Show that the cantor set contains no isolated points 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!!!
 A: $C_i$ is the union of a finite number of disjoint closed intervals of lengths $\left(\frac{1}{4}\right)^i$.  The trick is to realize that the endpoints of these interval at step $i$ are in $C$ - they never get removed.
Let $\epsilon>0$.  Then choose $n$ so that $\left(\frac{1}{4}\right)^n<\epsilon$.
Assume $x\in C$.  Then $x\in C_n$, so $x$ is in a closed interval of length $\left(\frac{1}{4}\right)^n$ contained in $C_n$.  Pick $y\neq x$ to be one of the endpoints of that interval. Then $|y-x|\leq\left(\frac{1}{4}\right)^n<\epsilon$.  But by the observation above, $y\in C$.
So no point $x\in C$ is isolated.
A: The argument below uses base $4$ expansions. We always use the "non-terminating" expansion when a number has two expansions.
The numbers that were removed are the ones that have a $1$ or a $2$ in their base $4$ expansion.  So what is left is the numbers that only have  $0$'s and/or $3$'s in their base $4$ expansion.  Now given such a number $x$, it is possible to produce an infinite sequence $(x_n)$ of such numbers, all different from $x$, but approaching $x$.  This is done simply by changing "digits" of $x$ ($0$ to $3$ or $3$ to $0$) further and further along in the base $4$ expansion of $x$.
Once one has a solid intuitive grasp of the process, base $4$ expansions can be stripped from the argument, and we end up with a concise elegant proof such as the one given by Thomas Andrews. 
