※ 引述《cckk3333 (皓月)》之铭言： : 2. Let f: R -> R be a bounded function. Prove that f is continuous if and : only if the graph = { (x,y) | y = f(x) } of f is a closed subset of R^2 Proof. Suppose f is a continuous function defined on |R, we define F＝(I,f):|R→|R^2, then F is continuous on |R. So, the graph is clearly a closed subset of |R^2. Conversely, if the graph G_f is closed, we will show that f is continuous at a, where a is given. Consider the graph which is contained in the compact region D＝{(x,y): x in [a-1,a+1], y in [-M,M]}, where M ＝ sup {|f(x)|: x in |R}. So, the restricted graph, called G'_f, is also compact. Let P:G'_f→[a-1,a+1] by P(x,f(x))＝x. It is obvious that P is continuous and one-to-one on the compact set G'_f. Hence, its inverse function P^-1 ＝ (I,f):[a-1,a+1] → G'_f is also continuous (＊). It implies that f is continuous on [a-1,a+1]. In particular, f is continuous at a. Since a is arbitrary, we have proved that f is continuous on |R. NOTE. (1) The assumption that f is bounded is needed, e.g., f(x) ＝ 1/x if x ＞ 0, ＝ 0 if x ＝ 0, ＝ 1/x if x ＜ 0. (2) For (＊), Walter Rudin, Principles of Mathematical Analysis, 3rd ed. McGraw-Hill, 1976. Theorem 4.17, Chap. 4, p. 90. (3) Reference: Walter Rudin, Principles of Mathematical Analysis, 3rd ed. McGraw-Hill, 1976. Exercise 6, Chap. 4, p. 99. (4) Related Exercise: Let f:(0,1)→|R. If the graph G_f is path-connected, then f is continuous on (0,1). (5) If you are interested with this, you can google "closed graph theorem" in Functional Analysis. -- Good taste, bad taste are fine, but you can't have no taste. -- ※ 编辑: math1209 来自: 140.113.25.169 (01/19 12:01)