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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.

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    $\begingroup$ I have no time to check your proof, but I am wondering if using the characterization of the Cantor set using numbers in base $3$ with no $1$ would not be easier. $\endgroup$
    – Taladris
    Commented Nov 7, 2022 at 4:03
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    $\begingroup$ As a second alternative to the comment above, consider a subset of $[0,1]$ not in the cantor set (so some middle third set) and note doubling it gives the same as adding the first and last third set (this is basically the same argument as the ternary sequences but using sets). It's nice to have your own proof checked, and perhaps someone will, but in some sense it seems your proof missed the forest for the trees... $\endgroup$
    – Brevan Ellefsen
    Commented Nov 7, 2022 at 4:25
  • $\begingroup$ @BrevanEllefsen How does the set approach you suggest show we cover adding a member of the first third and a member of the middle (excluded) third? $\endgroup$
    – SRobertJames
    Commented Nov 7, 2022 at 15:19
  • $\begingroup$ You don't need to ever add a member of a first and middle third to get all of $[0,2]$. You of course get this in the sum $C + C$, but all this means is that some point have multiple representations in the sum. All you need is to be able to double any given subset, and for the ones that are excluded we can replace the double by the sum of the first and last third. $\endgroup$
    – Brevan Ellefsen
    Commented Nov 7, 2022 at 16:37
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    $\begingroup$ It might help you to think about what would happen if you considered $2C$. This is a subset of $C + C$ given by "stretching" the cantor set to be in $[0,2]$ rather than $[0,1]$. The things that are missing in $2C$ are the stretched middle thirds, but these are exactly the sums of first and last thirds in $C$ so are already included in $C+C$ $\endgroup$
    – Brevan Ellefsen
    Commented Nov 7, 2022 at 16:42

1 Answer 1

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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.
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