Sticky Cantor sets

Importance: Medium ✭✭
Subject: Topology
Keywords: Cantor set
Recomm. for undergrads: no
Posted by: porton
on: February 6th, 2011

\begin{conjecture} Let $C$ be a \Def{Cantor set} embedded in $\mathbb{R}^n$. Is there a self-homeomorphism $f$ of $\mathbb{R}^n$ for every $\epsilon$ greater than $0$ so that $f$ moves every point by less than $\epsilon$ and $f(C)$ does not intersect $C$? Such an embedded Cantor set for which no such $f$ exists (for some $\epsilon$) is called "sticky". For what dimensions $n$ do sticky Cantor sets exist? \end{conjecture}

I borrowed this conjecture from \href [this forum thread]{}.

Certainly I understand this conjecture wrongly: $C$ is a subset of a line segment. Consider a homeomorphism which moves all points of $\mathbb{R}^n$ orthogonally to this line segment by $\epsilon/2$. This would be a solution of this problem. Obviously it is not what is meant.

Indeed I submit the problem to OPG as is in the hope that somebody will correct my wrong understanding and adjust the formulation to not be misunderstood as by me.

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* indicates original appearance(s) of problem.


Your misunderstanding comes from the definition of a Cantor set. A Cantor set is a set homeomorphic to the usual middle-thirds Cantor set. In general it does not have to lie on a line segment.


"embedded" does not imply that it is still a subset of the line. It just says that it's one-to-one and a homeomorphism with the image. The conjecture requires to prove that there exists a Cantor which cannot be separated from itself, so showing an example where it can be separated is not relevant.

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