The Cantor set + the Cantor set = $[0,2]$

Question

Let $C$ be the Cantor set. Prove $ C + C = [0,2]$. (Notation: If $S, R$ are sets, then $S + R$ is the set of all $s + r$ with $s \in S, r \in R$.)

Proofs of this are readily available. This question asks to help me complete my proof. I ask that you do not give away the entire answer, but simply help me take the next step.

Definitions:
Let $I$ be a closed interval $[a,b]$. Then $I'$, the cantorization of $I$, is $[a, a + \delta] \cup [b - \delta, b]$ with $\delta = \frac{b-a}{3}$. The interval $[a, a + \delta]$ is called the lower third, the interval $[b - \delta, b]$ is called the upper third, and $I \setminus I'$ is called the middle third.

Let $S$ be a finite union of disjoint, separated, closed intervals, each of uniform (finite) width. Then $S'$, the cantorization of $S$, is the union of $I'$ for all $I$ where $I$ is a closed interval, a subset of $S$, and separated from $S \setminus I$.

Lemma: $S + S = S' + S'.$

Proof:

Let $I_1$ and $I_2$ be two suitable intervals in $S$ such $I_1 = [a, a + \delta], I_2 = [b, b + \delta], a \leq b$. Then

$$(I_1 \cup I_2) + (I_1 \cup I_2) = [2a, 2a + 2\delta] \cup [a+b, a +b + 2\delta] \cup [2b, 2b + 2\delta] \\$$ and $$I_1' = [a, a + \frac{1}{3}\delta] \cup [a + \frac{2}{3}\delta, a + \delta] \\ (I_1' + I_1') = [2a, 2a + \frac{2}{3}\delta] \cup [2a + \frac{2}{3}\delta, 2a + \frac{4}{3}\delta] \cup [2a + \frac{4}{3}\delta, 2a + 2\delta] = [2a, 2a + 2\delta] \\ I_2' = [b, b + \frac{1}{3}\delta] \cup [b + \frac{2}{3}\delta, b + \delta] \\ (I_2' + I_2') = [2b, 2b + \frac{2}{3}\delta] \cup [2b + \frac{2}{3}\delta, 2b + \frac{4}{3}\delta] \cup [2b + \frac{4}{3}\delta, 2b + 2\delta] = [2b, 2b + 2\delta] \\ (I_1' + I_2') = [a + b, a + b + \frac{2}{3}\delta] \cup [a + b + \frac{2}{3}\delta, a + b + \frac{4}{3}\delta] \cup [a + b + \frac{4}{3}\delta, a + b + 2\delta] \\ (I_1' + I_2') = [a + b, a + b + 2\delta] \\ (I_1' \cup I_2') + (I_1' \cup I_2') = [2a, 2a + 2\delta] \cup [a+b, a +b + 2\delta] \cup [2b, 2b + 2\delta] \\ $$ so $$ (I_1 \cup I_2) + (I_1 \cup I_2) = (I_1' \cup I_2') + (I_1' \cup I_2')$$

Since this holds for any $I_1, I_2$ in $S$, $S + S = S' + S'.$

Main Proof:

By induction using the lemma, for any $n \in \mathbb{n}, C_n + C_n = [0,2].$ We now need to show this for $C$ itself, and here is where I need some help.

It is not enough that for any $n$, there exists $x_n, y_n \in C_n$ such that $x_n + y_n = t$. We need to show that the $x$ and $y$ are in $C$, that is, that the same $x_n, y_n$ are in all $C_n$. One way to do this is to apply that sequences within compact sets converge to limits within the compact set. Thus, $(x_n) \to x \in C_n, (y_n) \to y \in C_n$. But does that show that if for all $n$, $x_n + y_n = t$, then $x + y = t$?

Another approach might be to take for each $n$ the intersection of $C_n$ and the set of ($x, y \in [0,2]$ that sum to $t$). Since both of these are compact, their intersection is compact. But defining the set ($x, y \in [0,2]$ that sum to $t$) is tricky.

Answer 1

You have given the next step of your approach basically.

One way to do this is to apply that sequences within compact sets converge to limits within the compact set. Thus, $(x_n) \to x \in C_n, (y_n) \to y \in C_n$. But does that show that if for all $n$, $x_n + y_n = t$, then $x + y = t$?

Yes, it does. More specifically, you can show that

• There is subsequence $\{x_{i_n}\}_{i_n}$ of $\{x_n\}_n$ that converges.
• If $x_{i_n}\in C_{i_n}$ for $i_n\to\infty$ converges, it converges to a point in $C$. So does $t-x_{i_n}=y_{i_n}$.
• Taking limits preserves addition.