cannot stabilize modulo $I$ else if it does stabilize at some $\alpha(*)$, then define $F:\kappa\to ON$ by

$f(i)=h(i)$ if $i\in t_\alpha$ and $f(i)=g(i)$ otherwise.

It is straightforward to check that $F$ would then be an upper bound of $\vec f$ with $F\leq g$ and $F\neq_I g$, contradicting that $g$ is a lub. So as before, we can find a club $E\subseteq \lambda$ such that for all $\alpha<\beta$ in $E$, we have that $t_\alpha\neq t_\beta$ with $t_\beta\subseteq_I t_\alpha$. Letting $\delta\in S\cap acc(E)$, we can again find $\{\beta_\epsilon : \epsilon<\kappa^+\}$ such that

1. $\beta_\epsilon\cap C_\delta=C_{\beta_\epsilon}$;
2. $t_{\beta_\epsilon}\neq t_{\beta_{\epsilon +1}}$.

Let $i(\epsilon)\in t_{\beta_\epsilon}\setminus t_{\beta_{\epsilon+1}}$, and again find an unbounded subset of $\kappa^+$ such that $i(\epsilon)=i(*)$. If we let $\epsilon(1)<\epsilon(2)<\kappa^+$ be such that $i(\epsilon(1))=i(\epsilon(2))=i(*)$, we then get that

$h(i(*))\geq f_{\beta_{\epsilon(2)}}(i(*))>f_{\beta_{\epsilon(1)+1}} (i(*))> h(i(*))$.

This is silly, and so $g$ is an exact upper bound of $\vec f$.