Cantor Set is Closed in Real Number Space

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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.

$\blacksquare$


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