(1) $A$ is nonempty subset of $\mathbb{R}$.
(2) For all $x<y \in A$ there is $z \notin A$ such that $x<z<y$.
(3) $A$ is perfect.
Then is there homeomorphism between $A$ and cantor set?
Intuitively, I think this is right because $A$ is uncountable because of (3), and set is almost like distinct many points. Which is almost like Cantor set.
Without compactness this is false, take for example any subset $A$ of the standard Cantor set $C$ such that $A$ and $C-A$ are both dense in $C$. The subset $A$ can be countable, e.g. the endpoints of the deleted intervals of $C$.
You'll need to add compactness, then it is true. It is a standard theorem that every compact, totally disconnected, perfect, nonempty metrizable space is homeomorphic to the Cantor set; your hypothesis (2) provides the "totally disconnected" property. A proof can be found in the book by Moise, "Geometric topology in dimensions 2 and 3"; and a link to Brouwer's original paper is given in http://en.wikipedia.org/wiki/Cantor_space.
