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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: http://math.stackexchange.com/questions/33110/self-complete-set-in-square is there a relation between the two?2011-06-08
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    @Asaf Karagila It is changed a little bit. I found your question on universal sets and decided to reformulate my in more formal way.2011-06-08
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    @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.2011-06-08
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    @Gortaur: Also, you might want to add that this is a reformulation of a previous question. It took me a few minutes to find the question and see that you were the OP, and it is not a dupe.2011-06-08
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    @Asaf Karagila: dupe? Oh, you mean that I asked for help in question2 to then answer in question1 and get reputation? I'm not so tricky) Thanks anyway, edited it.2011-06-08
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    I think that the sets $X$ and $\bigcup_{x\in X} s(x)$ are trivially invariant. Am I missing something? (Do I understand correctly that you are working with one fixed set $G$?)2011-06-08
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    @Gortaur: I think you should mention (edit) that you have posted it to mathoverflow too: http://mathoverflow.net/questions/67275/sets-invariant-under-sections2011-06-08
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    @Martin: you're certainly right. I've edited, one should look for invariant subsets of $A\neq X$.2011-06-08
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    @Asaf: I have noticed you've discussed the problem at MO, so you might be interested to know that a new answer appeared here. Of course, it is possible that I misunderstood the problem. (Should I have posted it to MO instead? Is there a consensus on cases like this.)2011-06-09
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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