Characterization of Euclidean Borel Sigma-Algebra/Closed equals Compact

Theorem
Let $\mathcal C^n$ and $\mathcal K^n$ be the collections of closed and compact subsets of the Euclidean space $\left({\R^n, \tau}\right)$, respectively.

Then:


 * $\sigma \left({\mathcal C^n}\right) = \sigma \left({\mathcal K^n}\right)$

where $\sigma$ denotes generated $\sigma$-algebra.

Proof
By the Heine-Borel Theorem, $\mathcal K^n \subseteq \mathcal C^n$.

Thus from Generated Sigma-Algebra Preserves Subset, $\sigma \left({\mathcal K^n}\right) \subseteq \sigma \left({\mathcal C^n}\right)$.

Next, let, for all $n \in \N$, $B^- \left({\mathbf 0; n}\right)$ be the closed ball of radius $n$ around $\mathbf 0$ in $\R^n$.

Observe that $\R^n = \displaystyle \bigcup_{n \mathop \in \N} B^- \left({\mathbf 0; n}\right)$.

Now let $U \in \mathcal C^n$ be a closed subset of $\R^n$.

Then from Intersection with Subset is Subset and Intersection Distributes over Union:


 * $\displaystyle U = U \cap \R^n = U \cap \bigcup_{n \mathop \in \N} B^- \left({\mathbf 0; n}\right) = \bigcup_{n \mathop \in \N} \left({U \cap B^- \left({\mathbf 0; n}\right)}\right)$

From Intersection of Closed Sets is Closed, $U \cap B^- \left({\mathbf 0; n}\right)$ is closed for all $n \in \N$.

By definition, $B^- \left({\mathbf 0; n}\right)$ is bounded.

Thus, by the Heine-Borel Theorem, $U \cap B^- \left({\mathbf 0; n}\right)$ is compact.

Thus, any closed set is the countable union of compact sets.

By the third axiom for a $\sigma$-algebra, this means that $\mathcal C^n \subseteq \sigma \left({\mathcal K^n}\right)$.

Now the definition of generated $\sigma$-algebra ensures that $\sigma \left({\mathcal C^n}\right) \subseteq \sigma \left({\mathcal K^n}\right)$.

Hence the result, by definition of set equality.