Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
    
It seems very close to a previous question that you asked: math.stackexchange.com/questions/33110/… 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
1  
@Gortaur: I think you should mention (edit) that you have posted it to mathoverflow too: mathoverflow.net/questions/67275/sets-invariant-under-sections –  Martin Sleziak Jun 8 '11 at 17:12
5  
@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

1 Answer 1

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 $X$.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.