# Jech's proof of Silver's Theorem on SCH

Jech's textbook proves Silver's Theorem–that if the Singular Cardinals Hypothesis holds for all singular cardinals of countable cofinality, then it holds for all singular cardinals–by breaking up the question into several lemmas and considering the case $$\kappa=\aleph_{\omega_1}$$ to simplify notation. I have a question about a technical detail in the proof of one of these lemmas. Here are the relevant statements:

Lemma 8.15: Assume that $$\aleph_\alpha^{\aleph_1}<\aleph_{\omega_1}$$ for all $$\alpha<\omega_1$$. Let $$F$$ be an almost disjoint family of functions (i.e. if $$f \ne g$$ and $$f,g \in F$$, then there is some $$\beta<\omega_1$$ such that $$\gamma>\beta$$ implies $$f(\gamma)\ne g(\gamma)$$) contained in the product $$\prod_{\alpha<\omega_1}A_\alpha$$ such that the set $$\{\alpha<\omega_1:|A_\alpha|\le \aleph_{\alpha+1}\}$$ is stationary. Then $$|F|\le \aleph_{\omega_1+1}$$.

Lemma 8.16: Assume that $$\aleph_\alpha^{\aleph_1}<\aleph_{\omega_1}$$ for all $$\alpha<\omega_1$$. Let $$F$$ be an almost disjoint family of functions contained in the product $$\prod_{\alpha<\omega_1}A_\alpha$$ such that the set $$\{\alpha<\omega_1:|A_\alpha|\le \aleph_\alpha\}$$ is stationary. Then $$|F|\le \aleph_{\omega_1}$$.

We assume Lemma 8.16 (the proof of which makes sense to me as presented by Jech) to prove Lemma 8.15. The issue I have is with the following detail in the proof of 8.15:

For every $$f \in F$$, let $$F_f=\{g \in F:$$ for some stationary set $$T$$, $$g(\alpha) for all $$\alpha \in T\}$$. By Lemma 8.16, $$|F_f|\le \aleph_{\omega_1}$$.

In order to substantiate the use of 8.16, we need to argue that $$F_f \subset \prod_{\alpha<\omega_1}A_\alpha$$ where $$\{\alpha:|A_\alpha|\le\aleph_\alpha\}$$ (or rather $$\{\alpha:A_\alpha \subset \omega_\alpha\}$$) is stationary. This is intuitively clear, but I am having trouble cooking up a nice, crisp proof of it. Does anyone know how to do it?

Alternatively, I'm interested in any good writeup of Silver's Theorem. I tried googling around, but I didn't find anything. Thanks!

I should clarify my question a little bit. First, it is indeed an important detail that we assume $$f(\alpha) \in \omega_{\alpha+1}$$ and hence $$|f(\alpha)|\le \aleph_\alpha$$. Thanks to Andres for pointing that out.

Let's denote $$\prod_{\alpha<\omega_1}B_\alpha$$ to be the product containing $$F_f$$. We cannot technically have $$B_\alpha=f(\alpha)$$ because it's possible that $$g \in F_f$$ and yet $$f(\alpha) for some small index $$\alpha$$. It's more realistic to have $$B_\alpha=\sup\{g(\alpha):g \in F\}$$, though I'm not sure this would work. The more important concern is proving that $$S:=\{\alpha<\omega_1:|B_\alpha|=|\{g(\alpha):g \in F\}|\le \aleph_\alpha\}$$ is in fact stationary. We're done if we have this fact (because we appeal to 8.16), and I can see why it should be true, but it's not clear to me how to prove it.

There are more details to Jech's proof of 8.15, but the question of why the set $$S$$ is necessarily stationary is the only detail that is bothering me.

## 1 Answer

You left out an important detail in your outline, namely that before we define the sets $F_f$, we have identified (for $\alpha$ in the relevant stationary set $S$) each $A_\alpha$ with a subset of $\omega_{\alpha+1}$. The outcome of this is that, since $f(\alpha)\in A_\alpha$, then it is an ordinal of size at most $\omega_\alpha$, and therefore $A'_\alpha=f(\alpha)$ is a set as in Lemma 8.15. If this happens for a stationary set of indices, we are done since $\prod_{\alpha}A_\alpha$ can be replaced with $\prod_{\alpha}A'_\alpha$ when discussing $F_f$.

The other detail left out in the outline is that, before defining the sets $F_f$, the further simplifying assumption was made that all $A_\alpha$ have size at most $\omega_{\alpha+1}$. Without this assumption, the construction of the ultrafilter needs to be done relative to the stationary set $S$ of indices where this happens and, in particular, the set $T$ should have stationary intersection with $S$.

• We can't use $A'_\alpha=f(\alpha)$ here because it's possible that $f(\alpha)<g(\alpha)$ for some $\alpha$ and $g \in F_f$. It is also not obvious that $\{\alpha:|\{g(\alpha):g \in F_f\}| \le \aleph_\alpha\}$ is stationary. Why is this the case? Commented Dec 8, 2013 at 0:09
• @Maxwell You have an equivalence relation that identifies two functions if they agree in a lot of points, and you can use it to see that $\prod A'_\alpha$ is all you need to use. But perhaps a cleaner alternative is to use $A'_\alpha=A_\alpha$ if $f(\alpha)<g(\alpha)$ and $A'_\alpha$ as I mentioned otherwise. Commented Dec 8, 2013 at 0:13
• Okay, that makes sense. How can we show that the set of indices where $\{g(\alpha):\alpha \in F_f\} \le \aleph_\alpha$ is stationary? Commented Dec 8, 2013 at 1:40
• @Maxwell, you simply show that for each stationary set $T$ the set $\prod_{\alpha<\omega_1}A^T_\alpha$, where $A^T_\alpha=f(\alpha)$ if $\alpha\in T$, and $A^T_\alpha=A_\alpha$, then as $|f(\alpha)|\leq\omega_{\alpha}$ for all $\alpha$ in the stationary set $T$, thus by lemma 8.16 $\prod_{\alpha<\omega_1}A^T_\alpha\leq\aleph_{\omega_1}$, but there are less than $\aleph_{\omega_1}$ stationary subsets of $\omega_1$. This argument can't get any simpler... Commented Dec 8, 2013 at 2:59
• I see, you're considering each stationary set $T$ as a separate case. Thanks. Commented Dec 8, 2013 at 7:52