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I have recently rediscovered (after several years) the wonder of the Cantor set (so rich and so beautiful!). I have two questions that are unrelated, but they are both about Cantor sets.

  1. Let $K$ be a non-empty compact, perfect, metric space such that $K \simeq K \times K$. Is $K$ necessarily homeomorphic to the Cantor set, or the Hilbert cube or some combination of both?
  2. Let $C$ be the Cantor set, $K$ the set of points $\exp(i2\pi x)$ where $x\in C $, and $S$ the set of all chords between points of $K$. Is $S$ convex?
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  • $\begingroup$ I fixed the LaTeX to get the braces to work. $\endgroup$
    – David Roberts ♦
    Apr 19, 2011 at 1:08
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    $\begingroup$ What is a "combination" of two spaces? $\endgroup$
    – Qiaochu Yuan
    Apr 19, 2011 at 1:21
  • $\begingroup$ Qiaochu, this is rather loose but initially I thought "product of" but it might be too restrictive. Then I thought "constructed from", but that's too broad... $\endgroup$
    – Portland
    Apr 19, 2011 at 1:48

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  1. What about product of countably many copies of a circle?

  2. If $C$ is the standard Cantor set then No. One can find a point inside of a triangle with vertexes in your $K$ which does not lie on any chord. To see it consider hexagon $ABCDEF$ inscribed in a circle. Notice that its main diagonals intersect at one point iff $$AB{\cdot}CD{\cdot}EF=BC{\cdot}DE{\cdot}FA.$$ Moreover $$BCDE\cup DEFA \cup FABC= ABCDEF.$$ if and only if
    $$AB{\cdot}CD{\cdot}EF\le BC{\cdot}DE{\cdot}FA.\ \ \ \ \ (\star)$$ Now choose points $A$, $B$, $C$, $D$, $E$ and $F$, which correspond to the following points in the Cantor set: $\tfrac1{3^n}$, $\tfrac 2{3^n}$, $\tfrac 7{3^{n+1}}$, $\tfrac 8{3^{n+1}}$, $\tfrac 1{3^{n-1}}$ and $\tfrac 2{3^{n-1}}$. Set $\alpha=\tfrac\pi{3^{n+1}}$. If $S$ is convex then $(\star)$ holds; i.e. $$\sin (3{\cdot}\alpha){\cdot}\sin \alpha{\cdot}\sin (9{\cdot}\alpha)\le\sin (15{\cdot}\alpha){\cdot}\sin \alpha{\cdot}\sin \alpha,$$ which is not true for small $\alpha$. Here is how it looks: alt text

P.S. My original answer stated that the set $S$ has zero measure; but in fact $S$ has positive measure --- it was noticed by Tapio Rajala. All this made me to take this question seriously and to draw this nice picture...

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  • $\begingroup$ Thanks Anton, 1. I'm not familiar with Cantor torus, do you have a reference? 2. $K$ has zero measure, but $S$? $\endgroup$
    – Portland
    Apr 19, 2011 at 1:57
  • $\begingroup$ I edit 1, I did not get what you ask about 2. $\endgroup$
    – Anton Petrunin
    Apr 19, 2011 at 2:39
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    $\begingroup$ K has one-dimensional measure zero, S has two-dimensional measure zero? But perhaps this needs proof. $\endgroup$
    – Gerald Edgar
    Apr 19, 2011 at 4:07
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    $\begingroup$ I don't see the argument yet for $S$ having measure zero. $C \times C$ has dimension $\frac{\log 4}{\log 3} > 1$ so just taking a Lipschitz map does not tell you much. Also, if one would take the cords between two copies of $C$ which are on two different lines in $\mathbb{R}^2$, then $S$ would have positive $2$-dimensional measure by the Marstrand projection theorem (?) $\endgroup$
    – Tapio Rajala
    Apr 19, 2011 at 7:01
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    $\begingroup$ @Anton, I am not sure this is the case: Take $S$ to be (for example) the cords between $0 \times C$ and $1 \times C$. Then the intersection $S \cap (t\times\mathbb{R}) = tC + (1-t)C$ has positive $\mathcal{H}^1$-measure for $\mathcal{H}^1$-almost every $t \in [0,1]$. (This is because this collection is just a reparametrized and scaled version of a collection of projections of $C \times C$.) Thus, $\mathcal{H}^2(S) > 0$. $\endgroup$
    – Tapio Rajala
    Apr 19, 2011 at 13:35

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