[proofplan]
We verify that $\Phi$ is a bijection, that it is continuous, and that it is a closed map (equivalently, its inverse is continuous). Bijectivity follows from the uniqueness of $\{0,2\}$-ternary expansions. Continuity of $\Phi$ follows from the fact that agreement in the first $N$ coordinates forces proximity in $\mathbb{R}$. Since $\{0,2\}^{\mathbb{N}}$ is compact (by Tychonoff) and $\mathcal{C}$ is Hausdorff, a continuous bijection from a compact space to a Hausdorff space is automatically a homeomorphism.
[/proofplan]
[step:Verify that $\Phi$ is a well-defined bijection]
By the [Ternary Characterisation of the Cantor Set](/theorems/1196), $\mathcal{C}$ consists precisely of those $x \in [0,1]$ admitting a ternary expansion with all digits in $\{0, 2\}$. Hence $\Phi$ maps into $\mathcal{C}$.
**Surjectivity.** Given $x \in \mathcal{C}$, the [Ternary Characterisation](/theorems/1196) provides a sequence $(a_k)_{k \in \mathbb{N}}$ with $a_k \in \{0, 2\}$ and $x = \sum_{k=1}^{\infty} a_k / 3^k$. Then $\Phi((a_k)) = x$.
**Injectivity.** Suppose $\Phi((a_k)) = \Phi((a_k'))$ for two sequences $(a_k), (a_k') \in \{0,2\}^{\mathbb{N}}$. Then $\sum_{k=1}^{\infty} a_k / 3^k = \sum_{k=1}^{\infty} a_k' / 3^k$. If $(a_k) \ne (a_k')$, let $m$ be the first index with $a_m \ne a_m'$. Then $|a_m - a_m'| = 2$ (the only possibility since $a_m, a_m' \in \{0,2\}$ and $a_m \ne a_m'$), giving
\begin{align*}
\frac{2}{3^m} = \frac{|a_m - a_m'|}{3^m} \le \sum_{k=m+1}^{\infty} \frac{|a_k - a_k'|}{3^k} \le \sum_{k=m+1}^{\infty} \frac{2}{3^k} = \frac{2 \cdot 3^{-(m+1)}}{1 - 1/3} = \frac{1}{3^m}.
\end{align*}
This gives $2/3^m \le 1/3^m$, a contradiction. Hence $(a_k) = (a_k')$ and $\Phi$ is injective.
[guided]
The injectivity argument exploits the fact that in base $3$, the "carry capacity" of all digits beyond position $m$ is strictly less than a single unit at position $m$. Concretely, even if all subsequent digits differ maximally ($|a_k - a_k'| = 2$ for every $k > m$), the total discrepancy they can produce is $\sum_{k=m+1}^{\infty} 2/3^k = 3^{-m}$, which is strictly less than the discrepancy $2/3^m$ at position $m$. This is the same principle that makes positional numeral systems work: a digit difference at position $m$ cannot be compensated by digits at later positions.
[/guided]
[/step]
[step:Equip $\{0,2\}^{\mathbb{N}}$ with the product topology and verify compactness]
Give $\{0, 2\}$ the discrete topology (as a finite set with two elements). The product space $\{0,2\}^{\mathbb{N}} = \prod_{k=1}^{\infty} \{0,2\}$ carries the product topology, in which a basis of open sets consists of all **cylinder sets**
\begin{align*}
\mathcal{U}_{N; d_1, \ldots, d_N} := \{(a_k) \in \{0,2\}^{\mathbb{N}} : a_k = d_k \text{ for } 1 \le k \le N\}
\end{align*}
for $N \in \mathbb{N}$ and $d_1, \ldots, d_N \in \{0, 2\}$.
Each factor $\{0, 2\}$ is compact (finite discrete spaces are compact). By the Tychonoff theorem, the product $\{0,2\}^{\mathbb{N}}$ is compact.
[/step]
[step:Prove that $\Phi$ is continuous]
We show that the preimage of every open set in $\mathcal{C}$ (under the subspace topology from $\mathbb{R}$) is open in $\{0,2\}^{\mathbb{N}}$. It suffices to show that for every $(a_k) \in \{0,2\}^{\mathbb{N}}$ and every $\varepsilon > 0$, there exists $N \in \mathbb{N}$ such that the cylinder set $\mathcal{U}_{N; a_1, \ldots, a_N}$ maps into $B(\Phi((a_k)), \varepsilon) \cap \mathcal{C}$.
Let $(a_k') \in \mathcal{U}_{N; a_1, \ldots, a_N}$, so $a_k' = a_k$ for $1 \le k \le N$. Then
\begin{align*}
|\Phi((a_k)) - \Phi((a_k'))| = \left|\sum_{k=N+1}^{\infty} \frac{a_k - a_k'}{3^k}\right| \le \sum_{k=N+1}^{\infty} \frac{2}{3^k} = \frac{2 \cdot 3^{-(N+1)}}{1 - 1/3} = \frac{1}{3^N}.
\end{align*}
Choosing $N$ so that $3^{-N} < \varepsilon$ (any $N > \log(1/\varepsilon)/\log 3$ suffices), the entire cylinder maps into the $\varepsilon$-ball around $\Phi((a_k))$. Since cylinder sets form a basis, $\Phi$ is continuous.
[guided]
The estimate shows that agreement in the first $N$ coordinates forces the images to be within $3^{-N}$ of each other. This is the hallmark of continuity for maps defined by convergent series: the tail of the series can be made uniformly small by fixing sufficiently many initial terms. The key computation is the geometric series bound on the tail, which gives $3^{-N}$ independently of the particular digits $a_k'$ for $k > N$.
[/guided]
[/step]
[step:Conclude that $\Phi$ is a homeomorphism]
We have established that $\Phi: \{0,2\}^{\mathbb{N}} \to \mathcal{C}$ is a continuous bijection. The domain $\{0,2\}^{\mathbb{N}}$ is compact (by Tychonoff, as shown above). The codomain $\mathcal{C}$ is Hausdorff, being a subspace of $\mathbb{R}$.
A continuous bijection from a compact space to a Hausdorff space is a homeomorphism: the image of every closed set is closed (closed subsets of a compact space are compact, continuous images of compact sets are compact, and compact subsets of a Hausdorff space are closed), so $\Phi$ is a closed map, and hence $\Phi^{-1}$ is continuous. Therefore $\Phi$ is a homeomorphism.
[/step]
