5
$\begingroup$

I came across this club-guessing exercise on Cardinal Arithmetic by Abraham and Magidor in the Handbook of Set Theory.

Let $\kappa, \lambda$ be regular cardinals $\kappa^{++}<\lambda$ and let $F$ be a (partial) function $:\ \subseteq [\lambda]^{<\kappa}\to \lambda$. For each $\delta\in S^\lambda_{\kappa^{++}}$ (cardinals in $\lambda$ with cofinality $\kappa^{++}$) there exists a club $E_{\delta}$ such that $[E_{\delta}]^{<\kappa}\subseteq dom(F)$. Show $S=\{\alpha\in S^\lambda_\kappa: \exists club \ D\subseteq \alpha \forall a,b\in D (a<b\rightarrow F(\{d\in D: d\leq a\})<b)\}$ is stationary.

What I can do so far is assume that $F$ is total, and then sparse out the elements in the club by the function $F$. This is what I have tried so far: Assume $F$ is total. Fix a club $E\subset \lambda$. Fix a club-guessing sequence $\langle C_i: i\in S^{\kappa^{++}}_\kappa\rangle$. By induction on $\alpha<\kappa^{++}$ build a club set. Suppose we have already defined $F_\alpha\subset E$, if $\sup(F_\alpha)$ is not yet in $F_\alpha$, put it in and go to the next stage (in particular in the limit stage, this make sure the sequence of elements is continuous). Otherwise, list $F_\alpha$ as $\langle f_j: j<\alpha\rangle$. Pick $f_{\alpha+1}$ to be the least in $E$ such that it is greater than $\sup_{k<i}F(\{f_j: j\in C_i \wedge j<\min\{k,\alpha\} \})$ for all $i\in S^{\kappa^{++}}_\kappa$. In this way, we build a club $E_{\delta}\subset \delta\cap E$ of order type $\kappa^{++}$ for some $\delta$ of cofinality $\kappa^{++}$. List them as $\langle e_i: i\in \kappa^{++}\rangle$. Then by club-guessing, there exists some $i\in S^{\kappa^{++}}_\kappa$ such that $C_i\subseteq \{j\in \kappa^{++}: e_j\in E_{\delta}\}$. Then $e_i\in E\cap S$ as desired.

Now to incorporate $E_\delta$, I'm afraid it might mess up with the original indices if I do it as above. Any thoughts would be appreciated!

$\endgroup$
  • 1
    $\begingroup$ Show $S=\ldots$ is not a complete sentence (specifically since you didn't define $S$ before hand). $\endgroup$ – Asaf Karagila Feb 12 '15 at 17:02
  • 1
    $\begingroup$ Oops. My problem of typing. $\endgroup$ – Jing Zhang Feb 12 '15 at 17:12
  • 1
    $\begingroup$ Why are there three votes to close this question? It’s clear, and the OP has certainly put some effort into it. $\endgroup$ – Brian M. Scott Feb 12 '15 at 21:39
  • $\begingroup$ For the curious, this is Exercise $2.20$ in the cited article, which is available here [DVI]. $\endgroup$ – Brian M. Scott Feb 12 '15 at 21:52
  • 1
    $\begingroup$ @BrianM.Scott I was hoping someone would enlighten me on this as well... $\endgroup$ – Jing Zhang Feb 12 '15 at 23:55
1
$\begingroup$

After thinking for a while, I believe I could prove the original statement as follows.

The modification to the previous argument is when defining continuous $f_{\alpha+1}\in E$ we pick it such that it is greater than $\sup_{k<i} F({f_j:j\in C_i∧j\leq\min\{k,\alpha\}})$ for all $i\in S^{\kappa^{++}}_\kappa$ whenever the value F is defined (if it is not defined we just take whatever available which has not been picked before). Now again we have $\delta$ with cofinality $\kappa^{++}$ such that $F_{\delta}\subset E$ a club. Let $E_\delta\subset \delta$ a club given as in the assumption such that $[E_\delta]^{<\kappa}\subset dom(F)$. Let $D=E_\delta\cap F_{\delta}$ then $D$ is also a club in $\delta$. Also notice that $L=\{j<\kappa^{++}: f_j\in E_{\delta}\}$ then it is a club in $\kappa^{++}$. By club-guessing, for some $i\in S^{\kappa^{++}}_\kappa$we have a club $C_i\subset L$ (so $i \in L$). Verify the property for $P=\{f_k: k\in C_i\}$ club in $f_i$. Given $p<q\in C_i$, by construction we have $F(\{f_d\in F_\delta: d\in C_i \wedge d\leq p\}) < f_{p+1}\leq f_q$. Therefore $f_i\in E\cap S$ and $cf(f_i)=\kappa$.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.