The topology of Cantor set? How to prove $f:\{0,2\}^{\mathbb{N}}\rightarrow C$ the Cantor set, is a homeomorphism?

Question

$P=\{0,2\}^{\mathbb{N}}$ and $C$ is Cantor sets. Then $$f(\{a_n\})=\sum_{n=1}^\infty\frac{a_n}{3^n}$$ with $a_n\in\{0,2\}$ is a bijection from $P$ to $C$.

My first question: If I choose the discrete topology of $\{0,2\}$ and choose the product topology in $P$, does $P$ also has the discrete topology?

I think the answer is yes, because every point in $P$ as a set is an open set in $P$, then $P$ must has the discrete topology. Am I right?

My second question: If I choose $C$ as subspace topology of $\mathbb{R}$, then must the topology of $C$ not the discrete topology?

I also think it's correct and $C$ is not discrete topology. Because every point in $C$ can't be written as an intersection of open interval and $C$.

My third question: But I heard that $f$ is a homeomorphism of $P$ and $C$. How is it possilbe to have a homeomorphism from a discrete topology to a non-discrete topology? If this saying is right, how to prove the homeomorphism?

Answer 1

The product topology will not be the discrete topology. The product topology is generated by sets of the form $$ \prod_{j=1}^{\infty} U_j $$ where each set $U_j \in \{ \varnothing, \{0\}, \{2\}, \{0,2\}\}$, and only finitely many of the $U_j$ are singletons or empty. This second condition gives us something different from the discrete topology. Note that the box topology on $\{0,2\}^{\mathbb{N}}$ will give the discrete topology, but the box topology is not the product topology.

Answer 2

The product topology on $P = \{0, 2\}^{\Bbb{N}}$ induced from the discrete topology on $\{0, 2\}$ is not the discrete topology. The points in $P$ are not open. The product topology on $P$ is the smallest topology that makes each projection $\pi_i : P \to \{0, 2\}$ continuous. Each open set $X$ in the product topology is a union of sets such that $\pi_i[X] = \{0, 2\}$ for all but finitely many $i$ (so $X$ is defined by a disjunction of finite set of conditions on the values of the $\pi_i$). If this topology were discrete any singleton set would be open, but it takes an infinite set of conditions on the values of the $\pi_i$ to define a singleton set (and forming unions doesn't help). So the answer to your first question is no and you need to rethink your second and third questions.

Answer 3

To answer the first question the set $\{0,2\}$ is compact in the discrete topology and thus from Tychonov's theorem the set $X=\{0,2\}^{\mathbb{N}}$ is compact with respect to the product topology.

So $X$ cannot be a discrete topological space because a discrete topological space $(X,\mathcal{T})$ is compact if and only if the $X$ is finite.

For the second question ,the Cantor set is a compact space with the subspace topology of the real line.

So it cannot be a discrete space because it is infinite.