[proofplan] We prove both directions using the iterative construction of $\mathcal{C} = \bigcap_{n=0}^{\infty} C_n$. For the forward direction, we show that membership in $C_n$ forces the $n$-th ternary digit to lie in $\{0,2\}$, so membership in the full intersection forces all digits into $\{0,2\}$. For the converse, we show that any $x$ with all ternary digits in $\{0,2\}$ belongs to every $C_n$ and hence to $\mathcal{C}$. [/proofplan] [step:Show that $x \in C_n$ forces the first $n$ ternary digits into $\{0, 2\}$] We prove by induction on $n$ that if $x \in C_n$, then $x$ admits a ternary expansion $x = \sum_{k=1}^{\infty} a_k / 3^k$ with $a_k \in \{0, 2\}$ for all $1 \le k \le n$. **Base case.** $C_0 = [0,1]$, and every $x \in [0,1]$ has a ternary expansion, so the condition on zero digits is vacuous. **Inductive step.** Suppose the claim holds for $C_n$. The set $C_n$ consists of $2^n$ closed intervals, each of length $3^{-n}$. A generic constituent interval of $C_n$ has the form \begin{align*} I_{a_1, \ldots, a_n} := \left\{ \sum_{k=1}^{\infty} \frac{c_k}{3^k} : c_k = a_k \text{ for } 1 \le k \le n, \; c_k \in \{0,1,2\} \text{ for } k > n \right\} \end{align*} where each $a_k \in \{0,2\}$. This interval has left endpoint $\sum_{k=1}^{n} a_k / 3^k$ and right endpoint $\sum_{k=1}^{n} a_k / 3^k + 3^{-n}$. Passing from $C_n$ to $C_{n+1}$ removes the open middle third of each such interval. In terms of ternary digits, the left third of $I_{a_1, \ldots, a_n}$ consists of points with $c_{n+1} = 0$, the middle third consists of points with $c_{n+1} = 1$ (excluding endpoints that have alternative representations), and the right third consists of points with $c_{n+1} = 2$. Removing the open middle third retains exactly those points with $c_{n+1} \in \{0, 2\}$ (the endpoints $c_{n+1} = 1$ with $c_k = 0$ for all $k > n+1$, or $c_k = 2$ for all $k > n+1$, have alternative ternary representations with $c_{n+1} \in \{0, 2\}$). Hence $x \in C_{n+1}$ implies $a_k \in \{0,2\}$ for $1 \le k \le n+1$. [guided] The key idea is to track what happens to ternary digits at each removal step. Every real number $x \in [0,1]$ can be written in base $3$ as $x = \sum_{k=1}^{\infty} a_k / 3^k$ with $a_k \in \{0,1,2\}$ (with the standard convention that $1 = 0.222\ldots_3$). At stage $1$, we remove the open middle third $(1/3, 2/3)$. In base $3$, the interval $(1/3, 2/3)$ consists of numbers of the form $0.1 c_2 c_3 \ldots_3$ where the representation with $a_1 = 1$ is the unique one — except at the endpoints. The endpoint $1/3 = 0.1000\ldots_3$ has the alternative representation $0.0222\ldots_3$, and $2/3 = 0.2000\ldots_3$. Both endpoints belong to $C_1$ and have representations with $a_1 \in \{0,2\}$. So membership in $C_1$ means: $x$ has a ternary expansion with $a_1 \in \{0,2\}$. The inductive step repeats this reasoning within each constituent interval of $C_n$. The interval $I_{a_1, \ldots, a_n}$ is partitioned by the $(n+1)$-th digit: the left third has $c_{n+1} = 0$, the middle third has $c_{n+1} = 1$, the right third has $c_{n+1} = 2$. Removing the open middle third eliminates all interior points with $c_{n+1} = 1$, and the boundary points of the middle third belong to adjacent thirds under their alternative ternary representations. [/guided] [/step] [step:Conclude the forward direction: $x \in \mathcal{C}$ implies all digits in $\{0, 2\}$] If $x \in \mathcal{C} = \bigcap_{n=0}^{\infty} C_n$, then $x \in C_n$ for every $n \in \mathbb{N}$. By the inductive result of the previous step, $x$ has a ternary expansion with $a_k \in \{0,2\}$ for all $1 \le k \le n$. Since this holds for every $n$, we obtain $a_k \in \{0,2\}$ for all $k \in \mathbb{N}$. [/step] [step:Prove the converse: all digits in $\{0, 2\}$ implies $x \in \mathcal{C}$] Suppose $x = \sum_{k=1}^{\infty} a_k / 3^k$ with $a_k \in \{0,2\}$ for all $k \in \mathbb{N}$. We show $x \in C_n$ for every $n \ge 0$. The set $C_n$ is the union of the $2^n$ intervals $I_{d_1, \ldots, d_n}$ over all choices of $(d_1, \ldots, d_n) \in \{0,2\}^n$. Taking $(d_1, \ldots, d_n) = (a_1, \ldots, a_n)$, we have \begin{align*} x = \sum_{k=1}^{n} \frac{a_k}{3^k} + \sum_{k=n+1}^{\infty} \frac{a_k}{3^k}. \end{align*} The tail satisfies $0 \le \sum_{k=n+1}^{\infty} a_k / 3^k \le \sum_{k=n+1}^{\infty} 2/3^k = 3^{-n}$, so $x$ lies in the interval $[\sum_{k=1}^{n} a_k/3^k, \; \sum_{k=1}^{n} a_k/3^k + 3^{-n}] = I_{a_1, \ldots, a_n} \subset C_n$. Since $x \in C_n$ for every $n$, we conclude $x \in \bigcap_{n=0}^{\infty} C_n = \mathcal{C}$. [/step]