When does $f:X \to X$ give $L_f(A) \subseteq R_f(A)$ for all $A \subseteq X$?

Question

Let $f:X \rightarrow X$ be a continuous function on a topological space $X$. Under what conditions is it the case for every subset $A \subseteq X$ that $$L_f(A)=A \cap \bigcap_{i=1}^{\infty} f^{-i}((X-A)\cup f^{-1}(A) ) \subseteq \bigcap_{i=1}^{\infty} f^{-i}(A)=R_f(A)?$$ This appears to be equivalent to the condition that $$A \cap \bigcap_{i=1}^{\infty}[(X - f^{-i}(A)) \cup f^{-i}(A)] \subseteq \bigcap_{i=1}^{\infty} f^{-i}(A),$$ if that helps. This is a question equivalent to when the dynamic topological logic induction axiom $A,*\circ(A \supset \circ A) \rightarrow *A$ holds in topological semantics.

Answer 1

Let $f:X \to X$ be continuous. The condition that for every $A \subseteq X$ that $$L(A)=A \cap \bigcap_{i=1}^{\infty} f^{-i}((X \setminus A)\cup f^{-1}(A) ) \subseteq \bigcap_{i=1}^{\infty} f^{-i}(A)=R(A)$$ is equivalent to $f$ having both of the following properties:

1. $f:X \to X$ is a bijection,
2. $Orb(f,x):= \{f^i(x) \mid i \in \mathbb{Z}\}$ is finite for all $x \in X$

Let $f:X \to X$ be a continuous function such that for all $A \subseteq X$ \begin{align} L(A)=A \cap \bigcap_{i=1}^\infty f^{-i}\left( (X\setminus A) \cup f^{-1}(A)\right) \subseteq \bigcap_{i=1}^\infty f^{-i}(A)=R(A) \quad (*) \end{align} and in particular this is true for all singelton $A$. Suppose that $A=\{a\}$, $a \not\in f^{-1}(A).$ So $a$ is not in the right hand side of $(*)$, hence not the left hand side, assuming $f$ has this property. So for some $n$ we have that $$a \not\in f^{-n}\left( (X\setminus A) \cup f^{-1}(A) \right).$$ So we have that $$f^{n+1}(a) \not\in f\left( (X \setminus A) \cup f^{-1}(A) \right)= f(X \setminus A) \cup \{a\} $$ and that gives us that $f^n(a)=a$ as otherwise $f^{n+1}(a) \in f(X \setminus A)$. Suppose that $B=\{b\} $ and that $b,c \in f^{-1}(B)$. So $f(c)=b$ and if $c \neq b$, but this contradicts the fact that there would be an $n>0$ such that $f^n(c) =c$ by our last claim since $f^n(c)=f^{n-1}(b)=b$, hence $f^{-1}(B)=B$. So at this moment we have that for all $x$ that there is an $n>0$ so that $f^n(x)=x$ and that $f$ is a bijections since every element has exactly one preimage. We have now completed showing that $L(A) \subseteq R(A)$ means that $f$ has property 1. and 2..

Let $f:X \to X$ be a bijection and suppose that $Orb(f,x)$ is finite for each $x \in X$. Suppose that $a \not \in R(A)$ and $a \in A$ (the case $a \not \in A$ is trivial), so there is an $i>0$ such that $a \not \in f^{-i}(A)$, and we can assume $i$ to be the least such. Also suppose that $|Orb(f,a)|=n \geq 1$. Let $i>1$, so in particular we have that $a \in f^{1-i}(A)$. Notice that $$f^{1-i} \left( (X \setminus A) \cup f^{-1}(A) \right)= (X \setminus f^{1-i}(A)) \cup f^{-i}(A)$$ does not contain $a$ so $x \not\in L(A)$.

Say $i=1$, so $a \not \in f^{-1}(A)$, and $n \neq 1$ as otherwise $a \in f^{-1}(A)$ since $a$ would be fixed by $f$. We have that, by our assumption of $Orb(f,a)=n$, $a \not\in f^{-1-n}(a)$ and $a \in f^{-n}(A)$, $n>1$. This reduces $i=1$ to the case of $i>1$ since $$(X \setminus f^{-n}(A)) \cup f^{-1-n}(A)$$ does not contain $a$. So $a \not \in L(A)$. We have proven that $L(A) \subseteq R(A)$ hence property 1. and 2. give us $L(A) \subseteq R(A)$.

A question I thought a bit about is whether $f^{-1}$ is continuous (hence $f$ a homeomorphism)? I am not sure, I am guessing if it were continuous it would come from a fairly basic fact in general topology that I just don't know and I can't think of a proof at the moment, or it is false in general.

(edit:I have now asked the question about whether $f^{-1}$ is continuous)