 |
 |
|
|
[ Pobierz caÅ‚ość w formacie PDF ] REMARKS: (a) In the definition, δ is allowed to depend on and x. (b) When E = E = R with the usual metric, the preceding is the classical defini- tion of continuity. (c) The condition for T to be continuous at x can be rephrased in more geometric terms as follows: for every 0 there is a δ 0 such that T maps the open ball B(x, δ) of E into the open ball B (T x, ) of E . Here, B(x, δ) = {y ∈ E : d(x, y) δ}, B (T x, ) = {y ∈ E : d (T x, y) }. 45 -1 -1 46 FUNCTIONS ON METRIC SPACES Continuity and Open Sets 12.2 THEOREM. A mapping T : E → E is continuous if and only if T B is an open subset of E for every open subset B of E . PROOF. Suppose that T is continuous. Let B ⊂ E be open. We want to show that, then, A = T B is open, that is, for every x in A there is δ 0 such that B(x, δ) ⊂ A. To this end, fix x in A, note that y = T x is in B, and therefore, there is 0 such that B (y, ) ⊂ B (since B is open). By the continuity of T , for that , there is a δ 0 such that T maps B(x, δ) into B (y, ). Since B (y, ) ⊂ B, we have B(x, δ) ⊂ A as needed. Suppose that T B is open in E for every open subset B of E . Let x in E be arbitrary. We want to show that, then, T is continuous at x. To this end, fix 0. Since B (T x, ) is open, its inverse image is open, that is A = T B (T x, ) is an open subset of E. Note that x is in A; therefore, there is a δ 0 such that B(x, δ) ⊂ A, and then T maps B(x, δ) into B (T x, ). So, T is continuous at x. Continuity and Convergence If (xn) is a sequence in E, we write xn → x to mean that (xn) converges to x in E in the metric d, that is, d(xn, x) → 0. Similarly, we write yn → y to mean that the sequence (yn) in E converges to y in the metric d . The following is probably the most useful characterization of continuity. 12.3 THEOREM. A mapping T : E → E is continuous at the point x of E if and only if (xn) ⊂ E, xn → x ⇒ T xn → T x. PROOF. Suppose that T is continuous at x. Let (xn) ⊂ E be such that xn → x. We want to show that, then, T xn → T x, which is equivalent to showing that for every 0 the ball B (T x, ) contains all but finitely many of the points T xn. To this end, fix 0. By the continuity of T at x, there is δ 0 such that T maps B(x, δ) into B (T x, ). Since xn ∈ B(x, δ) for all but finitely many n, it follows that T xn ∈ B (T x, ) for all but finitely many n, which is as desired. Suppose that T is not continuous at x. Then, there is 0 such that for every δ 0 there is y in E such that d(x, y) δ and d (T x, T y) ≥ . Thus, for that , -1 -1 -1 -1 d d d d d d 12. CONTINUOUS MAPPINGS taking δ = 1, 1/2, 1/3, . . . we can pick y = x1, x2, x3, . . . such that d(xn, x) 1/n and d (T xn, T x) ≥ . Hence, there is a sequence (xn) ⊂ E such that xn → x but (T xn) does not converge to T x. Compositions The following result is recalled best by the phrase “a continuous function of a continu- ous function is continuous”. 12.4 THEOREM. If T : E → E is continuous at x ∈ E and S : E → E is continuous at T x ∈ E , then S ◦ T : E → E is continuous at x ∈ E. If T is continuous and S is continuous, then S ◦ T is continuous. PROOF. The second assertion is immediate from the first. To show the first, let (xn) ⊂ E be such that xn → x. If T is continuous at x, the T xn → T x by the last theorem; d 47 d d d and if S is continuous at T x, this in turn implies that S(T xn) → S(T x) by the last theorem again, which means that S ◦ T is continuous at x. EXAMPLES. 12.5 Constants. Let T : E → E be defined by T x = b where b in E is fixed. This T is continuous. 12.6 Identity. Let T : E → E be defined by T x = x. This T is continuous, as is easy to see from Theorem 12.2 or 12.3. 12.7 Restrictions. Let T : E → E be continuous. For D ⊂ E, the restriction of T to D is the mapping S : D → E defined by putting Sx = T x for each x ∈ D. Obviously, the continuity of T implies that of S. 12.8 Discontinuity. Let f : R → R be defined by setting f(x) = 1 if x is rational and f(x) = 0 if x is irrational. This function is discontinuous at every x ∈ R. To see it, fix x in R. For every δ 0, the ball B(x, δ) has infinitely many rationals and infinitely many irrationals. Thus, it is impossible to satisfy the condition for continuity at x (for any 1). 12.9 Lipschitz continuity. A mapping T : E → E is said to satisfy a Lipschitz [ Pobierz caÅ‚ość w formacie PDF ] |