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It is a quite famous fact that $C+C=[0,2]$, what about $ C \cdot C $ ?

Is it measurable? Yes

It's clearly measurable because it's compact, being a continuous image of the compact set C×C.

Is it possible to list all the intervals of $[0,1]\setminus C\cdot C$ ?

What is its Lebesgue measure ?

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  • $\begingroup$ It's clearly measurable because it's compact, being a continuous image of the compact set $C\times C$. $\endgroup$ – bof Mar 20 '14 at 9:58
  • $\begingroup$ It looks like an interesting question. Unless $C\cdot C$ is measure zero or an interval, however, it seems likely there is no simple expression for it. So first determine whether it is measure zero, or is an interval. $\endgroup$ – GEdgar Mar 20 '14 at 14:15
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    $\begingroup$ It can't be an interval because surely $C\cdot C$ doesn't take values in $(\frac{1}{3},\frac{4}{9})$ $\endgroup$ – user136725 Mar 20 '14 at 19:01
  • $\begingroup$ Inasmuch as $C\cdot C$ is closed, its complement in $[0,1]$ is the union of countably many disjoint open intervals. One of those intervals is $(\frac13,\frac49)$ as remarked by @Febo; others are $(\frac19,\frac4{27})$ and $(\frac79,\frac{64}{81})$, etc. $\endgroup$ – bof Mar 22 '14 at 0:29
  • $\begingroup$ Well, how to find all of them? $\endgroup$ – user136725 Mar 22 '14 at 7:46
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Your question is already treated in mathoverflow. See Arithmetic products of Cantor Sets with interesting, nearly up-to-date information. The question treats self-similar sets on $\mathbb R$ in general. Results with respect to Cantor sets are e.g. that the Hausdorff Dimension $$\dim_H(C\cdot C) = \min(2 \cdot \dim_H(C),1)=\min\left(2\frac{\ln 2}{\ln 3},1\right) = 1.$$ Another interesting aspect is, that according to the experts there, an answer to the question if $C\cdot C$ has positive measure seems to be out of reach at the time.

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