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Math Exam 508
October 12, 2006
Jerry L. Kazdan
12:00 1:20
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 innite set in a metric
space (perhaps R ) with the specied property.
A1. Bounded with exactly two limit points.
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Solution: The set {(1)n (1 + n ), n = 1, 2, 3, . . .} in R .
A2. Containing all of its limit points.
Solution: Lots of exmples: 1). The empty set.
4). The closed interval {0 x 1 in R} .
2). All of R .
3). The point {0} R .
A3. Distinct points {xj , j = 1, 2, 3, . . .} with xi = xj for i = j that is compact.
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Solution: The following subset of the real numbers: {0} { n , n = 1, 2, 3, . . .} .
A4. Closed and bounded but not compact.
Solution: The closed unit ball x 1 in 2 . The standard basis vectors e1 = (1, 0, 0, . . .),
e2 = (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 = 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 +
1
n
R, n = 1, 2, 3, . . .}
closed
bounded
compact
countable
open
closed
bounded
compact
countable
c) {(x, y ) R2 : 0 < y 1}
open
closed
bounded
compact
countable
d) {(x, y ) R2 : x = 0}
open
closed
bounded
compact
countable
b) {1} {1 +
1
n
open
R, n = 1, 2, 3, . . .}
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e) {(x, y ) R2 : x2 + y 2 = 1}
open
closed
bounded
compact
countable
f) {(x, y ) R2 : x2 + y 2 1}
open
closed
bounded
compact
countable
g) {(x, y ) R2 : y > x2 }
open
closed
bounded
compact
countable
closed
bounded
compact
countable
h) {(k, n) R2 : k, n any positive integers}
open
Part C: Traditional Problems (4 problems, 20 points each)
C1. In R , if an A and bn B , show that the product an bn AB .
Solution: Let pn = an A 0, qn = bn B 0. Then
an bn = (pn + A)(qn + B ) = pn qn + 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 pn qn 0. Given > 0 (which we
may assume satises < 1), pick N so that if n > N then |pn | < and |qn | < . Consequently
|pn qn | < 2 < .
a1 + + an
be the sequence averages of (arithmetic
n
mean). If ak converges to A , show that the averages Cn also converge to A .
C2. Given a real sequence {ak } , let Cn =
Solution: Letting Bn = an A 0, I could reduce to the case A = 0. Instead, for variety I
proceed directly. Note that
Cn A =
Given any
(a1 A) + + (an A)
a1 + + an
A=
n
n
> 0, pick N so that if n > N then |an A| < . Then write
Cn A =
(a1 A) + + (aN A)
(aN +1 A) + + (an A)
+
.
n
n
In
Jn
Now
[n (N + 1)]
n
=
for any n > N.
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 sucientnly large
|Jn | <
|In | <
Consequently, |Cn A| |In | + |Jn | < 2 .
2
NM
<.
n
C3. Let Kj , j = 1, 2, . . . be compact sets in a metric space. Give a proof or counterexample for
each of the following assertions.
a) K1 K2 is compact.
Solution: True. Since compact sets are closed, then K1 K2 is a closed subset of the
compact set K1 , and hence compact.
b) K1 K2 is compact.
Solution: True. Let {U } be any open cover of K1 K2 . A nite number of these, say
{V1 , . . . , Vk } , cover K1 , and {W1 , . . . , Wn } , cover K2 . Then {V1 . . . Vk W1 . . . Wn }
is the desired nite cover of K1 K2 .
c)
j =1 Kj is compact.
Solution: Counterexample. The non-negative real numbers {x 0} is the union of the
compact sets (closed intervals) Kj = {j 1 x j ; j = 1, 2, . . .} . Since this set is not
bounded, it is not compact.
C4. In a complete metric space M , let d(x, y ) denote the distance. Assume there is a constant
0 < c < 1 so that the sequence xk satises
d(xn+1 , xn ) < cd(xn , xn1 ) for all
n = 1, 2, . . . .
a) Show that d(xn+1 , xn ) < cn d(x1 , x0 ).
Solution: Since d(x2 , x1 ) < cd(x1 , x0 ), then
d(x3 , x2 ) < cd(x2 , x1 ) < c2 d(x1 , x0 ).
Using this,
d(x4 , x3 ) < cd(x3 , x2 ) < c3 d(x1 , x0 ).
The induction to the general case is obvious.
b) Show that the {xk } is a Cauchy sequence.
Solution: Say n > k . Then using the previous part and that 0 < c < 1
d(xn , xk ) d(xn , xn1 ) + . . . + d(xk+1 , xk )
(cn1 + cn2 + + ck ) d(x1 , x0 )
(ck (1 + c + c2 + c3 + . . .) d(x1 , x0 ) =
ck
d(x1 , x0 ).
1c
Pick N so that cN < . If n > k > N then
d(xn , xk ) <
1c
d(x1 , x0 ).
c) Show that there is some p M so that limn xk = p .
Solution: Since the metric space is complete, there is a point p in the metric space to
which the Cauchy sequence xk converges.
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