# Is the cantor set a connected set?

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.

• No, of course not.
– bof
Jan 16, 2016 at 23:14
• No. It's intersection with $[0, 1/3]$ is nonempty, and it's intersection with $[2/3, 1]$ is nonempty. But $[0, 1/3] \cup [2/3, 1]$ is disconnected. Jan 16, 2016 at 23:15
• Far from it: it’s totally disconnected, even zero-dimensional. Jan 16, 2016 at 23:16
• Have you ever seen a picture of the Cantor set? Jan 16, 2016 at 23:18
• @angryavian $[0,1]$ is not a subset of $[0, 1/3] \cup [2/3, 1]$. The relevant theorem is: if $B$ is disconnected and $A$ is a connected subset of $B$, then $A$ must lie in a single connected component of $B$. Jan 16, 2016 at 23:22

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.
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.