Math 501
Introduction to Real Analysis
Instructor: Alex Roitershtein
Iowa State University
Department of Mathematics
Summer 2015
Exam #2
Solutions to the sample
This is a take-home examination. The exam includes 6 questions. The total mark is 100
points. Please show all the work, not only the answers.
1. [17 points]
(a) Solve Exercise 40(c) in Chapter 2 of the textbook.
(b) Solve Exercise 40(d) in Chapter 2 of the textbook.
Solution:
(a) Suppose that the graph Gf = {(p, y) ∈ M × R : y = f p} is a compact. If f is not
continuous at p ∈ M, then there exists ε > 0 and pn ∈ M such that for all n ∈ N,
d(pn , p) < 1/n
but
|f (pn ) − f (p)| > ε.
(1)
In particular,
lim pn = p.
n→∞
Since Gf is compact, there is a converging subsequence of points (pnk , f pnk ) ∈ Gf such
that
lim (pnk , f pnk ) = (p, f p) ∈ Gf .
k→∞
Thus
lim f pnk = f p.
n→∞
However, this is a contradiction to the assertion in (1). Thus the assumption that f is
not continuous at p is false.
(b) For instance,
f (x) =
0
if x ≤ 0
1/x if x > 0.
The graph of this function consists of two infinite branches (corresponding to, respectively, negative and positive values of x), each of them is a closed subset of R2 .
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2. [17 points]
(a) Solve Exercise 78(a) in Chapter 2 of the textbook.
(b) Solve Exercise 78(c) in Chapter 2 of the textbook.
Solution:
(a) Let p and q be two arbitrary points in M. Fix any ε > 0. Let p1 = p, and d1 = d(p, q).
If d1 ≤ ε/2 then {p1 , q} is an ε-chain which connects p and q. Suppose that d1 > ε/2.
Since the set M is connected, the intersection of the open balls M 2ε (p1 ) and Md1 − 2ε (q)
is non-empty. Pick any
p2 ∈ M 2ε (p1 ) ∩ Md1 − 2ε (q)
and let d2 = d(p2 , q). If d2 ≤ ε/2 then {p1 , p2 , q} is an ε-chain which connects p and
q. Suppose that d2 > ε/2. Since the set M is connected, the intersection of the open
balls M 2ε (p2 ) and Md2 − 2ε (q) is non-empty. Pick any
p3 ∈ M 2ε (p1 ) ∩ Md2 − 2ε (q)
And so on, until dk = d(pk , q) < ε/2 and hence {p1 , p2 , . . . , pk , q} is an ε-chain which
connects p and q.
(b) The set R\Z is chain-connected because R is chain-connected and, furthermore, the
restriction of R\Z on any finite interval [p, q] differ form the chain-connected [p, q] by,
perhaps, only finitely many points. In words, by replacing integer points with close
to them non-integer points, any ε-chain in [p, q] can be altered to a (ε + δ)-chain with
arbitrary small δ > 0 avoiding integer points. Since ε > 0 and δ > 0 are arbitrary,
R\Z is chain-connected.
3. [17 points] Miscellaneous (and not necessarily related each to other problems)
(a) Construct a compact set of real numbers such that its limit points form a countable
set.
(b) Let (pn )n∈N be a Cauchy sequence of points in a metric space (X, d). Suppose that
some subsequence (pnk )k∈N of pn converges. Prove that then pn converges to the same
limit.
(c) Solve Exercise 54 in Chapter 2 of the textbook.
(d) Solve Exercise 71 in Chapter 2 of the textbook.
Solution:
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(a) For instance, define for any n, m ∈ N,
an,m =
n
X
2−k +
k=1
1
.
m
Then the limit points of the set S = {an,m : n, m ∈ N} are
an,∞ =
n
X
2−k +
k=1
1
,
m
n, m ∈ N ∪ {∞}.
Thus the set S∞ = an,m : n, m ∈ N ∪ {+∞} contains all its limit points. That is,
S∞ is closed. Clearly, S∞ is bounded. Hence it is compact.
(b) Fix any ε > 0 and let Nε ∈ N be such that n, m > Nε implies |pn − pm | < ε. Consider
now any pn with n > Nε . Let τε ∈ N be such that t > τε implies |pnt − p| < ε. Then, if
t > τε and nt > Nε , we have
|pn − p| ≤ |pn − pnt | + |pnt − p| < 2ε.
Therefore, p = limn→∞ pn .
(c) Consider the following counterexample. A is a union of two disjoint disks in R2 and
a segment outside of both the disks connecting two points on their boundary. For
instance, A = A1 ∪ A2 ∪ A3 , where (draw the picture)
A1 = {(x, y) ∈ R2 : (x − 2)2 + y 2 = 1},
A2 = {(x, y) ∈ R2 : (x + 2)2 + y 2 = 1},
A3 = {(x, y) ∈ R2 : −1 ≤ x ≤ 1, y = 0}.
Then int A = int A1 ∪ int A2 is a union of two disjoint open disks, and hence is a
disconnected set.
(d) Let
s=
inf
x∈A,y∈B
d(x, y).
(2)
Then there exist sequences xn ∈ A and yn ∈ B such that
lim d(xn , yn ) = s.
n→∞
(3)
Since both A and B compact, we can choose a subsequences xnk and ymk such that
lim xnk = x ∈ A
k→∞
and
lim ymk = y ∈ B.
k→∞
It follows from
d(x, y) ≤ d(x, xnk ) + d(xnk , ymk ) + d(yyk , y)
along with (3) and (2) that d(x, y) = s.
It is worth to notice that the claim becomes false if the word “compact” is replaced
by “closed” in its statement. For instance, the planar curves defined by the equations
y = x and y = x + (1 + |x|)−1 may serve as a counterexample.
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4. [16 points] Let A1 , A2 , . . . be subsets of a metric space. Let
Un = ∪nk=1 Ak
and
U = ∪∞
k=1 Ak .
(a) Prove that Un = ∪nk=1 Ak , n ∈ N.
(b) Prove that U ⊃ ∪∞
k=1 Ak .
(c) Show, by example, that the inclusion U ⊃ ∪∞
k=1 Ak might be proper.
Solution: Throughout the solution we will use the identity S = lim S and think of the
closure of S as the limit set of S.
(a) We have:
U n = {x : Mr (x) ∩ Un 6= ∅ for all r > 0} = ∩k∈N {x : M1/k (x) ∩ Un 6= ∅} .
Since Un = ∪nj=1 Aj , there is at least one set Ai(x) for every x ∈ U n such that
M1/k (x) ∩ Ai(x) 6= ∅ for infinitely many values of k ∈ N.
But then x ∈ Ai(x) and hence Un ⊂ ∪nj=1 Aj . The inverse inclusion ∪nj=1 Aj ⊂ Un is
trivial (see the solution to (b) given below). Therefore,
U n = ∪nj=1 Aj ,
as desired.
(b) If p is a limit point of Ai for some i ∈ N, then clearly p is a limit point of the union
U = ∪∞
k=1 Ak . Thus
U ⊃ ∪∞
k=1 Ak .
(c) A counterexample to the inverse inclusion is, for instance, the following setting:
An = (1/n, ∞).
Then U = (0, ∞), and hence U = [0, ∞). On the other hand, An = [1/n, ∞), and
hence
∞
∪∞
n=1 An = ∪n=1 An = U 6= U .
5. [17 points] Solve Exercise 55 in Chapter 2 of the textbook.
Solution:
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(a) Counterexample showing that the closure in R of a disconnected subset of R may be
connected:
M = (0, 1) ∪ (0, 2).
M = [0, 2] is a (connected) interval in R.
(b) Counterexample showing that the interior of a disconnected subset of R may be connected:
M = (0, 1) ∪ {2}.
int M = (0, 1) is a (connected) interval in R.
6. [16 points] Let (R, d) be a metric space where R = (0, +∞] and the metrics d is defined
as follows:
1 1
d(x, y) = − x y
Let N∞ = {+∞, 1, 2, . . .} be the set including all positive integers together with +∞.
(a) Describe geometrically open balls Mr (x) in (R, d).
(b) Describe geometrically bounded sets in (R, d).
(c) Describe all interior, all cluster, and all isolated points of N∞ in (R, d) (a point p of a
subset S of a metric space is called isolated if it belongs to S and there exists r > 0
such that Mr (p) ∩ S contains only one point, namely p itself).
(d) Is N∞ open in (R, d)? Closed? Perfect (a set S in a metric space is called perfect if
any point of S is its cluster point)? Dense? Bounded?
Solution:
(a) Neighborhoods Mr (x) in (R, d) are open intervals either in the form (a, b) (where a
might be 0 to include neighborhoods of +∞) or in the form (a, +∞). Indeed, if r < x−1 ,
then
1 1
− <r
x y
is equivalent to
1
1
1
− r < < + r,
x
y
x
5
which is an open interval
1
x
+r
−1
<y<
1
x
−r
−1
.
If if r ≥ x−1 , then
1 1
− <r
x y
is equivalent to
0<
1
1
< + r,
y
x
which is an open interval
1
x
+r
−1
< y < +∞.
On the other hand, any interval (a, b) is Mr (x) if
1
1
= +r
a
x
and
1
1
= − r,
b
x
which yields
1 −1
x=2·
+
a b
1
and
1 1 1
−
.
r= ·
2
a b
In above computations, a−1 = +∞ if a = 0. Finally, (a, ∞) for a > 0, is Mr (x) with
x = 2a
and
r=
1
.
2a
(b) A set is S bounded in (R, d) if it contained entirely in a open neighborhood in (R, d).
It is not hard to verify that this implies that S is bounded if and only if there is a
usual Euclidean ball B centered at the origin, such that B ∩ S = ∅.
(c) N∞ does not have interior points, the only cluster point is +∞, all its points are
isolated. In fact, any neighborhood of n ∈ N∞ in (R, d) of radius less than one has an
empty intersection with N∞ .
(d) N∞ is not open, is closed (notice that +∞ ∈ R), is not perfect (its only cluster point
is +∞), is not dense, and is bounded in (R, d).
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