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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:

Cheenta College Mathematics Program

## 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.