cofinality of $\prod_{n \in \omega} \aleph_n$ under everywhere dominance

Question

In Don Monk's notes on set theory (http://euclid.colorado.edu/~monkd/full.pdf), it is written on page 784 (the second item after Proposition 33.6) that the cofinality of $\prod_{n \in \omega} \aleph_n$ under everywhere dominance is singular. How do we know this? From his argument I can only see that the cofinality is not less than $\aleph_\omega$.
Also, what else do we know about the cofinality of the poset?

Answer 1

I think the claim as stated is consistently false. First, one can run a similar diagonal argument as in the case of the dominating number for sets of reals to show that the cofinality is at least $(\aleph_\omega)^+$.

Let $\{f_\alpha\mid \alpha\in \aleph_\omega\}$ be a family of functions in $\prod_{n\in\omega}\aleph_n$. We show that there is a function $h\in \prod_{n\in\omega}\aleph_n$ that is not everywhere dominated by any of the $f_\alpha$.

For $k\neq 0$, set $h(k+1)=(\sup_{i<\aleph_k}f_i(k+1))+1$. For definiteness, set $h(0)=0$. Observe that this is well-defined, because $\sup_{i<\aleph_k}f_i(k+1)$ is a $\aleph_k$-sequence in $\aleph_{k+1}$, and therefore bounded. This $h$ is not dominated by any $f_\alpha$.

Suppose for contradiction that it is dominated by some $f_\alpha$, let $n$ be least such that $\alpha<\aleph_n$ then $h(n+1)=(\sup_{i<\aleph_n}f_i(n+1))+1 >f_{\alpha}(n+1)$. Contradiction!

So the cofinality must be at least $(\aleph_{\omega})^+$. But under GCH, This is the only possibility, because the whole set $\prod_{n\in\omega}\aleph_n$ has cardinality $(\aleph_{\omega})^+$.

An easier example of a poset with singular cofinality is (say) $(\mathcal{P}(\aleph_\omega)\smallsetminus\{\emptyset\}, \leq)$, where $X\leq Y$ iff $X\supseteq Y$. The set of singletons is cofinal, and any set $A$ of cardinality less than $\aleph_\omega$ won't be cofinal, by considering the least ordinal whose singleton is not in $A$