Proposition 9.2.1🔗

Proposition 9.2.1. Let X be a compact metric space and \{V_\alpha\} an open cover. Then there are a finite index set A_0, functions \varpi_\alpha : X \to [0,\infty), and a constant M, such that: each \varpi_\alpha is M-Lipschitz and nonnegative; \varpi_\alpha \equiv 0 for \alpha \notin A_0; \sum_{\alpha \in A_0} \varpi_\alpha \equiv 1; and \varpi_\alpha = 0 outside V_\alpha for every \alpha.

Proof. If X = \emptyset there is nothing to prove. If some V_{\alpha_0} = X, take \varpi_{\alpha_0} \equiv 1 and all others 0. Otherwise every complement V_\alpha^{\mathrm c} is nonempty and closed. By compactness choose a finite subcover, indexed by A_0. For \alpha \in A_0 let \varphi_\alpha(x) := \operatorname{dist}(x, V_\alpha^{\mathrm c}): nonnegative, 1-Lipschitz, vanishing exactly off V_\alpha (the complement is closed), and positive on V_\alpha. The sum \Sigma := \sum_{\alpha \in A_0}\varphi_\alpha is continuous and everywhere positive (each point is in some V_\alpha, \alpha \in A_0), hence on the compact X bounded below by some m > 0 and above by some B. Set \varpi_\alpha := \varphi_\alpha / \Sigma for \alpha \in A_0 (and 0 otherwise). These are nonnegative, sum to 1, vanish off V_\alpha, and are Lipschitz: a quotient of a bounded Lipschitz function by a Lipschitz function bounded below is Lipschitz, with constant expressible in terms of the Lipschitz constants, B and m (numerator: |\varphi(x)\Sigma(y) - \varphi(y)\Sigma(x)| \le (B + B\,|A_0|)\, d(x,y) by adding and subtracting \varphi(y)\Sigma(y); denominator \ge m^2). \blacksquare

In Lean, the finite subcover, the Lipschitz quotient bound, the normalization step, and the assembled partition of unity are: