Let \((M,d)\) be a metric space, let \(C\subseteq M\) be compact, and let \((U_\alpha)_{\alpha\in I}\) be an open cover of \(C\). Show that there exists \(\varepsilon>0\) such that for every \(p\in C\), the open ball \[ B(p,\varepsilon) \] is contained in at least one \(U_\alpha\).
Solution
1. For every \(p\in C\), there are an index \(\alpha(p)\in I\) and a number\(r_p>0\) such that\[ B(p,2r_p)\subseteq U_{\alpha(p)}. \]
Since the family \((U_\alpha)\) covers \(C\), choose \(\alpha(p)\) with \[ p\in U_{\alpha(p)}. \] The set \(U_{\alpha(p)}\) is open in the ambient metric space \(M\), so there is \(\rho_p>0\) such that \[ B(p,\rho_p)\subseteq U_{\alpha(p)}. \] Take \[ r_p=\frac{\rho_p}{2}. \]
2. There are points\[ p_1,\ldots,p_m\in C \]such that\[ C \subseteq \bigcup_{j=1}^m B(p_j,r_{p_j}). \]
The balls \[ B(p,r_p), \qquad p\in C, \] form an open cover of the compact set \(C\). Compactness gives a finite subcover.
3. Define\[ \varepsilon \coloneqq \min_{1\leq j\leq m}r_{p_j}. \]Then\[ \varepsilon>0. \]
Every \(r_{p_j}\) is positive, and the minimum is taken over finitely many numbers.
4. For every \(p\in C\), there is some \(j\) such that\[ B(p,\varepsilon) \subseteq B(p_j,2r_{p_j}). \]
Fix \(p\in C\). By step 2, choose \(j\) such that \[ d(p,p_j)<r_{p_j}. \] If \[ q\in B(p,\varepsilon), \] then step 3 gives \[ d(q,p)<\varepsilon\leq r_{p_j}. \] The triangle inequality therefore gives \[ d(q,p_j) \leq d(q,p)+d(p,p_j) < r_{p_j}+r_{p_j} = 2r_{p_j}. \] Thus \(q\in B(p_j,2r_{p_j})\).
5. For every \(p\in C\), the ambient ball\[ B(p,\varepsilon) \]is contained in one member of the original cover.
4. By step 1,\[ B(p_j,2r_{p_j}) \subseteq U_{\alpha(p_j)}. \]Hence\[ B(p,\varepsilon) \subseteq U_{\alpha(p_j)}. \]
6. Q.E.D.
The positive number \(\varepsilon\) from step 3 has the required property by step 5.