Let \(S \subseteq \mathbf{R}^n\) and suppose \(\mathcal{A} = {A_i | i \in I}\) is a family of open subsets of \(\mathbf{R}^n\) such that

\[

S \subseteq \bigcup_{i \in I} A_i.

\]

We call such a family \(\mathcal{A}\) an **open cover** of \(S\). We call a subset \(K \subseteq \mathbf{R}^n\) **topologically compact** if every open cover \(\mathcal{A}\) of \(K\) admits a finite subcover. That is for every open cover \(\mathcal{A}\), there exists \(k \in \mathbf{N}\) such that there exist \(A_1, A_2, \ldots, A_k \in \mathcal{A}\) with

\[

K \subseteq \bigcup_{i = 1}^k A_i.

\]

**Theorem**. If \(K \subseteq \mathbf{R}^n\) is topologically compact, then it is closed and bounded.

**Proof**. We first show that \(K\) is bounded. To this end, consider the family of open sets given by \(A_i = {x \in \mathbf{R}^n \big| |x| 1/i}.

\]

Note that \(\bigcup A_i = \mathrm{R} \setminus{x}\), thus (since \(x \notin K\)) \({A_i}\) is an open cover of \(K\). Since this family admits an open subcover and the \(A_i\) are nested (\(i Heine-Borel Theorem.