## Thursday, June 30, 2016

### On Weak Compactness II [edited]

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

Lingering a little on the material from the last post just for curiosity's sake:

Theorem

The following are equivalent for an inaccessible cardinal $\kappa$:

1.  $\kappa$ is weakly compact.

2.  For every sequence of functions $\langle f_\alpha:\alpha<\kappa\rangle$ with $f_\alpha:\alpha\rightarrow\alpha$, there is an $f:\kappa\rightarrow\kappa$ such that for every $\alpha<\kappa$ there is a $\beta\geq\alpha$ with $f_\beta\restr\alpha= f\restr\alpha$.

3. For every sequence of sets $\langle A_\alpha:\alpha<\kappa\rangle$ with $A_\alpha\subseteq\alpha$, there is a set $A\subseteq\kappa$ such that for every $\alpha<\kappa$ there is a $\beta\geq\alpha$ with $A_\beta\cap\alpha = A\cap\alpha$.

4. For every $C$-sequence $\langle C_\alpha:\alpha<\kappa\rangle$ there is a closed unbounded $C\subseteq\kappa$ such that for all $\alpha<\kappa$ there is a $\beta\geq\alpha$ with $C_\beta\cap\alpha= C\cap\alpha$.

We'll work this out next time to make sure, but (1) implies (2) by Shelah's result used last time,  and (4) implies (1) by a result of Todorcevic  (Theorem 6.3.5 of his book "Walks on Ordinals and their Characteristics").

One thing to note is that subtle cardinals, weakly ineffable cardinals, and ineffable cardinals all are defined using something that looks like condition (3).  It might be worthwhile to see if the corresponding modifications of (2) and (4) give equivalent characterizations.

--------------

Edit:  (2) follows from (3) immediately, as we just set $f_\alpha$ to be the characteristic function of $A_\alpha$, and then $f$ is the characteristic function of the $A$ we need.

(4) follows from (3) immediately, as it is easily seen that if we apply (3) to a $C$-sequence $\langle C_\alpha:\alpha<\kappa\rangle$ then the resulting $A$ must be closed.