Cantor set as an intersection I've seen that the Cantor set $C$ can be expressed as a countable intersection of $C_n$'s where $C_n=C_{n-1}-\bigcup_{k=0}^{3^{n-1}-1}(\frac{3k+1}{3^n},\frac{3k+2}{3^n})$. 
Taking $C_1,C_2$ into consideration, this is clear, however, when trying to come up with this specific equation on my own, I cannot. I mean, how can we be sure that inductively, this formulation gives the correct $C_n$'s as we know. 

What might help is to see that the formula in the question is a piece of the inductive proof of a closed form description of $C_n$, expressed using trinary expansions of integers.
For each $n$, each integer $k \in \{0,...,3^n-1\}$ can be written as a trinary number $a_{n-1}a_{n-2}...a_0$ with entries in the set of trigits $\{0,1,2\}$, such that 
$$k = a_{n-1}3^{n-1} + ... + a_1 3^1 + a_0 3^0
$$
Let $T_n$ be the set of such integers whose trinary expansion contains no $1$'s. For example, $T_1 = \{0,2\}$. Also, $T_2 = \{0,2,7,9\}$ which in trinary is $T_2=\{00,02,20,22\}$. 
Let me sketch the proof that
$$(*) \qquad\qquad C_n = \bigcup_{k\in T_n} \biggl[ \frac{3k}{3^{n+1}}, \frac{3k+3}{3^{n+1}} \biggr]
$$
You can check easily enough that $C_0=[0,1]$, and $C_1=[0,\frac{1}{3}] \cup [\frac{2}{3},1]$, and 
$$C_2=\biggl[0,\frac{1}{9}\biggr] \cup \biggl[\frac{2}{9},\frac{1}{3}\biggr] \cup \biggl[\frac{2}{3},\frac{7}{9}\biggr] \cup \biggl[\frac{8}{9},1\biggr]
$$
and that this agrees with $(*)$.
So, let's assume as an induction hypothesis that
$$C_{n-1} = \bigcup_{k=T_{n-1}} \biggl[ \frac{3k}{3^{n}}, \frac{3k+3}{3^{n}} \biggr]
$$
Next, we use that $C_n$ equals what you get by removing the open middle third of each of the component intervals of $C_{n-1}$. Thus, for each $k \in T_{n-1}$, we want to remove the open middle third of the interval 
$$\biggl[ \frac{3k}{3^{n}}, \frac{3k+3}{3^{n}} \biggr]
$$
which is open interval
$$\biggl( \frac{3k+1}{3^{n}}, \frac{3k+2}{3^{n}} \biggr)
$$
leaving
$$\biggl[ \frac{3k}{3^{n}}, \frac{3k+3}{3^{n}} \biggr] - \biggl( \frac{3k+1}{3^{n}}, \frac{3k+2}{3^{n}} \biggr)
$$
and taking their union gives us $C_n$ and the formula
$$C_n = \bigcup_{k \in T_{n-1}} \biggl[ \frac{3k}{3^{n}}, \frac{3k+3}{3^{n}} \biggr] - \biggl( \frac{3k+1}{3^{n}}, \frac{3k+2}{3^{n}} \biggr)
$$
Collecting all the first terms of each summand we get the formula in your question:
$$C_n = C_{n-1} - \bigcup_{k=1}^{3^{n-1}-1}\biggl( \frac{3k+1}{3^{n}}, \frac{3k+2}{3^{n}} \biggr)
$$
Thus, your formula is the exact expression of the statement that $C_n$ is equal to $C_{n-1}$ with the open middle thirds removed from each component interval of $C_{n-1}$.
To complete the induction, we have to do a little bit of rewriting, expressing 
$$\biggl[ \frac{3k}{3^{n}}, \frac{3k+3}{3^{n}} \biggr] - \biggl( \frac{3k+1}{3^{n}}, \frac{3k+2}{3^{n}} \biggr) = \biggl[ \frac{3k}{3^{n}}, \frac{3k+1}{3^{n}} \biggr] \bigcup \biggl[\frac{3k+2}{3^{n}}, \frac{3k+3}{3^{n}} \biggr]  
$$
and taking the union we get
$$C_n =  \bigcup_{k\in T_{n-1}} \biggl[ \frac{3k}{3^{n}}, \frac{3k+1}{3^{n}} \biggr] \bigcup \biggl[\frac{3k+2}{3^{n}}, \frac{3k+3}{3^{n}} \biggr] 
$$
Now with a bit more rewriting of the union, you can finish the induction and get the formula for $C_n$ at the beginning of this answer, using the trinary arithmetic fact that 
$$T_n = \bigcup_{k \in T_{n-1}} \{3k,3k+2\}
$$
