Heine-Borel Theorem/Euclidean Space/Necessary Condition/Proof 1

Theorem

For any natural number $n \ge 1$, a closed and bounded subspace of the Euclidean space $\R^n$ is compact.

Proof

Let $C \subseteq \R^n$ be closed and bounded.

Since $C$ is bounded, $C \subseteq \closedint a b^n = B$ for some $a, b \in \R$.

By the Heine-Borel Theorem: Real Line and by Topological Product of Compact Spaces, it follows that $B$ is compact.

From Euclidean Topology is Product Topology, it follows that $B$ is compact in the usual Euclidean topology induced by the Euclidean metric.

By the Corollary to Closed Set in Topological Subspace, we have that $C$ is closed in $B$.

By Closed Subspace of Compact Space is Compact, $C$ is compact.

$\blacksquare$

Source of Name

This entry was named for Heinrich Eduard Heine and Émile Borel.