The set $A_n$ consists of closed intervals of length $1/3^n$ and the distance between them at least $1/3^n$ . Indeed, this is true for $A_1$ . At every step a closed interval of $1/3^n$ is divided into three parts with two parts in $A_{n+1}$ such that the distance between them is $1/3^{n+1}$ .(a) For any two points $x\neq y$ there is $n$ such that they cannot lie in the same closed interval of $A_n$ , and, therefore, there is a point $z$ between them not in the Cantor set, and the set can be separated by $(-\infty,z)$ and $(z,+\infty)$ .(b) It’s a closed subspace of a compact space.(c) We have already shown by induction that it is the union of $2^n$ closed intervals each of length $1/3^n$ . By the same argument endpoints of the intervals are never excluded (only interior points are excluded).(d) Consider a sequence of closed intervals containing a given point of $C$ , for each one choose an endpoint which does not equal the point, and all the endpoints are in $C$ , by (c), and the sequence converges to the point. Therefore, it is a limit point of the set of all other points in $C$ .(e) It is nonempty (by (c), for example, or by the fact that it is the intersection of a nested sequence of closed intervals), compact and Hausdorff with no isolated points. By Theorem 27.7, it is uncountable.