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:
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.
