Can Cantor set constructed on non compact or disconnected set, such as $(0,1)$ or $[0,1]$\ $\mathbb{Q}$ The construction of Cantor set is always start from a compact set. I wonder what if we have $[0,1)$ at first, and then construct Cantor set?
Edit:My original thought is we must construct Cantor set from compact set, so the complement of Cantor set will give an open interval in $[0,1)$. Which doesn't sound right now. What about $(0,1)$ instead? What is the case of $[0,1]$\ $\mathbb{Q}$?Can we still construct Cantor set from them?
 A: I think I know what you are after.
The first iteration $D_1 = [0,1)$.
The second iteration $D_2 = [0,\frac13) \cup [\frac23,1)$.
The second iteration $D_3 = [0,\frac19) \cup [\frac29,\frac13) \cup [\frac23,\frac79) \cup [\frac89,1)$.
The $k$th iteration $D_k = \frac13 C_{k-1} \cup (\frac23 + \frac13 C_{k-1})$.
Then $D = \bigcap_k D_k$.
We can express $D$ using base 3 expansions.  The Cantor Set is
$$ C = \{0.a_1a_2a_3\dots (\text{base $3$}):a_i=0\text{ or }2\} ,$$
and $D$ is those elements of the Cantor set whose base 3 expansion does not end in recurring 2s.
If you start with $(0,1)$, you will end up with those elements of the Cantor set whose base 3 expansion neither terminate, nor end up in recurring 2s.
If you start with $[0,1] \setminus \mathbb Q$, you will simply end up with $C \setminus \mathbb Q$.
A: If you follow the exact prescription of THE Cantor set but starting with $[0,1)$ then, as the other answer shows, you will be missing the point $1$ and so in the end you will not get THE Cantor set, and in fact what you will get is not even compact.
However, one can use the same procedure to construct A Cantor set starting with any closed interval $C_0=[a,b]$: first remove the middle third open interval of $C_0$ which in this case is the open interval $(a + \frac{b-a}{3},b - \frac{b-a}{3})$, and what's left is
$$C_1 = [a, a + \frac{b-a}{3}] \cup [\frac{b-(b-a)}{3},b]
$$
then remove the middle third open intervals of each of the two closed intervals comprising $C_1$ and what's left is $C_2$, and so on; and then intersect them all. The result will be a compact set which is very much like the original Cantor set, in fact it is similar to the original Cantor set in the strict geometrical sense of "similarity" which is used in studying "similar triangles" in the plane: what you get will be a linearly scaled copy of the original Cantor set.
So, if you are happy to locate a linearly scaled copy of the original Cantor set inside the interval $[0,1)$, you can start with any closed subinterval $C_0=[a,b] \subset [0,1)$, for example $C_0=[1/4,3/4]$ or $C_0=[0,1/3]$.
