Cantor set is closed

Statement

The Cantor set $C = \bigcap_{n \ge 0} C_n$ is a closed subset of $\mathbb{R}$.

Remarks

Proof: each $C_n$ is a union of $2^n$ closed intervals, so it is closed. An intersection of closed sets is closed. Since $C \subseteq [0, 1]$, the set $C$ is also bounded, and thus compact by the Heine–Borel theorem. Each endpoint of an interval of some $C_n$ lies in $C$. For example, $\tfrac{1}{3}$ and $\tfrac{2}{3}$ are in $C$.

Explanations

L1 human

At each step we keep a finite union of closed intervals, which is a closed set. Closed sets stay closed under intersections, so the limit set is closed.