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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.

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    $\begingroup$ No, of course not. $\endgroup$
    – bof
    Jan 16, 2016 at 23:14
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    $\begingroup$ 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. $\endgroup$ Jan 16, 2016 at 23:15
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    $\begingroup$ Far from it: it’s totally disconnected, even zero-dimensional. $\endgroup$ Jan 16, 2016 at 23:16
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    $\begingroup$ Have you ever seen a picture of the Cantor set? $\endgroup$ Jan 16, 2016 at 23:18
  • $\begingroup$ @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 $B$. $\endgroup$ Jan 16, 2016 at 23:22

2 Answers 2

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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.

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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.

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