Why isn't the Cantor set a discrete topology? 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?
 A: 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.
A: Here's the confusing thing about infinite products: The product topology on an $\prod A_i$ has as a subbasis (a collection of sets from which one can obtain any open set from finite intersection and arbitrary union) things of the form $\prod U_i$, where $U_i$ is open in $A_i$, and for all but one $i$, $U_i=A_i$.
So the topology is not discrete, as any open set containing the origin contains every point that agrees with the origin except on some finite set of coordinates.
The box topology is obtained by allowing $U_i$ to be an arbitrary open set for every $A_i$. This gives much finer topologies (stronger to analysts, weaker to topologists — because it's harder for sequences to converge, but easier for sets to be open).
For example, the function $f: \Bbb R\to\Bbb R^\infty$ defined by $f(x) = (x,x,x,\ldots)$ is continuous if $\Bbb R^\infty$ is given the product topology, but is horrendously discontinuous if it is given the box topology, as in the box topology, the set $\prod\limits_i(-\frac1i,\frac1i)$ is an open neighborhood of $f(0)$, but its preimage is just $\{0\}$.
