Chapter 4: Problem 57
An open cover \(\mathcal{U}\) of a topological space \(X\) is called locally finite if each \(x \in X\) has a neighborhood that intersects only finitely many members of \(U\). If \(U, \mathcal{V}\) are open covers of \(X, V\) is a refinement of \(U\) if for each \(V \in V\) there exists \(U \in \mathcal{U}\) with \(V \subset U . X\) is called paracompact if every open cover of \(X\) has a locally finite refinement. a. If \(X\) is a \(\sigma\)-compact LCH space, then \(X\) is paracompact. In fact, every open cover \(U\) has locally finite refinements \(\left\\{V_{\alpha}\right\\},\left\\{W_{\alpha}\right\\}\) such that \(\bar{V}_{\alpha}\) is compact and \(\bar{W}_{\alpha} \subset V_{\alpha}\) for all \(\alpha\). (Let \(\left\\{U_{n}\right\\}_{1}^{\infty}\) be as in Proposition 4.39. For each \(n\), \(\left\\{E \cap\left(U_{n+2} \backslash \bar{U}_{n-1}\right): E \subset U\right\\}\) is an open cover of \(\bar{U}_{n+1} \backslash U_{n}\). Choose a finite
Short Answer
Step by step solution
Key Concepts
These are the key concepts you need to understand to accurately answer the question.