Compactness and Open Covers

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.