## 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!