Context
There is a well-known result on the Generalized Continuum Hypothesis for singular cardinals:
- Silver's Theorem: Let $\aleph _{\lambda }$ be a singular cardinal such that its cofinality is greater than $\omega$. If for every $\alpha < \lambda$, $2^{\aleph _{\alpha }}=\aleph _{\alpha +1}$, then $2^{\aleph _{\lambda }}=\aleph _{\lambda +1}$.
Proof Sketch: For simplicity, let us assume that $\lambda = \omega _1$, the first uncountable ordinal number.
For each $\alpha < \omega _1$, let $A_{\alpha }= \mathcal{P}(\omega _{\alpha })$, so $|A_{\alpha }|=2^{\aleph _{\alpha }}=\aleph _{\alpha +1}$. For every set $X \subseteq \omega _{\omega _1}$, let $f_X \in \prod _{\alpha < \omega _1}A_ {\alpha }$ be the function defined by$$f_X (\alpha )= X \cap \omega _{\alpha }.$$The set $F= \{ f_X \mid X \in \mathcal{P} (\omega _{\omega _1}) \}$ is a family of almost disjoint functions, meaning that for any $f_X$ and $f_Y$ in $F$ there is some $\alpha < \omega _1$ such that $f(\beta ) \neq g(\beta )$ for all $\beta \ge \alpha$.
Now, let $U$ be an ultrafilter on $\omega _1$ that extends the closed unbounded filter. Thus every set $S \in U$ is stationary.
Let us define a relation $<$ on $F$ as follows:$$f < g \quad \text{if and only if} \quad \{ \alpha < \omega _1 \mid f(\alpha ) < g(\alpha ) \} \in U.$$ It can be shown that $<$ is a linear ordering on $F$.
Now, if $f,g \in F$ and $g <f$, then $g \in F_f$, where$$F_f=\{ g \in F \mid \text{for some stationary } T, g(\alpha ) < f( \alpha ) \text{ for all } \alpha \in T \}.$$By a slight modification of the following lemma, $|F_f| \le \aleph _{\omega _1}$. Thus for every $f \in F$, $|\{ g \in F \mid g < f \}| \le \aleph _{\omega _1}$.
As $<$ is a linear ordering of the set $F$, it follows that $|F| \le \aleph _{\omega _1 +1}$. Therefore, $2^{\aleph _{\omega _1}}=\aleph _{\omega _1 +1}$.
Lemma: Let $\{ A_{\alpha } \mid \alpha < \omega _1 \}$ be a family of sets such that $|A_{\alpha }| \le \aleph _{\alpha }$ for every $\alpha < \omega _1$, and let $F$ be a family of almost disjoint functions,$$F \subset \prod _{\alpha < \omega _1} A_{\alpha }.$$Then $|F| \le \aleph _{\omega _1}$.
This lemma can be proved as follows.
Let $S_0$ be the set of all limit ordinals $0 < \alpha < \omega _1$. For $f\in F$ and and $\alpha \in S_0$, let $f^{*}(\alpha )$ denote the least $\beta$ such that $f( \alpha ) < \omega _{\beta }$. As $f^{*}(\alpha ) < \alpha$ for every $\alpha \in S_0$, there exists, by Fodor's Lemma , a stationary set $S \subset S_0$ such that $f^{*}$ is constant on $S$. Therefore $f$ restricted to $S$ is a function from $S$ into $\omega _{\beta }$, for some $\beta < \omega _1$, which is denoted by $\varphi (f)$.
It can be shown that $\varphi$ is a one-to-one mapping from $F$ into the set of functions defined on a subset $S$ of $\omega _1$ into some $\omega _{beta} < \omega _{\omega _1}$. Thus we have$$|F| \le 2^{\aleph _1} \cdot \sum _{\beta < \omega _1} \aleph _{\beta } ^{\aleph _1} \le \aleph _{\omega _1}.$$
Question
I have no question about details of the proof. However, I cannot understand how such a proof can come into one's mind from scratch; this proof does not seem intuitive and natural to me at all. What background and idea leads a one to such a proof? What is the intuition behind the proof? What were being thought when someone went to prove this?
I have the fish; please teach me how to fish.
P.S. Please note that my math knowledge is very low, so please answer this post in a way that I can understand.