# Sticky Cantor sets

\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]{http://www.mathkb.com/Uwe/Forum.aspx/math/16972/Current-Status-of-Topology}.

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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## Bibliography

% Example: %*[B] Claude Berge, Farbung von Graphen, deren samtliche bzw. deren ungerade Kreise starr sind, Wiss. Z. Martin-Luther-Univ. Halle-Wittenberg Math.-Natur. Reihe 10 (1961), 114. % %[CRS] Maria Chudnovsky, Neil Robertson, Paul Seymour, Robin Thomas: \arxiv[The strong perfect graph theorem]{math.CO/0212070}, % Ann. of Math. (2) 164 (2006), no. 1, 51--229. \MRhref{MR2233847} % % (Put an empty line between individual entries)

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### M

"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.

## Misunderstanding

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.