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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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It seems very close to a previous question that you asked:… is there a relation between the two? – Asaf Karagila Jun 8 '11 at 15:36
@Asaf Karagila It is changed a little bit. I found your question on universal sets and decided to reformulate my in more formal way. – Ilya Jun 8 '11 at 15:49
@Gortaur: I see. You might want to consult the MathOverflow link the comments to my question, I cross-posted and got some interesting answers from Clinton Conley there. – Asaf Karagila Jun 8 '11 at 15:58
@Gortaur: I think you should mention (edit) that you have posted it to mathoverflow too: – Martin Sleziak Jun 8 '11 at 17:12
@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. :-) – Asaf Karagila Jun 9 '11 at 18:40
up vote 1 down vote accepted

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. – Ilya Jun 10 '11 at 8:08

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