It is well known that Cantor set $C$ contains no intervals, hence for every $p,q\in C$ such that (without loss of generality) $p<q$, there exists $r\in \mathbb{R}\setminus C$ such that $p<r<q$. You can always find a point between two different points in $C$ that is not in $C$, since $C$ is disconnected.
Thus for each point $p\in C$, there exits and open interval $(a,b)\subset\mathbb{R}$ such that $C\cap (a,b)=\{p\}$. So every singleton in $C$ is an open set.
It easily follows that $C$ must be a discrete topology.
But I keep hearing that $C$ is NOT a discrete topology, because it is homeomorphic to an infinite product space of discrete spaces consisting of $\{0,1\}$. My argument would be that they are BOTH discrete. Am I wrong?
Best Answer
This is not true: For every $p$, $\exists\ a,b$ such that $C \cap (a,b) = \{p\}$.
For example, $\frac 14 \in C$. In fact, there exists a non-constant sequence of endpoints in $C$ that converges to $\frac 14$. That sequence is: $$a_n = \sum_{k=1}^n \frac{2}{3^{2k}} \quad \implies \quad a_n \to \frac{1}{4}$$
This shows that $C$ does not have the discrete topology, because it has a limit point.