prove that the ternary cantor set is compact and a perfect set.
My trial:
I know that I should prove that it is closed and bounded, for proving that it is closed because finite union of closed sets (the cantor set definition is this ) are closed and also using that arbitrary intersection of closed sets are closed.
Now to prove that it is bounded shall I prove that its lenght is zero?
How can I prove that it has no isolated points?
Let $C_n=\frac{C_{n-1}}{3}\cup \left(\frac{2}{3}+\frac{C_{n-1}}{3}\right)$ with $C_0=[0,1]$.
$C_n$ is closed
$C_n$ is contained in $[0,1]$
Every interval in $C_n$ is at maximum $(\frac{1}{3})^n$ in length
Cantor set is defined as:
$C:=\cap_{n=0}^{\infty}C_n$
By definition it is clear that
it is bounded (is contained in $[0,1]$)
it is closed (is the infinite intersection of closed sets)
Every neighborhood of a point in $C$ must contain another point of $C$at least: you can prove it using the third property of $C_n$
Note that this property is noteworthy, since the set is totally disconnected
Rigorous proof: By simple computations, you can note that the Cantor set consists of all the real numbers of the unit interval that do not require 1 in their ternary expansion. Thus, for every $c\in C=0.c_1c_2\dots$, there exists, for all $n$, a number $c_n \in C$ such that $|c-c_n|<3^{-n}$, constructed in this way: truncate $c$ to the $n+1$-th digit, and substitute it with $2$ if it is $0$ and viceversa. The number so constructed is still in $C$, and has a distance from $c$ of $\frac{2}{3}\cdot 3^{-n}$.