# 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 $\vec f$.