314 – C+C=[0,2]

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


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



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.)


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),


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


\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.


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 0s and 2s (so a\in C), and the expansion of b has only 0s and 1s: If


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 0s and 1s), we can set


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


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


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