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This question is about a sort of "weak topological $k$-mixing" where the $k$-point set is replaced by a (topological) segment.

Let $f:M\rightarrow M$ be a homeomorphism on a (compact) topological manifold $M$. I am looking for a system with the following property: There is a topological segment $C\subset M$ such that for any connected set $A\subset M$ and any $\varepsilon>0$ there is $n$ so that $d(f^{n}(C),A)<\varepsilon~$, where $d$ is the Hausdorff distance (with respect to some fixed metric on $M$). Such $C$ is a kind of a 'ninja' as it approximates any connected region of $M$ with arbitrary precision.

The reason for this question is that for a class of systems I have an idea how to construct a 'ninja set' $C$ with respect to all subsets of $M$, but this $C$ is not connected, it is zero dimensional. So it is a natural question to ask whether connected ninjas exist and the most simple connected set is the segment. (Note that a connected ninja may approximate only connected sets.) I don't believe that 'ninja segments' exist in dimension 2 as there is no enough space for a ninja to operate.

In case the answer is affirmative, one may ask about smooth systems with this property or to look for 'ninjas' in some known systems.

s::l

share|improve this question
    
what is a topological segment? –  Nishant Chandgotia Apr 16 '11 at 19:39
    
@Nishant: I'd assume that "topological segment" means arc, i.e., homeomorphic copy of the interval $[0,1]$ --- unless and until someone tells me otherwise. –  Andreas Blass Apr 16 '11 at 21:55
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