The Cantor set is homeomorphic to $\{0,1\}^{\mathbb{N}}$ with the product of discrete topologies, the ternary digits being the coordinates

For every $b \in K$ the terms $2b_j 3^{-(j+1)}$ lie between $0$ and $2 \cdot 3^{-(j+1)}$, and $\sum_j 2 \cdot 3^{-(j+1)}$ converges with sum $1$, since it is $2 \cdot 3^{-1}\sum_j 3^{-j} = (2/3)(3/2)$ by [L1] and [L3]. So $\Phi(b)$ is defined and lies in $[0,1]$.

$3^{-k} \to 0$, since $\sum_k 3^{-k}$ converges by [L1]; hence for every real $\varepsilon > 0$ there is $n \in \mathbb{N}$ with $3^{-n} < \varepsilon$, a positive rational below $\varepsilon$ serving as the tolerance.

Each cylinder $Z(b,n)$ is open in $K$: it is the box with factor $\{b_j\}$ at each $j<n$ and $D$ elsewhere, and every subset of $D$ is open. Moreover $Z(c,n) = Z(b,n)$ whenever $c \in Z(b,n)$, the defining condition being agreement of the first $n$ coordinates.

The cylinders form a basis of $K$: given a basic box $\prod_j U_j$ with $U_j = D$ off a list $j_0,\dots,j_{r-1}$ and a point $b$ in it, put $n := 1 + j_{m_0}$ where $j_{m_0}$ is the largest of the listed indices, available by [L5] for $r \ge 1$, and $n := 0$ for $r = 0$; then $b \in Z(b,n) \subseteq \prod_j U_j$, since every listed index is $< n$.

$\Phi(b) = \Phi_0\big((2b_k)_k\big)$ for every $b \in K$, the two series having the same terms $2b_k 3^{-k-1}$; so by [A1] the map $\Phi$ is a bijection of $K$ onto $C$. This is claim 1.

For every $n \in \mathbb{N}$: $\sum_{j \ge n} 2 \cdot 3^{-(j+1)} = 3^{-n}$, by shifting the index and applying [L1] and [L3] as in step 1.1.

Suppose $b_j = c_j$ for every $j < n$. For $N \ge n$ the finite sum $\sum_{j<N} 2(b_j - c_j)3^{-(j+1)}$ has vanishing terms for $j < n$, and each remaining term lies between $-2 \cdot 3^{-(j+1)}$ and $2 \cdot 3^{-(j+1)}$, so by [L2] the finite sum lies between $-3^{-n}$ and $3^{-n}$, using step 2.2 and [L1]. Letting $N$ grow and applying [L1] and [L2] gives $|\Phi(b) - \Phi(c)| \le 3^{-n}$.

Suppose $b_j = c_j$ for every $j < m$ and $b_m \ne c_m$; interchanging $b$ and $c$ if necessary, take $b_m = 1$ and $c_m = 0$. For $N > m$ the finite sum $\sum_{j<N} 2(b_j-c_j)3^{-(j+1)}$ equals $2 \cdot 3^{-(m+1)}$ plus a term bounded below by $-\sum_{j \ge m+1} 2 \cdot 3^{-(j+1)} = -3^{-(m+1)}$, by step 2.2, [L1] and [L2]; so it is at least $3^{-(m+1)}$. Letting $N$ grow gives $\Phi(b) - \Phi(c) \ge 3^{-(m+1)}$, hence $|\Phi(b)-\Phi(c)| \ge 3^{-(m+1)}$.

Steps 3.1 and 3.2 are claim 2.

$\Phi$ is continuous: let $V$ be open in $C$ and $b \in \Phi^{-1}[V]$; by [A3] there is $\varepsilon > 0$ with $B_C(\Phi(b),\varepsilon) \subseteq V$, by step 1.2 there is $n$ with $3^{-n} < \varepsilon$, and by step 3.1 every $c \in Z(b,n)$ has $|\Phi(c)-\Phi(b)| \le 3^{-n} < \varepsilon$, so $Z(b,n) \subseteq \Phi^{-1}[V]$; and $Z(b,n)$ is open by step 1.3.

For $b \in K$ and $n \in \mathbb{N}$: $B_C(\Phi(b), 3^{-n}) \subseteq \Phi[Z(b,n)]$. Indeed such a point is $\Phi(c)$ for a unique $c \in K$ by step 2.1; if $c \notin Z(b,n)$, let $m$ be the least index with $b_m \ne c_m$, which exists by [L5] and satisfies $m < n$, and then step 3.2 gives $|\Phi(c)-\Phi(b)| \ge 3^{-(m+1)} \ge 3^{-n}$ by [L3], contradicting the choice of $\Phi(c)$.

$\Phi$ is an open map onto $C$: by step 1.3 the set $\Phi[Z(b,n)]$ contains, around each of its points $\Phi(c)$ with $c \in Z(b,n)$, the ball $B_C(\Phi(c),3^{-n}) \subseteq \Phi[Z(c,n)] = \Phi[Z(b,n)]$ by step 4.3; so each $\Phi[Z(b,n)]$ is open in $C$ by [A3], and by step 1.4 every open subset of $K$ is a union of cylinders, whose image is the union of their images.

By step 2.1 the map $\Phi$ is a bijection onto $C$, by step 4.2 it is continuous and by step 5.1 it is open, so it is a homeomorphism by [L6]. This is claim 3, and with steps 2.1 and 4.1 all three claims are proved.