Is the cantor set a connected set?

Question

The Cantor set is created by deleting the open middle third from each of a set of line segments repeatedly.

Is the cantor set a connected set?

Thank you.

Answer 1

Let $C$ denote the cantor set. Assuming the usual topology, the open(inside of $C$) sets $(-1,1/2)\cap C$ and $(1/2,2)\cap C$ partition $C$ showing that $C$ is indeed disconnected.

Answer 2

Let $C \subseteq [0,1]$ be the Cantor set. For every $x,y \in C$ with $x \neq y$ there exists by the construction of $C$ some $z \in [0,1]$ with $z \notin C$ and $x < z < y$. Then both $(-1,z) \cap C = [-1,z] \cap C$ and $(z,2) \cap C = [z,2] \cap C$ are clopen in $C$ with respect to the subspace topology.

So every two points of $C$ are respectively contained in two disjoint clopen subsets of $C$. Because every connected component of $C$ is contained in a clopen subset of $C$ it follows that no two distinct points of $C$ are contained in the same connected component. So $C$ is totally disconnected, i.e. all connected components are singletons.