[proofplan] We construct an explicit surjection from $\mathcal{C}$ onto $[0,1]$, which establishes $|\mathcal{C}| \ge |[0,1]| = 2^{\aleph_0}$. Since $\mathcal{C} \subset [0,1]$, we also have $|\mathcal{C}| \le 2^{\aleph_0}$, giving equality. The surjection is the Cantor function restricted to $\mathcal{C}$, which maps each ternary expansion in $\{0,2\}^{\mathbb{N}}$ to a binary expansion in $\{0,1\}^{\mathbb{N}}$. [/proofplan]

[step:Construct a surjection from $\mathcal{C}$ onto $[0,1]$ via digit substitution] By the [Ternary Characterisation of the Cantor Set](/theorems/1196), every $x \in \mathcal{C}$ has a ternary expansion $x = \sum_{k=1}^{\infty} a_k / 3^k$ with $a_k \in \{0, 2\}$ for all $k \in \mathbb{N}$. Define \begin{align*} \psi: \mathcal{C} &\to [0,1] \\ x = \sum_{k=1}^{\infty} \frac{a_k}{3^k} &\mapsto \sum_{k=1}^{\infty} \frac{a_k/2}{2^k}. \end{align*} Since $a_k/2 \in \{0, 1\}$, the image $\psi(x) = \sum_{k=1}^{\infty} b_k / 2^k$ with $b_k := a_k/2 \in \{0,1\}$ is a binary expansion of a number in $[0,1]$.[/step]

[guided]The map $\psi$ is the restriction of the Cantor function $\varphi$ to $\mathcal{C}$, but we do not need any properties of $\varphi$ beyond this explicit formula. The idea is to exploit the bijection between $\{0,2\}$ and $\{0,1\}$: dividing each ternary digit by $2$ converts a base-$3$ expansion using digits $\{0,2\}$ into a base-$2$ expansion using digits $\{0,1\}$. We must verify that $\psi$ is well-defined. A point $x \in \mathcal{C}$ can have at most two ternary expansions using digits in $\{0,2\}$ — but the only ambiguity in ternary expansions is between a terminating expansion and one ending in repeating $2$'s. For instance, $1/3 = 0.0222\ldots_3 = 0.1000\ldots_3$, but the second representation uses the digit $1$ and is therefore not available in $\{0,2\}^{\mathbb{N}}$. Hence each $x \in \mathcal{C}$ has a **unique** ternary representation using only digits $\{0,2\}$, and $\psi$ is well-defined.[/guided]

[step:Verify that $\psi$ is surjective] Let $y \in [0,1]$. Then $y$ has a binary expansion $y = \sum_{k=1}^{\infty} b_k / 2^k$ with $b_k \in \{0, 1\}$ (choosing the expansion ending in repeating $1$'s when $y$ has two binary representations). Set $a_k := 2 b_k \in \{0, 2\}$ for each $k$, and define $x := \sum_{k=1}^{\infty} a_k / 3^k$. Since all digits $a_k$ lie in $\{0,2\}$, the [Ternary Characterisation](/theorems/1196) gives $x \in \mathcal{C}$. By construction, \begin{align*} \psi(x) = \sum_{k=1}^{\infty} \frac{a_k/2}{2^k} = \sum_{k=1}^{\infty} \frac{b_k}{2^k} = y. \end{align*} Hence $\psi$ is surjective. [/step]

[step:Conclude $|\mathcal{C}| = 2^{\aleph_0}$] The surjection $\psi: \mathcal{C} \to [0,1]$ gives $|\mathcal{C}| \ge |[0,1]|$. Since $\mathcal{C} \subset [0,1]$, the inclusion gives $|\mathcal{C}| \le |[0,1]|$. By the Cantor-Bernstein theorem, $|\mathcal{C}| = |[0,1]| = |\mathbb{R}| = 2^{\aleph_0}$. In particular, $\mathcal{C}$ is uncountable. [/step]