## 314 – C+C=[0,2]

Recall that the Cantor set $C$ is defined as the intersection $\bigcap_n C_n$ where

$C_0=[0,1]$

and $C_{n+1}$ is obtained by removing from each closed interval that makes up $C_n$ its open middle third, so

$C_1=[0,1/3]\cup[2/3,1]$,

$C_2=[0,1/9]\cup[2/9,1/3]\cup[2/3,7/9]\cup[8/9,1]$,

etc. Each $C_n$ is the union of $2^n$ closed intervals, each of length $1/3^n$.

Let’s prove that $C+C=\{x+y\mid x,y\in C\}$ is the interval ${}[0,2]$. (Cf. Abbott, Understanding analysis, Exercise 3.3.6.)

1.

The usual proof consists in showing inductively that $C_n+C_n=[0,2]$ for all $n$. This is easy: Note first that

$\displaystyle C_{n+1}=\frac13 C_n+\left(\frac13C_n+\frac23\right)$,

where

$\displaystyle \frac13C_n=\left\{\frac x3\mid x\in C_n\right\}$

and

$\displaystyle\frac13 C_n+\frac23=\left\{\frac{x+2}3\mid x\in C_n\right\}$.

This equality is verified by induction. Using this, we can use induction again to verify that, indeed, $C_n+C_n=[0,2]$ for all $n$.

We clearly have that $C+C\subseteq \bigcap_n C_n+C_n=[0,2]$. To prove the converse, for each $z\in[0,2]$ and each $n$, pick $x_n,y_n\in C_n$ such that $x_n+y_n=z_n$. The sequence of $x_n$ is bounded, so it has a convergent subsequence $x_{n_k}$. The corresponding subsequence $y_{n_k}$ has itself a convergent subsequence $y_{n_{k_m}}$. One argues that their limit values $x,y$ belong to $C$, because they belong to each $C_n$, since these sets are nested and closed. Finally, it follows immediately that $x+y=z$ as well.

2.

A very elegant different argument is obtained by using an alternative characterization of $C$: Note that each $x\in[0,1]$ can be written in base three as

$\displaystyle x=0.x_1x_2x_3\dots=\sum_{n=1}^\infty\frac {x_n}{3^n}$

where each $x_i$ is $0$, $1$, or $2$. By induction, one easily verifies that $x\in C_n$ iff it admits such an expansion with $x_n\ne1$. It follows that $x\in C$ iff it admits an expansion where no $x_i$ is $1$.

Given $z\in[0,2]$, we have $z/2\in[0,1]$, so we can write $z/2=a+b$ where the ternary expansion of $a$ has only $0$s and $2$s (so $a\in C$), and the expansion of $b$ has only $0$s and $1$s: If

$z/2=0.t_1t_2\dots$,

we can set $a=0.a_1a_2\dots$ where $a_i=0$ unless $t_i=2$, in which case $a_i=2$ as well, and similarly $b=0.b_1b_2\dots$ where $b_i=0$ unless $t_i=1$, in which case $b_i=1$ as well.

We then have that $z=2(a+b)=(a+2b)+a$, and both $a+2b$ and $a$ are in $C$.

This construction has the further advantage of making clear that the typical $z$ admits continuum many ($=|\mathbb R|$) representations as sum of two members of $C$: If we can split $b=c+d$ (where the expansions of $c,d$ only have $0$s and $1$s), we can set

$z=(a+2c)+(a+2d)$.

This gives us as many representations as subsets of $\{n\in\mathbb N\mid b_n=1\}$.

3.

The related problem of describing $C\cdot C$ appears to be much more complicated. See here and here.