In the course of trying to prove a larger claim, I want to demonstrate directly that $A = P \cup H_0 \cup H_1 \cup H_2 ... \cup \ K_0 \cup K_1 \cup K_2 ...$
where $P = \bigcap_{n \lt \omega}f^n[A] = \bigcap_{n \lt \omega}f^n[B]$,
$H_n = f^n[A] \setminus f^n[B]$, and $K_n = f^n[B] \setminus f^{n+1}[A]$.
Here is the extra information that is relevant:
$f: A \xrightarrow{1-1} B$, and the notation $f^n$ simply refers to $n$ iterations of the function $f$ (with $f^0$ representing $I_A$)
$A \supseteq B \supseteq f[A] \supseteq f[B] \supseteq f^2[A] \supseteq f^2[B] \supseteq...$
Previously, I have also proven that $P$, $H_m$ and $K_n$ are all pairwise disjoint.
Seeing as the one direction is fairly trivial, I have condensed the desired proof into the following form:
$A \subseteq \bigcap_{n \lt \omega}f^n[A] \ \cup \ \bigcup_{n\lt \omega}f^n[A] \setminus f^n[B] \ \cup \ \bigcup_{n\lt \omega} f^n[B] \setminus f^{n+1}[A] $
I have successfully demonstrated $A \subseteq \bigcap_{n \lt \omega}f^n[A] \ \cup \ \bigcup_{n\lt \omega}f^n[A] \setminus f^n[B] \ \cup \ \bigcup_{n\lt \omega} f^n[B] \setminus f^{n+1}[A] $. However, I could only do so using a proof by contradiction.
Specifically, I assumed $A \nsubseteq \bigcap_{n \lt \omega}f^n[A] \ \cup \ \bigcup_{n\lt \omega}f^n[A] \setminus f^n[B] \ \cup \ \bigcup_{n\lt \omega} f^n[B] \setminus f^{n+1}[A] $ and arrived at a contradiction.
My set up for this was $\exists x \in A \Big(x \notin P \land x \notin \bigcup_{n\lt \omega} H_n \land x \notin \bigcup_{n\lt \omega} K_n \Big )$
$x \notin P \leftrightarrow \exists x \in \omega \Big (x \notin f^n[A] \Big) \land \exists n \in \omega \Big (x \notin f^n[B]\Big)$
$x \notin \bigcup_{n\lt \omega} H_n \leftrightarrow \forall n \in \omega \Big ( x \notin f^n[A] \Big ) \lor \forall n \in \omega \Big (x \in f^n[B] \Big) $
$x \notin \bigcup_{n\lt \omega} K_n \leftrightarrow \forall n \in \omega \Big ( x \notin f^n[B] \Big ) \lor \forall n \in \omega \Big (x \in f^{n+1}[A] \Big) $
The contradictions that arise from these three simultaneous statements are straightforward.
I am really at loss on how to prove this claim directly.
I've played around with it quite a bit using a working example of $A=\mathbb N$, $B=\mathbb N \setminus \{2,3\}$ and $f$ being the natural number square function (where $n \mapsto n^2$)...where, for example, $P$ equals $\{0,1\}$
I have some visual intuition about how these sets partition $A$:
If anyone could offer some input on how to formulate a direct proof, I would greatly appreciate it. Cheers~
Let
$$C_n=\begin{cases} f^{n/2}[A],&\text{if }n\text{ is even}\\ f^{(n-1)/2}[B],&\text{if }n\text{ is odd}\,, \end{cases}$$
so that $C_0=A$, $C_1=B$, and $C_n\supseteq C_{n+1}$ for all $n\in\omega$. For each $x\in A\setminus P$ let $n(x)=\min\{k\in\omega:x\notin C_k\}$; clearly $n(x)>0$, and
$$\begin{align*} x\in C_{n(x)-1}\setminus C_{n(x)}&=\begin{cases} f^{k-1}[B]\setminus f^k[A],&\text{if }n(x)=2k\\ f^k[A]\setminus f^k[B],&\text{if }n(x)=2k+1 \end{cases}\\ &=\begin{cases} K_{k-1},&\text{if }n(x)=2k\\ H_k,&\text{if }n(x)=2k+1\,. \end{cases} \end{align*}$$
Thus,
$$A\setminus P\subseteq\bigcup_{n\in\omega}H_n\cup\bigcup_{n\in\omega}K_n\,,$$
and the result follows.