# Cantor set + Cantor set =$[0,2]$

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?

• 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).
– Did
Commented Feb 20, 2013 at 13:48
• Relevant
– MJD
Commented Feb 20, 2013 at 14:40
• There are two proofs here. Commented Feb 21, 2013 at 15:24

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 $[0,1] \subseteq \dfrac{1}{2}C+\dfrac{1}{2}C$. $\square$

• How to you know that if $x$ belongs to the interval $0$ to $1$ then It is of the form of that infinite sum ? Commented Feb 24, 2019 at 22:00

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 $[0,2]\subset C+C$.

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.

• Your answer is more of a comment, there is no reason to bump an old question, that already has two perfectly fine answers.
– user29123
Commented Apr 4, 2015 at 20:29
• 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. Commented Apr 5, 2015 at 19:42
• 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. Commented May 2, 2022 at 4:37

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: $$2 = 1 + 1\\ 1 = 1 + 0\\ 0 = 0 + 0$$

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

• set = number ?! Commented Apr 4, 2015 at 20:26
• @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. Commented Apr 4, 2015 at 22:46

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 $$C \displaystyle=\lim_{n\to\infty}C_n=\cap_{n\in\mathbb N}C_n$$.

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.

$$[0,1]=\frac{1}{2}[0,2]$$. 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 $$z'=x+y\in C+C$$.