# Cantor Set is Closed in Real Number Space

## Theorem

Let $\mathcal C$ be the Cantor set.

Let $\left({\R, \tau_d}\right)$ be the real number space $\R$ under the Euclidean topology $\tau_d$.

Then $\mathcal C$ is a closed subset of $\left({\R, \tau_d}\right)$.

## Proof

By definition, the Cantor set is the complement of a union of open sets relative to the closed interval $\left[{0 \,.\,.\, 1}\right]$.

By the definition of a topology, that union is itself open in $\R$.

The closed interval $\left[{0 \,.\,.\, 1}\right]$ is itself the complement of a union of open sets $\left({-\infty \,.\,.\, 0}\right) \cup \left({1 \,.\,.\, \infty}\right)$.

Hence the result.

$\blacksquare$