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.