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Let $X$ be a compact in the Polish space (metric, complete, separable) and $G\subseteq X\times X$ is open. For $x\in X$ we define the section of $G$: $ s(x) = \{y\in X|\langle x,y \rangle \in \bar{G}\}. $

The set $A'\subseteq X$ is invariant if for all $x\in A'$ holds $s(x)\subset A'$. How to verify if there are non-empty invariant subsets of given compact $A$? Maybe there are known equivalent problems?

It will be even helpful in the case $X = [0,1]$.

This is reformulated and changed a little bit problem from my previous question: Self-complete set in square

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    @Matrtin: I check both the sites more often than teenage girls update their Facebook status. So I'm not sure what you mean by "new answer". I had seen it when it was first posted. :-)2011-06-09

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The sets $X$ and $\bigcup_{x\in X} s(x)$ are always invariant.

For any $X$, if $G=X\times X$ then the only invariant sets are $X$ and $\emptyset$.

If $X=[0,1]$ and $G=\{(x,y); |x-y|<\varepsilon\}$ for some given $\varepsilon>0$ then the only invariants sets are $X$ and $\emptyset$. (A similar set $G$ works for $X=\mathbb R$.)

So without some additional assumptions you cannot avoid the situation that the only non-empty invariant subset is $X$.

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    Yes, it was clear in fact. The problem is the followin: if for $G\subset X\times X$ and $A\subset X \exists A'\subset A:A'$ is invariant? I am looking for the procedure which can for *any* $G$ give an answer on this question. I even know how to find $A'$ with $A_n\to A'$, but there are no bounds on convergence of this sets and this procedure will work only if such $A'$ exists.2011-06-10