Cantor set and Hilbert cube, or anything else? 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.


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*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?

*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?

 A: *

*What about product of countably many copies of a circle?

*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: 
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... 
