I have to construct a few things before I get to my question: it's possible we don't need all of it, but I am stuck on the very last step so I should recap everything so far.

Let $\kappa$ be a singular cardinal with $\operatorname{cf}\kappa = \lambda > \omega$, and let $T \subseteq \mathcal{P}(\kappa)$. Define $T\upharpoonright_{\alpha} := \{a \cap \alpha \mid a \in T \}$. Let $C = \langle \alpha_{\xi} \mid \xi < \lambda \rangle$ be a strictly increasing continuous sequence of cardinals with limit $\kappa$.

Suppose furthermore that $\alpha < \kappa \implies \alpha^{\lambda} < \kappa$, and that the set $\{ \alpha < \kappa : | T\upharpoonright_{\alpha} | \le \alpha\}$ is stationary in $\kappa$.

Now, the set $S = \{ \xi < \lambda : | T\upharpoonright_{\alpha_{\xi}} | \le \alpha_{\xi} \}$ is stationary in $\lambda$, essentially by hypothesis. So for each $\xi \in S$, we may define a one-to-one function $f_{\xi} : T\upharpoonright_{\alpha_{\xi} } \to \alpha_{\xi}$, and for each $a \in T$, define $g_a (\xi) = f_{\xi}(a \cap \alpha_{\xi} )$.

If we define $S_0$ to be the elements of $S$ which are limit ordinals, then $g_a(\xi) < \alpha_{\eta}$ for some $\eta < \xi$ by continuity of $C$, so let $h_a(\xi)$ be the least such $\eta$. Since $S_0$ is stationary, and $h_a: S_0 \rightarrow \lambda$ is regressive, by Fodor's Theorem $h_a$ is constant on a stationary set. In other words, for each $a \in T$, there is some $S_a$ stationary $\subset S_0$ and $\eta(a) < \lambda$ such that $h_a$ is constantly $\eta(a)$ on $S_a$.

Let $(S',\eta')$ be a pair with $S' \subseteq S_0$ stationary in $\lambda$, and $\eta' < \lambda$. Define $T' = \{ a \in T \mid S_a = S' \wedge\ \eta(a) = \eta'\}$. My goal is to show that $|T'| \le \kappa$. Because $\large a \in T' \implies g_a ( \xi) < \alpha_{\eta '}$ for all $\xi \in S'$, we must have that $\large| T' \upharpoonright_{\alpha_{\xi}}|\le \alpha_{\eta '}$ for all $\xi \in S$. Now here is the part I do not understand.

"So for each $a,b \in T$ with $a \not = b$, then the sequences $\langle a \cap \alpha_{\xi} \mid \xi \in S' \rangle$ and $\langle b \cap \alpha_{\xi} \mid \xi \in S' \rangle$ are distinct, and hence $\large |T ' | \le (\alpha_{\eta '} ) ^{\lambda}$."

Now I am not sure how to verify anything in that sentence. How do I show the sequences are distinct? And how do I work out the final statement? I think that it has something to do with the cardinality of the subsets of $\large\alpha_{\eta '}$ of size $\lambda$ , because I think that set has cardinality $\large( a_{η'} )^λ$ . Any help would be appreciated, very much (because it means I can finish proving Silver's Theorem :).

  • $\begingroup$ I have a sneaking suspicion that in my actual question we should take $a,b \in T'$, not $T$. But I can't be sure. $\endgroup$ – Paul Slevin Feb 29 '12 at 17:08
  • $\begingroup$ You only need $a,b\in T'$, but the statement is true as written. $\endgroup$ – Brian M. Scott Mar 1 '12 at 1:51
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    $\begingroup$ By the way, for a nice PCF-y proof of Silver's theorem you can check this chapter of The Handbook of Set Theory by Uri Abraham and Menachem Magidor. The proof of Silver's theorem is relatively simple and both of them are excellent writers. $\endgroup$ – Asaf Karagila Mar 1 '12 at 1:56

If $a,b\in T$ with $a\ne b$, there is some $\zeta\in\kappa$ that belongs to exactly one of $a$ and $b$. $C$ is cofinal in $\kappa$, so there is some $\xi\in\lambda$ such that $\zeta<\alpha_\xi$, and it follows that $a\cap\alpha_\xi\ne b\cap\alpha_\xi$. Any $\xi'\in\lambda\setminus\xi$ works equally well, $\lambda\setminus\xi$ is cofinal in $\lambda$, and $S'$ is stationary in $\lambda$, so we may as well assume that $\xi\in S'$.

In particular this is true for distinct $a,b\in T\,'$: if $a,b\in T\,'$, and $a\ne b$, there is a $\xi\in S'$ such that, and therefore the sequences $\langle a\cap\alpha_\xi:\xi\in S'\rangle$ and $\langle b\cap\alpha_\xi:\xi\in S'\rangle$ are distinct. Moreover, each of the functions $f_\xi$ is injective, so $f_\xi(a\cap\alpha_\xi)\ne f_\xi(b\cap\alpha_\xi)$ whenever $a\cap\alpha_\xi\ne b\cap\alpha_\xi$, and therefore the sequences $$\Big\langle f_\xi(a\cap\alpha_\xi):\xi\in S'\Big\rangle=\Big\langle g_a(\xi):\xi\in S'\Big\rangle$$ and $$\Big\langle f_\xi(b\cap\alpha_\xi):\xi\in S'\Big\rangle=\Big\langle g_b(\xi):\xi\in S'\Big\rangle$$ are distinct. Finally, $g_a(\xi)<\alpha_{\eta\,'}$ for all $a\in T\,'$ and $\xi\in S'$, so the map

$$T\,'\to{^{S'}\alpha_{\eta\,'}}:a\mapsto\Big\langle g_a(\xi):\xi\in S'\Big\rangle$$ is injective, and $|T\,'|\le \left|{^{S'}\alpha_{\eta\,'}}\right|=(\alpha_{\eta\,'})^{\,|S'|}=(\alpha_{\eta\,'})^\lambda$.

  • $\begingroup$ What does your notation $|^{S'} \alpha_{\eta '}|$ mean? Sequences in $\alpha_{\eta '}$ of length $|S'|$ ? $\endgroup$ – Paul Slevin Mar 1 '12 at 12:01
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    $\begingroup$ ${^A}B$ is a fairly standard notation for the set of functions from $A$ to $B$, so it’s the cardinality of the set of functions from $S'$ to $\alpha_{\eta\,'}$. $\endgroup$ – Brian M. Scott Mar 1 '12 at 13:09

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