Thursday, May 30, 2013

Easy Parts

$\DeclareMathOperator{\pp}{pp} \def\pcf{\rm{pcf}} \DeclareMathOperator{\cov}{cov} \def\cf{\rm{cf}} \def\REG{\sf {REG}} \def\restr{\upharpoonright} \def\bd{\rm{bd}} \def\subs{\subseteq} \def\cof{\rm{cof}} \def\ran{\rm{ran}} \DeclareMathOperator{\PP}{pp}$

This is a direct continuation of the previous post.  I want to refresh your memory about some definitions, and then proceed to a proof of the easy parts of the chain of inequalities we were investigating.

We have assumed $\cf(\lambda)<\theta=\cf(\theta)<\lambda$.

Definition
1. $\pp_{<\theta}(\lambda)$ is defined to be the supremum of all cardinals of the form $\cof(\prod\mathfrak{a}/\mathcal{U})$, where
• $\mathfrak{a}$ is cofinal in $\lambda\cap\REG$,
• $|\mathfrak{a}|<\theta$, and
• $\mathcal{U}$ is an ultrafilter on $\mathfrak{a}$ disjoint to the ideal of bounded subsets of $\mathfrak{a}$.
2. $\cf_{<\theta}(\prod(\lambda\cap\REG), <_{J^\bd_\lambda})$ is the minimum cardinality of a family of functions $\mathcal{F}\subseteq\prod(\lambda\cap\REG)$ such that for any cofinal $\mathfrak{a}\subseteq\lambda\cap\REG$ of cardinality $\lambda$ and any $g\in\prod\mathfrak{a}$, there is an $f\in \mathcal{F}$ such that $g<f\restr\mathfrak{a}$ modulo the bounded ideal on $\mathfrak{a}$
3. $\cov(\lambda,\lambda,\theta,2)$ is the minimum cardinality of a family $\mathcal{P}\subseteq [\lambda]^{<\lambda}$ such that for any $B\in [\lambda]^{<\theta}$ there is a $A\in\mathcal{P}$ with $B\subseteq A$

We will show

$\pp_{<\theta}(\lambda)\leq\cf_{<\theta}(\prod(\lambda\cap\REG), <_{J^\bd_\lambda})\leq\cov(\lambda,\lambda,\theta,2).$

This gives us the first two inequalities in our chain, and both are easy.

For the first, suppose $\mathcal{F}$ is as described, and let $\mathfrak{a}$ and $\mathcal{U}$ be as in the definition of $\pp_{<\theta}(\lambda)$

The set $$\{\mathcal{f}\restr\mathfrak{a}:f\in\mathcal{F}\}$$
(or rather the set of equivalence classes of these functions modulo $\mathcal{U}$) is cofinal in $\prod\mathfrak{a}/\mathcal{U}$ because $U$ is disjoint to the bounded ideal on $\mathfrak{a}$ and the inequality follows immediately.

For the second, suppose $\mathcal{P}\subseteq[\lambda]^{<\lambda}$ is as described above.
For $A\in\mathcal{P}$, let
us define a function $f_A\in\prod(\lambda\cap \REG$) by
$$f_A(\kappa)= \begin{cases} \sup(A\cap \kappa)+1 & \text{if this is less than \kappa,}\\ 0 &\text{otherwise}. \end{cases}$$

Does this collection of functions work? Suppose $B$ is a cofinal subset of $\lambda\cap\REG$ of cardinality $<\theta$, and let $g\in\prod B$. Choose $A\in\mathcal{P}$ such that $\ran(g)\subseteq A$.

If $\kappa\in B$ is sufficiently large (say $|A|<\kappa$), then
$$g(\kappa)\leq\sup(A\cap\kappa)<f_A(\kappa)<\kappa,$$
and we are done.

Tuesday, May 28, 2013

A chain of inequalities

$\def\pp{\rm{pp}} \def\pcf{\rm{pcf}} \def\cov{\rm{cov}} \def\cf{\rm{cf}} \def\REG{\sf {REG}} \def\restr{\upharpoonright} \def\bd{\rm{bd}} \def\subs{\subseteq} \def\cof{\rm{cof}} \def\ran{\rm{ran}}$

I spent the weekend reading the very last section of Shelah's Cardinal Arithmetic, and I want to try my hand at writing a reasonable exposition of it.

The main concern he has is the following problem::

Question:
Suppose $\lambda$ is a singular cardinal.  Is $\pp(\lambda)=\cov(\lambda,\lambda,(\cf(\lambda))^+,2)$?

He discusses this question quite a bit in the Analytical Guide for Cardinal Arithmetic.  The "interesting" case is when $\lambda$ is a fixed point of cofinality $\aleph_0$.

In this post, I want to gather together a few inequalities culled from the book and try to build a coherent picture of what's going on.  I'm going to leave some things undefined for now just to get the statement out.
Over my next few posts I'll try to define everything and provide proofs for each of the inequalities.

__________
Theorem
Suppose $\cf(\lambda)<\theta=\cf(\theta)<\lambda$.

Define
$$\lambda(1)=\cf_{<\theta}(\prod(\lambda\cap\REG),<_{J^\bd_{\lambda}})$$

and let $\lambda(2)$ be the minimum cardinality of a family $\mathcal{P}\subseteq[\lambda(1)]^{<\lambda}$ such that for any infinite $B\in[\lambda(1)]^{\theta}$, there is an $A\in \mathcal{P}$ with $A\cap B$ infinite.

Then

$$\pp_{<\theta}(\lambda)\leq\lambda(1)\leq\cov(\lambda,\lambda,\theta, 2)\leq\lambda(2)\leq\cov(\lambda(1),\lambda,\theta^+,\theta).$$
__________

The main case of interest is when $\theta=\cf(\lambda)=\aleph_0$.  In this situation, we get:
$$\lambda(1)=\cf_{\aleph_0}(\prod(\lambda,\cap\REG, <_{J^\bd_{\lambda}}),$$

and

$$\lambda(2)=\min\{|\mathcal{P}|:\mathcal{P}\subseteq[\lambda(1)]^{<\lambda}\wedge(\forall B\in [\lambda(1)]^{\aleph_1})(\exists A\in\mathcal{P})[|A\cap B|\geq\aleph_0]\},$$

while the conclusion is

$$\pp(\lambda)\leq\lambda(1)\leq\cov(\lambda,\lambda,\aleph_1, 2)\leq\lambda(2)\leq\cov(\lambda(1),\lambda,\aleph_2,\aleph_1).$$

Friday, May 24, 2013

Note on [Sh:371] Claim 3.6

$\def\pp{\rm{pp}} \def\pcf{\rm{pcf}} \def\cov{\rm{cov}} \def\cf{\rm{cf}} \def\REG{\sf {REG}} \def\restr{\upharpoonright} \def\bd{\rm{bd}} \def\subs{\subseteq} \def\cof{\rm{cof}}$

I wanted to write out a proof of Claim 3.6 on page 340 of Cardinal Arithmetic for my own benefit.  Be warned that the proof will be clunky as I haven't tried to smooth it out yet.

Theorem (Shelah)
Assume the following:
• $\mu$ is a singular cardinal.
• $\cf(\mu)\leq\theta<\mu$, and
• $\pp_\theta(\mu)<\mu^{+\theta^+}$.
Then $\pp_{<\mu}(\mu)=\pp_\theta(\mu)$.

Proof:

It suffices to prove that whenever $\mathfrak{a}$ is an unbounded subset of $\mu\cap\REG$ of cardinality less than $\mu$, then there is a $\sigma<\mu$ such that
$$\max\pcf(\mathfrak{a}\setminus\sigma)\leq\pp_\theta(\mu).$$
Note that if $|\mathfrak{a}|<\sigma<\mu$, then $\mathfrak{a}\setminus\sigma$ is progressive. This means that $\max\pcf(\mathfrak{a}\setminus\sigma)$ is defined for all sufficiently large $\sigma<\mu$, and furthermore we need only verify the above for progressive $\mathfrak{a}$.

Let $\mathfrak{a}$ be a progressive cofinal subset of $\mu\cap\REG$, and assume by way of contradiction that

$$\sigma<\mu\Longrightarrow \pp_\theta(\mu)<\max\pcf(\mathfrak{a}\setminus\sigma).$$

Let $\langle \mathfrak{b}_\lambda:\lambda\in\pcf(\mathfrak{a})\rangle$ be a generating sequence for $\pcf(\mathfrak{a})$, and let us define
$$X:=\pcf(\mathfrak{a})\cap (\mu, \pp_\theta(\mu)].$$

Given a finite subset $x$ of $X$ and $\sigma<\mu$, we know
$$(\mathfrak{a}\setminus\sigma)\setminus\bigcup_{\lambda\in x}\mathfrak{b}_\lambda\neq\emptyset$$
as otherwise we would have
$$\max\pcf(\mathfrak{a}\setminus\sigma)\leq\max\{\lambda:\lambda\in x\}\leq\pp_\theta(\mu),$$

Since we assume $\pp_\theta(\mu)<\mu^{+\theta^+}$, we know $|X|\leq\theta$, so let $\langle x_i:i<\theta\rangle$ enumerate all finite subsets of $X$ (possibly with repetition).  Also fix an increasing
sequence $\langle \mu_j:j<\cf(\mu)\rangle$ of cardinals cofinal in $\mu$ with $\theta<\mu_0$.

For each $i<\theta$ and $j<\cf(\mu)$, choose

$$\theta(i, j)\in (\mathfrak{a}\setminus\mu_j)\setminus\bigcup_{\lambda\in x_i}\mathfrak{b}_\lambda$$

and define
$$\mathfrak{b}=\{\theta(i,j):i<\theta, j<\cf(\mu)\}.$$

Clearly $\mathfrak{b}$ satisfies the following:
• $\mathfrak{b}\subseteq\mathfrak{a}$ (hence $\mathfrak{b}$ is a progressive set of regular cardinals)
• $\mathfrak{b}$ is cofinal in $\mu$, and
• $|\mathfrak{b}|\leq\theta$.
----------------------------
Claim 1: There is a $\sigma<\mu$ and a finite $x\subseteq X$ such that $\mathfrak{b}\setminus\sigma\subseteq\bigcup\{\mathfrak{b}_\lambda:\lambda\in x\}.$

Proof. Given the definition of $\pp_\theta(\mu)$, we know there is a $\sigma_0<\mu$ such that
$$\max\pcf(\mathfrak{b}\setminus\sigma_0)\leq\pp_\theta(\mu),$$
hence
$$\mathfrak{b}\setminus\sigma_0\in J_{\leq\pp_\theta(\mu)}[\mathfrak{a}].$$

This means we can find $\mathfrak{c}\in J_{<\mu}[\mathfrak{a}]$ and finite $x\subseteq X$ such that
$$\mathfrak{b}\setminus\sigma_0\subseteq \mathfrak{c}\cup\bigcup\{\mathfrak{b}_\lambda:\lambda\in X\}.$$

Since sets in $J_{<\mu}[\mathfrak{a}]$ are bounded below $\mu$, the result follows immediately.
--------------------------

We finish by noting that our construction of $\mathfrak{b}$ guarantees that
$\mathfrak{b}\setminus\bigcup\{\mathfrak{b}_\lambda:\lambda\in x\}$ is unbounded in $\mu$, and this contradicts the conclusion of Claim 1.

Tuesday, May 21, 2013

Let's try this again

This is a test post to see how well I can get Mathjax to work here.

When I started my blog on wordpress  a couple of years ago, I used it as an experiment in exposition.  The first project I worked on was an explication of the first theorem in Shelah's paper [Sh:430], which states

$\def\pp{\rm{pp}} \def\cov{\rm{cov}} \def\cf{\rm{cf}} \def\REG{\sf {REG}} \def\restr{\upharpoonright} \def\bd{\rm{bd}} \def\subs{\subseteq}$

Theorem
If  $\mu$ is singular of uncountable cofinality and $\pp(\mu)=\mu^+$, then $\cov(\mu,\mu,\cf(\mu)^+,2)=\mu^+$.

(Here $\cov(\mu,\mu,\cf(\mu)^+,2)$ is the minimum cardinality of a family $\mathcal{F}\subseteq[\mu]^{<\mu}$ with the property that for any $B\in [\mu]^{\cf(\mu)}$ there is an $A\in\mathcal{F}$ such that $B\subseteq A$. This is potentially smaller than the cofinality of $[\mu]^{\cf(\mu)}$.)

I've been interested in seeing what occurs if we try to remove the "uncountable cofinality" assumption.  What I can show is that the statement

$$\cov(\mu,\mu,\cf(\mu)^+, 2)=\mu^+$$

is equivalent to the existence of a family $\langle f_\alpha:\alpha<\mu^+\rangle$  in $\prod(\mu\cap\REG)$ such that

• $\alpha<\beta\Longrightarrow f_\alpha<f_\beta$ modulo the bounded ideal on $\mu\cap\REG$, and
• whenever $\mathfrak{a}$ is a cofinal  subset of $\mu\cap\REG$ of cardinality $\cf(\mu)$, the restrictions $\langle f_\alpha\restr\mathfrak{a}:\alpha<\mu^+\rangle$ are cofinal in $\prod\mathfrak{a}/J^\bd[\mathfrak{a}]$
The family $\langle f_\alpha:\alpha<\mu^+\rangle$ could rightly be called a "master scale" for $\mu$:
the assumption $\pp(\mu)=\mu^+$ is equivalent to "every cofinal $\mathfrak{a}\subseteq\mu\cap\REG$ of cardinality $\cf(\mu)$ carries a scale'', while the ostensibly stronger assumption about $\cov$ gives us a single "master scale'' that works uniformly for all such $\mathfrak{a}$.

This concludes my test post.  This was certainly much much easier than at wordpress because of the ability to define macros!