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Hi,

doing my research I found the following problem and I´ll be glad if someone could give a reference.

We say that a compact connected subset $K$ of the plane is psuedo laminated if the following hold:

  • $K$ has a partition by $C^1$ curves without boundary. For each $x$ in $K$ denote by $W(x)$ the leaf through $x.$

  • For any $x\in K$ and any $y\in W(x)$ there exist a neighborhood $U_x$ and a continuous map $\phi:U_x\times [0,1]\to\mathbb{R}^2$ such that: 1) $\phi(x,0)=x$ 2) $\phi(x,1)=y$ 3) for any $z\in K\cap U_x$ we have that $\phi(z,t)\in W(z)$ for any $t\in[0,1].$

I´m looking for the following results:

1) Let $K$ be a compact connected subset of $\mathbb{R}^2$ and pseudo laminated. Then $K$ separates the plane in at least two connected components

2) If the complement of $K$ has just two connected components, then $K$ is just a circle.

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  • $\begingroup$ Does U_x contain all of W(x), or just the two points x,y ? Also, I may be missing something in 2), because I have the impression that if you take K to be the union of circles of radii in [1,2], then K is pseudo laminated (you may even take U_x to be the entire plane and phi to be a family of rotations indexed by t), its complement has two connected components, but K is not homeomorphic to a circle. $\endgroup$ – Mathieu Baillif Mar 11 '13 at 21:57
  • $\begingroup$ Couldn´t $K$ in 2) be a "closed ring"? $\endgroup$ – Ramiro de la Vega Mar 11 '13 at 21:58
  • $\begingroup$ I´m sorry, $K$ has emtpy interior. U_x ia an open neighbourhood of $x.$ $\endgroup$ – Martin Mar 11 '13 at 22:07
  • $\begingroup$ I do not know a ref, but it does not seem to be hard to prove. BTW, do you know that the staement does not hold if you drop compactness? $\endgroup$ – Anton Petrunin Mar 12 '13 at 2:49
  • $\begingroup$ I agree, it seems not very hard to be proved. It is a bit faraway from i was doing and it was better to include a reference than a proof. If you drop compactness the statemement does not hold. For instance if you take just a curve which lool likes the graph of $\sin (1/x)$ at both "ends" does not separate the plane. Also, if you don´t have the second part of the definition of pseudo-laminatios the statement does not hold (the stable manifold of a Smale´s horshoe does not separates the plane) $\endgroup$ – Martin Mar 12 '13 at 12:47

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