The generalized continuum hypothesis can't fail first at $\omega_{\omega_{1}}$ I am willing to prove that the GCH cannot first fail at a singular cardinal. For this purpouse I am following the strategy outlined by Kunen in his 2013 book (see Exercises III.6.16-6.17-6.18-6.19).
I have not any problems regarding to the first exercise (III.6.16) however I am not able to take advantage of the hints from the exercises 6.17-6.18. I write down both.
A family of functions $\mathcal{F}\subset \lambda^\kappa$ is a eventually disjoint family (edf) if $|\{\alpha<\kappa:\,f(\alpha)= g(\alpha)\}|<\kappa$ for every $f,g\in \mathcal{F}$ and $f\neq g$
Exercise III 6.17 Assume that $\lambda=\omega_{\omega_1}$, $\forall\alpha<\omega_1\,[2^{\aleph_\alpha}<\lambda]$, $\mathcal{F}\subset \lambda^{\omega_1}$ an eventually disjoint family, and $\{\alpha<\omega_1:\,f(\alpha)<g(\alpha)\}$ is stationary for every $f\in\mathcal{F}$, where $g:\omega_1\rightarrow\lambda$ and $g(\alpha)<\omega_{\alpha+1}$ for all $\alpha$. Prove that $|\mathcal{F}|\leq \lambda$.
My Attempt: Let's denote by $S_f=\{\alpha<\omega_1:\,f(\alpha)<g(\alpha)\}$. For every $\alpha\in S_f$ we have $|f(\alpha)|\leq |g(\alpha)|\leq \omega_\alpha$ since $g(\alpha)<\omega_{\alpha+1}$ and therefor $|f(\alpha)|=\omega_{f'(\alpha)}$ where $f'(\alpha)<\alpha$. Now applying Fodor's Lemma to $f'$ and $S_f$ we can find an stationary set $S'_f\subset S_f$ such that $S'_f=\{\alpha<\omega_1: |f(\alpha)|=\omega_{\beta_0}\}$. It is clear by the hypothesis $\forall\alpha<\omega_1\,[2^{\aleph_\alpha}<\lambda]$ that $|Club(\omega_1)|\leq \lambda$. So my question is, is the map $f\mapsto S'_f$ injective? If it would not be the case, how can we prove that $|\mathcal{F}|\leq \lambda$?
Also, if the answer to the first question would be affirmative, I would like to know why it is necessary to assume $2^{\aleph_\alpha}<\lambda$ further than $\alpha>1$.
Exercise III 6.18 Assume that $\lambda=\omega_{\omega_1}$, $\forall\alpha<\omega_1\,[2^{\aleph_\alpha}=\aleph_{\alpha+1}]$. Prove that $2^\lambda=\lambda^+$.
My Attempt By the hypothesis and the Exercise III-6.16 we can take an edf family $\mathcal{F}\subset \lambda^{\omega_1}$ of cardinality $2^\lambda$ such that $f(\alpha)<\omega_{\alpha+1}$ for all $f\in\mathcal{F}$ and $\alpha<\omega_1$. Given $f,g\in\mathcal{F}$ two different functions, it can be shown that $f<^* g\leftrightarrow \{\alpha\in\omega_1:\,f(\alpha)<g(\alpha)\}\in Club(\omega_1)$ is a totally order. Now given $g\in\mathcal{F}$ consider the edf $\mathcal{F}_g=\{f\in \mathcal{F}: f<^* g\}$. Applying Exercise III.6.17 we conclude that $|\mathcal{F}_g|\leq \lambda$. Since $<^*$ is a totally order given two diferent functions $g,g'\in\mathcal{F}$ then $\mathcal{F}_g\neq \mathcal{F}_{g'}$. Is there any way to take profit of this information to prove the inequality $|\mathcal{F}|\leq \lambda^+$?
Any help or hint would be much appreciated.
 A: Exercise III 6.17:
For $f \in \mathcal F$ let
$S_{f} := \{ \alpha < \omega_{1} \mid f(\alpha) < g(\alpha) \}$. First
consider the case that $g \colon \omega_{1} \to \lambda$ satisfies
$g(\alpha) = \omega_{\alpha}$ for every $\alpha < \omega_{1}$. By
Fodor, we may, for each $f \in F$, choose some stationary
$T_{f} \subseteq S_{f}$ and some cardinal $\kappa_{f} < \lambda$ such
that $f(\alpha) < \kappa_{f}$ for all $\alpha \in T_{f}$. For each
$f \in \mathcal F$ let
$f^{*} := (\kappa_{f}, T_{f}, \{ (\alpha,f(\alpha)) \mid \alpha \in
T_{f}\})$. Let $f,g \in \mathcal F$ be distinct. Since $\mathcal F$ is
an eventually disjoint family, there is some $\gamma < \omega_{1}$ such
that
$\{\alpha < \omega_{1} \mid f(\alpha) = g(\alpha) \} \subseteq
\gamma$. Since $T_{f}, T_{g}$ are stationary and thus unbounded in
$\omega_{1}$, this implies
$\{(\alpha,f(\alpha)) \mid \alpha \in T_{f} \} \neq \{ (\alpha,
g(\alpha)) \mid \alpha \in T_{g} \}$, i.e. $f^{*} \neq g^{*}$. In
particular, the map $\mathcal F \ni f \mapsto f^{*}$ is injective. Now
$f^{*} \in \lambda \times \mathcal P(\omega_{1}), \bigcup_{\alpha <
  \omega_{1}} \mathcal P(\omega_{1} \times \omega_{\alpha})$ and thus
$| \mathcal F | \le | \lambda \times \omega_{1} \times \bigcup_{\alpha
  < \omega_{1}} \mathcal P(\omega_{1} \times \omega_{\alpha}) | \le
\sup_{\alpha < \omega_{1}} \lambda \cdot \omega_{1} \cdot
2^{\omega_{\alpha}} \le \lambda.$ (Here we use that
$2^{\omega_{\alpha}} \le \lambda$ for all $\alpha < \omega_{1}$.)
The proof for general $g \colon \omega_{1} \to \lambda$ such that
$g(\alpha) < \omega_{\alpha+1}$ for all $\alpha < \omega_{1}$ is
similar. For each $\alpha < \omega_{1}$ fix an injection
$\Gamma_{\alpha} \colon g(\alpha) \to \omega_{\alpha}$. Let $S_{f}$ be
as above. For each $\alpha \in S_{f}$, we have that
$\Gamma_{\alpha}(f(\alpha)) < \Gamma_{\alpha}(g(\alpha)) <
\omega_{\alpha}$ and there is hence some $\kappa_{f} < \lambda$ and
some stationary $T_{f} \subseteq S_{f}$ such that for all $\alpha \in
T_{f} \colon \Gamma_{\alpha}(f(\alpha)) < \kappa_{f}$. Now let $f^{*}
= (\kappa_{f}, T_{f}, \{ (\alpha, \Gamma_{\alpha}(f(\alpha))) \mid
\alpha \in T_{f} \})$. As before, the map $\mathcal F \ni f \mapsto
f^{*}$ is injective and thus implies that $| \mathcal F | \le \lambda$.
Exercise III 6.18:  Let $\mathcal F \subseteq ^{\omega_{1}}\lambda$ be an eventually
  disjoint family of size $2^{\lambda}$ such that for all $f \in
  \mathcal F \forall \alpha < \omega_{1} \colon f(\alpha) <
  \omega_{\alpha +1}$. For $f,g \in \mathcal F$ write $f <^{*} g$ iff $\{
  \alpha < \omega_{1} \mid f(\alpha) < g(\alpha) \}$ contains a club and
  let $<^{*} \subseteq \prec$  be a strict total order on $\mathcal
  F$. For $g \in \mathcal F$ let $F_{g} := \{ f \in \mathcal F \mid f
  \prec g \}$. Let $f \prec g$. I claim that $\{ \alpha < \omega_{1} \mid f(\alpha) <
  g(\alpha) \}$ is stationary in $\omega_{1}$:
Suppose not. Then there is some club $C \subseteq \omega_{1}$ such
  that for all $\alpha \in C \colon f(\alpha) \ge g(\alpha)$. Since $f,g
  \in \mathcal F$ are distinct and $\mathcal F$ is an eventually
  disjoint family, the set $\{ \alpha  \in C \mid g(\alpha) < f(\alpha)
  \}$ contains some club $D \subseteq \omega_{1}$, witnessing that $g
  \prec f$. Contradiction.
We may now apply Exercise III 6.17: For each $g \in \mathcal F$, we
  now know that $F_{g}$ has size $\le \lambda$. Now, recursively
  construct a sequence $(g_{\alpha} \mid \alpha < \lambda^{+})$ as
  follows: Pick any $g_{0} \in \mathcal F$. Given $(g_{\alpha} \mid
  \alpha < \gamma)$ for some $\gamma < \lambda$, we have that $\bigcup
  \{ F_{g_{a}} \mid \alpha < \gamma \}$ has size at most $| \lambda | \cdot
  \gamma = \lambda < 2^{\lambda}$. There is hence some $g_{\gamma} \in
  \mathcal F$ such that for all $\alpha < \gamma \colon g_{\alpha} \prec
  g_{\lambda}$. This finishes our construction.
I now claim that $\bigcup \{ \mathcal F_{g_{\alpha}} \mid \alpha < \lambda^{+} \}=
  \mathcal F$.
Suppose not. Then there is some $f \in \mathcal F$ such that
  $g_{\alpha} \prec f$ for all $\alpha < \lambda^{+}$ (by the totality of $\prec$). Hence $\mathcal
  F_{f}$ has size $\ge \lambda^{+}$. Contradiction.
We thus have that $\mathcal F = \bigcup \{g_{\alpha} \mid \alpha <
  \lambda^{+}\}$ has size at most $\lambda \cdot \lambda^{+} =
  \lambda^{+}$. In particular, we have $2^{\lambda} = \lambda^{+}$.
