- Show by example that the union of infinitely many compact subsets of a metric space need not be compact. (c) If \((X,d)\) is a metric space and \(K\subset X\) is compact, define \(d(x_0,K)=\inf_{y\in K} d(x_0,y)\). Prove that there exists a point \(y_0\in K\) such that \(d(x_0,K)=d(x_0,y_0)\).
Solution
1. (b) The union of infinitely many compact sets need not be compact.
in \(\RR\) (a metric space), the singletons \(K_n = \{n\}\) are compact (finite, hence compact). Their union \(\cup_n K_n = \NN\) is not compact: the open cover \(\{(n - 1/2, n + 1/2) : n \in \NN\}\) has no finite subcover, since each open interval contains exactly one point of \(\NN\). (Equivalently, \(\NN\) is closed but unbounded, hence not compact.)
2. (c) Setup: the function \(x \mapsto d(x_0, x)\) is continuous.
by the triangle inequality, \(|d(x_0,x) - d(x_0,y)| \le d(x,y)\), so the map is \(1\)-Lipschitz.
3. (c) The distance to \(K\) is attained.
\(K\) is compact and \(d(x_0, \cdot)\) is continuous (<1>2), so \(d(x_0,\cdot)\) attains its minimum on \(K\): there is \(y_0 \in K\) with \(d(x_0, y_0) = \min_{y \in K} d(x_0, y) = d(x_0, K)\), where the last equality is the definition of \(d(x_0,K)\) as an infimum, attained by compactness.
4. Q.E.D.