Euclidean Spaces have a very nice property. In \( \mathbb{R}^n \) (equipped with standard Euclidean metric), every closed and bounded set is a compact set. The converse is also true. Every compact set is closed and bounded). This property is known as **Heine Borel Theorem**.

**Recall that: **A set V in a topological space X is compact iff every open cover of V has a finite subcover.

Sometimes, we want *other *spaces to have this property. Consider any metric space (X, d) with the property: **every closed ball is compact. **This type of space is known as **proper metric space. **

We will prove a simple theorem related to proper spaces to illustrate their properties.

**Theorem: **Closed and Bounded Annulus in Proper Metric Spaces is Compact

**Proof: **Fix a point \( x_0 \in X \). Suppose \( B [ x_0 , r] = \{ x \in X | d(x_0, x) \leq r \} \). In simpler terms, \( B[x_0, r] \) is a closed ball.

Let** int (A)** denote the interior points of the set A. (Recall that a point \( a\in A \) is called an interior point if we can find a neighborhood of **a** that is contained in **A**).

If a < b then \(A = B[x_0, b] – int (B[x_0, a] ) \) is a *honest-to-goodness* closed and bounded annulus. We will show that **A is compact**.

Suppose \( \{U_{\alpha} \}_{\alpha \in \Lambda} \) is an arbitrary open cover of A. Then \( \{ \{U_{\alpha} \}_{\alpha \in \Lambda} ,int (B[x_0, a] ) \} \) is an open cover for \( B[x_0, b] \).

Since (X, d) is **proper**, by definition, \( B[x_0, b] \) is compact.

Hence a finite subclass of \( \{ \{U_{\alpha} \}_{\alpha \in \Lambda} ,int (B[x_0, a] ) \} \) covers \( B[x_0, b] \). Since A is a subset of \( B[x_0, b] \), hence this finite subclass also covers A. Therefore we have found a finite subcover of A (for an arbitrary open cover of it).

This implies A is compact.

Also see:

## Moral

This is a standard strategy. To show some set is compact, start with an arbitrary open cover of that set and find a finite subcover of it.