Let \(K\subset\mathbb R^n\) be compact, and let \(\{B_j\}_{j=1}^\infty\) be a sequence of open balls covering \(K\). Prove that there is \(\varepsilon>0\) such that every open \(\varepsilon\)-ball centered at a point of \(K\) is contained in one of the balls \(B_j\).
Solution
For \(x\in\RR^n\) and \(r>0\), write \[ B(x,r)\coloneqq\{y\in\RR^n:\norm{y-x}<r\}. \]
1. For every \(x\in K\), there are an index \(j(x)\) and a number\(r_x>0\) such that\[ B(x,2r_x)\subseteq B_{j(x)}. \]
Because the balls \(B_j\) cover \(K\), choose \(j(x)\) with \[ x\in B_{j(x)}. \] Write \[ B_{j(x)}=B(c_x,R_x). \] Since \(x\) lies in this open ball, \[ d_x\coloneqq R_x-\norm{x-c_x}>0. \] Set \[ r_x\coloneqq\frac{d_x}{2}. \] If \(y\in B(x,2r_x)\), then by the triangle inequality, \[ \norm{y-c_x} \leq \norm{y-x}+\norm{x-c_x} < 2r_x+\norm{x-c_x} = R_x. \] Hence \(y\in B_{j(x)}\).
2. There are points \(x_1,\ldots,x_N\in K\) such that\[ K\subseteq\bigcup_{i=1}^N B(x_i,r_{x_i}). \]
The family \[ \{B(x,r_x):x\in K\} \] is an open cover of \(K\). Since \(K\) is compact, it has a finite subcover.
3. Define\[ \varepsilon\coloneqq \min_{1\leq i\leq N}r_{x_i}. \]Then\[ \varepsilon>0. \]
Every \(r_{x_i}\) is positive by step 1, and the minimum is taken over finitely many such numbers.
4. For every \(y\in K\), the ball \(B(y,\varepsilon)\) is contained inone of the original covering balls \(B_j\).
3. Thus\[ z\in B(x_i,2r_{x_i}), \]and step 1 gives\[ z\in B_{j(x_i)}. \]Therefore\[ B(y,\varepsilon)\subseteq B_{j(x_i)}. \]
5. Hence there exists a uniform radius\[ \boxed{\varepsilon>0} \]such that every \(\varepsilon\)-ball centered at a point of \(K\) iscontained in one member of the given cover.
Step 3 provides a positive \(\varepsilon\), and step 4 proves the required containment for every center \(y\in K\).
6. Q.E.D.
Step 5 is exactly the required conclusion.