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
lim ykjl = y ∈ B. Note also that the subsequence {xkjl } still converges to x.
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
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.
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 is positive. By Theorem 11.27, F is uniformly continuous.