I am trying to prove that the Cantor set (C) is closed without using the fact "the intersection of closed sets is closed". My proof is as follows.
Proof: Let $ \{x_{n}\} $ be a sequence of elements of $ C $ such that $ \{x_{n}\} $ converges for some $ x\in [0,1] $.
Notice that for each $ n\in \mathbb{N} $, $ x_{n} $ can be written as $ \sum \limits_{k=1}^{\infty}\frac{x_{n,k}}{3^{k}} $ where $ x_{n,k}\in \{0,2\} $ for each $ k\in \mathbb{N} $.
Since $ x\in [0,1] $ we have that $ x=\sum \limits_{k=1}^{\infty}\frac{x_{k}}{3^{k}} $ where $ x_{k}\in \{0,1,2\} $ for each $ k\in \mathbb{N} $.
Since $ \{x_{n}\} $ converges to $ x $, for each $ k\in \mathbb{N} $, there exists $ n_{k}\in \mathbb{N} $ such that for each $ n\geq n_{k} $, $ |x_{n}-x|<\frac{1}{3^{k}} $.
This implies for each $ k\in \mathbb{N} $, there exists $ n_{k}\in \mathbb{N} $ such that for each $ n\geq n_{k} $, $ x_{k}=x_{n,k} $.
Therefore for each $ k\in \mathbb{N} $, $ x_{k}\in \{0,2\} $ and hence $ x\in C $. Thus the Cantor set (C) is closed.
Can someone verify my proof? Is there something missing? Is this proof depend on the consideration of two different ternary expansion of some numbers(end points of the removed middle third intervals in the construction of C) of $[0,1]$?
Thanks for the feedback.
Best Answer
We are told to use the following definition of the Cantor set $C$: A point $x\in[0,1]$ is in $C$ if it has a ternary expansion $x=\sum_{k=1}^\infty{d_k\over 3^k}$ with all $d_k\in\{0,2\}$.
Let $x$ be an accumulation point of $C$. We have to prove that $x\in C$. Assume to the contrary. Then $x$ has a ternary expansion containing a first digit $d_r=1$, and not all subsequent digits $=0$ or all subsequent digits $=2$. It follows that there are positive numbers $s$ and $t$ with $$x=\sum_{k=1}^{r-1}{d_k\over 3^k}+3^{-r}+s=\sum_{k=1}^{r-1}{d_k\over 3^k}+2\cdot 3^{-r}-t\ .$$ Consider now a sequence $n\to x_n\in C$ with $\lim_{n\to\infty} x_n=x$. If $x_n<x$ then $x_n\leq \sum_{k=1}^{r-1}{d_k\over 3^k}+3^{-r}$, whence $x-x_n\geq s$, and similarly, if $x_n>x$ then $x_n-x\geq t$. It follows that the sequence $(x_n)_{n\geq1}$ cannot converge to $x$.