Say we take the set of infinite binary codes $\{0,1\}^\mathbb{N}$, which is often written as $2^\mathbb{N}$, mapped to the Cantor set defined previously as $C_n=\frac{c_{n-1}}{3} \cup \left( \frac{2}{3} + \frac{C_{n-1}}{3}\right)$ or it can be described as the set of real numbers in $[0,1]$ having ternary expansions using only the digits $0$ and $2$, i.e., real numbers of the form $$\sum_{n=1}^\infty \frac{a_n}{3^n},$$ where each $a_n$ is either $0$ or $2$.
For each positive integer $n$ let $D_n = \{0,1\}$ with the discrete topology, and let $$X = \prod_{n=1}^\infty D_n$$ with the product topology. \begin{align*} &f: 2^\mathbb{N} \to C\space \mbox{defined as for a sequence}\space a=(a_i)_{i=0}^\infty. \\ &\Rightarrow f(a) = \sum_{i=0}^\infty \frac{2a_i}{3^{i+1}}. \end{align*}
A topological space is a Cantor space if and only if it is non-empty, perfect, compact, totally disconnected, and metrizable.
Then $f$ is a homeomorphism of metric spaces, meaning that all the properties in the above definition hold for the space $2^\mathbb{N}$ and thus this is a Cantor space.I am assuming $C$ is written in ternary and thus that is why the summation is as shown in the first summation. Now how would one go by proving something like closed/compact/disconnected/perfect? I am usually given a set and told to prove these properties but I do not truly understand where to start with just a homeomorphism. I want to prove that $f(a) = \sum_{i=0}^\infty \frac{2a_i}{3^{i+1}}$ would have those properties.