Consider an infinite sequence of subsets of the interval $[0,1]$ obtained in the following way; set $C_{0}=[0,1]$. $C_{1}$ is obtained by removing the middle open half of $C_{0}$, that is
$C_{1}= C_{0} -(1/4,3/4)=[0,1/4] \cup [3/4,1]$
C2 is obtained by removing the middle open half of the interval of C1, that is ,C2=C1-(1/16,3/16)-(13/16,15,16). The set $C=\cap C_{n}$ for nϵ{0,1,2…∞} is called cantor set. Consider C as a subspace of [0,1]. Show that C is closed, compact and contains no isolated points.
I think for $C$ is closed because $C$ is an intersection of closed sets $C_n$ , looking $[0,1]$ is a closed interval. but any closed subset of $\mathbb R$ is compact, hence compactness of $C$ comes because it is a closed subset of $[0,1]$. I have no idea how to show that $C$ contains no isolated point! How may I get to this? Thank you!!!