I have a question about the following statement:
Let $X$ be a set and $f:X\to X$ a map, then for any subset $A\subset X$ and all $n\in\mathbb{N}$
$$A\triangle f^{-n}(A)\subseteq\bigcup_{i=0}^{n-1}f^{-i}(A)\triangle f^{-(i+1)}(A).\tag{$\ast$}$$
I think I have a proof for this statement but it seems a bit convoluted:
Proceeding via induction, the case $n=1$ is clear; now, assume $(\ast)$ holds for all $1\leq k\leq n$ for some $n\in\mathbb{N}$, then
\begin{align*} A\triangle f^{-(n+1)}(A) &= \underbrace{[A\cap f^{-(n+1)}(A)^c]}_{:=(1)}\cup\underbrace{[f^{-(n+1)}(A)\cap A^c]}_{=:(2)}. \end{align*} We investigate (1) and (2) separately. \begin{align*} (1) = A\cap f^{-(n+1)}(A)^c&= [A\cap(f^{-n}(A)\cup f^{-n}(A)^c)]\cap f^{-(n+1)}(A)^c\\ &= [(A\cap f^{-n}(A))\cup (A\cap f^{-n}(A)^c)]\cap f^{-(n+1)}(A)^c\\ &=(A\cap f^{-n}(A)\cap f^{-(n+1)}(A)^c)\cup (A\cap f^{-n}(A)^c\cap f^{-(n+1)}(A)^c)\\ &\subseteq (f^{-n}(A)\cap f^{-(n+1)}(A)^c)\cup(A\cap f^{-n}(A)^c). \end{align*} With an anologous argument for (2) (writing $A^c=A^c\cap(f^{-n}(A)\cup f^{-n}(A)^c)$) we find \begin{align*} (2)\subseteq(A^c\cap f^{-n}(A))\cup (f^{-(n+1)}\cap f^{-n}(A)^c). \end{align*} Hence \begin{align*} A\triangle f^{-(n+1)}(A)&\subseteq [(f^{-n}(A)\cap f^{-(n+1)}(A)^c)\cup(A\cap f^{-n}(A)^c)]\cup[(A^c\cap f^{-n}(A))\cup (f^{-(n+1)}\cap f^{-n}(A)^c)]\\ &= A\triangle f^{-n}(A)\cup f^{-n}(A)\triangle f^{-(n+1)}(A)\\ &\overset{(\ast)}{\subseteq}\bigcup_{i=0}^{n-1}f^{-i}(A)\triangle f^{-(i+1)}(A)\cup f^{-n}(A)\triangle f^{-(n+1)}(A)\\ &=\bigcup_{i=0}^{n}f^{-i}(A)\triangle f^{-(i+1)}(A), \end{align*} which proves the claim.
Now, I think the proof feels a bit convoluted. This proof is just a small supplementary result for a 90-minute seminar that I'm going to give and I don't want to spend that much time on it, since there's a lot more material to cover. I was wondering if there is a slicker way of getting the result. If you have any ideas, let me know. Any help is appreciated.
Thank you.
A simple proof might show that for any sets $A,B,C$ we have $A \triangle B \subseteq A \triangle C \cup C \triangle B$. Your statement then follows immediately by adding in all intermediate sets and does not require any properties of the pre-image: $$A \triangle f^{-n}(A) \subseteq A \triangle f^{-1} (A) \cup f^{-1} (A)\triangle f^{-n} (A) \dots \subseteq \bigcup_{i=0}^{n-1} f^{-i} (A)\triangle f^{-(i+1)} (A)$$