maximum with respect inclusion of a function whose output are sets Suppose that $S(x)$ is a function from a compact space $A$ to a space
of sets $S$. Suppose that there exists a map $W: A\to A$ and
$S(x)\subseteq S(W(x)).$ Does there exist a point $x$ such
that the equality holds
$$S(x)= S(W(x))$$
 A: I believe that the answer is "no". Let $A := S^1$ be the unit circle and let $W \colon S^1 \to S^1$ be a rotation by an irrational angle. Thus $W$ defines an action of $\mathbb{Z}$ on $S^1$ all of whose orbits are countable. For each point $p \in S^1$ define $S(p)$ to be the "backwards orbit of $p$", namely $S(p) := \{ W^n(p) | n \leq 0 \}$. Applying $W$ to a point $p$ we find that the set $S(W(p)) = S(p) \cup \{W(p)\} \supsetneq W(p)$. Since this holds for every point of $p$, we find that there is no point with the required property.
A: If $W$ admits a periodic point, then, of course, $S(x)$ is constant along its orbit. But, on the contrary, whatever nice and well-structured $A$ and $W$ you have, if $W$ has no periodic points, there is always the bad map $S(x):=$ the negative invariant set generated by $x$, that is $\cup_{k\in\mathbb{N}}W^{-k}(x),$ and this map fails to satisfy the thesis while $S(x)\subset S(W(x))$ holds for all $x$.
This is to convince you that some assumption on the map $S(x)$ is in order. Here, a mild and natural assumption, to be coupled with the compactness of $A\neq\emptyset$, is weak upper semicontinuity, that is,
$$ S:A\to2^X$$
is continuous w.r.to the product topology on $2^X$ where the two-point space $2:=\{0,1\}$ is endowed with the left-order topology (whose only proper open subset is $\{0\}$). As a consequence, the continuous image of $A$ is a compact subset of $2^X$, therefore it has a maximal element with respect to inclusion. Due to your assumption on $W$, the equality necessarily holds for any maximal set $S(x),$ proving your thesis (incidentally, note that no further assumption on $W$ is needed).
Rmk 1. The upper semicontinuity of $S$ introduced above may be equivalently stated as:


*

*$\operatorname{graph(S)}:=\{(a,x)\in A\times X\\ : a\in S(x)\}$ is closed in $A\times X$, where $X$ has the discrete topology;

*$S^* :X\to 2^A$ is a closed map, that is, for any $a\in A$ the set $S^* (x):=\{a\in A\\,:\\, x\in S(a) \}$ is a closed subset of $A$;

*for any  $a\in A$ (denoting $\mathcal{N}_ a$ the family of nbd's of $a$), there holds 
$$\limsup_{b\to a}  S(b) := \cup_ {U\in\mathcal{N}_ a} \cap_ {b\in U} S(b) \subset S(a).$$
Rmk 2. The fact that a compact subset $K$ of $2^X$, where $2$ has the left-order topology, admits a  maximal element, is, of course, a consequence of the Zorn lemma. Indeed, if $\Gamma$ is an infinite chain in $K$, it has a limit point in $K$, which turns out to be an upper bound of $\Gamma$.  
