The Cantor set is exactly the set of $\sum_{k \ge 1} a_k 3^{-k}$ with every $a_k \in \{0,2\}$, and this gives a bijection with $\{0,1\}^{\mathbb{N}}$

$\Phi$ is well defined and takes values in $[0,1]$. For $a \in D$ every term $a_k 3^{-k-1}$ is $\ge 0$ by [L1] and [L9], and for every $n$ the partial sum satisfies $\sum_{k<n} a_k 3^{-k-1} \le \sum_{k<n} 2 \cdot 3^{-1} \cdot 3^{-k} = 2 \cdot 3^{-1} \sum_{k<n} 3^{-k} \le 2 \cdot 3^{-1} \cdot 3 \cdot 2^{-1} = 1$, by [L3], [L4] and [L9]. So by [L3] the series converges, its sum $\Phi(a)$ satisfies $0 \le \Phi(a) \le 1$, and $\Phi(a) \in [0,1]$ by [L1].

Shift identity: $\Phi(a) = a_0 \cdot 3^{-1} + 3^{-1}\Phi(\sigma a)$ for every $a \in D$. Indeed by [L2] the partial sums satisfy $\sum_{k<n+1} a_k 3^{-k-1} = a_0 3^{-1} + \sum_{j<n} a_{j+1} 3^{-j-2} = a_0 3^{-1} + 3^{-1}\sum_{j<n} a_{j+1}3^{-j-1}$, using $3^{-j-2} = 3^{-1}\cdot 3^{-j-1}$ from [L1] and [L9]; letting $n$ grow and using [L5] and [L2] gives the identity.

Self-similarity of $C$, claim 4. If $y \in C$ then $y \in C_n$ for every $n$, so $y \cdot 3^{-1} \in \tfrac13 C_n \subseteq C_{n+1}$ and $2 \cdot 3^{-1} + y \cdot 3^{-1} \in \tfrac23 + \tfrac13 C_n \subseteq C_{n+1}$ for every $n$, whence both lie in $\bigcap_n C_{n+1} = C$ by [L1]; this gives the inclusion $\supseteq$. Conversely let $x \in C$, so $x \in C_{n+1}$ for every $n$. By [L1] the first half of $C_{n+1}$ lies in $[0,\tfrac13]$ and the second in $[\tfrac23,1]$, and $\tfrac13 < \tfrac23$ by [L9]. If $x \le \tfrac13$ then $x \notin [\tfrac23,1]$, so for every $n$ one has $x \in \tfrac13 C_n$, that is $3x \in C_n$; hence $3x \in C$ and $x \in \tfrac13 C$. If $x > \tfrac13$ then $x \notin [0,\tfrac13]$, so for every $n$ one has $x \in \tfrac23 + \tfrac13 C_n$, that is $3x - 2 \in C_n$; hence $3x - 2 \in C$ and $x \in \tfrac23 + \tfrac13 C$. Disjointness is [L1] and [L9], since $\tfrac13 C \subseteq [0,\tfrac13]$ and $\tfrac23 + \tfrac13 C \subseteq [\tfrac23,1]$.

$\Phi(a) \in C$ for every $a \in D$. By induction on $n$ ([L6]) the statement "for every $a \in D$, $\Phi(a) \in C_n$" holds for every $n$: at $n = 0$ it is step 1.1 and [L1]; and if it holds at $n$, then for $a \in D$ the value $a_0$ is $0$ or $2$, so step 1.2 gives $\Phi(a) = 3^{-1}\Phi(\sigma a) \in \tfrac13 C_n$ in the first case and $\Phi(a) = 2 \cdot 3^{-1} + 3^{-1}\Phi(\sigma a) \in \tfrac23 + \tfrac13 C_n$ in the second, so $\Phi(a) \in C_{n+1}$ by [L1]. Hence $\Phi(a) \in \bigcap_n C_n = C$.

The digit recursion. Fix $x \in C$ and let $T : \mathbb{R} \to \mathbb{R}$ be $T(y) := 3y$ for $y \le 3^{-1}$ and $T(y) := 3y - 2$ for $y > 3^{-1}$, a definition by cases on the total order ([L9]) and so a genuine function. By [L6] there is $y : \mathbb{N} \to \mathbb{R}$ with $y_0 = x$ and $y_{n+1} = T(y_n)$; put $a_n := 0$ when $y_n \le 3^{-1}$ and $a_n := 2$ otherwise, so that $a \in D$ and $y_{n+1} = 3 y_n - a_n$ for every $n$. Every $y_n$ lies in $C$, by induction on $n$: $y_0 = x \in C$; and if $y_n \in C$ then, by step 1.3, either $y_n \in \tfrac13 C \subseteq [0,\tfrac13]$ or $y_n \in \tfrac23 + \tfrac13 C \subseteq [\tfrac23,1]$, and these two cases are exactly $y_n \le \tfrac13$ and $y_n > \tfrac13$ by [L9]; in the first $y_n = z \cdot 3^{-1}$ with $z \in C$ and $y_{n+1} = 3y_n = z \in C$, in the second $y_n = 2 \cdot 3^{-1} + z \cdot 3^{-1}$ with $z \in C$ and $y_{n+1} = 3y_n - 2 = z \in C$.

$\Phi$ is injective. Let $a, b \in D$ with $a \ne b$; the set of $k$ with $a_k \ne b_k$ is a nonempty subset of $\mathbb{N}$, so by [L7] it has a least element $k$, and by symmetry we may take $a_k = 0$ and $b_k = 2$. By [L5], $\Phi(b) - \Phi(a) = \sum_{j \ge 0}(b_j - a_j)3^{-j-1}$, and the terms with $j < k$ vanish, so by [L2] this equals $2 \cdot 3^{-k-1} + R$ with $R := \sum_{j \ge k+1}(b_j - a_j)3^{-j-1}$. Every $b_j - a_j$ is at least $-2$, so the series $\sum_{j \ge k+1}\big((b_j - a_j) + 2\big)3^{-j-1}$ has nonnegative terms and hence nonnegative sum by [L3], giving $R \ge -\sum_{j \ge k+1} 2 \cdot 3^{-j-1} = -2 \cdot 3^{-k-2} \cdot 3 \cdot 2^{-1} = -3^{-k-1}$ by [L2], [L4], [L5] and [L9]. Therefore $\Phi(b) - \Phi(a) \ge 2 \cdot 3^{-k-1} - 3^{-k-1} = 3^{-k-1} > 0$ and $\Phi(a) \ne \Phi(b)$.

The value is recovered from the digits. With $x$, $(y_n)$ and $a$ as in step 2.2, put $s_n := \sum_{k<n} a_k 3^{-k-1}$. Then $x = s_n + 3^{-n} y_n$ for every $n$, by induction on $n$ ([L6]): at $n = 0$ both sides are $x$, since $s_0 = 0$ by [L2] and $3^{0} = 1$; and if $x = s_n + 3^{-n}y_n$ then $s_{n+1} + 3^{-n-1}y_{n+1} = s_n + a_n 3^{-n-1} + 3^{-n-1}(3y_n - a_n) = s_n + 3^{-n}y_n = x$, using [L1], [L2] and [L9].

Hence $x = \Phi(a)$, so $C \subseteq \Phi[D]$. Every $y_n$ lies in $C \subseteq [0,1]$ by step 2.2 and [L1], so $0 \le x - s_n = 3^{-n}y_n \le 3^{-n}$ by step 3.1 and [L9]. Given a rational $\varepsilon > 0$, [L8] supplies $N$ with $3^{-n} < \varepsilon$ for all $n \ge N$, and then $|s_n - x| = x - s_n \le 3^{-n} < \varepsilon$ by [L8]; so $s_n \to x$. But $s_n \to \Phi(a)$ by [L2], since $(s_n)$ is the sequence of partial sums of the series defining $\Phi(a)$, and limits are unique by [L8]; therefore $x = \Phi(a)$ with $a \in D$.

By steps 2.1 and 4.1 the image of $D$ under $\Phi$ is exactly $C$, which with step 1.1 is claim 1; step 2.3 is claim 2, so $\Phi$ is a surjection from $D$ onto $C$ that is injective, that is, a bijection (Injection, surjection, bijection); the map $b \mapsto (2b_k)_k$ is a bijection from $\{0,1\}^{\mathbb{N}}$ onto $D$, with inverse $a \mapsto (a_k \cdot 2^{-1})_k$ by [L9], and a composition of bijections is a bijection, which is claim 3; and step 1.3 is claim 4.