Regressive injective function in a set of infinite cardinals This is also from Kunen, Set Theory, ch. II:

Let $A$ be a set of infinite cardinals such that for each $\lambda$ regular $A\cap\lambda$ is not stationary in $\lambda$. Show that there is an injective function $g$ with domain $A$ such that $\forall\alpha\in A(g(\alpha)<\alpha)$. 

I tried using $\lambda=\sup A^+$, took a closed unbounded set $C$ disjoint from $A\cap\lambda=A$ and defined $g(\alpha)=\sup(C\cap\alpha)$ if $\sup(C\cap\alpha)>g(\eta)\forall\eta<\alpha$ and $g(\alpha)=\eta$ if $\sup(C\cap\alpha)=g(\eta),\eta<\alpha$, but then realized it could happen that $\alpha>\eta=\sup(C\cap\beta)=g(\beta),\eta<\beta<\alpha$, so g wouldn't be injective. 
Help, please!
 A: Apparently, this problem can be viewed as a problem of counting weakly inaccessibles, first of all we will deal with the "0" case:

Theorem $1$: If $\sup A$ is less than the first weakly inaccessible, then there exists an injective regressive function from $A$ to $\sup A$.

The method we are using is basically counting $\aleph$-fixed points:

Lemma $1$: if $A$ is a set of cardinals, and there are no $\aleph$-fixed points in $A$, then there is an injective regressive function $f_A:A\to\sup A$

Proof: a simple counting function will work. (The minimal element of $A$ will be send to $0$, the second smallest will be $1$, and so on).
This is trivially injective, and clearly $f_A(x)≤x$ for all $x$, notice that $f_A(\aleph_β)≤β$, so if $f_A(\aleph_β)=\aleph_β$ we get $\aleph_β=β$ contradiction.

Observation $1$: if $A$ contains a single $\aleph$-fixed point, it has an injective regressive function

We can simply send the $\aleph$-fixed point to $0$, and then use Lemma $1$ but start counting from $1$.
We can take a similar idea to get

Observation $ω$: if $A$ contains countably many $\aleph$-fixed point, it has an injective regressive function

Simply send the $\aleph$-fixed points into the odd numbers, and use Lemma $1$ but count the evens the first $ω$ cardinals.
We can continue this kind of observations going up and up till $\aleph$-fixed point many $\aleph$-fixed points (as an exercise, try to see how to do "Observation  $ω_1$"), but in fact the method for Observation $ω$ is enough for us.

From here on out, $f_X$, for a set of cardinals $X$, means an injective regressive function.

Fact $1$: $κ$ is weakly inaccessible if and only if it is regular, $\aleph$-fixed point, and there are $κ$-many $\aleph$-fixed points bellow $κ$


Proof of theorem $1$:

Now, let $A$ be as I described above, WLOG assume $A=λ∩Card$ for some cardinal $λ<$ the first inaccessible, we will prove it by induction on $A$.
For $x\in A$ let $A_x=\{y\in A\mid y<x\}$, and let $A'=\{y\in A\mid \text{ $y$ is an $\aleph$-fixed point}\}$ and $A^∘=A\setminus A'$.
Assume that there is an $f_{A_α}$ for each $α<x$.

Lemma $2$: there exists $f_{A_x'}$

If $A_x'=∅$, it is trivial.
If $\operatorname{otp} A_x'$ is a successor, then there exists $y<x$ such that $A_x'\setminus\{\max A_x'\}⊆A_y$, so let $g_{A_x'}=f_{A_y}\restriction A_x'\setminus\{\max A_x'\}$ and then $f_{A_x'}(x)=1+g_{A_x'}(x)$ for $x\ne \max A_x'$ and $f_{A_x'}(\max A_x')=0$ will work.
Otherwise, let $(β_i\mid i\in \operatorname{cof}|A_x'|)$ be a cofinal sequence that starts with $0$.
Define $g_{i}:A_{β_{i+1}}\setminus A_{β_i}→\kappa$ as $x↦β_i+f_{A_{β_i}}(x)$ (as ordinal addition).
Let $g=\bigcup g_i$ and let $G=f_{\{β_i\mid i\in\operatorname{cof}|A_x'|\}}$ (Try to show why it exists, hint: use fact 1).
Define $o(ν+n)=ν+2n,e(ν+n)=ν+2n+1$ for $ν$ limit and $n$ finite.
Then $f_{A_x'}=o\circ g\cup e\circ G$ $\square$
Let $h=f_{A_x^∘}$(it exists by Lemma 1).
Then $f_{A_x}=o\circ h\cup e\circ f_{A_x'}$.
Then $f_{A_λ}=f_A$.$\qquad\square$


Theorem $2$: if $κ$ is the first (weakly) inaccessible, and $A\subseteq κ$ satisfy the conditions in the question, then there exists $f_A:A\to\kappa$ that is regressive and injective.

Proof:
Let $κ$ be the first (weakly) inaccessible.
Let $A=\{a_i\mid i\in\kappa\}\subseteq κ$ (we always assume that the indexes is the natural counting of the set) be set of cardinals such that $A∩κ=A$ is not stationary in $κ$, and let $C=\{c_i\mid i\in\kappa\}$ be a club set that is disjoint from $A$.
Define recursively $\{b_i\mid i\in\kappa\}=B⊆A$ as follows:

*

*$b_0=a_0$

*assume $b_α$ is defined for all $α<β∈κ$, then $b_β$ will be the first $a_γ$ such that $\sup(a_γ∩C)>b_α$ for all $α<β$
It is easy to see that $F:B\to\kappa$ defined by $b_i\mapsto \sup(b_i∩C)$ is injective regressive function.
For each $i∈κ$ let $F_i:b_{i+1}\setminus b_i→b_{i+1}\setminus b_i$ be injective regressive function (to see that it exists, see the definition of $g_i$ in theorem 1).
I claim that $F\cup\bigcup_{i\in\kappa}F_i$ is an injective function.
It is enough to check that $F\cup F_i$ is injective for each $i$, which is the same as saying that $F(b_{i+1})∉\operatorname{Range}(F_i)$, but it can't be, because if it were, then for some $F(b_{i+1})<a_γ<b_{i+1}$ we would have $\sup(α_γ∩C)>b_i$, contradiction to the definition of $b_{i+1}$.
So: $f_A=F\cup\bigcup_{i\in\kappa}F_i$ is injective regressive function on $A$.$\qquad\square$

Theorem $3$: Let $A$ be a set of infinite cardinals that satisfy the condition of the question, then it has an $f_A:A\to\sup A$ that is injective and regressive

Proof:
By induction on $\sup A$, if $\sup A$ is less than the first inaccessible, then by theorem $1$ we are done, if it is equal, then by theorem $2$ we are done.
Assume that $κ=\sup A$ is greater than the first inaccessible, and that the theorem is true for all $λ<κ$, if $κ$ is not inaccessible, then let $λ'$ be the supremum of the inaccessibles bellow $κ$, let $F:A\cap λ'→λ'$ be injective regressive function, let $G':A\setminus λ'→κ$ be injective regressive function (the proof $G'$ exists is basically identical to the proof of theorem $1$), and let $G$ be $x\mapsto λ'+G'(x)$(ordinal addition). $G\cup F$ is the wanted function.
If $κ$ is inaccessible, repeat the proof of theorem $2$. (The only difference is that in theorem $2$ we had used theorem $1$ for the existence of the $F_i$'s, and here we will use the (strong) induction step)$\qquad\blacksquare$
