Cantor set

Example

Let $C_0 = [0, 1]$. For $n \ge 0$, let $C_{n+1}$ be the set that we get from $C_n$ when we remove the open middle third of each of its closed intervals. Thus $C_1 = [0, \tfrac{1}{3}] \cup [\tfrac{2}{3}, 1]$. The Cantor set is $C = \bigcap_{n \ge 0} C_n$, with the subspace topology of $\mathbb{R}$.

Remarks

Equivalently, $C$ is the set of $x \in [0, 1]$ that have a base-3 expansion $x = \sum_{n \ge 1} a_n 3^{-n}$ with all $a_n \in \{0, 2\}$. The set $C$ is closed and bounded, so it is compact. It is totally disconnected and has no isolated points. It is uncountable, and it is homeomorphic to the space of binary sequences. Its Lebesgue measure is $0$, because $C_n$ has total length $(2/3)^n$. The Cantor set is nowhere dense in $\mathbb{R}$.

Explanations

L1 human

Start with $[0, 1]$, remove the middle third, then remove the middle third of each piece that is left, and go on forever. What remains is a set with no length, but with as many points as $\mathbb{R}$.