exercise VII.G5 in kunen

Question

Recall $Fn(\alpha,\beta,\gamma)$ is the set of partial functions with domain contained in $\alpha$, range contained in $\beta$, size $< \! \gamma$. How do you prove that $Fn(\aleph_\omega,2,\aleph_\omega)$ collapses $\aleph_{\omega+1}$? I see why it collapses $\aleph_\omega$ but I don't see why it kills the successor too.

Answer 1

It's wrong! Not again Ken!

Assume $\aleph_\omega$ is a strong limit. Let $\lbrace p_\alpha : \alpha < \aleph_{\omega+1} \rbrace$ be a set of conditions in $\mathbb{P} = Fn(\aleph_\omega,2,\aleph_\omega)$. Since $\aleph_{\omega+1}$ is regular, there is some $n < \omega$ such that $A = \lbrace \alpha : |p_\alpha| = \aleph_n \rbrace$ has size $\aleph_{\omega+1}$. Let $2^{\aleph_n} = \aleph_m < \aleph_\omega$. Take the first $\aleph_{m+1}$ many members of $A$, and call this set $B$.

Consider the set of domains $\lbrace dom(p) : p \in B \rbrace$. Since $\aleph_m^{\aleph_n} = \aleph_m$, the delta-system lemma applies, and there is a set $C \subseteq B$ of size $\aleph_{m+1}$ and a fixed set $r$ such that for all $p,q \in C$, $dom(p) \cap dom(q) = r$. Since there are only $\aleph_m$ functions from $r$ to 2, there are $p,q \in C$ which are compatible. Therefore, $\mathbb{P}$ has the $\aleph_{\omega+1}$ chain condition.

On the other hand, it is not hard to show that if $\aleph_\omega$ is not a strong limit, then $\mathbb{P}$ collapses $\aleph_\omega^{<\aleph_\omega}$.

It's very hard to prove something that is consistently false.