Cantor set's endpoints. Prove that:
If $[a,b]$ is one of the closed intervals that makes up one the approximation $C_k$ of the Cantor set then the endpoints $\{a,b\}\subset C$ where $C$ is the cantor set.
I should prove that each endpoint of $C_k$ belongs to all $C_m$ where $m\geq k$. 
For example if we take $C_0$ I would prove that $\{0,1\}\subset C$. But $[0,1/3]\subset C_1$ and $[2/3,1]\subset C_1$; $[0,1/3^2]\subset C_2$, $[0,1/3^2]$, $[8/3^2,1]\subset C_3$... in general $[0,1/3^m]\subset C_m$, $[(3^m-1)/3, 1]\subset C_m$. Then ${0,1}\subset C$. Am I right?
If this is true, can I use the same argument to $[a,b]$?
Thanks!
 A: Probably it is easier if we recall, more formally, how the construction of $C$ is done:
First, let $C_0=[0,1]$ (which is a union of $2^0=1$ disjoint intervals of length $3^0=1$).
Suppose that we are given $C_k$, which is as a union of $2^k$ disjoint closed intervals of length $3^{-k}$, say $C_k=\bigcup_{j=1}^{2^k}[a_{k,j},b_{k,j}]$, where $a_{k,j}<b_{k,j}<a_{k,j+1}$ (this choice of $a_{k,j}$ and $b_{k,j}$ is unique). Then we remove the middle third of each interval $[a_{k,j},b_{k,j}]$, that is, let $$c_{k,j}=a_{k,j}+3^{-1}(b_{k,j}-a_{k,j})\quad \text{and}\quad d_{k,j}=b_{k,j}-3^{-1}(b_{k,j}-a_{k,j}).$$
Then let $C_{k+1}=\bigcup_{j=1}^k[a_{k,j},c_{k,j}]\cup[d_{k,j},b_{k,j}]$, which is a union of $2^{k+1}$ disjoint closed intervals of length $3^{-(k+1)}$, so the induction process continues. Note, in particular, that $a_{k,j}$ is an endpoint of an interval which makes up $C_{k+1}$, so in particular $a_{k,j}\in C_{k+1}$.
Finally, the Cantor set is defined as $C=\bigcap_{k=1}^\infty C_k$. We also know that $C_0\supseteq C_1\supseteq C_2\supseteq\cdots$.
Let's use the notation above to prove the result. Suppose $[a,b]$ is one of the intervals which make up $C_k$, that is, $[a,b]=[a_{k,j},b_{k,j}]$ for a certain $j$. Let's show that $a=a_{k,j}\in C$. First, note that if $n\leq k$, then $a\in C_k\subseteq C_n$.
On the other hand, as we noted above, $a=a_{k,j}$ is an endpoint of some of the intervals which make up $C_{k+1}$, so $a\in C_{k+1}$. The same fact implies that $a\in C_{k+2}$, $C_{k+3}$, etc..., that is, $a\in C_m$ for all $m>k$.
Therefore, $a\in C_n$ for all $n=0,1,2,\ldots$, so $a\in\bigcap_n C_n=C$. Similarly, $b\in C$.
