Possible troubles in Shelah's book "Cardinal Arithmetic"

Question

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 $$ \operatorname{cov} \left( \aleph_{\omega + \omega} , \aleph_{\omega + 1} , \aleph_{\omega + 1} , \aleph_0 \right) ? $$

Answer 1

I found a way to prove in detail a slightly weaker version of the Observation 5.3(7) cited above:

Definition Given an ordinal $\alpha$, and any cardinals $\mu$, $\eta$, $\theta$ and $\sigma$, with $\eta \geq \theta$, define $$ \operatorname{s}_{\mu , \eta , \theta , \sigma} = \sup \left\lbrace \operatorname{cov} \left( \nu , \nu , {\left( \operatorname{cf} (\nu) \right)}^{+} , \operatorname{cf} (\nu) \right) : \eta \leq \nu \leq \mu , \sigma \leq \operatorname{cf} (\nu) < \theta \right\rbrace $$ and $$ \operatorname{i}_{\mu , \eta , \theta , \sigma} \left( \alpha \right) = \left\lbrace \begin{array}{ll} \mu , & \textrm{if } \alpha = 0 ; \\ \sup \left\lbrace \operatorname{i}_{\mu , \eta , \theta , \sigma} \left( \beta \right) : \beta < \alpha \right\rbrace , & \textrm{if } \alpha \textrm{ is a limit ordinal;} \\ \operatorname{i}_{\mu , \eta , \theta , \sigma} \left( \beta \right) + \operatorname{s}_{\operatorname{i}_{\mu , \eta , \theta , \sigma} \left( \beta \right) , \eta , \theta , \sigma} , & \textrm{if } \alpha = \beta + 1 . \end{array} \right. $$

Proposition Let $\mu$, $\eta$, $\theta$ and $\sigma$ be cardinals such that $$ \mu \geq \eta = \operatorname{cf} (\eta) \geq \theta > \sigma = \operatorname{cf} (\sigma) \geq \aleph_{0} . $$ Then, $$ \operatorname{cov} \left( \mu , \eta , \theta , \sigma \right) \leq \operatorname{i}_{\mu , \eta , \theta , \sigma} \left( \omega^{2} \right) . $$

I needed this proposition in the following paper, where I included, for the reader's convenience, this proposition as Proposition 3.17, with a detailed proof.

A.M.E. Levi. Reflection for lindelöf degree and inaccessible cardinals, Acta Mathematica Hungarica 144(1) (2014), 182-195. doi:10.1007/s10474-014-0442-0

A manuscript version can be viewed here.

When $\sigma = \aleph_{0}$, we show that $$ \operatorname{i}_{\mu , \eta , \theta , \sigma} \left( \omega \right) < \operatorname{i}_{\mu , \eta , \theta , \sigma} \left( \omega + 1 \right) \leq \operatorname{i}_{\mu , \eta , \theta , \sigma} \left( \omega^{2} \right) ; $$ but when $\sigma > \aleph_{0}$, we prove that $$ \operatorname{i}_{\mu , \eta , \theta , \sigma} \left( \omega \right) = \operatorname{i}_{\mu , \eta , \theta , \sigma} \left( \alpha \right) $$ for every ordinal $\alpha > \omega$. Hence, we can prove the following result, which is essentially the book's one, with the additional hypotheses $\kappa = \operatorname{cf} (\kappa)$ and $\sigma > \aleph_{0}$:

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