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
$$\max\pcf(\mathfrak{a}\setminus\sigma)\leq\max\{\lambda:\lambda\in x\}\leq\pp_\theta(\mu),$$
contradicting our assumption.

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.








No comments: