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Math 508 Exam 1 Jerry L. Kazdan
October 12, 2006 12:00 – 1:
Directions This exam has three parts, Part A has 4 problems asking for Examples (20 points, 5 points each), Part B asks you to describe some sets (20 points), Part C has 4 traditional problems (60 points, 15 points each). Closed book, no calculators – but you may use one 3′′^ × 5 ′′^ card with notes.
Part A: Examples (4 problems, 5 points each). Give an example of an infinite set in a metric space (perhaps R) with the specified property.
A–1. Bounded with exactly two limit points.
Solution: The set {(−1)n(1 + (^1) n ), n = 1, 2 , 3 ,.. .} in R.
A–2. Containing all of its limit points.
Solution: Lots of exmples: 1). The empty set. 2). All of R. 3). The point { 0 } ∈ R. 4). The closed interval { 0 ≤ x ≤ 1 in R}.
A–3. Distinct points {xj , j = 1, 2 , 3 ,.. .} with xi 6 = xj for i 6 = j that is compact.
Solution: The following subset of the real numbers: { 0 } ∪ { (^) n^1 , n = 1, 2 , 3 ,.. .}.
A–4. Closed and bounded but not compact.
Solution: The closed unit ball ‖x‖ ≤ 1 in ` 2. The standard basis vectors e 1 = (1, 0 , 0 ,.. .), e 2 = (0, 1 , 0 , 0 ,.. .), etc have no convergent subsequence. Another example: the real numbers {x ∈ R | 0 ≤ x ≤ 1 } with the discrete metric: d(x, y) = 1 for x 6 = y , d(x, x) = 0.
Part B: Classify sets (20 points) For each of the following sets, circle the listed properties it has:
a) {1 + (^) n^1 ∈ R, n = 1, 2 , 3 ,.. .} open closed bounded compact countable
b) { 1 } ∪ {1 + (^1) n ∈ R, n = 1, 2 , 3 ,.. .}
open closed bounded compact countable
c) {(x, y) ∈ R^2 : 0 < y ≤ 1 } open closed bounded compact countable
d) {(x, y) ∈ R^2 : x = 0} open closed bounded compact countable
e) {(x, y) ∈ R^2 : x^2 + y^2 = 1} open closed bounded compact countable
f) {(x, y) ∈ R^2 : x^2 + y^2 ≤ 1 } open closed bounded compact countable
g) {(x, y) ∈ R^2 : y > x^2 } open closed bounded compact countable
h) {(k, n) ∈ R^2 : k, n any positive integers}
open closed bounded compact countable
Part C: Traditional Problems (4 problems, 20 points each)
C–1. In R, if an → A and bn → B , show that the product anbn → AB.
Solution: Let pn = an − A → 0, qn = bn − B → 0. Then
anbn = (pn + A)(qn + B) = pnqn + Aqn + Bpn + AB.
Using that for convergent sequences xn and yn we know lim(xn + yn) = lim xn + lim yn and lim(cxn) = c lim xn , we see that it is enough to show that pnqn → 0. Given > 0 (which we may assume satisfies < 1), pick N so that if n > N then |pn| < and |qn| < . Consequently |pnqn| < ^2 < .
C–2. Given a real sequence {ak}, let Cn =
a 1 + · · · + an n
be the sequence of averages (arithmetic mean). If ak converges to A, show that the averages Cn also converge to A.
Solution: Letting Bn = an − A → 0, I could reduce to the case A = 0. Instead, for variety I proceed directly. Note that
Cn − A =
a 1 + · · · + an n
− A =
(a 1 − A) + · · · + (an − A) n Given any > 0, pick N so that if n > N then |an − A| < . Then write
Cn − A =
(a 1 − A) + · · · + (aN − A) ︸ ︷︷^ n ︸ In
(aN +1 − A) + · · · + (an − A) ︸ ︷︷^ n ︸ Jn
Now |Jn| <
[n − (N + 1)] n
n n
= for any n > N.
We will show that by choosing n even larger, we can make |In| < . Since the sequence an − A converges, it is bounded, so for some M we have |an − A| < M. Thus for n sufficientnly large
|In| <
N M
n
Consequently, |Cn − A| ≤ |In| + |Jn| < 2 .
