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
$$S(x)= S(W(x))$$
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6$\begingroup$ I notice that A is assumed to be compact, but W is not assumed to be continuous and S is also not assumed to have any continuity properties (such as continuity in the sense of the Hausdorff metric, or Kuratowski semicontinuity). Presumably this is an error. $\endgroup$– Ian MorrisCommented Aug 25, 2010 at 16:43
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$\begingroup$ I agree to Ian Morris: as is written, the question does not refer to the topology of A except for the assumption that A is compact. This means that A can be any set. If we want to turn a set A to a compact topological space, we can just give it the trivial topology. Something is wrong with the current question as it is, but I do not know what the correct question is. $\endgroup$– Tsuyoshi ItoCommented Aug 30, 2010 at 15:22
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$\begingroup$ However it's quite an interesting question; it seems to me that what is lacking is a semicontinuity assumption on S(x). $\endgroup$– Pietro MajerCommented Aug 30, 2010 at 21:39
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1$\begingroup$ I think it's a bad idea to use the same symbol, $S$, for both a function and the codomain of that function. $\endgroup$– Gerry MyersonCommented Aug 31, 2010 at 0:39
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2 Answers
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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.
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$\begingroup$ Btw, I realize now that compactness was not used in the argument above: this is suspicious... Presumably you also wanted some property of the "space of sets $S$" and of the function $A \to S$ so that compactness would play a role, right? $\endgroup$– damianoCommented Aug 25, 2010 at 16:31
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$\begingroup$ In reality no, there is no required property for the space of sets S. $\endgroup$– albertoCommented Aug 25, 2010 at 17:26
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$\begingroup$ So compactness of $A$ was not relevant? In any case, is the example above fine for you? I was trying to find a "natural" example for which $W$ was continuous and iterations of $W$ had no finite orbits, since a finite orbit gives you immediately the consequence you were asking to avoid. Obviously there are easier examples: for instance, just do the same as in my answer, but only considering a single orbit, so that $A=\mathbb{Z}$, the function $W$ is the shift by 1 and $S(n) = \left\{m \in \mathbb{Z} \,|\, m \leq n \right\}$. Topologize $\mathbb{Z}$ as you want, if you need to. $\endgroup$– damianoCommented Aug 25, 2010 at 17:52
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1$\begingroup$ I could add another hypothesis. I assume that the space of sets S is a topological space and A->S is continuous $\endgroup$– albertoCommented Aug 25, 2010 at 18:20
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3$\begingroup$ @alberto: this is not going to be enough to make the question interesting, as the only open sets of $S$ could be $S$ itself and the empty set. I suspect that, if the question were properly formulated, then the answer would be "no", and compactness would play a key role in finding a "fixed point". Unfortunately, I am not able to guess what the right question should be. $\endgroup$– damianoCommented Aug 25, 2010 at 19:40
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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 $\Gamma$.
answered Aug 30, 2010 at 13:38