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Let us have the well-known Cantor set. We know that the Cantor set has continuum amount of numbers in the interval $[0,1]$.

Here is the thing that I need to prove: Prove, that every $x \in [0,2]$ can be written in the form $a+b$, where $a$ and $b$ are part of the Cantor set.

Edit: If I prove that Cantor set + Cantor set $= [0,2]$, is that a good solution as well?

My idea:

We can write all the elements of Cantor set instead of decimal, in 3-based number system. In this way, every number of the Cantor set has the form $0,abc...$ but it doesn't contain the digit 1, since I deleted that part of the numbers. Now, with this form, I am able to make the Cantor set $ \Rightarrow\ [0,1]$ bijection, which means I am able to write down every $x \in [0,1]$ number with the elements of Cantor set. From this part, I can pick any $a,b\in[0,1]$ and with them I can cover the interval $[0,2]$ as well.

Is my proof correct, or what you think about this task? Anything can help! :)

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Atvin
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    might be of possible interest: http://math.stackexchange.com/questions/309080/cantor-set-cantor-set-0-2?rq=1 – Sujaan Kunalan Mar 18 '15 at 19:44
  • Yeah, I just asked that, if that is true, is my task also true? :) – Atvin Mar 18 '15 at 19:44
  • If I understand right, your solution sounds suspicious: You're taking a bijection of $C$ with $[0,1]$, then saying $[0, 1] + [0, 1] = [0, 2]$. But that's not what your question stipulated. In other words, you're "not allowed" to map $C$ bijectively to $[0,1]$, you have to use elements of $C$ itself. – Andrew D. Hwang Mar 18 '15 at 19:53

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Let $C$ denote the Cantor set, then $\frac12C$ makes up of all the element in $[0,\frac12]$ of which the ternary expansion doesn't contains the digit 2. It's easy to show that $\frac12C+\frac12C=[0,1]$

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