Show that if \(X\) is a metric space and \(A\subseteq X\) is compact then \(A\) is bounded.
Solution
Goal: Show that if \(X\) is a metric space and \(A \subseteq X\) is compact, then \(A\) is bounded.
1. Fix a point \(p \in X\); the open balls \(\theset{B(p, n)}_{n \in \NN}\) form an open cover of \(X\).
Every point of \(X\) is at finite distance from \(p\), so it lies in some \(B(p, n)\); balls are open in a metric space.
2. The restriction to \(A\) is an open cover of \(A\), so it has a finite subcover \(B(p, n_1), \ldots, B(p, n_k)\).
\(A\) is compact.
3. Let \(N := \max\theset{n_1, \ldots, n_k}\); then \(A \subseteq B(p, N)\).
Each \(a \in A\) lies in some \(B(p, n_j) \subseteq B(p, N)\) since \(n_j \leq N\).
4. \(A\) is bounded.
\(A\) is contained in a ball of finite radius \(N\) centered at \(p\), which is the definition of boundedness in a metric space.
5. Q.E.D.
<1>2–<1>4 establish the claim.