Heine-Borel Theorem/Euclidean Space/Necessary Condition

Theorem

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

Proof 1

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.

$\blacksquare$

Proof 2

The proof holds for $n = 1$, as follows.

Suppose $C$ is a closed, bounded subspace of $\R$.

Then $C \subseteq \closedint a b$ for some $a, b \in \R$.

Moreover, $C$ is closed in $\closedint a b$ by the Corollary to Closed Set in Topological Subspace.

Hence $C$ is compact, by Closed Subspace of Compact Space is Compact.

Now suppose $C \subseteq \R^n$ is closed and bounded.

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

Now $B$ is compact by Topological Product of Compact Spaces.

Also, $C$ is closed in $B$ by the Corollary to Closed Set in Topological Subspace.

Hence $C$ is compact, by Closed Subspace of Compact Space is Compact.

$\blacksquare$

Source of Name

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