Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I found some possible troubles in Observation 5.3(7) in the Chapter II of the Shelah's book "Cardinal Arithmetic" (page 86). For convenience, I quote the result and the proof in the book here (together with Observation 5.3(4) that is used in the proof of 5.3(7)):

Observation 5.3

(4) If $\lambda > \kappa \left( \geq \theta > \sigma \right)$, $\sigma$ regular then $$ \operatorname{cov} \left( \lambda , \kappa , \theta , \sigma \right) = \sum_{\mu \in \left[ \kappa , \lambda \right]} \operatorname{cov} \left( \mu , \mu , \theta , \sigma \right) . $$

(7) If $\lambda \geq \kappa \geq \theta > \sigma = \operatorname{cf} (\sigma)$, $\operatorname{cf} (\kappa) \geq \theta$, $\lambda_{0} = \lambda$, $$ \lambda_{n+1} = \sup \left\lbrace \operatorname{cov} \left( \mu , \mu , \tau^{+} , \tau \right) : \kappa \leq \mu \leq \lambda_{n} , \operatorname{cf} (\mu) = \tau \in \left[ \sigma, \theta \right) \right\rbrace $$ then $$ \operatorname{cov} \left( \lambda , \kappa , \theta , \sigma \right) \leq \bigcup_{n < \omega} \lambda_{n} . $$

Proof: 7) Let $\chi$ be regular large enough, by induction on $n$ choose $N_{n} \prec \left( H(\chi) , \in \right)$ of cardinality $\lambda_{n}$ such that $$ \left\lbrace N_{0}, \ldots , N_{n-1}, \lambda , \kappa , \theta , \sigma \right\rbrace \cup \left( \lambda_{n} + 1 \right) \subseteq N_{n} , $$ and $$ \mathcal{P}_{n} = \left\lbrace A \in N_{n} : \left| A \right| < \kappa , A \subseteq \lambda \right\rbrace $$ and $\mathcal{P}_{\omega} = \bigcup_{n < \omega} \mathcal{P}_{n}$. Suppose $X \subseteq \lambda$, $\left| X \right| < \theta$ and for no $\mathcal{P} \subseteq \mathcal{P}_{\omega}$, $\left| \mathcal{P} \right| < \sigma$ is $X \subseteq \bigcup_{A \in \mathcal{P}} A$; let $I$ be the $\sigma$-complete ideal on $X$ generated by $\left\lbrace X \cap A : A \in \mathcal{P}_{\omega} \right\rbrace$, so $X \notin I$. Let $$ \theta_{n} = \min \left\lbrace \left| \mathcal{P} \right| : \mathcal{P} \subseteq \mathcal{P}_{n} , \bigcup_{A \in \mathcal{P}} A \cap X \notin I \right\rbrace ; $$ now $\theta_{n} \leq \left| X \right| < \theta$ and $\operatorname{cf} \left( \theta_{n} \right) \geq \sigma$ and $\theta_{n+1} < \theta_{n}$ (use 5.3(4) applied to $\operatorname{cov} \left( \lambda_{n} , \kappa , \theta_{n} , \theta_{n} \right)$), contradiction.
$\square$

First, it is easy to show that $\theta_{n} \leq \left| X \right| < \theta$, $\operatorname{cf} \left( \theta_{n} \right) \geq \sigma$ and $\theta_{n+1} \leq \theta_{n}$. To show that $\theta_{n+1} < \theta_{n}$, we must apply Observation 5.3(4) to $\operatorname{cov} \left( \lambda_{n} , \kappa , {\left( \theta_{n} \right)}^{+} , \theta_{n} \right)$ ($\operatorname{cov} \left( \lambda_{n} , \kappa , \theta_{n} , \theta_{n} \right)$ is an "uninteresting" covering number: $\operatorname{cov} \left( \lambda_{n} , \kappa , \theta_{n} , \theta_{n} \right) \leq \lambda_{n}$ - consider the family ${[\lambda_{n}]}^1$).

Now, the major difficulty is that we need $\operatorname{cf} (\theta_{n}) = \theta_{n}$ to use 5.3(4), but I don't see how to prove this.

Supposing $\operatorname{cf} (\theta_{n}) = \theta_{n}$, I wrote a detailed proof for 5.3(7): applying 5.3(4), we can show that $$ \operatorname{cov} \left( \left| \mathcal{P}_{n} \right| , \kappa , {\left( \theta_{n} \right)}^{+} , \theta_{n} \right) \leq \operatorname{cov} \left( \lambda_{n} , \kappa , {\left( \theta_{n} \right)}^{+} , \theta_{n} \right) \leq \lambda_{n+1} = \left| N_{n+1} \right| , $$ and then use this to "cover" the set in ${[\mathcal{P}_{n}]}^{\theta_{n}}$ that testifies the definition of $\theta_{n}$, with a set in ${[\mathcal{P}_{n+1}]}^{< \theta_{n}}$.

My questions are: (answers specific to the case $\sigma = \aleph_0$ are welcome too)

1) Is possible to prove that $\operatorname{cf} (\theta_{n}) = \theta_{n}$?

2) Is possible to prove that $\operatorname{cov} \left( \lambda_{n} , \kappa , {\left( \theta_{n} \right)}^{+} , \theta_{n} \right) \leq \lambda_{n+1}$ when $\operatorname{cf} (\theta_{n}) < \theta_{n}$?

3) Is possible to prove that the sequence $(\theta_{n})$ is not eventually constant?

Some observations:

i) This is my proof that $\operatorname{cov} \left( \lambda_{n} , \kappa , {\left( \theta_{n} \right)}^{+} , \theta_{n} \right) \leq \lambda_{n+1}$, if $\operatorname{cf} (\theta_{n}) = \theta_{n}$:

$$ \operatorname{cov} \left( \lambda_{n} , \kappa , {\left( \theta_{n} \right)}^{+} , \theta_{n} \right) = \sum_{\mu \in \left[ \kappa , \lambda_{n} \right]} \operatorname{cov} \left( \mu , \mu , {\left( \theta_{n} \right)}^{+} , \theta_{n} \right) \leq $$ $$ \lambda_{n} \cdot \sup \left\lbrace \operatorname{cov} \left( \mu , \mu , {\left( \theta_{n} \right)}^{+} , \theta_{n} \right) : \kappa \leq \mu \leq \lambda_{n} \right\rbrace . $$

Now, it is easy to show that $\operatorname{cov} \left( \mu , \mu , {\left( \theta_{n} \right)}^{+} , \theta_{n} \right) = \operatorname{cf} (\mu)$ when $\operatorname{cf} (\mu) \neq \theta_{n}$. Thus,

$$ \sup \left\lbrace \operatorname{cov} \left( \mu , \mu , {\left( \theta_{n} \right)}^{+} , \theta_{n} \right) : \kappa \leq \mu \leq \lambda_{n} \right\rbrace \leq $$ $$ \lambda_{n} + \sup \left\lbrace \operatorname{cov} \left( \mu , \mu , {\left( \theta_{n} \right)}^{+} , \theta_{n} \right) : \kappa \leq \mu \leq \lambda_{n} , \operatorname{cf} (\mu) = \theta_{n} \right\rbrace = $$ $$ \lambda_{n} + \sup \left\lbrace \operatorname{cov} \left( \mu , \mu , {\left( \operatorname{cf} (\mu) \right)}^{+} , \operatorname{cf} (\mu) \right) : \kappa \leq \mu \leq \lambda_{n} , \operatorname{cf} (\mu) = \theta_{n} \right\rbrace \leq $$ $$ \lambda_{n} + \sup \left\lbrace \operatorname{cov} \left( \mu , \mu , {\left( \operatorname{cf} (\mu) \right)}^{+} , \operatorname{cf} (\mu) \right) : \kappa \leq \mu \leq \lambda_{n} , \sigma \leq \operatorname{cf} (\mu) < \theta \right\rbrace = $$ $$ \lambda_{n} + \lambda_{n+1} = \lambda_{n+1}, $$ hence $$ \operatorname{cov} \left( \lambda_{n} , \kappa , {\left( \theta_{n} \right)}^{+} , \theta_{n} \right) \leq \lambda_{n} \cdot \lambda_{n+1} = \lambda_{n+1} . $$ $\square$

ii) It is more convenient to define $$ \lambda_{n+1} = \lambda_{n} + \sup \left\lbrace \operatorname{cov} \left( \mu , \mu , {\left( \operatorname{cf} (\mu) \right)}^{+} , \operatorname{cf} (\mu) \right) : \kappa \leq \mu \leq \lambda_{n} , \sigma \leq \operatorname{cf} (\mu) < \theta \right\rbrace . $$

iii) If $\eta$ is any cardinal with $\sigma \leq \operatorname{cf} (\eta) = \eta < \theta$, then we can show that $$ \operatorname{cov} \left( \lambda_{n} , \kappa , \eta^+ , \eta \right) \leq \lambda_{n+1} $$ (same argument of my observation (i)).

iv) The proof works when $\sigma < \theta \leq \aleph_{\sigma}$, since $\sigma \leq \operatorname{cf} (\xi) \leq \xi < \aleph_{\sigma}$ implies $\operatorname{cf} (\xi) = \xi$.

v) When $\sigma = \aleph_0$ (the case that interests me): if we define $$ \theta_{\omega} = \min \left\lbrace \left| \mathcal{P} \right| : \mathcal{P} \subseteq \mathcal{P}_{\omega} , \bigcup_{A \in \mathcal{P}} A \cap X \notin I \right\rbrace , $$ then we can show that $\theta_{\omega}$ is regular. If the sequence $(\theta_{n})$ is eventually constant, then there is $k \in \omega$ such that $\theta_{n} = \theta_{k}$ for any $n \geq k$, and $$ \aleph_0 = \sigma \leq \operatorname{cf} (\theta_{\omega}) = \theta_{\omega} \leq \operatorname{cf} (\theta_{k}) < \theta_{k} < \theta . $$

A more elaborated argument shows that $\operatorname{cf} (\theta_{k}) = \aleph_0$. Hence, $$ \aleph_0 = \sigma = \operatorname{cf} (\theta_{\omega}) = \theta_{\omega} = \operatorname{cf} (\theta_{k}) < \theta_{k} < \theta . $$

vi) Considering observation (iv), everything works for $$ \operatorname{cov} \left( \aleph_{\omega + \omega} , \aleph_{\omega + 1} , \aleph_{\omega} , \aleph_0 \right) . $$ Does the same occur with $$ \operatorname{cov} \left( \aleph_{\omega + \omega} , \aleph_{\omega + 1} , \aleph_{\omega + 1} , \aleph_0 \right) ? $$

share|improve this question
    
Could you please check that my separation of the book material and your material is correct? –  David Roberts Oct 25 '13 at 2:57
    
Yes, thank you, David Roberts. –  Alberto Levi Oct 25 '13 at 4:18
5  
@Alberto: Dear Alberto. There are some informal discussions among my colleagues those who are working in pcf theory that there are many incorrect lemmas in Shelah's book "Cardinal Arithmetic" even they said me that some of the theorems have contradictory assumptions! Perhaps your question is one of them. –  Ali Sadegh Daghighi Oct 25 '13 at 19:02
1  
@AliSadeghDaghighi is the set of mistakes in Shelah's book countable or uncountable? –  Norbert Nov 5 '13 at 6:15
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.