PCF Theory and The Tukey Spectrum

We begin with some set-up. Since $A$ is progressive, we may apply Proposition 2.17 to fix a sequence $\left<B_{\lambda}:\lambda\in\operatorname{pcf}(A)\right>$ of generators for $\operatorname{pcf}(A)$. For each $\lambda\in\operatorname{pcf}(A)$, $B_{\lambda}$ is progressive, being a subset of the progressive set $A$. Thus $\lambda=\operatorname{maxpcf}(B_{\lambda})=\operatorname{cf}\left(\prod B_{\lambda},<\right)$, using $B_{\lambda}\in J_{<\lambda^{+}}\setminus J_{<\lambda}$ for the first equality and applying Proposition 2.19 for the second equality. Therefore, for each $\lambda\in\operatorname{pcf}(A)$, we may choose a sequence $\vec{f}^{\lambda}=\left<f^{\lambda}_{\alpha}:\alpha<\lambda\right>$ of functions in $\prod A$ so that $\left<f^{\lambda}_{\alpha}\upharpoonright B_{\lambda}:\alpha<\lambda\right>$ is cofinal in $\left(\prod B_{\lambda},<\right)$.

Fix a cardinal $\kappa\in\operatorname{spec}(A)$, suppose that $\kappa$ is not a limit point of $\operatorname{pcf}(A)$, and we will show that $\kappa\in\operatorname{pcf}(A)$. Because $\kappa$ is not a limit point of $\operatorname{pcf}(A)$, we know that $|\operatorname{pcf}(A)\cap\kappa|<\kappa$; this will permit us to apply a pigeonhole argument at a crucial step later in the proof.

Since $\kappa\in\operatorname{spec}(A)$, let $\left<f_{\gamma}:\gamma<\kappa\right>$ enumerate a set $\cal{F}$ of $\kappa$-many functions in $\prod A$ so that every $\kappa$-sized subset of $\cal{F}$ is unbounded in $\left(\prod A,<\right)$. To show that $\kappa\in\operatorname{pcf}(A)$, it suffices to show that $J_{<\kappa}\neq J_{<\kappa^{+}}$. Since $\cal{F}$ is bounded in $\prod A/J_{<\kappa^{+}}$ (because this reduced product is $<\kappa^{+}$-directed, by Proposition 2.15), it in turn suffices to show that $\cal{F}$ is unbounded in $\prod A/J_{<\kappa}$.

We work by contradiction and suppose that $\cal{F}$ is bounded in $\prod A/J_{<\kappa}$. Our goal is to define a set $\cal{X}$ of fewer than $\kappa$-many functions in $\prod A$ so that every $f\in\cal{F}$ is pointwise below some $g\in\cal{X}$. Supposing we can define such an $\cal{X}$, we obtain a contradiction as follows: since $\kappa$ is regular, there is a single $g\in\cal{X}$ which is pointwise above $\kappa$-many elements of $\cal{F}$, and this contradicts the fact that $\cal{F}$ witnesses that $\kappa\in\operatorname{spec}(A)$.

We begin the construction of $\cal{X}$ as follows. Since $\cal{F}$ is bounded in $\prod A/J_{<\kappa}$, let $g$ be a bound. For each $n\in\omega$, each sequence $\vec{\lambda}=\lambda_{0}>\dots>\lambda_{n}$ in $\operatorname{pcf}(A)\cap\kappa$, and each sequence $\vec{\alpha}=\left<\alpha_{0},\dots,\alpha_{n}\right>$ with $\alpha_{i}\in\lambda_{i}$ for all $i\leq n$, we let $h(\vec{\alpha},\vec{\lambda})$ be the function $\max(f^{\lambda_{0}}_{\alpha_{0}},\dots,f^{\lambda_{n}}_{\alpha_{n}},g)$; see Notation 3.1. We let $\cal{X}$ be the set of such $h$. Since by assumption $\operatorname{pcf}(A)\cap\kappa$ is bounded below $\kappa$, we see that $\cal{X}$ consists of fewer than $\kappa$-many functions.

Now we verify that $\cal{X}$ has the desired property. Recalling that $\left<f_{\alpha}:\alpha<\kappa\right>$ enumerates $\cal{F}$, fix some $\alpha<\kappa$. Since $f_{\alpha}<_{J_{<\kappa}}g$, we know that

is in $J_{<\kappa}$. Thus $\max\operatorname{pcf}(z_{\alpha})<\kappa$, and so by Proposition 2.18, we may find a sequence $\lambda_{0}>\dots>\lambda_{n}$ in $\operatorname{pcf}(A)\cap\kappa$ so that $z_{\alpha}\subseteq\bigcup_{i\leq n}B_{\lambda_{i}}$. Then, for each $i\leq n$, pick $\alpha_{i}<\lambda_{i}$ so that

(i.e., this holds pointwise on $B_{\lambda_{i}}$). Then we see that

Indeed, if $a\notin z_{\alpha}$, then $f_{\alpha}(a)<g(a)$. On the other hand, if $a\in z_{\alpha}$, there is some $i\leq n$ so that $a\in B_{\lambda_{i}}$, and hence $f_{\alpha}(a)<f^{\lambda_{i}}_{\alpha_{i}}(a)$. Since the function $\max(f^{\lambda_{0}}_{\alpha_{0}},\dots,f^{\lambda_{n}}_{\alpha_{n}},g)$ is in $\cal{X}$, this completes the main part of the proof.

For the “in particular” part of the theorem, observe that if $\operatorname{pcf}(A)$ is progressive, then there are no regular limit $\kappa$ in $\lim(\operatorname{pcf}(A))$. Otherwise, $|\operatorname{pcf}(A)|\geq\kappa>\min(\operatorname{pcf}(A))$, contradicting that $\operatorname{pcf}(A)$ is progressive. ∎

Set $\mu:=\sup(A)$; note that we are making no assumption about whether or not $\mu\in A$. Since $\kappa\in\operatorname{spec}(A)$, let $\cal{F}$ be a set of $\kappa$-many functions in $\prod A$ so that for all $\cal{F}_{0}\in[\cal{F}]^{\kappa}$, $\cal{F}_{0}$ is unbounded in $(\prod A,<)$. Let

and

Note that since $\kappa<\mu=\sup(A)$, $A\setminus(\kappa+1)$ is a non-empty set of regular cardinals. Also, note that $\cal{F}_{>\kappa}$ is bounded in $\prod(A\setminus(\kappa+1),<)$, since $\cal{F}_{>\kappa}$ consists of at most $\kappa$-many functions, and we can take a sup of $\kappa$-many elements on each coordinate in $A\setminus(\kappa+1)$.

We next argue that $\cal{F}_{\leq\kappa}$ has size $\kappa$ and that every $\kappa$-sized subset is unbounded in $(\prod A\cap(\kappa+1),<)$; this will show the desired result. First suppose for a contradiction that $\cal{F}_{\leq\kappa}$ has size $<\kappa$. Then there is a single $\bar{f}\in\prod A\cap(\kappa+1)$ so that for $\kappa$-many $g\in\cal{F}$, $g\upharpoonright(\kappa+1)=\bar{f}$. Let $\cal{F}_{0}$ be this set of $g\in\cal{F}$. But then $\cal{F}_{0}$ is bounded in the entire product $(\prod A,<)$, using the observation from the previous paragraph to bound the elements of $\cal{F}_{0}$ on coordinates in $A$ above $\kappa$. A nearly identical argument shows that $\cal{F}_{\leq\kappa}$ must satisfy that every $\kappa$-sized subset is unbounded in $(\prod A\cap(\kappa+1),<)$. ∎

To begin, let $\left<\mu_{\nu}:\nu<\kappa\right>$ be the increasing enumeration of $A$, and let $\left<\zeta_{i}:i<\kappa\right>$ enumerate a club $C\subseteq\kappa$ with $C\cap A=\emptyset$. By relabeling if necessary, we take $\zeta_{0}=0$.

Suppose for a contradiction that $\operatorname{cf}(\prod A/D)\leq\kappa$; then the cofinality must equal exactly $\kappa$ since $D$ extends the tail filter on $A$. Let $\vec{f}=\left<f_{\alpha}:\alpha<\kappa\right>$ be a sequence of functions in $\prod A$ which is increasing and cofinal modulo $D$. We obtain our contradiction by showing that $\vec{f}$ is bounded in $\prod A$, modulo the tail filter on $A$, and hence modulo $D$.

Given $\nu<\kappa$ (corresponding to $\mu_{\nu}$), let $i<\kappa$ so that $\nu\in[\zeta_{i},\zeta_{i+1})$ (such an $i$ exists since $C$ is club and $\zeta_{0}=0$). The following observation, though simple, is crucial: for each $i<\kappa$, $\mu_{\zeta_{i}}>\zeta_{i}$: indeed, $\mu_{\zeta_{i}}\geq\zeta_{i}$ since the enumeration of $A$ is increasing. But $\zeta_{i}\in C$ and $\mu_{\zeta_{i}}\in A\subseteq\kappa\setminus C$, so $\mu_{\zeta_{i}}>\zeta_{i}$. Therefore $\mu_{\nu}$ is also strictly above $\zeta_{i}$, and so

is below $\mu_{\nu}$. We let $g(\mu_{\nu})<\mu_{\nu}$ be above this sup. $g$ is then a member of $\prod A$.

We finish the proof by showing that $g$ bounds each of the $f_{\xi}$ on a tail. To this end, fix $\alpha<\kappa$, and let $i<\kappa$ so that $\alpha\leq\zeta_{i}$. We claim that for all $\nu\geq\zeta_{i}$, $g(\mu_{\nu})>f_{\alpha}(\mu_{\nu})$. Fix $\nu\geq\zeta_{i}$, and let $j\geq i$ so that $\nu\in[\zeta_{j},\zeta_{j+1})$. Then

where the last inequality follows since $\alpha\leq\zeta_{i}\leq\zeta_{j}$. ∎

Fix such a $V$ (for example, work in $L$ up to the first $\lambda$ which is Mahlo in $L$, if such a $\lambda$ exists, and otherwise work in all of $L$). Let $A$ be a set of regular cardinals, and we will prove the result by induction on the order type of $A$.

We first dispense with the case when $\max(A)$ exists. Let $\lambda:=\max(A)$. Then, applying Lemma 2.9 and Fact 2.11, as well as our inductive assumption, we get

Now we suppose that $A$ does not have a max. Let $\mu:=\sup(A)$, and fix $\kappa\in\operatorname{spec}(A)$. We have a few cases on $\kappa$.

First suppose that $\kappa<\mu$. Then by Lemma 3.3, $\kappa\in\operatorname{spec}(A\cap(\kappa+1))$. Since the order type of $\bar{A}:=A\cap(\kappa+1)$ is less than the order type of $A$, we apply the induction hypothesis to conclude that

Now suppose that $\kappa\geq\mu$. If $\kappa=\mu$ (and in particular, $\mu$ is a regular limit cardinal), then because $A$ is unbounded in $\mu$, $\kappa=\mu\in\lim(A)$.

The final case is that $\kappa>\mu$ (here $\mu$ may be either regular or singular). Since $\mu$ is a limit cardinal, our cardinal arithmetic assumption implies that $\prod A$ has size $\mu^{\mu}=\mu^{+}$. Hence no cardinal greater than $\mu^{+}$ is in $\operatorname{spec}(A)$ or in $\operatorname{pcf}(A)$.

To finish the proof in this final case, we will argue that $\mu^{+}\in\operatorname{pcf}(A)$. Towards this end, let $D$ be an ultrafilter on $A$ which extends the tail filter on $A$. Then $\operatorname{cf}(\prod A/D)\geq\mu=\sup(A)$. If $\mu$ is singular, then the regular cardinal $\operatorname{cf}(\prod A/D)$ is greater than $\mu$. On the other hand, if $\mu$ is regular, then $\mu$ is not a Mahlo cardinal by assumption. This in turn, with the help of Lemma 3.4, implies that $\operatorname{cf}(\prod A/D)>\mu$. Thus in either case on $\mu$, $\operatorname{cf}(\prod A/D)>\mu$. This cofinality must be exactly $\mu^{+}$, however, since $\prod A$ has size $\mu^{+}$.

For the “consequently” part of the theorem, recall that $\operatorname{spec}(A)$ consists, by definition, of regular cardinals. Thus if there are no regular limit cardinals, $\lim(A)$ contains no regular cardinals, and this implies that $\operatorname{spec}(A)\subseteq\operatorname{pcf}(A)$. By Lemma 2.12, we conclude that $\operatorname{pcf}(A)=\operatorname{spec}(A)$. ∎

We show that functions which are constant on a tail witness the result. For each $\alpha<\kappa$, let $f_{\alpha}$ be the function in $\prod A$ which takes value $0$ on all $a\in A$ with $a\leq\alpha$, and which takes value $\alpha$ on all $a\in A$ with $\alpha<a$. We claim that the sequence $\left<f_{\alpha}:\alpha<\kappa\right>$ enumerates a set which witnesses that $\kappa\in\operatorname{spec}(A)$.

Towards this end, let $X\in[\kappa]^{\kappa}$, and we will show that $\left<f_{\alpha}:\alpha\in X\right>$ is unbounded in $(\prod A,<)$. Since $A$ is stationary, we may find some $a\in A\cap\lim(X)$. Then for all $\alpha\in X\cap a$, $f_{\alpha}(a)=\alpha$. Since $a$ is a limit point of $X$, we have that $\left\{f_{\alpha}(a):\alpha\in X\cap a\right\}$ is cofinal in $a$. This completes the proof. ∎

Let $C\subseteq\kappa$ be a club with $C\cap A=\emptyset$. Enumerate $C$ in increasing order as $\left<\zeta_{i}:i<\kappa\right>$, where we assume $\zeta_{0}=0$. Also, enumerate $A$ in increasing order as $\left<\mu_{i}:i<\kappa\right>$. As in the proof of Lemma 3.4, we have that for each $i<\kappa$, $\mu_{\zeta_{i}}>\zeta_{i}$.

Next, let $\left<f_{\alpha}:\alpha<\kappa\right>$ be an enumeration of a set $\cal{F}$ of $\kappa$-many functions in $\prod A$, and we will show that $\cal{F}$ does not witness that $\kappa\in\operatorname{spec}(A)$. Fix a $\kappa$-model $M$ (see Definition 2.20) which contains $A$, $C$, and $\left<f_{\alpha}:\alpha<\kappa\right>$ as elements. By the weak compactness of $\kappa$, let $\mathcal{U}$ be an $M$-normal ultrafilter on $\cal{P}(\kappa)\cap M$.

Our strategy is to use $\cal{U}$ to successively freeze out longer and longer initial segments of many functions on the sequence $\left<f_{\alpha}:\alpha<\kappa\right>$. We will then bound their tails using the non-stationarity of $A$.

For each $i<\kappa$, the product

is a member of $M$, and hence a subset of $M$. Since $\kappa$ is strongly inaccessible, this product has size $<\kappa$. Applying the fact that $\cal{U}$ is a $\kappa$-complete ultrafilter on $\cal{P}(\kappa)\cap M$, we may find a function $\varphi_{i}$ in $\prod_{\nu\in[\zeta_{i},\zeta_{i+1})}\mu_{\nu}$ so that

Next, define an increasing sequence $\left<\beta_{j}:j<\kappa\right>$ below $\kappa$ so that for all $j<\kappa$,

This also uses the completeness of $\cal{U}$ to see that for each $j<\kappa$, $\bigcap_{i<j}Z_{i}$ is in $\cal{U}$ and has size $\kappa$.

As a result of this freezing out, we have the following: for a fixed $i<\kappa$, and all $j>i$, $\beta_{j}\in Z_{i}$. Hence, for all $\nu\in[\zeta_{i},\zeta_{i+1})$,

Now consider an $i<\kappa$ and $\nu\in[\zeta_{i},\zeta_{i+1})$. Let

i.e., all values of all of the $f_{\beta_{j}}$ on the column $\mu_{\nu}\in A$. By applying the argument in the previous paragraph, we conclude that $R(\nu)$ in fact equals

Now fix an arbitrary $\nu<\kappa$. Since $C$ is a club and $\zeta_{0}=0$, there is an $i$ so that $\nu\in[\zeta_{i},\zeta_{i+1})$. We claim that $R(\nu)=\left\{f_{\beta_{j}}(\mu_{\nu}):j\leq i\right\}\cup\left\{\varphi_{i}(\mu_{\nu})\right\}$ has size less than $\mu_{\nu}$. If $i$ is finite, then $R(\nu)$ is finite, and hence has size smaller than $\mu_{\nu}$ (which is, after all, an infinite cardinal). On the other hand, if $i$ is infinite, then

Consequently, for each $\nu<\kappa$, $R(\nu)$ is bounded in $\mu_{\nu}$, as $\mu_{\nu}$ is regular. But by definition of $R(\nu)$, this means that for all $\nu<\kappa$, $\left\{f_{\beta_{j}}(\mu_{\nu}):j<\kappa\right\}$ is bounded in $\mu_{\nu}$. Thus $\left\{f_{\beta_{j}}:j<\kappa\right\}$ enumerates a set of $\kappa$-many functions from $\cal{F}$ which is bounded in $(\prod A,<)$. Since $\cal{F}$ was arbitrary, this shows that $\kappa\notin\operatorname{spec}(A)$.

For the final statement of the theorem, note that if $A\subseteq\kappa$ is non-stationary, then since $\kappa$ is Mahlo, Proposition 4.1 shows that $\kappa\notin\operatorname{spec}(A)$. ∎

Let $\left<\dot{f}_{\alpha}:\alpha<\kappa\right>$ be a sequence of $\mathbb{P}$-names for elements of $\prod A$. We will find an $X\in[\kappa]^{\kappa}$ in $V$ so that $\mathbb{P}$ forces that $\left<\dot{f}_{\alpha}:\alpha\in X\right>$ is bounded.

Indeed, using the c.c.c. of $\mathbb{P}$, for each $\alpha<\kappa$, we may find a function $\varphi_{\alpha}\in\prod A$ so that $\mathbb{P}\Vdash(\forall a\in A)\,[\dot{f}_{\alpha}(a)<\varphi_{\alpha}(a)]$. Namely, let $\varphi_{\alpha}(a)$ be above the sup of the countably-many $\gamma\in a$ so that $\gamma$ is forced to be the value of $\dot{f}_{\alpha}(a)$ by some condition in $\mathbb{P}$.

By the Theorem 4.2, let $X\in[\kappa]^{\kappa}$ so that $\left<\varphi_{\alpha}:\alpha\in X\right>$ is bounded in the product $(\prod A,<)$, say with $h$ as a bound. Then $\mathbb{P}$ forces that for each $\alpha\in X$, $\dot{f}_{\alpha}$ is pointwise below $h$. ∎

If $D$ is an ultrafilter on $A$ that concentrates on a bounded subset of $A$, then the strong inaccessibility of $\kappa$ implies that $\operatorname{cf}(\prod A/D)<\kappa$. On the other hand, if $D$ extends the tail filter on $\kappa$, then by Lemma 3.4, $\operatorname{cf}(\prod A/D)>\kappa$. ∎

Suppose otherwise, with $\cal{F}_{0}$ as a counterexample. Then since $\operatorname{ub}(\cal{F}_{0})$ is finite and has a max below $\kappa$, $\prod\operatorname{ub}(\cal{F}_{0})$ has size below $\kappa$. Let $\cal{F}_{1}\in[\cal{F}_{0}]^{\kappa}$ so that the function $f\in\cal{F}_{1}\mapsto f\upharpoonright\prod\operatorname{ub}(\cal{F}_{0})$ is constant, say with value $\bar{f}$. Then we can bound all of $\cal{F}_{1}$ in the entire product $\prod A$ using $\bar{f}$ on the coordinates in $\operatorname{ub}(\cal{F}_{0})\supseteq\operatorname{ub}(\cal{F}_{1})$. ∎