[proofplan] We show $\mathcal{C} + \mathcal{C} = [0,2]$ by proving both inclusions. The inclusion $\mathcal{C} + \mathcal{C} \subset [0,2]$ is immediate from $\mathcal{C} \subset [0,1]$. For the reverse, given $z \in [0,2]$, we construct $x, y \in \mathcal{C}$ with $x + y = z$ by a digit-by-digit greedy algorithm in base $3$, choosing at each position $k$ digits $a_k, b_k \in \{0,2\}$ whose sum matches the required digit of $z$ (after accounting for carries from lower-order positions). [/proofplan] [step:Establish the inclusion $\mathcal{C} + \mathcal{C} \subset [0, 2]$] Since $\mathcal{C} \subset [0,1]$, for any $x, y \in \mathcal{C}$ we have $0 \le x + y \le 2$. Hence $\mathcal{C} + \mathcal{C} \subset [0,2]$. [/step] [step:Represent $z \in [0, 2]$ in a form compatible with digit decomposition] Let $z \in [0,2]$. We construct sequences $(a_k)_{k \in \mathbb{N}}$ and $(b_k)_{k \in \mathbb{N}}$ with $a_k, b_k \in \{0,2\}$ such that \begin{align*} z = \sum_{k=1}^{\infty} \frac{a_k + b_k}{3^k}. \end{align*} At each position $k$, the sum $a_k + b_k$ takes one of three values: $0 + 0 = 0$, $0 + 2 = 2 + 0 = 2$, or $2 + 2 = 4$. So we need to express $z$ in the form $\sum_{k=1}^{\infty} s_k / 3^k$ where each $s_k \in \{0, 2, 4\}$, then split each $s_k$ into $a_k + b_k$. [guided] The challenge is that $z \in [0,2]$ has a standard ternary expansion using digits in $\{0, 1, 2\}$ (or, accounting for the range $[0,2]$, the first digit may be as large as $2$). We need an expansion using digits in $\{0, 2, 4\}$ instead. The idea is to build this expansion greedily, handling carries that arise from the conversion. [/guided] [/step] [step:Construct the digit sequences $(a_k)$ and $(b_k)$ by a greedy algorithm with carries] We construct the digits inductively. Define a sequence of **remainders** $r_0, r_1, r_2, \ldots$ by $r_0 := z$ and, at each stage $k \ge 1$, we select $s_k \in \{0, 2, 4\}$ and set $r_k := r_{k-1} - s_k / 3^k$, ensuring $r_k \in [0, 2 \cdot 3^{-k}]$. At stage $k$, the remainder $r_{k-1}$ satisfies $0 \le r_{k-1} \le 2 \cdot 3^{-(k-1)}$ (verified below). We choose $s_k$ as follows: - If $0 \le r_{k-1} \le \frac{2}{3^k}$, set $s_k := 0$. Then $r_k = r_{k-1} \in [0, 2 \cdot 3^{-k}]$. - If $\frac{2}{3^k} < r_{k-1} \le \frac{4}{3^k}$, set $s_k := 2$. Then $r_k = r_{k-1} - \frac{2}{3^k} \in [0, 2 \cdot 3^{-k}]$. - If $\frac{4}{3^k} < r_{k-1} \le \frac{6}{3^k} = \frac{2}{3^{k-1}}$, set $s_k := 4$. Then $r_k = r_{k-1} - \frac{4}{3^k} \in [0, 2 \cdot 3^{-k}]$. These three cases cover the entire interval $[0, 2 \cdot 3^{-(k-1)}]$ (noting $6/3^k = 2/3^{k-1}$), so the construction always succeeds. The base case $r_0 = z \in [0, 2] = [0, 2 \cdot 3^0]$ holds, and at each step $r_k \in [0, 2 \cdot 3^{-k}]$, so the induction is valid. After all stages, $r_k \to 0$ (since $0 \le r_k \le 2 \cdot 3^{-k}$), and by telescoping \begin{align*} z = r_0 = \sum_{k=1}^{n} \frac{s_k}{3^k} + r_n \to \sum_{k=1}^{\infty} \frac{s_k}{3^k}. \end{align*} Now decompose each $s_k \in \{0, 2, 4\}$: \begin{align*} s_k = 0 &\implies a_k := 0, \; b_k := 0, \\ s_k = 2 &\implies a_k := 0, \; b_k := 2, \\ s_k = 4 &\implies a_k := 2, \; b_k := 2. \end{align*} (The choice for $s_k = 2$ is not unique — $a_k = 2, b_k = 0$ also works — but any consistent choice suffices.) [guided] The greedy algorithm works because the range $[0, 2 \cdot 3^{-(k-1)}]$ of the remainder at stage $k$ is exactly three times the desired range $[0, 2 \cdot 3^{-k}]$ at stage $k+1$, and the three choices $s_k \in \{0, 2, 4\}$ tile this range into three subintervals of length $2 \cdot 3^{-k}$ each. This is precisely the "base-$3$ expansion with digits $\{0, 2, 4\}$" — a non-standard positional system that encodes the arithmetic sum of two standard $\{0, 2\}$ expansions. Why can we always find a decomposition $s_k = a_k + b_k$ with $a_k, b_k \in \{0,2\}$? Because $\{0, 2, 4\} = \{a + b : a, b \in \{0,2\}\}$. The Minkowski sum of $\{0,2\}$ with itself is exactly $\{0, 2, 4\}$, and this is the algebraic core of the entire argument. [/guided] [/step] [step:Verify that $x, y \in \mathcal{C}$ and $x + y = z$] Define $x := \sum_{k=1}^{\infty} a_k / 3^k$ and $y := \sum_{k=1}^{\infty} b_k / 3^k$. Since $a_k \in \{0, 2\}$ and $b_k \in \{0, 2\}$ for all $k$, the [Ternary Characterisation of the Cantor Set](/theorems/1196) gives $x, y \in \mathcal{C}$. Moreover, \begin{align*} x + y = \sum_{k=1}^{\infty} \frac{a_k}{3^k} + \sum_{k=1}^{\infty} \frac{b_k}{3^k} = \sum_{k=1}^{\infty} \frac{a_k + b_k}{3^k} = \sum_{k=1}^{\infty} \frac{s_k}{3^k} = z. \end{align*} Since $z \in [0,2]$ was arbitrary, $[0,2] \subset \mathcal{C} + \mathcal{C}$, completing the proof that $\mathcal{C} + \mathcal{C} = [0,2]$. [/step]