Heine-Borel Theorem/Euclidean Space/Necessary Condition/Proof 2

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

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

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

Then $C \subseteq \left[{a \,.\,.\, b}\right]$ for some $a, b \in \R$.

Moreover, $C$ is closed in $\left[{a \,.\,.\, b}\right]$ by the Corollary to Closed Sets 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, $C \subseteq \left[{a \,.\,.\, b}\right] \times \left[{a \,.\,.\, b}\right] \times \cdots \times \left[{a \,.\,.\, b}\right] = B$ 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 Sets in Topological Subspace.

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