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I am trying to prove that

$C+C =[0,2]$ ,where $C$ is the Cantor set.

My attempt:

If $x\in C,$ then $x= \sum_{n=1}^{\infty}\frac{a_n}{3^n}$ where $a_n=0,2$

so any element of $C+C $ is of the form $$\sum_{n=1}^{\infty}\frac{a_n}{3^n} +\sum_{n=1}^{\infty}\frac{b_n}{3^n}= \sum_{n=1}^{\infty}\frac{a_n+b_n}{3^n}=2\sum_{n=1}^{\infty}\frac{(a_n+b_n)/2}{3^n}=2\sum_{n=1}^{\infty}\frac{x_n}{3^n}$$

where $x_n=0,1,2, \ \forall n\geq 1$.

Is this correct?

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    $\begingroup$ This shows that $C+C\subseteq[0,2]$ (which is kind of obvious since $C\subset[0,1]$) and one also needs to show that $[0,2]\subseteq C+C$ (fortunately, roughly the same idea, only reversed, works). $\endgroup$
    – Did
    Commented Feb 20, 2013 at 13:48
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    $\begingroup$ Relevant $\endgroup$
    – MJD
    Commented Feb 20, 2013 at 14:40
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    $\begingroup$ There are two proofs here. $\endgroup$ Commented Feb 21, 2013 at 15:24

7 Answers 7

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Short answer for this question is "your argument is correct ". To justify the answer consider some particular $n_{0}\in \mathbb{N}$. Since $$\sum_{n=1}^{\infty}\dfrac{a_{n}}{3^{n}},\sum_{n=1}^{\infty}\dfrac{b_{n}}{3^{n}}\in C$$ we have that $ x_{n_{0}}=\dfrac{a_{n_{0}}+b_{n_{0}}}{2}\in \{0,1,2\} $. ($ x_{n_{0}}=0 $ if $ a_{n_{0}}=b_{n_{0}}=0 $. $ x_{n_{0}}=2 $ if $ a_{n_{0}}=b_{n_{0}}=2 $. Otherwise $ x_{n_{0}}=1 $. )

Then clearly $$\sum_{n=1}^{\infty}\dfrac{x_{n}}{3^{n}}\in [0,1].$$ So $$2\sum_{n=1}^{\infty}\dfrac{x_{n}}{3^{n}}=\sum_{n=1}^{\infty}\dfrac{a_{n}}{3^{n}}+\sum_{n=1}^{\infty}\dfrac{b_{n}}{3^{n}}\in [0,2].$$ Hence $ C+C\subseteq [0,2] $. To complete the proof you must show that the other direction as well. To show $ [0,2]\subseteq C+C $ it is enough to show $ [0,1]\subseteq \dfrac{1}{2}C+\dfrac{1}{2}C $.

Observe that $ b\in \dfrac{1}{2}C $ if and only if there exists $ t\in C $ such that $ b=\dfrac{1}{2}t $.

Hence $$ b\in \dfrac{1}{2}C\text{ if and only if }b=\sum\limits_{n = 1}^\infty \frac{b_n}{3^n}\text{ ; where }b_{n}=0\text{ or }1. $$ Now let $ x\in [0,1] $. Then $$ x= \sum\limits_{n = 1}^\infty \frac{x_n}{3^n}\text{ ; where }x_{n}=0,1\text{ or }2. $$ Here we need to find $ y,z\in \dfrac{1}{2}C $ such that $ x=y+z $. Let's define $ y=\sum\limits_{n = 1}^\infty \frac{y_n}{3^n} $ and $ z=\sum\limits_{n = 1}^\infty \frac{z_n}{3^n} $ as follows.

For each $n\in \mathbb{N}$, $ y_{n}=0 $ if $ x_{n}=0 $ and $ y_{n}=1 $ if $ x_{n}=1,2 $.

For each $n\in \mathbb{N}$, $ z_{n}=0 $ if $ x_{n}=0,1 $ and $ z_{n}=1 $ if $ x_{n}=2 $.

Thus $y,z\in \dfrac{1}{2}C $ and for each $n\in \mathbb{N}$, $ y_{n}+z_{n}=0 $ if $ x_{n}=0 $ , $ y_{n}+z_{n}=1 $ if $ x_{n}=1 $ and $ y_{n}+z_{n}=2 $ if $ x_{n}=2 $.

Therefore $x=y+z\in \dfrac{1}{2}C+\dfrac{1}{2}C$ and hence $[0,1] \subseteq \dfrac{1}{2}C+\dfrac{1}{2}C$. $\square $

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  • $\begingroup$ How to you know that if $x$ belongs to the interval $0$ to $1$ then It is of the form of that infinite sum ? $\endgroup$
    – Someone
    Commented Feb 24, 2019 at 22:00
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You can easily show that every number in $[0,2]$ of the form $m/3^n$, $0\leq m\leq 2\times 3^m$ can be write with a sum $x+y$, where $x,y\in C$. Since $A=\{m/3^n;0\leq m\leq 2\times 3^m \}$ is dense in $[0,2]$ and $C$ is compact we have that $[0,2]\subset C+C$.

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We need to show both inclusions. $C+C\subseteq [0,2]$ is pretty obvious, however $[0,2]\subseteq C+C$ is no longer. For the proof of the second statement I recommend the book written by Steven G. Krantz - "A Guide to Topology", which has a really good analytic. You can find it online http://books.google.com/books?id=O3tyezxgv28C&printsec=frontcover&hl=pl&redir_esc=y#v=onepage&q=cantor&f=false look at page 35.

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    $\begingroup$ Your answer is more of a comment, there is no reason to bump an old question, that already has two perfectly fine answers. $\endgroup$
    – user29123
    Commented Apr 4, 2015 at 20:29
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    $\begingroup$ Unfortunately thanks to the reputation system here, I can't comment because I don't have 50 reputation points. And for me, these 2 answers here were not as good as that found in the book I gave. May be useful for someone in the future, so I don't consider this as a bump. $\endgroup$
    – Andrew
    Commented Apr 5, 2015 at 19:42
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    $\begingroup$ It would be more fruitful to give the full details and highlight why exactly you prefer one proof over another. I would also look for a related post to which this could he aďded. $\endgroup$
    – Kurt G.
    Commented May 2, 2022 at 4:37
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Since $C \subset [0,1]$ we have $1/2(C+C) \subset [0,1]$ by the convexity of $[0,1]$.

The inclusion $[0,1] \subset \frac{1}{2}C + \frac{1}{2}C$ was explained by @Tgymasb, anyways, the way I see it

$$\frac{1}{2}C = \sum_{n \ge 1} \frac{a_n}{3^n}$$ with $a_n \in \{0,1\}$ while $$[0,1] = \sum_{n \ge 1} \frac{c_n}{3^n}$$ with $c_n \in \{0,1,2\}$ and the point is that any function from some domain to $\{0,1,2\}$ is a sum of two functions mapping to $\{0,1\}$, and this is because we have the obvious decompositions: $$2 = 1 + 1\\ 1 = 1 + 0\\ 0 = 0 + 0$$

$\tiny{ \text{( pointwise addition so no carries)}}$

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    $\begingroup$ set = number ?! $\endgroup$
    – Rasmus
    Commented Apr 4, 2015 at 20:26
  • $\begingroup$ @Rasmus: Hi, the meaning of $C+C$ is the set of all sums $c+c'$ with $c$, $c'$ in $C$. So the sum of two sets is another set. In the particular case that the sets have exactly one element their addition is like the addition of the corresponding numbers. $\endgroup$
    – orangeskid
    Commented Apr 4, 2015 at 22:46
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The set $C+C$ is a non-empty compact subset of $\mathbb R$. Since $C=\frac13C+\big\{0,\frac23\big\}$, we have $(C+C)=\frac13(C+C)+ \big\{0,\frac23,\frac43\big\}$. So both $C+C$ and the interval $[0,2]$ are fixed points of the map $F\mapsto \frac13 F+ \big\{0,\frac23,\frac43\big\}$. The latter is a contraction on the set $\mathcal K$ of all non-empty compact subsets of $\mathbb R$ w.r.to the Hausdorff distance $d_H$, so they coincide. $\square$

rmk. Note that since $(\mathcal K,d_H)$ is a complete metric space, you can also use the contraction principle to define $C$ as the unique non-empty compact $C\subset \mathbb R$ such that $C=\frac13C+\big\{0,\frac23\big\}$; the iteration $C_{n+1}:=\frac13C_n+\big\{0,\frac23\big\}$ starting from $C_0:=[0,1]$ produces the usual construction $C \displaystyle=\lim_{n\to\infty}C_n=\cap_{n\in\mathbb N}C_n$.

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By $1-C=C$ we can just do it for all $x\in [0,1]$. Then denote $x=x_1x_2...x_m...$(where $x=\sum_n3^{-n}x_n$ with $x_n=0,1,2$) and we then construct as $a=\sum_n3^{-n}a_n$,$b=\sum_n3^{-n}b_n$ with $a_n,b_n=0,2$ and $a+b=x$.

We do it iteratively, suppose $x_1=0$, we set $a_1=b_1=0$; and suppose $x_1=1$, we set $a_1=b_1=0$; and suppose $x_1=2$, we set $a_1=2,b_1=0$. Then we see $x-a_1-b_1=0x_2x_3...$(first case) or $x-a_1-b_1=1x_2x_3...$(second case); and then in the first case, and if $x_2=2$, we set $a_2=2,b_2=0$; and if $x_2=1$, we set $a_2=0,b_2=0$; and if $x_2=0$, we set $a_2=0,b_2=0$. And in the second case, if $x_2=1$, we set $a_2=b_2=2$; if $x_2=2$, we set $a_2=b_2=2$; and if $x_2=0$, we set $a_2=2,b_2=0$. Then by this construction, we see $x-a_1a_2-b_1b_2=00x_3x_4...$ or $01x_3x_4...$.

And iterate this process again and again to construct a $a=a_1...a_n...$ and $b=b_1b_2...b_n...$ such that $x-a_1...a_n-b_1b_2...b_n=0...0x_{n+1}x_{n+2}....$ or $0...1x_{n+1}x_{n+2}...$. Then we can see $x=a+b$ indeed.

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$[0,1]=\frac{1}{2}[0,2]$. Suppose $z\in[0,1]$ and set $z'=\frac12 z\in [0,1]$. Let $z'=\sum^\infty_{n=0}\frac{a_n}{3^n}$, where $a_n\in\{0,1,2\}$. Let $w=\sum_{n:a_n=2}\frac{a_n}{3^n}=2\sum_{n:a_n=2}\frac{1}{3^n}$. Notice that $$z'=\frac12w+(\frac12w+(z'-w))$$

$x'=\frac12w\leq 1/2$ has ternary expansion consisting of $0$'s and $1$'s only; hence $x=2x'\in C$. Similarly, $y'=\frac12w+(z-w)$ has ternary expansion consisting of $0$'s and $1$'s only; hence $y'\leq\sum^\infty_{n=1}3^{-n}=\frac12$, and $2y'\in C$.

Then $z'=x+y\in C+C$.

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