Math 342 Homework 8 Daniel Walton January 22, 2014 11.2 : 8, 9, 10 Exercise (8). Let A and B be sequentially compact subsets of R. Define K = {(x, y) ∈ R2 : x ∈ A, y ∈ B}. Prove that K is sequentially compact. Solution (8). Let {uk }∞ k=1 ⊆ K be a sequence. We can express the k-th term as uk = (xk , yk ). The terms xk and yk for sequences in A and B, respectively. Since {xk } ⊆ A and A is sequentially compact, ∃ a subsequence xkj such that lim xkj = x ∈ A and since {ykj } ⊆ B, ∃ a subsequence ykjl such that j→∞ lim ykjl = y ∈ B. Note also that the subsequence {xkjl } still converges to x. l→∞ Then uklj converges componentwise to u = (x, y) ∈ K, and by the componentwise convergence criterion, uklj converges to u ∈ A. Exercise (9). Let A be a subset of Rn that is sequentially compact and let v be a point in Rn \A. Prove that there is a point u0 in the set A such that dist(u0 , v) ≤ dist(u, v) for all points u ∈ A. Solution (9). We first need the fact that the distance function as we defined it is a continuous function. Well, pi (u) is a continuous function by Proposition 11.1, and by repeatedly applying Theorem 11.1, we have that (p1 (u) − p1 (v))2 + . . . + (pn (u) − pn (v))2 is a continuous function. From Math 341, the square root p function from R to R is also a continuous function. By Theorem 11.5, (p1 (u) − p1 (v))2 + . . . + (pn (u) − pn (v))2 = ||u − v|| is continuous. Now consider the set {||u − v|| : u ∈ A}, which is the image of the distance function. By Theorem 11.22, this set is sequentially compact and attains a smallest value d, which is greater than 0 since v ∈ / A. Then we know that there exists some u0 ∈ A where ||u0 − v|| attains this value d. Since d is the minimum distance, ||u0 − v|| ≤ ||u − v|| for all u ∈ A. We now show that this point is not necessarily unique with the following example. Let A be the circle of radius 1 centered at the origin in R2 . Let v = (0, 0). Then for all u ∈ A, dist(u, v) = 1. Exercise (10). A mapping F : Rn → Rm is Lipschitz if there is a number C such that dist(F(u), F(v)) ≤ Cdist(u, v) for all points u, v ∈ Rn . The number C is called a Lipschitz constant for the mapping. Show that a Lipschitz mapping is uniformly continuous. 1 Solution (10). Let > 0 and choose δ = /C. Then for u, v ∈ Rn satisfying ||u − v|| < δ, we have that ||F (u) − F (v)|| ≤ C||u − v|| < Cδ = C = since C C is positive. By Theorem 11.27, F is uniformly continuous. 2

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