Cantor set is homeomorphic to the space of binary sequences

Statement

Let $\{0, 1\}$ have the discrete topology and let $\{0, 1\}^{\mathbb{N}}$ have the product topology. Then the map $$\varphi \colon \{0, 1\}^{\mathbb{N}} \to C, \qquad \varphi((a_n)_{n \ge 1}) = \sum_{n \ge 1} \frac{2 a_n}{3^n}$$ is a homeomorphism onto the Cantor set $C$.

Remarks

The map is a bijection, because each point of $C$ has exactly one base-3 expansion with digits in $\{0, 2\}$. It is continuous: if two sequences agree in their first $N$ terms, their images differ by at most $3^{-N}$. The space $\{0, 1\}^{\mathbb{N}}$ is compact by Tychonoff's theorem, and $C$ is Hausdorff. Thus $\varphi$ is a homeomorphism, by this theorem. A space homeomorphic to $C$ is a Cantor space. By a theorem of Brouwer, these are exactly the nonempty compact metrizable spaces that are totally disconnected and have no isolated points.

Explanations

L1 human

Each point of the Cantor set is described by an infinite list of choices: left or right third, at each step. These lists are the binary sequences, and the topology of the Cantor set matches the product topology on them.