Proving the Cantor set is closed (without using the fact "the intersection of closed sets is closed") 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.

Suppose (in ternary) $x=0.1=0.0222...,$ a point of the cantor set. This point $x$ is approached by the sequence $x_1,x_2,...=0.022,0.0222,0.02222,...$ and even though each term of the approaching sequence is all 0's or 2's, it is not immediate that the approached number $x$ must be all 0's or 2's.
This kind of thing can happen only at points for which there are two ternary representations, but IMO these cases need to be handled. Otherwise the idea looks like a proof to me.
