Lebesgue_Integration_on_Euclidean_Space__solutions_

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Lebesgue Integration on Euclidean Space
Solutions to Exercises
David Malison
November 2, 2017
Chapter 1
Section A
Problem 1
Suppose x ∈
⋂
i∈I A
c
i . Then x /∈ Ai for all i ∈ I. But this means x /∈ ∪i∈IAi, so x ∈ (∪i∈IAi)
c
.
Conversely, suppose x ∈ (∪i∈IAi)c. Then x /∈ ∪i∈IAi, which implies that x /∈ Ai for all i ∈ I. But this
means x ∈
⋂
i∈I A
c
i . We conclude that (∪i∈IAi)
c
=
⋂
i∈I A
c
i . Take complements of both sides to obtain
∪i∈IAi =
(⋂
i∈I A
c
i
)c
. The second law follows by defining Bi = A
c
i for all i ∈ I and substituting to obtain
∪i∈IBci =
(⋂
i∈I Bi
)c
.
Problem 2
(a) Suppose x ∈
⋂∞
j=1
(⋃∞
k=j Ak
)
. This means that for any j, there exists k ≥ j such that x ∈ Ak. But this
implies x ∈ Ak infinitely often.
Conversely, suppose x ∈ Ak for infinitely many k. This means that for any j, there must exist a k ≥ j
such that x ∈ Ak. But this implies x ∈
⋂∞
j=1
(⋃∞
k=j Ak
)
.
(b) Suppose x ∈
⋃∞
j=1
(⋂∞
k=j Ak
)
. This means there exists some j such that x ∈ Ak for all k ≥ j. But this
means x ∈ Ak for all but finitely many k.
Now suppose x ∈ Ak for all but finitely many k. That means there exists some j such that x ∈ Ak for
all k ≥ j. But this implies x ∈
⋃∞
j=1
(⋂∞
k=j Ak
)
.
(c) If x is in Ak for all but finitely many k, x ∈ Ak infinitely often.
(d) Method 1: (lim supAk)
c =
(⋂∞
j=1
(⋃∞
k=j Ak
))c
=
⋃∞
j=1
((⋃∞
k=j Ak
)c)
=
⋃∞
j=1
(⋂∞
k=j A
c
k
)
= lim inf Ack.
Method 2: If x is not in Ak infinitely often, then there exists some j such that x /∈ Ak for k ≥ j. But
this implies x ∈ Ack for all but finitely many k. Conversely, if x ∈ Ack for all but finitely many k, then x
cannot be in Ak infinitely often.
1
(e) Suppose x ∈ (lim sup(Ak ∪Bk))c. By part (d), we know that this implies x ∈ lim inf(Ak ∪ Bk)c. But
this implies there exists j such that x is not in Ak or Bk for all k ≥ j, which means that x is not in Ak
or Bk infinitely often. We conclude that x ∈ (lim supAk ∪ lim supBk)c.
Conversely, suppose x ∈ (lim supAk∪lim supBk)c. By part (d), we know that this implies x ∈ lim inf Ack∩
lim inf Bck. This means that there exists j such that x ∈ Ack for all k ≥ j and there exists l such that
x ∈ Bck for all k ≥ l. But this implies x ∈ Ack∩Bck for all k ≥ max{j, l}, which implies x ∈ lim inf Ack∩Bck.
Part (d) then implies x ∈ (lim sup(Ak ∪Bk))c.
(f)
lim inf(Ak ∩Bk) = (lim sup(Ack ∪Bck))
c
by part (d)
= (lim supAck ∪ lim supBck)
c
by part (e)
= (lim supAck)
c ∩ (lim supBck)c by Problem 1
= lim inf Ak ∩ lim inf Bk by part (d)
(g) If Ak is decreasing, then ∪j=kAk = Aj . It is then immediate from the definition of lim supAk given in
part (a) that lim supAk = ∩∞j=1Aj . From part (c), we know that lim inf Ak ⊂ lim supAk, so all that
remains is to show that ∩∞k=1Ak ⊂ lim inf Ak. But this follow directly from the definition of lim inf Ak
given in part (b).
(h) If Ak is increasing, then A
c
k is decreasing. By part (g), we have that lim inf A
c
k = lim supA
c
k = ∩∞k=1Ack.
Taking complements yields (lim inf Ack)
c = (lim supAck)
c = ∪∞k=1Ak. Applying the result in part (d), we
have lim supAk = lim inf Ak = ∪∞k=1Ak.
Problem 3
Consider the sequence of open intervals An =
(
0, 1n
)
. The An are decreasing and
1
n+1 ∈ An for all n. Now
fix x ∈ R. If x ≤ 0, x /∈ An for all n. If x > 0, there exists k such that x > 1k . This implies x /∈ An for all
n ≥ k. Thus no x ∈ R is in An for all n.
Section B
Problem 4
Define the equivalence relation ∼ over N by i ∼ j ⇐⇒ f(i) = f(j). By the Axiom of Choice, we can pick
a representative of each equivalence class to form a set B ⊂ N. We know that B is countable by Problem
6, so either B is finite or there exists a bijection g : N → B. If B is finite, A is finite and we are done.
If B is infinite, define the function h : N → A by f ◦ g. If i 6= j, then h(i) = f(g(i)) 6= f(g(j)) = h(j)
because g(i) and g(j) belong to different equivalence classes. Thus h is injective. If x ∈ A, then there
exists i ∈ B such that f(i) = x. Because g is surjective, there exists j ∈ N such that g(j) = i. But then
h(j) = f(g(j)) = f(i) = x, so that h is surjective.
2
Problem 5
Let B = {n ∈ N : f(a) = n for some a ∈ A}. Since B ⊂ N, it is countable by Problem 6. Thus either B
is finite or there exists a bijection g : N → B. If B is finite, then A is finite and we are done. Otherwise,
define the function h : N→ A by f−1 ◦ g. h is injective because it is the composition of injective functions.
If a ∈ A, then there exists n ∈ B such that f(a) = n. Because g is onto, there exists m ∈ N such that
g(m) = n. Then h(m) = f−1(g(m)) = a, so h is onto.
Problem 6
Suppose A is countable and B ⊂ A. If B is finite, there is nothing to prove. Therefore suppose B is
infinite. Since A is countable, there exists an injection f : N → A. Define ai = f(i), so that we can
write A = {ai : i ∈ N}. We define a function g : N → B as follows. Pick n1 large enough so that
I1 = {1 ≤ i ≤ n1 : ai ∈ B} is non-empty (this is possible since B is non-empty). Define i1 = min I1 and
g(1) = ai1 . Next pick n2 large enough so that I2 = {1 ≤ i ≤ n2 : ai ∈ B − {ai1}} is non-empty. Define
i2 = min I2 and g(2) = ai2 . Continue in this fashion by choosing nk such that
Ik =
{
{1 ≤ i ≤ nk : ai ∈ B − {ai1 , ai2 , · · · , aik−1}
}
is non-empty and then defining ik = min Ik and g(k) = aik . If an ∈ B, then there exists k ≤ n such that
g(k) = an. Thus g is surjective. If j < k, then ij < ik by construction. Thus g(j) 6= g(k), which implies g is
injective.
Problem 7
Suppose A is an infinite set. Let g be a choice function on the power set of A (such a choice function exists
by the Axiom of Choice). Define a function f : N→ A by
f(1) = g(A)
f(2) = g(A− {f(1)})
...
f(k) = g(A− {f(1), f(2), · · · , f(k − 1)})
...
The set B = {f(i) : i = 1, 2, · · · } is countable by construction.
3
Problem 8
Arrange the elements of N× N in an array
(1, 1) (1, 2) (1, 3) · · ·
(2, 1) (2, 2) · · ·
(3, 1) · · ·
...
Define the function f : N→ N× N by moving along the anti-diagonals of this matrix:
f(1) = (1, 1)
f(2) = (2, 1)
f(3) = (1, 2)
f(4) = (3, 1)
f(5) = (2, 2)
f(6) = (1, 3)
...
This is called the Cantor pairing function. It should be apparent that f is a bijection, although the rigorous
proof of this fact is somewhat tedious.
Problem 9
A real number q is a positive rational iff there exists natural numbers N and M such that q = NM . Thus
f(n,m) = nm defines a surjection f : N× N → Q+. By Problem 8, we know that N× N is countable. Thus
there exists a bijection g : N→ N×N. Define the function h : N→ Q+ by h = f ◦g. Then h is a composition
of surjective functions and therefore is surjective. Then Problem 4 implies that Q+ is countable. An identical
argument can be used to show that the negative rationals Q− are countable. Since Q = Q− ∪ {0} ∪Q+, by
Problem 10 we know that Q is countable.
Problem 10
Let {An} denote a countable collection of countable sets. We can list the elements of the union of these sets
as
a11 a12 a13 · · ·
a21 a22 · · ·
a31 · · ·
...
where Ai = {ai1, ai2, · · · }, i = 1, 2, · · · . Each horizontal list is allowed to terminate and there may be only a
finite number of horizontal lists. Let B ⊂ N×N denote the collection of indicies in this array. If B is finite,
then ∪nAn is finite and we are done. If B is infinite, Problems 6 and 8 imply the existence of a bijection
4
f : N→ B. Define the bijection g : B → ∪nAn by g(i, j) = aij and the function h : N→ ∪nAn by h = g ◦ f .
h is a composition of bijective functions and therefore is bijective. Thus ∪nAn is countable.
Problem 11
It is obvious that y ∈ (0, 1) by construction. y 6= xk for all k because by construction y and xk differ in the
kth digit of the decimal expansion.
Section C
Problem 12
Square both sides to obtain
|x+ y| ≤ |x|+ |y| ⇐
N∑
i=1
(xi + yi)
2 ≤
N∑
i=1
x2i + 2|x||y|+
N∑
i=1
y2i
⇐
N∑
i=1
xiyi ≤ |x||y|
⇐
∣∣∣∣∣
N∑
i=1
xiyi
∣∣∣∣∣ ≤ |x||y|
Square both sides again to obtain
⇐
∣∣∣∣∣∣
N∑
i=1
N∑
j=1
xiyixjyj
∣∣∣∣∣∣ ≤
N∑
i=1
x2i
N∑
j=1
y2j
⇐ 0 ≤
N∑
i=1
N∑
j=1(x2i y
2
j − |xi||yi||xj ||yj |)
⇐ 0 ≤ 1
2
N∑
i=1
N∑
j=1
(x2i y
2
j − 2|xi||yi||xj ||yj |+ x2jy2i )
⇐ 0 ≤ 1
2
N∑
i=1
N∑
j=1
(|xi||yj | − |xj ||yi|)2
⇐ 0 ≤
∑
1≤i<j≤N
(|xi||yj | − |xj ||yi|)2
Problem 13
If y = cx for some c ∈ R+, then
|x+ y| = |(1 + c)x| = (1 + c)|x| = |x|+ c|x| = |x|+ |y|
5
Conversely, suppose |x+ y| = |x|+ |y|. Then we have
|x+ y| = |x|+ |y| ⇒
N∑
i=1
(xi + yi)
2 =
N∑
i=1
x2i + 2|x||y|+
N∑
i=1
y2i
⇒
N∑
i=1
xiyi = |x||y|
⇒
N∑
i=1
N∑
j=1
xiyixjyj =
N∑
i=1
x2i
N∑
j=1
y2j
⇒ 0 =
∑
1≤i<j≤N
(xiyj − xjyi)2
⇒ xiyj = xjyi for all i 6= j
If x = 0, then x = 0 · y. Suppose x 6= 0 and pick xi 6= 0. Then we must have yj = yixixj for all j. But this
implies y = cx, where c = yixi .
Problem 14
(a) This follows from the definition of square root.
(b) If x = y, then d(x, y) =
√
0 + · · ·+ 0 = 0. If d(x, y) = 0, then (xi − yi)2 = 0 for all i, which implies
xi = yi for all i.
(c) d(x, y) =
√∑N
i=1(xi − yi)2 =
√∑N
i=1(yi − xi)2 = d(y, x).
(d) d(x, y) = |x− y| = |x− z + z − y| ≤ |x− z|+ |z − y| = d(x, z) + d(z, y)
Problem 15
From Problem 14, we have
d(x, y) ≤ d(x, z) + d(y, z)
d(x, z) ≤ d(x, y) + d(y, z)
which implies
d(x, y)− d(x, z) ≤ d(y, z)
d(x, z)− d(x, y) ≤ d(y, z)
These inequalities then imply |d(x, y)− d(x, z)| ≤ d(y, z).
Problem 16
Suppose r+ r′ ≤ d(x, x′). Then by Problem 14(d), r+ r′ ≤ d(x, y) + d(x′, y) for all y ∈ Rn. But this means
no y can satisfy d(x, y) < r and d(x, y) < r′, so B(x, r) and B(x, r′) are disjoint.
6
Conversely, suppose r + r′ > d(x, x′). Define y = rr+r′x+
r′
r+r′x
′. Then we have
d(x, y) = |x− y| d(x′, y) = |x′ − y|
=
∣∣∣∣ r′r + r′ (x− x′)
∣∣∣∣ = ∣∣∣∣ rr + r′ (x′ − x)
∣∣∣∣
=
|x− x′|
r + r′
r′ =
|x′ − x|
r + r′
r
=
d(x, x′)
r + r′
r′ =
d(x, x′)
r + r′
r
< r′ < r
Thus y ∈ B(x, r) ∩B(x′, r′), so the two sets are not disjoint.
Problem 17
Suppose d(x, x′) ≤ r′ − r. Then for any y ∈ B(x, r), we have d(x′, y) ≤ d(x′, x) + d(x, y) < r′ − r+ r = r′ so
that y ∈ B(x′, r′).
Conversely, suppose B(x, r) ⊂ B(x′, r′). If x = x′, then we must have r ≤ r′. Now suppose x 6= x′. Define
yn = x+
(
1− 1n
)
r
d(x,x′) (x− x
′) for n ≥ 1. Now d(x, yn) =
(
1− 1n
)
r < r, which implies by assumption that
d(x′, yn) < r
′. But d(x′, yn) = d(x, x
′) +
(
1− 1n
)
r. Combining these results, we have d(x, x′) < r′ − r + 1nr.
Since this holds for all n, we can take the limit as n goes to infinity to obtain d(x, x′) ≤ r′ − r.
Problem 18
(a) This follows since there are no points in ∅.
(b) This follows since B(x, r) ⊂ Rn for all x ∈ Rn and r > 0.
(c) Let {Si : i ∈ I} be a collection of open sets and suppose x ∈
⋃
i∈I Si. Then there exists i ∈ I such that
x ∈ Si. Since Si is open, there exists r > 0 such that B(x, r) ⊂ Si. But this implies B(x, r) ⊂
⋃
i∈I Si.
(d) Let {Si : i = 1, · · · , n} be a finite collection of open sets and suppose x ∈
⋂n
i=1 Si. Since x ∈ Si
for all i, for each i there exists ri > 0 such that B(x, ri) ⊂ Si. Define r = mini ri > 0. Since
B(x, r) ⊂ B(x, ri) ⊂ Si for i = 1, · · · , n, we have that B(x, r) ⊂
⋂n
i=1 Si.
Problem 19
Suppose Sn =
(
− 1n ,
1
n
)
, n = 1, 2, · · · ,. Then ∩∞n=1Sn = {0}, which is not open.
Problem 20
Fix x ∈ R and r > 0 and suppose x′ ∈ B(x, r). Then d(x′, x) < r, so we can find an r′ such that
0 < r′ ≤ r − d(x′, x). By Problem 17, we have that B(x′, r′) ⊂ B(x, r).
7
Problem 21
(a) This follows directly from the definition of an open set.
(b) Suppose x ∈ Ao. Then there exists r > 0 such that B(x, r) ⊂ A. Since B(x, r) is open, for any
x′ ∈ B(x, r) we can find r′ > 0 such that B(x′, r′) ⊂ B(x, r). But this implies B(x′, r′) ⊂ A, so x′ is an
interior point of A. This means B(x, r) ⊂ Ao, so x is an interior point of Ao.
(c) This follows from parts (a) and (b).
(d) Suppose x ∈ (A ∩B)o. Then there exists r > 0 such that B(x, r) ⊂ A ∩B. But this means B(x, r) ⊂ A
and B(x, r) ⊂ B, so x ∈ Ao and x ∈ Bo.
Conversely, suppose x ∈ Ao ∩Bo. Then there exists r1 and r2 such that B(x, r1) ⊂ A and B(x, r2) ⊂ B.
Define r = min{r1, r2}. Then B(x, r) ⊂ A and B(x, r) ⊂ B, so B(x, r) ⊂ A ∩B. Thus x ∈ (A ∩B)o.
(e) Suppose A = (0, 1) and B = [1, 2). Then A∪B = (0, 2) but Ao ∪Bo = (0, 1)∪ (1, 2). Since 1 ∈ (A∪B)o
but 1 /∈ Ao ∪Bo, the two sets are not equal.
(f) If x ∈ Ao, then there exists r > 0 such that B(x, r) ⊂ A. But B(x, r) is open, so x is in the union of all
open subsets of A.
Now suppose x ∈ B ⊂ A where B is open. Then there exists r > 0 such that B(x, r) ⊂ B ⊂ A, so x is
an interior point of A.
(g) This follows directly from part (f).
Problem 22
Parts (a) and (b) follow immediately from Problem 18(a) and Problem 18(b). Suppose {Ai : i ∈ I} is a
collection of closed sets. Then Aci is open for all i ∈ I. Problem 18(c) then implies that ∪i∈IAci is open. But
this means (∪i∈IAci )c = ∩i∈IAi is closed. Problem 18(d) implies that if I is finite, ∩i∈IAci is open. This
then implies (∩i∈IAci )c = ∪i∈IAi if I is finite.
Problem 23
If for every r > 0 the set A ∩B(x, r) is infinite, then by definition x is a limit point of A.
Conversely, suppose x is a limit point of A and assume there exists an r such that B(x, r) ∩A is finite. Let
{x1, · · · , xn} denote the collection of points in B(x, r)∩A such that xk 6= x. Define r′ = min1≤k≤n d(xk, x).
Then B(x, r′) ∩A does not contain any elements not equal to x, so x cannot be a limit point.
Problem 24
Suppose A is closed. Let x be a limit point of A. If x ∈ Ac, then there exists r > 0 such that B(x, r) ⊂ Ac.
But this implies B(x, r) ∩A = ∅, which means that x cannot be a limit point. Thus x must be in A.
Conversely, suppose A contains all its limit points. Suppose x ∈ Ac. Since x is not a limit point, there exists
r > 0 such that A ∩B(x, r) = ∅. But this implies B(x, r) ⊂ Ac, which means that Ac is open.
8
Problem 25
Suppose x ∈ A− and fix r > 0. If x ∈ A, then x ∈ A∩B(x, r). If x is a limit point of A, then A∩B(x, r) 6= ∅
by definition.
Conversely suppose that the set A∩B(x, r) 6= ∅ for all r > 0. If x is not a limit point of A, then there exists
r̄ > 0 such that B(x, r̄)− {x} ⊂ Ac. But since A ∩B(x, r̄) is non-empty, we must have x ∈ A.
Problem 26
Suppose x ∈ A−. If x ∈ A, then choose xk = x for all k. If x /∈ A, then x is a limit point of A. Define rk = 1k .
For each k, there exists xk ∈ B(x, rk)∩A. But 0 ≤ d(xk, x) < 1k for all k, which implies limk→∞ d(xk, x) = 0.
Conversely, suppose there exists a sequence {xk}∞k=1 with xk ∈ A for all k such that xk → x. Suppose
x ∈ Ac. For any r > 0 there exists n ≥ 1 such that d(xk, x) < r for all k ≥ n. Since xn ∈ B(x, r) ∩ A and
xn 6= x, x is a limit point of A.
Problem 27
Suppose x ∈ Aoc. Then for any r > 0, there exists x′ ∈ B(x, r) ∩ Ac. If x ∈ Ac, then x ∈ Ac−. If x ∈ A,
then x′ 6= x for all r so x is a limit point of Ac. Thus x ∈ Ac−.
Conversely, suppose x ∈ Ac−. If x ∈ Ac, then x ∈ Aoc since Ac ⊂ Aoc. If x ∈ A, then x is a limit point of
Ac. This means that for any r > 0 there exists x′ ∈ B(x, r) ∩Ac. But this implies x ∈ Aoc.
Problem 28
(a) Statement: A is closed ⇐⇒ A− = A.
Proof: This follows directly from Problem 24.
(b) Statement: A− is closed.
Proof: The result in Problem 27 implies that that A−c = Aco. Since Aco is open, A− is closed.
(c) Statement: A−− = A−.
Proof: This follows from parts (a) and (b).
(d) Statement: (A ∪B)− = A− ∪B−.
Proof: From Problem 21(d), we know that (Ac ∩ Bc)o = Aco ∩ Bco. Taking complements and applying
the result in Problem 27, we obtain
(A ∪B)− = (Ac ∩Bc)c− = (Ac ∩Bc)oc = Acoc ∪Bcoc = Acc− ∪Bcc− = A− ∪B−
(e) Statement: (A ∩B)− does not in general equal A− ∩B−.
Proof: Consider A = (0, 1) and B = [1, 2). Then (A ∩B)− = ∅ but A− ∩B− = {1}.
9
(f) Statement: A− = the intersection of all closed sets containing A.
Proof: Let B denote the intersection of all closed sets containing A. Since A− is closed and contains A,
B ⊂ A−. Suppose x ∈ Bc. Then there exists a set closed G containing A such that x ∈ Gc. But this
implies x ∈ Ac and there exists r > 0 such that B(r, x) ⊂ Gc ⊂ Ac. But this means x is not a limit
point of A, sox ∈ A−c.
(g) Statement: A− is the smallest closed set containing A (i.e. A− is a closed set containing A and if B is
any closed set containing A, then A− ⊂ B.)
Proof: This follows directly from part (f).
Problem 29
Define B = {y : d(x, y) ≤ r}. Suppose y ∈ B(x, r)−. By Problem 26, there exists a sequence yk ∈ B(x, r)
for all k with yk → y. But d(y, x) ≤ d(y, yk) + d(yk, x) < d(y, yk) + r. Since this inequality holds for all k
and d(y, yk)→ 0, we must have d(y, x) ≤ r. This implies y ∈ B, so we must have B(x, r)− ⊂ B.
Next suppose that y ∈ B. Fix r′ > 0. If r′ > r, then x ∈ B(x, r) ∩ B(y, r′). If r′ ≤ r, define x′ =
r′
r+1x +
r+1−r′
r+1 y. Then d(x
′, x) = r+1−r
′
r+1 d(x, y) < r and d(x
′, y) = d(x,y)r+1 r
′ < r′, so x′ ∈ B(x, r) ∩ B(y, r′).
From Problem 25, we know that this implies y ∈ B(x, r)−. Thus B ⊂ B(x, r)−.
Notice that B(x, r) is an open subset of B (see problem 20), so by Problem 21(f) we must have B(x, r) ⊂ Bo.
Now suppose y ∈ Bo. Then there exists r′ such that B(y, r′) ⊂ B. Pick any z ∈ B(y, r′) with z 6= y. Then
d(x, y) ≤ d(x, z) + d(y, z) ≤ r + d(y, z) < r, which implies y ∈ B(x, r). Thus Bo ⊂ B(x, r).
Problem 30
Suppose A = (0, 1) ∪ (1, 2). Then A− = [0, 2] so that A−o = (0, 2) 6= A.
Problem 31
Suppose A is both open and closed and that there exists x ∈ A and y ∈ Ac. Define zθ = θx + (1 − θ)y for
θ ∈ [0, 1]. Let θ∗ = sup{θ ∈ [0, 1] : zθ ∈ A}. Then there exists a sequence θn ∈ {θ ∈ [0, 1] : zθ ∈ A} such
that θn → θ∗. Since zθn ∈ A and zθn → zθ∗ , zθ∗ is in A− (Problem 26). Since A is closed, zθ∗ ∈ A (Problem
28(a)). But notice that for any θ > θ∗, we must have zθ ∈ Ac. Let θ̃n be a sequence in (θ∗, 1] converging
downward to θ∗. But zθ̃n ∈ A
c for all n and zθ̃n → zθ∗ , so zθ∗ is in A
c−. But Ac is closed, so zθ∗ ∈ Ac.
Since zθ∗ cannot be in both A and A
c, we have a contradiction. Thus if A is both open and closed, either
A is empty or Ac is empty. This implies A = ∅ or A = Rn.
Section D
Problem 32
(a) Let {Gi, i ∈ I} be an open covering of ∅. Then Gi contains ∅ for all i, so Gi is a finite subcovering.
10
(b) Let {Gi, i ∈ I} be an open covering of the finite set A = {a1, · · · , an}. For each j ∈ {1, 2, · · · , n}, there
exists ij such that aj ∈ Gij . But then {Gij : j = 1, 2, · · · , n} is a finite subcover of A.
(c) Suppose A and B are compact and {Gi, i ∈ I} is an open covering of A ∪ B. Since {Gi, i ∈ I} is an
open cover of A and of B, there exists finite subcovers of A and B. Let IA and IB denote the index sets
of these finite subcovers. Then {Gi : i ∈ IA ∪ IB} is a finite subcover of A ∪B.
(d) Suppose {Ak, k = 1, 2 · · · } is a collection of compact sets. Suppose ∪n−1k=1Ak is compact for some n ≥ 2.
Then ∪nk=1Ak = ∪
n−1
k=1Ak∪An is compact by part (c). It therefore follows by induction that that ∪nk=1Ak
is compact for n = 2, 3, · · · .
(e) Fix r > 0 and consider the collection of sets
{
B
(
x, r − �n
)
, n = 1, 2, · · ·
}
where 0 < � < r. If d(x, y) < r,
then there exists n large enough such that d(x, y) < r − �n . This implies
{
B
(
x, r − �n
)
, n = 1, 2, · · ·
}
is
an open cover of B(x, r). Let N ⊂ N denote a finite index set. Since N is finite, it contains a largest
number N . Then any y such that r − �N < d(x, y) < r is in B(x, r) but not in ∪n∈NB
(
x, r − �n
)
. Thus{
B
(
x, r − �n
)
: n ∈ N
}
is not a cover of B(x, r).
(f) Consider the collection of sets {B(0, n), n = 1, 2, · · · }. This is an open cover of Rn. Let N ⊂ N denote
a finite index set. Since N is finite, it contains a largest number N . Then any y such that N < d(0, y)
is in Rn but not in ∪n∈NB (x, n). Thus {B (0, n) : n ∈ N} is not a cover of Rn.
Problem 33
Suppose A is compact. Consider the collection of open sets
Gn = {B(0, n) : n = 1, 2, · · · }
Since this collection covers all of Rn, it must form an open cover of A. By compactness, there must exist
a finite set N ⊂ N such that A ⊂ ∪n∈NGn. Since N is finite, it contains a largest number N . But then
A ⊂ B(0, N), so A is bounded.
Problem 34
Consider the sets A = N and B = ∪∞n=1{n− 12n}. A and B are closed and disjoint. However, for any � > 0,
there exists n such that 12n < �. But then n ∈ A and n−
1
2n ∈ B but d(n, n−
1
2n ) < �.
Problem 35
For any non-empty set A ⊂ Rn, define d(x,A) = inf{d(x, y) : y ∈ A} (see Chapter 1, Section F). If F1
is empty, take G1 = ∅ and G2 = Rn. Likewise, if F2 is empty take G1 = Rn and G2 = ∅. Thus we
can assume without loss that F1 and F2 are nonempty. Let G1 = {x ∈ Rn : d(x, F1) < d(x, F2)} and
G2 = {x ∈ Rn : d(x, F1) > d(x, F2)}. G1 and G2 are disjoint by construction.
To see that F1 ⊂ G1, fix x ∈ F1. Since F2 is closed and F1 ⊂ F c2 , there exists r > 0 such that B(x, r) ⊂ F c2 .
This implies d(x, F2) ≥ r > 0. But since d(x, F1) = 0, we must have x ∈ G1. It can likewise be shown that
F2 ⊂ G2.
11
We next show that G1 is open. To see this, fix r > 0. Define Hr = {x : d(x, F1) < r} and fix x ∈ Hr. Define
r′ = r − d(x, F1) > 0. For any y ∈ B(x, r′) and any z ∈ F1, we must have
d(y, F1) ≤ d(y, z) ≤ d(y, x) + d(x, z)
Since this holds for all z ∈ F1, we must have
d(y, F1) ≤ d(y, x) + d(x, F1) < r′ + d(x, F1) = r
which implies y ∈ Hr. Thus Hr is open. Now define H̃r = {x : d(x, F2) > r}. Define r′ = d(x, F2) − r > 0
and fix x ∈ H̃r. Then for any y ∈ B(x, r′) and z ∈ F2, we have
d(y, z) ≥ d(x, z)− d(y, x) ≥ d(x, F2)− d(y, x)
Since this holds for all z ∈ F2, we have
d(y, F2) ≥ d(x, F2)− d(y, x) > d(x, F2)− r′ = r
This implies y ∈ H̃r, so H̃r is open. Now for any x ∈ G1, there exists r such that d(x, F1) < r < d(x, F2).
But this implies x ∈ Hr ∩ H̃r. Since Hr ∩ H̃r is open, there exists r′ > 0 such that B(x, r′) ⊂ Hr ∩ H̃r. But
Hr ∩ H̃r ⊂ G1, so B(x, r′) ⊂ G1. Thus G1 is open. A symmetric argument can be used to show that G2 is
open.
Problem 36
Let x1, x2, · · · be a bounded sequence in Rn. Define the set A = {x : x = xk for some k = 1, 2, · · · }. If
A is finite, then there exists x ∈ A such that xk = x infinitely often. Thus we can construct a convergent
subsequence using the index set {k ∈ N : xk = x}. If A is infinite, then by the Bolzano-Weierstrass Theorem
there exists a limit point x of A. Construct a subsequence as follows. Let n1 = 1. Pick nk+1 > nk such that
d(x, xnk+1) <
1
k . This is possible by Problem 23. Then xnk → x.
Section E
Problem 37
Suppose f is continuous at x0. Fix � > 0. Then there exists δ > 0 such that d(x, x0) < δ implies
d(f(x), f(x0)) = (
∑m
k=1(fk(x)− fk(x0))2)1/2 < �. But this implies |fk(x)− fk(x0)| < � for k = 1, · · · ,m.
Conversely, suppose f1, · · · , fm are continuous at x0 = (x01, · · · , x0m)′. Fix � > 0. For each k ∈ {1, 2, · · · ,m},
there exists δk > 0 such that d(x, x0) < δk implies d(fk(x), f(x)) <
�√
m
. Define δ = mink δk. Then if
d(x, x0) < δ, we have
d(f(x), f(x0)) =
(
m∑
k=1
(fk(x)− fk(x0))2
)1/2
<
(
m∑
k=1
�2
m
)1/2
= �
12
so that f is continuous at x0.
Problem 38
I first prove a useful lemma:
Lemma: Suppose f1 : Rk → R and f2 : Rl → R are continuous at x0 and y0, respectively. Let f : Rk×Rl → R,
g : Rk × Rl → R, and h : Rk × Rl → R be defined as f(x, y) = f1(x) + f2(y), g(x, y) = f1(x)f2(y) and
h(x, y) = f1(x)/f2(y), respectively. Then (i) f is continuous at (x0, y0), (ii) g is continuous at (x0, y0), and
(iii) h is continuous at (x0, y0) if f2(y0) 6= 0.
Proof: (i) Fix � > 0. There exists δ1 > 0 and δ2 > 0 such that |x − x0| < δ1 implies |f1(x) − f1(x0)| < �/2
and |y − y0| < δ2 implies |f2(y)− f2(y0)| < �/2. Define δ = min(δ1, δ2). Then d((x, y), (x0, y0)) < δ implies
that
d(f(x, y), f(x0, y0)) = |f1(x) + f2(y)− f1(x0)− f2(y0)| ≤ |f1(x)− f1(x0)|+ |f2(y)− f2(y0)| < �
(ii) Observe that
d(g(x, y), g(x0, y0)) = |f1(x)f2(y)− f1(x0)f2(y0)|
= |f1(x)f2(y)− f1(x0)f2(y) + f1(x0)f2(y)− f1(x0)f2(y0)|
≤ |f2(y)||f1(x)− f1(x0)|+ |f1(x0)||f2(y)− f2(y0)|
≤ |f2(y)− f2(y0)||f1(x)− f1(x0)|+ |f2(y0)||f1(x)− f1(x0)|+ |f1(x0)||f2(y)− f2(y0)|
There exists δ1 > 0 such that |y−y0| < δ1 implies |f2(y)−f2(y0)| < 1. There also exists δ2 > 0 and δ3 > 0 such
that |x− x0| < δ2 implies |f1(x)− f1(x0)| < �2(1+|f2(y0)|)and |y− y0| < δ3 implies |f2(y)− f2(y0)| <
�
2|f1(x0)|
(if f1(x0) = 0, pick δ3 = 1). Define δ = min(δ1, δ2, δ3). Then if d((x, y), (x0, y0)) < δ, we have
d(g(x, y), g(x0, y0)) < (1 + |f2(y0)|)|f1(x)− f1(x0)|+ |f1(x0)||f2(y)− f2(y0)| <
�
2
+
�
2
= �
(iii) Part (ii) will imply that h(x, y) is continuous at (x0, y0) if 1/f2(y) is continuous at y0 whenever f2(y0) 6= 0.
To see that this is indeed true, assume that f2(y0) 6= 0. Then there exists η > 0 such that 0 < η < |f2(y0)|.
Observe that
d(1/f2(y), 1/f2(y0)) =
∣∣∣∣f2(y0)− f2(y)f2(y)f2(y0)
∣∣∣∣ = |f2(y0)− f2(y)||f2(y)||f2(y0)|
for y such that f2(y) 6= 0. There exists δ1 > 0 such that |y−y0| < δ1 implies |f2(y0)−f2(y)| < η. There also
exists δ2 > 0 such that |y− y0| < δ2 implies |f2(y0)− f2(y)| < �(|f2(y0)| − η)|f2(y0)|. Define δ = min(δ1, δ2).
Then for |y − y0| < δ, we have
d(1/f2(y), 1/f2(y0)) <
|f2(y0)− f2(y)|
(|f2(y0)| − η)|f2(y0)|
< �
Corollary: Suppose f1 : Rk → R and f2 : Rk → R are continuous at x0. Let f : Rk → R, g : Rk → R, and
h : Rk → R be defined as f(x) = f1(x) + f2(x), g(x) = f1(x)f2(x) and h(x) = f1(x)/f2(x), respectively.
13
Then (i) f is continuous at x0, (ii) g is continuous at x0, and (iii) h is continuous at x0 if f2(x0) 6= 0.
Proof: Follow the same steps as in the proof of the lemma, replacing y and y0 with x and x0, respectively.
(a) Observe that f(tx) = 0 if x1x2 = 0, which is continuous for all t ∈ R. If x1x2 6= 0, we have
f(tx) =

t3x21x2
t4x41+t
2x22
if t 6= 0
0 if t = 0
Repeated application of parts (i) and (ii) of the lemma and its corollary can be used to show that all
multivariate polynomials are continuous. Part (iii) of the lemma then implies that f(tx) function is
continuous everywhere except at t = 0. To check continuity at t = 0, fix � > 0 and let δ = |x2|
x21
�. Then
if 0 < |t| < δ, we have
|f(tx)| =
∣∣∣∣ tx21x2t2x41 + x22
∣∣∣∣ < ∣∣∣∣ tx21x2x22
∣∣∣∣ = |t| x21|x2| < �
(b) Fix � ∈ (0, 12 ) and pick any δ > 0. Choose x1 6= 0 such that x
2
1 + x
4
1 < δ
2 and let x2 = x
2
1. Then
f(x) = 12 . Thus for any δ > 0, we can find x ∈ R
2 such that d(x, 0) < δ but d(f(x), f(0)) > �.
Problem 39
Fix � > 0 and let δ = �. Then if d(y, z) < δ, we have |d(x, y)− d(x, z)| ≤ d(y, z) < δ = �.
Problem 40
(a) x ∈ f−1
(⋃
i∈I Yi
)
⇐⇒ x ∈ A and there exists i ∈ I such that f(x) ∈ Yi ⇐⇒ x ∈
⋃
i∈I f
−1(Yi).
(b) x ∈ f−1
(⋂
i∈I Yi
)
⇐⇒ x ∈ A and f(x) ∈ Yi for all i ∈ I ⇐⇒ x ∈
⋂
i∈I f
−1(Yi).
(c) Suppose y ∈ f
(⋃
i∈I Xi
)
. Then there exists x ∈
⋃
i∈I Xi such that f(x) = y. But since x ∈ Xi for some
i ∈ I, f(x) ∈ f(Xi) for some i ∈ I. This implies y ∈
⋃
i∈I f(Xi).
Conversely, suppose y ∈ f(Xi) for some i ∈ I. Then there exists x ∈ Xi such that f(x) = y. But since
x ∈
⋃
i∈I Xi, y ∈ f
(⋃
i∈I Xi
)
.
(d) Suppose y ∈ f
(⋂
i∈I Xi
)
. Then there exists x ∈
⋂
i∈I Xi such that f(x) = y. But since x ∈ Xi for all
i ∈ I, f(x) ∈ f(Xi) for all i ∈ I. This implies y ∈
⋂
i∈I f(Xi).
(e) Suppose f : {0, 1} → {0} (i.e. f(0) = f(1) = 0). Let X1 = {0} and X2 = {1}. Then f(X1 ∩X2) = ∅ but
f(X1) ∩ f(X2) = {0}.
Problem 41
(a) Suppose y ∈ f(f−1(Y )). Then there exists x ∈ f−1(Y ) such that f(x) = y. This implies x ∈ A and
f(x) ∈ Y , so y ∈ Y ∩ f(A).
14
Conversely, suppose y ∈ Y ∩ f(A). Then there exists x ∈ A such that y = f(x). But since y ∈ Y , we
must have x ∈ f−1(Y ). This implies y ∈ f(f−1(Y )).
(b) Pick x ∈ X ⊂ A. Then f(x) ∈ f(X), so x ∈ f−1(f(X)).
(c) Suppose x ∈ f−1(f(X)) and f is one-to-one. Then x ∈ A and f(x) ∈ f(X). Thus there must exist
x′ ∈ X such that f(x′) = f(x). Because f is one-to-one, we must have x = x′ ∈ X.
Problem 42
Suppose f : A→ Rm is continuous on A ⊂ Rn. Let F ⊂ Rm be a closed set. Then F c is open, which implies
there exists an open set G ⊂ Rn such that f−1(F c) = A ∩G. But observe that
f−1(F c) = {x ∈ A : f(x) ∈ F c} = A ∩ f−1(F )c (1)
Taking complements of both sides gives us
Ac ∪Gc = Ac ∪ f−1(F )
Intersecting both sides with A, we obtain
f−1(F ) = A ∩Gc
But Gc is closed, yielding the desired result.
For the converse, pick an open set G ⊂ Rm. Then Gc is closed, so by assumption there exists a closed set
H ⊂ Rn such that f−1(Gc) = A∩H. But from equation (1), we know that f−1(Gc) = A∩ f−1(G)c. Taking
complements and intersecting both sides with A gives us f−1(G) = A∩Hc. Since Hc is open, f is continuous
on A.
Problem 43
Take f(x) = x2 and G = (−1, 1). Then f(G) = [0, 1) which is not open since 0 is not an interior point.
Problem 44
Take f(x) = 11+x2 and G = R. Then f(G) = (0, 1] which is not closed since 0 is a limit point of f(G) but is
not in f(G).
Problem 45
Let A and B denote the open subsets of A and B, respectively. Let g : B → A denote the continuous inverse
of the continuous bijection f : A→ B. Fix C ∈ A. Then we have
f(C) = {f(x) : x ∈ C} = {y ∈ B : g(y) ∈ C} = g−1(C)
15
But g−1(C) is an open set by continuity of g, so f(C) ∈ B. Thus f defines a mapping from A to B.
To see that this mapping is one-to-one, consider two non-equal sets C1 and C2 in A. Suppose without
loss that there exists x ∈ C1 ∼ C2. By construction, f(x) ∈ f(C1). If f(x) ∈ f(C2), there would exist
x′ ∈ C2 with f(x′) = f(x). But since x ∈ Cc2, this would violate the assumption that f is injective. Thus
f(x) ∈ f(C1) ∼ f(C2), so f(C1) and f(C2) are not equal. To see that the mapping is onto, fix D ∈ B. Then
by continuity of f , f−1(D) ∈ A. By Problem 41(a), we know that f(f−1(D)) = D ∩ f(A). But since f is
onto, f(A) = B so that f(f−1(D)) = D.
Problem 46
Using the result in Problem 42, we can follow the same proof in Problem 45 with the word “open” replaced
with “closed.”
Problem 47
Consider the function f(x) = 11+x2 and let K = [0, 1]. Then f
−1(K) = R, which is not compact.
Problem 48
Consider the function f(x) = sin(x2). Then f is bounded between −1 and 1. To see that f is continuous,
note that it is the composition of continuous functions.1 Thus f is continuous by the following result:
Lemma: Suppose f : Rk → Rl and g : Rl → Rm are continuous. Then h : Rk → Rm defined as h(x) = g(f(x))
is continuous.
Proof: Let A ⊂ Rm be open. Then
h−1(A) = {x ∈ Rk : g(f(x)) ∈ A}
= {x ∈ Rk : f(x) ∈ g−1(A)}
= f−1(g−1(A))
g−1(A) is open by continuity of g, which implies that f−1(g−1(A)) is open by continuity of f . We conclude
that h−1(A) is open, so h is continuous.
To show that f is not uniformly continuous on R, pick � ∈ (0, 1) and fix δ > 0. For any positive integer n
we have
∣∣∣f(√nπ)− f(√nπ + π/2)∣∣∣ = 1 > �. But we also have
∣∣∣√nπ −√nπ + π/2∣∣∣ = ∣∣∣∣∣ π/2√nπ +√nπ + π/2
∣∣∣∣∣ < 1√n
Therefore
∣∣∣√nπ −√nπ + π/2∣∣∣ < δ if we pick n ≥ 1δ2 . Thus for any δ > 0, we can find x, y ∈ R with
d(x, y) < δ but d(f(x), f(y)) > �.
1Rigorously proving the continuity of sinx is beyond the scope of these solutions. See, e.g., Rudin (1964) Chapter 8.
16
Problem 49
Fix � > 0. Let δ = �C . Then if d(x, y) < δ, we have
d(f(x), f(y)) ≤ Cd(x, y) < Cδ = �
Problem 50
Suppose x ∈ A−. Then there exists xk ∈ A with xk → x. But by Problem 39, this implies
lim
k→∞
d(xk, x)→ d(x, x) = 0
so that d(x,A) ≤ 0. But since d(x, y) ≥ 0 for all y ∈ A, we also must have d(x,A) ≥ 0. Thus d(x,A) = 0.
Conversely, suppose d(x,A) = 0. Then there exists x0 ∈ A− such that d(x,A) = d(x, x0) = 0. But by
Problem 14(b), this implies that x = x0.
Problem 51
If B ⊂ A, then
{d(x, y) : y ∈ B} ⊂ {d(x, y) : y ∈ A}
Thus if B 6= ∅, we must have
d(x,B) = inf{d(x, y) : y ∈ B} ≥ inf{d(x, y) : y ∈ A} = d(x,A)
Problem 52
By Problem 51, we must have d(x,A) ≥ d(x,A−). To show the reverse inequality, observe that there exists
x0 ∈ A− with d(x, x0) = d(x,A−). But then there must exist a sequence {xk} with xk ∈ A for all k such
that xk → x0. By Problem 39, this implies d(x, xk)→ d(x,A−). But this implies d(x,A−) ≥ d(x,A).
Problem 53
If A− = B−, Problem 52 implies
d(x,A) = d(x,A−) = d(x,B−) = d(x,B)
for all x ∈ Rn. Conversely, suppose d(x,A) = d(x,B) for all x ∈ R. Then from Problem 50 we have
x ∈ A− ⇐⇒ d(x,A) = 0 ⇐⇒ d(x,B) = 0 ⇐⇒ x ∈ B−
so that A− = B−.
17
Problem 54
Suppose F1 and F2 are disjoint closed subsetsof Rn. Let f : Rn → R denote an Urysohn function for the
pair F1 ⊂ F c2 . Consider the sets
G1 = {x ∈ Rn : f(x) > .5}, G2 = {x ∈ Rn : f(x) < .5}
If x ∈ F1, then f(x) = 1 so that x ∈ G1. If x ∈ F2, then f(x) = 0 so that x ∈ G2. Thus F1 ⊂ G1
and F2 ⊂ G2. G1 and G2 are disjoint by construction. To see that G1 and G2 are open, notice that
G1 = f
−1(−∞, .5) and G2 = f−1(.5,∞). The result then follows by the continuity of f .
Problem 55
Suppose A = {x}. Then {d(x, y)|x, y ∈ A} = {0}, so diam(A) = 0.
Suppose diamA = 0. Since d(x, y) ≥ 0 for all x, y ∈ A, we must have d(x, y) = 0 for all x, y ∈ A. But then
Problem 14(b) implies x = y for all x, y ∈ A, so that A must consist of a single point.
Problem 56
If A is bounded, there exists x0 ∈ Rn and r ∈ R such that A ⊂ B(x0, r). But then for any x, y ∈ A, we must
have
d(x, y) ≤ d(x, x0) + d(y, x0) ≤ 2r
This implies diamA ≤ 2r <∞.
Suppose diam(A) < ∞. Fix x0 ∈ A. For any x ∈ A, we must have d(x0, x) ≤ diam(A). Thus A ⊂
B(x0,diam(A) + 1).
Problem 57
Since A ⊂ A−, we must have diamA ≤ diamA−. To show the reverse inequality, first suppose diamA− =∞.
Fix M < ∞. For any � > 0, there exists x0, y0 ∈ A− such that d(x0, y0) ≥ M + �. Moreover, there exists
x, y ∈ A such that d(x, x0) < �/2 and d(y, y0) < �/2. This implies
M + � ≤ d(x0, y0) ≤ d(x0, x) + d(x, y) + d(y, y0) < d(x, y) + �
so that M < d(x, y). Thus for any M we can find x, y ∈ A with d(x, y) > M , so diamA =∞.
Next suppose diamA− < ∞. By Problem 59, we know there exists x0, y0 ∈ A− such that diamA− =
d(x0, y0). We this means there exists sequences {xn} and {yn} in A such that xn → x0 and yn → y0. Fix
� > 0 and choose n large enough that d(yn, y0) < �/2 and d(xn, x0) < �/2. Then we have
|d(xn, yn)− d(x0, y0)| ≤ |d(xn, yn)− d(xn, y0)|+ |d(xn, y0)− d(x0, y0)| ≤ d(yn, y0) + d(xn, x0) < �
18
where the second inequality follows from Problem 15. But this implies for any � > 0, there exists xn, yn ∈ A
with d(xn, yn) > diamA
− − �. Since this is true for all � > 0, we must have diamA ≥ diamA−.
Problem 58
For any y, z ∈ B(x, r), we have
d(y, z) ≤ d(x, y) + d(x, z) < 2r
Thus diamB(x, r) ≤ 2r.
To show the reverse inequality, fix � > 0. If � ≥ 2r, pick y = z = x. Otherwise, let y =
(
1− 2r − �
2|x|
)
x and
z =
(
1 +
2r − �
2|x|
)
x. Then
d(x, y) = d(x, z) =
0 if � ≥ 2rr − �2 otherwise
d(y, z) =
0 if � ≥ 2r2r − � otherwise
Thus for any � > 0, we can find y, z ∈ B(x, r) such that d(y, z) ≥ 2r − �. This implies diamB(x, r) ≥ 2r.
Problem 59
There exists a sequence (xn, yn) ∈ A × A such that d(xn, yn) → diam(A). Fix x, y ∈ A. Then d(xn, x) +
d(yn, y) ≤ 2 diamA <∞ for all n. This implies
d((xn, yn), (x, y)) =
√
d(xn, x)2 + d(yn, y)2 ≤
√
(d(xn, x) + d(yn, y))2 ≤ 2 diamA <∞
so that {xn, yn} is a bounded sequence. By the Bolzano-Weierstrass Theorem, there exists a convergent
subsequence (xnk , ynk)→ (x0, y0). Problem 26 then implies x0, y0 ∈ A−.
Next, we show d(xnk , ynk) → d(x0, y0). Fix � > 0 and define δ = �2 . Then for k large enough such that
d((xnk , ynk), (x0, y0)) < δ, we have
|d(xnk , ynk)− d(x0, y0)| ≤ |d(xnk , ynk)− d(xnk , y0)|+ |d(xnk , y0)− d(x0, y0)|
≤ d(ynk , y0) + d(xnk , x0)
≤ 2 ·max(d(ynk , y0), d(xnk , x0))
≤ 2 · d((xnk , ynk), (x0, y0))
< �
Thus, we have
d(x0, y0) = lim
k→∞
d(xnk , ynk) = diam(A)
19
Problem 60
By Problem 59, there exists x0, y0 ∈ A− such that d(x0, y0) = d. By definition, we must have that A ⊂
B(x0, d)
− ∩B(y0, d)−. Define x = x0+y02 . Then for any y ∈ B(x0, d)
− ∩B(y0, d)−, we must have
d(x, y) =
∣∣∣∣y − x02 + y − y02
∣∣∣∣ ≤ ∣∣∣∣y − x02
∣∣∣∣+ ∣∣∣∣y − y02
∣∣∣∣ ≤ d
Thus y ∈ B(x, d)−. This implies B(x0, d)− ∩B(y0, d)− ⊂ B(x, d)−, so that A ⊂ B(x, d)−.
Problem 61
Let x1, x2, and x3 be equidistant points in R2. Without loss, assume that x1 + x2 + x3 = 0 (otherwise,
work with x′i = xi − ( 13x1 +
1
3x2 +
1
3x3) for i = 1, 2, 3). The equilateral triangle with vertices at x1, x2, x3 is
defined as
A = {θ1x1 + θ2x2 + θ3x3 : θi ≥ 0, i = 1, 2, 3, θ1 + θ2 + θ3 = 1}
Notice that |x1| is given by
|x1| =
∣∣∣∣13(x1 − x2) + 13(x1 − x3)
∣∣∣∣
=
1
3
(√
2d2 + 2(x1 − x2) · (x1 − x3)
)
(2)
where d = d(x1, x2) = d(x1, x3) = d(x2, x3). We can use the law of cosines identity
2(x1 − x2) · (x1 − x3) = d(x1, x2)2 − d(x2, x3)2 + d(x1, x3)2 (3)
to obtain 2(x1 − x2) · (x1 − x3) = d2. Plugging into equation (2), we see that |x1| = d√3 . By symmetry, we
also have |x2| = |x3| = d√3 . Now consider any point y = θ1x1 + θ2x2 + θ3x3 ∈ A. The distance between 0
and y is given by
d(0, y) = |θ1x1 + θ2x2 + θ3x3|
≤ θ1|x1|+ θ2|x2|+ θ3|x3|
=
d√
3
Thus all points in A are contained in B
(
0, d√
3
)−
.
To see that this is the smallest ball which can contain A, pick a point x ∈ R2. If d(x, xi) ≤ r for i = 1, 2, 3,
we must have
3r2 ≥ d(x, x1)2 + d(x, x2)2 + d(x, x3)2 = |x1|2 + |x2|2 + |x3|2 + 3|x|2 ≥ d2
This implies r ≥ d√
3
. Thus any (closed) ball which contains the vertices of the triangle must have radius
greater than or equal to d√
3
.
(Received help from achille hui http://math.stackexchange.com/q/1490944)
20
Problem 62
Follow the same procedure as in Problem 61 using four equidistant points in R3.
Chapter 2
Section A
Problem 1
(a) ⇒ (c): Suppose λ(I) = 0 for a special rectangle I =×ni=1[ai, bi]. Then there exists i ∈ {1, · · · , n} such
that ai = bi. Define E = {x ∈ Rn : xi = 0}. It is easy to check that E is a subspace of Rn with dimension
n− 1:
• 0 ∈ E
• Suppose x and y are in E, so that xi = yi = 0. Then for any scalars a, b in R, we have axi + byi = 0.
This implies ax+ by ∈ E.
• Let ej denote the jth standard basis vector. Then e1, · · · , ei−1, ei+1, · · · , en are n− 1 linearly indepen-
dent vectors which span E, so the dimension of E is n− 1.
Fix any x0 ∈ I (by construction, special rectangles are non-empty). For any y ∈ I, define x = y − x0. Since
x0 and y are in I, we must have x0i = yi = ai. But this implies xi = 0, so x ∈ E. Thus for any y ∈ I, we
can find x ∈ E such that y = x0 + x. This implies I ⊂ {x0 + x : x ∈ E}.
(c) ⇒ (b): Suppose I ⊂ {x0 + x : x ∈ E} for some subspace E with dimension less than n. Since E has
dimension less than n, we can pick z ∈ Ec such that |z| = 1. Now pick y ∈ I. By assumption, there exists
x ∈ E such that y = x0 + x. For any r > 0, choose � ∈ (0, r) and let y′ = y + �z. Suppose y′ ∈ I, so that
there exists x′ ∈ E such that y′ = x0 + x′. But this would imply
x0 + x
′ = x0 + x+ �z ⇒ z =
1
�
(x′ − x)
Since E is a subspace, this would imply z ∈ E. Since z was chosen in Ec, this is a contradiction. We
conclude that y′ /∈ I. Thus for any y ∈ I and any r > 0, B(y, r) 6⊂ I. This means I has no interior points.
(b) ⇒ (a): Suppose λ(I) 6= 0 for a special rectangle I = ×ni=1[ai, bi]. Then ai < bi for all i. Define
xi =
ai+bi
2 , x = (x1, · · · , xn)
′, and r = min{ b1−a12 , · · · ,
bn−an
2 }. For any x
′ ∈ B(x, r), we have
|x′ − x| < r ⇒ |x′i − xi| < r, i = 1, · · · , n
⇒
∣∣∣∣x′i − ai + bi2
∣∣∣∣ < bi − ai2 i = 1, · · · , n
⇒ ai ≤ x′i ≤ bi i = 1, · · · , n
Thus x′ ∈ I, so B(x, r) ⊂ I. This means that x ∈ Io, so Io is not empty.
21
Problem 2
After step N , the Lebesgue measure of each non-overlapping square is
1
9N
. Since eight of nine squares within
a partition are kept at each step, the number of squares remaining after N steps is equal to 8N . It follows
that I ∼ ∪Nk=1Gk =
(
8
9
)N
.
Problem 3
Suppose G ⊂ Rn is open and P is a special polygon with P ⊂ G. Let G′ = G ∼ P . If G′ is empty, then
G = P . But since G is open and P is closed, Problem 31 of Chapter 1 implies that P = Rn or P = ∅.
Either is a contradiction with the definition of P , so G′ must be non-empty. Since G′ is a non-empty, open
set, we can find a special polygon P ′ ⊂ G′. Take P ′′ = P ′ ∪ P , which by construction satisfies P ′′ ⊂ G and
λ(P ′′) > λ(P ).
Problem 4
(a) Suppose G ⊂ Rn is a bounded open set. Then there exists x ∈ Rn and r ∈ R such that G ⊂ B(x, r).
Let I =×ni=1[xi − r, xi + r]. Then G ⊂ I, which implies λ(P ) ≤ λ(I) = (2r)n for all special polygons
P satisfying P ⊂ G. This implies λ(G) is boundedabove by (2r)n.
(b) Fix 0 < M <∞ and consider the sequence of special rectangles
In = [2
n−1, 2n]× [2−(n+2), 2−(n+1)]
Then In ⊂ G for n ≥ 2, In are non-overlapping, and λ(In) = 18 for each n. Pick N > 8M + 1 and
consider the special polygon P = ∪Nn=2In ⊂ G. Then λ(P ) = (N−1)/8 > M . Thus λ(G) is not bounded
above, so λ(G) =∞.
(c) Define the special rectangles
Ikn =
[
k
n
,
k + 1
n
]
×
[
0, e−k/n
]
and the special polygons
Pn =
n2−1⋃
k=0
Ikn
Then we have
λ(Pn) =
n2−1∑
k=0
1
n
(
e−
1
n
)k
=
1
n
· 1− e
−n
1− e− 1n
It can be shown that limn→∞ λ(Pn) = 1. For any special polygon P ⊂ G, there exists an N such that
P ⊂ Pn for all n ≥ N . Thus λ(P ) ≤ limn λ(Pn) = 1 for all P ⊂ G, which implies λ(G) ≤ 1.
To show the reverse inequality, define the special rectangles
Jkn =
[
k
n
,
k + 1
n
]
×
[
0, e−
k+1
n
]
22
and the special polygons
Pn =
n2−1⋃
k=1
Jkn
Then
λ(Pn) =
1
n
n2∑
k=1
e−
k+1
n = e−
2
n
n2−1∑
k=0
1
n
(
e−
1
n
)k
=
e−
2
n
n
1− e−n
1− e− 1n
so that limn→∞ λ(Pn) = 1. Now fix � > 0. There exists N such that |λ(Pn)− 1| < �2 for all n ≥ N . For
each n ≥ N , there also exists P ′n such that P ′n ⊂ G ∩ Pn and |λ(P ′n)− λ(Pn)| < �2 (consider shaving off
just a sliver from the top and bottom of each Jkn). Thus for any � > 0, we can find P
′
n ⊂ G such that
|λ(P ′n)− 1| ≤ |λ(P ′n)− λ(Pn)|+ |λ(Pn)− 1| < �. This implies λ(G) ≥ 1.
(d) Define the special rectangles
Ikn =
[
k
n
,
k + 1
n
]
×
[
0,
(
k
n
)−a]
for k = n, n+ 1, · · · and the sets
Pn =
n2⋃
k=n
Ikn
Then we have
λ(Pn) =
1
n
n2∑
k=n
(
k
n
)−a
= na−1
n2∑
k=n
k−a
From the (generalized) binomial theorem, we know that
(k + 1)−(a−1) =
∞∑
j=0
(
−(a− 1)
j
)
k−(a−1+j)
= k−(a−1) − (a− 1)k−a + k−(a+1)
∞∑
j=0
(
−(a− 1)
j + 2
)
k−j
= k−(a−1) − (a− 1)k−a +O
(
k−(a+1)
)
for k > 1. This implies
n2∑
k=n
k−a =
1
a− 1
(
n−(a−1) − n−2(a−1)
)
+O
 n2∑
k=n
k−(a+1)

=
1
a− 1
(
n−(a−1) − n−2(a−1)
)
+O
(
n−a
)
which implies
λ(Pn) =
1
a− 1
(
1− 1
na−1
+O
(
n−1
))
Thus limn→∞ λ(Pn) =
1
a−1 . For any special polygon satisfying P ⊂ G, there exists N large enough that
P ⊂ Pn for all n ≥ N . This implies λ(P ) ≤ 1a−1 for all P ⊂ G, so that λ(G) ≤
1
a−1 .
23
To show the reverse inequality, define the special rectangles
Jkn =
[
k
n
,
k + 1
n
]
×
[
0,
(
k + 1
n
)−a]
for k = n, n+ 1, · · · and the sets
Pn =
n2−1⋃
k=n
Ikn
Then we have
λ(Pn) =
1
n
n2−1∑
k=n
(
k + 1
n
)−a
= na−1
n2∑
k=n+1
k−a
=
1
a− 1
((
n
n+ 1
)a−1
− 1
na−1
+O(n−1)
)
so that limn→∞ λ(Pn) = 1. Now fix � > 0. There exists N such that |λ(Pn)− 1| < �2 for all n ≥ N . For
each n ≥ N , there also exists P ′n such that P ′n ⊂ G ∩ Pn and |λ(P ′n)− λ(Pn)| < �2 (consider shaving off
just a sliver from the top and bottom of each Jkn and a sliver off the left of Jnn). Thus for any � > 0,
we can find P ′n ⊂ G such that |λ(P ′n)− 1| ≤ |λ(P ′n)− λ(Pn)|+ |λ(Pn)− 1| < �. This implies λ(G) ≥ 1.
(Received help from Jack D’Aurizio http://math.stackexchange.com/q/1500518)
Problem 5
Let Gi, i ∈ I denote a collection of disjoint open sets in Rn. Define I ′ = {i ∈ I : Gi is non-empty}. Suppose
I ′ is not finite. For each i ∈ I ′, we can find xi = (xi1, · · · , xin)′ ∈ Gi and ri > 0 such that B(xi, ri) ⊂ Gi.
Pick � ∈ (0, ri√
n
) and construct the rectangle Ii = ×nj=1[xij − �, xij + �]. Then Ii ⊂ B(xi, ri). Since Q is
dense in R, we can find a rational qij between xij − � and xij + � for j = 1, · · · , n (see Theorem 1.20, Rudin
(1964)). Let qi = (qi1, · · · , qin)′ ∈ Qn. Define a mapping q : I ′ → Qn by q(i) = qi. Since the Gi sets are
disjoint, q is injective. Since Qn is countable, this implies the existence of an injective mapping between I ′
and N. Problem 5 of Chapter 1 then implies that I ′ is countable.
Problem 6
Let G be a non-empty, open set in R. Define the equivalence relation x ∼ y ⇐⇒ [min{x, y},max{x, y}] ⊂ G.
We can write each equivalence class as
{y ∈ G : x ∼ y} = (inf{y ∈ G : x ∼ y}, sup{y ∈ G : x ∼ y})
Thus each equivalence class is either empty or a non-empty, open interval (where sets of the form (−∞, a)
and (a,∞) are considered open intervals in this problem). Let Gi, i ∈ I denote the collection of non-empty
equivalence classes. We can write G = ∪i∈IGi, where we know I is either finite or countable by Problem 5.
24
To see that this representation is unique, suppose G = ∪i′∈I′G′i′ for some countable or finite collection of
non-empty disjoint open intervals G′i′ . Pick xi′ ∈ G′i′ for each i′ ∈ I ′. Then xi′ ∈ Gi for a unique i ∈ I, so
we can define the mapping f : I ′ → I by f(i′) = {i : xi′ ∈ Gi}. Suppose f(i′0) = f(i′1), so that xi′0 and xi′1
are both in Gi. But this would imply that xi′0 and xi′1 are in the same interval. By construction, we must
have i′0 = i
′
1. Thus f is injective. Now fix any i ∈ I and pick xi ∈ Gi. Then xi ∈ Gi′ for some i′ ∈ I ′. But
this means xi ∼ xi′ , so xi′ ∈ Gi. Thus f(i′) = i, so f is surjective.
Problem 7
We have that λ(G) = λ(∪k(ak, bk)) =
∑
k λ(ak, bk) =
∑
k(bk − ak) where the second equality follows from
O6 and the third from O7.
Problem 8
It is easy to check that B(0, 1) is convex, and therefore connected. By definition, a connected space cannot
be written as the union of two disjoint, non-empty open sets. Since any open rectangle contained in B(0, 1)
cannot contain all elements of B(0, 1), the result follows.
Problem 9
For each x ∈ G, there exists r > 0 such that B(x, r) ⊂ G. For each x, construct a closed rectangle with
rational vertices containing x and contained in B(x, r). Since there are only a countable number of such
rectangles and the union of all the rectangles equals G, we can write G = ∪∞k=1Ik where each Ik is a rectangle
with rational vertices.
Now define P1 = I1 and Pk = Ik ∼ int(∪k−1j=1Ij) for k = 2, 3, · · · . Then the Pk are non-overlapping special
polygons, so each Pk can be written as the finite union of non-overlapping special rectangles.
Alternatively, begin by defining G0 as
G0 = {[i1, i1 + 1]× · · · × [in, in + 1] : (i1, · · · , in)′ ∈ Zn}
Let H0 denote the family of cubes in G0 which are contained in G:
H0 = {G0 ∈ G0 : G0 ⊂ G}
Bisect the sides of the cubes in G0 ∼ H0 to obtain cubes with side 12 . Let G1 denote the family of these cubes
and let H1 denote the cubes in G1 which are contained in G. Continue recursively by letting Gk+1 denote
the cubes in Gk ∼ Hk bisected to have side length 2−(k+1) and defining Hk+1 as the cubes in Gk+1 which
are contained in G.
Define G′ as
G′ =
∞⋃
k=1
⋃
H∈Hk
H = {x : x is contained in H ∈ Hk for some k}
25
By construction, G′ is a countable union of non-overlapping rectangles contained in G. To see that G ⊂ G′,
first define the lattices
Λk = {2−k
n∑
i=1
aiei : ai ∈ Z}
where ei is the i-th coordinate vector in Rn. The vertices of all the cubes in Hk are contained in Λk. Now
pick any x ∈ G. Since G is open, there exists � > 0 such that B(x, �) ⊂ G. Pick k such that 2−k < �/
√
n.
Then there exists a cube of side length 2−k with vertices contained in Λk which contains x and which is
contained in B(x, �). But since B(x, �) ⊂ G, this cube must either be in Hk or be contained in some cube
which is in ∪k−1j=1Hj . But since x is in this cube, we must have x ∈ G′.
Now suppose G could be written as the finite union of special rectangles. Then G would be closed (Problem
22, Chapter 1). But by assumption G is open, so either G = ∅ or G = Rn (Problem 31, Chapter 1). Since
Rn cannot be written as a finite union of special rectangles and G is non-empty by assumption, we have a
contradiction.
Problem 10
From O5, we have
λ(G) = λ(∪∞k=1Ik) ≤
∞∑
k=1
λ(Ik)
We also have
∞∑
k=1
λ(Ik) =
∞∑
k=1
λ(Iok) = λ(∪∞k=1Iok) ≤ λ(G)
where the first equality follows from O7, the second from O6, and the inequality from O4.
Problem 11
Gk is the finite union of 8
k−1 disjoint open squares. By O7, each open square has Lebesgue measure 1
9k
.
Thus λ(Gk) =
8k−1
9k
. From O6, we then have
λ
( ∞⋃
k=1
Gk
)
=
∞∑
k=1
λ(Gk) =
∞∑
k=1
8k−1
9k
= 1
Problem 12Fix � > 0. Let Q = {qk}∞k=0 and let rk = �2−k. Construct the set G = ∪∞k=1B(qk, rk/2). Then G is open
(Problem 18, Chapter 1) and Q ⊂ G by construction. O5 implies
λ(G) ≤
∞∑
k=1
λ(B(qk, rk)) = �
∞∑
k=1
rk = �
∞∑
k=1
2−k = �
Since � was arbitrary, the result follows.
26
Problem 13
If R were countable, we could use the construction in Problem 12 to conclude that λ(R) ≤ � for all � > 0.
This would imply that λ(R) = 0. But λ(R) =∞, a contradiction. Thus R cannot be countable.
Problem 14
Suppose x ∈ C. At stage k in the construction, an interval of length 1
3k−1
containing x is divided into thirds
and the middle third is removed. Since x ∈ C, x must always lie in the bottom third or upper third of this
interval. If x is in the bottom third, assign αk = 0 and if x is in the upper third, assign αk = 2. For example,
since x ∈ Gc1 we must have x ∈ [0, 13 ] ∪ [
2
3 , 1]. Set α1 = 0 if x ∈ [0,
1
3 ] and α2 = 2 if x ∈ [
2
3 , 1]. Likewise,
since x ∈ Gc2 we must also have x ∈ [0, 19 ] ∪ [
2
9 ,
1
3 ] ∪ [
2
3 ,
7
9 ] ∪ [
8
9 , 1]. Set α2 = 0 if x ∈ [0,
1
9 ] ∪ [
2
3 ,
7
9 ] and α2 = 2
if x ∈ [ 29 ,
1
3 ] ∪ [
8
9 , 1]. Continuing in this way, we can construct a ternary expansion that consists of only 0’s
and 2’s. At stage k, the difference between the expansion and x is less than or equal to 1
3k
(i.e. the length
of the remaining intervals). Thus this ternary expansion converges to x.
Conversely, suppose x ∈ [0, 1] and x has a ternary expansion consisting only of 0’s and 2’s. Notice that
x ∈ Gk only if αk = 1 for all ternary expansions of x. Thus x ∈ Gck for all k, implying that x ∈ C.
Problem 15
Suppose x is an end point of an extracted interval belonging to Gk. Then x must have a terminating ternary
expansion
x = .α1α2 · · ·αk (ternary)
where α1, · · · , αk−1 ∈ {0, 2} and αk = {1, 2}. If αk = 1, we can write
x =
α1
3
+
α2
9
+ · · ·+ αk−1
3k−1
+
∞∑
i=1
2
3k+i
while if ak = 2 we can write
x =
α1
3
+
α2
9
+ · · ·+ αk−1
3k−1
+
1
3k
+
∞∑
i=1
2
3k+i
These latter two expressions give different ternary expansions of x.
Conversely, suppose x ∈ C. Let .α1α2 · · · denote a ternary expansion of x consisting of only 0’s and 2’s,
which we know exists by Problem 14. Let .α′1α
′
2 · · · denote a different ternary expansion of x and let k
denote the first digit such that αk 6= α′k. Then we must have
αk − α′k
3k
=
1
3k
∞∑
i=1
α′k+i − αk+i
3i
⇒ ak − a′k =
∞∑
i=1
α′k+i − αk+i
3i
The maximum value the right-hand side can attain is 1, which occurs when a′k+i− ak+i = 2 for i = 1, 2, · · · .
Likewise, the minimum value the right-hand side can attain is −1, which occurs when a′k+i − ak+i = −2 for
i = 1, 2, · · · . Thus ak − a′k must equal 1 or −1. If ak − a′k = 1, then ak = 2, ak+i = 0, a′k = 1, and a′k+i = 2
27
for i = 1, 2, · · · . Likewise, if ak − a′k = −1, then ak = 0, ak+i = 2, a′k = 1, and a′k+i = 0 for i = 1, 2, · · · . In
either case, there exists a ternary expansion of x which terminates after k terms. This expansion consists
only of 0’s and 2’s for the first k − 1 digits, and then contains a 1 or 2 in the k-th digit. But this implies x
is an end point of an extracted interval belonging to Gk.
Problem 16
Observe that
1
4
= 2
∞∑
k=1
1
9k
= .020202 · · · (ternary)
Thus 14 has a ternary expansion containing only 0’s and 2’s, so
1
4 ∈ C by Problem 14.
Likewise, observe that
1
10
= 8
∞∑
k=1
1
81k
= .00220022 · · · (ternary)
Thus 110 has a ternary expansion containing only 0’s and 2’s, so
1
10 ∈ C by Problem 14.
The end-points of the extracted intervals are of the form a
3k
for some positive integers a and k. If there
exists positive integers a and k such that 14 =
a
3k
, then 3k would be divisible by 4. Likewise, if there exists
positive integers a and k such that 110 =
a
3k
, then 3k would be divisible by 10. Since this is not possible, 14
and 110 cannot be endpoints.
Problem 17
Intuitively, x ∈ C should imply 1−x ∈ C since the sets are removed symmetrically. We can write the Cantor
set explicitly as
C = [0, 1] ∼
∞⋃
m=1
3m−1−1⋃
k=0
(
3k + 1
3m
,
3k + 2
3m
)
(1)
Suppose that 1 − x ∈ Cc, where x ∈ [0, 1]. Then for some m ≥ 1 and 0 ≤ k ≤ 3m−1 − 1, we must have
3k+1
3m < 1 − x <
3k+2
3m . This implies that
3k′+1
3m < x <
3k′+2
3m for k
′ = 3m−1 − (k + 1), which means x ∈ Cc.
We can conclude x ∈ C ⇒ 1− x ∈ C.
Alternatively, if x ∈ C then x =
∑∞
k=1
αk
3k
where αk ∈ {0, 2} for all k. Then 1− x =
∑∞
k=1
2
3k
−
∑∞
k=1
αk
3k
=∑∞
k=1
2−αk
3k
. But since 2 − αk ∈ {0, 2}, 1 − x has a ternary expansion consisting only of 0’s and 2’s. This
implies 1− x ∈ C.
Problem 18
It can be shown by direct computation that
3 · 26 + 1
39
<
1
249
<
3 · 26 + 2
39
which implies from equation (1) that 1249 ∈ C
c.
28
One can also show that
1
252
= .00000222000222000 · · · (ternary)
31
121
= .0202202022 · · · (ternary)
which implies that 1252 and
31
121 are in C.
Problem 19
Suppose x and 2x are in C. Then x =
∑∞
i=1
αi
3i and 2x =
∑∞
i=1
βi
3i for sequences αi and βi only taking values
in {0, 2}. Let k denote the first digit such that 2αi 6= βi (if no such digit exists, x = 2x = 0). Notice that
we must have αi = βi = 0 for i < k. We also have
2αk − βk =
∞∑
i=1
βk+i − 2αk+i
3i
By construction, the left-hand side is either -2, 2, or 4. The maximum value the right-hand side can attain
is 1, so the left-hand side must equal −2. The right-hand side equals −2 when βk+i = 0 and αk+i = 2
(βk+i = 2) for i = 1, 2, · · · . Thus if x and 2x are in C, there must exist a ternary expansion .α1α2 · · · of x
such that αi = 0 for i < k and αi = 2 for αi ≥ k for some positive integer k. But this means that the only
x such that both x and 2x are in C are of the form x = 1
3k
for k = 1, 2, · · · .
Problem 20
Suppose x and 2x − 1 are in C. Then by Problem 17, 1 − x and 2(1 − x) are also in C. Problem 19 then
implies 1− x = 1
3k
for some k, so that x = 1− 1
3k
for some k.
Problem 21
Since z2 ∈ [0, 1], we can write its ternary expansion as
z
2
= .γ1γ2 · · · (ternary)
Define the sequences
αi =
0 if γi = 02 if γi = 1 or γi = 2 βi =
0 if γi = 0 or γi = 12 if γi = 2
and define
x = .α1α2 · · · (ternary) y = .β1β2 · · · (ternary)
Notice that x and y have ternary expansions consisting only of 0’s and 2’s, and therefore are in C. Since
γi =
αi+βi
2 for all i, we also have
z
2 =
x+y
2 ⇒ z = x+ y.
29
Problem 22
If w ∈ [−1, 1], then w + 1 ∈ [0, 2]. By Problem 21, this implies w + 1 = x + y for some x, y ∈ C. But this
implies w = x− (1− y). By Problem 17, we know that 1− y ∈ C, so the result follows.
Problem 23
Suppose C were countable. Let ci denote the i-th element of C. By Problem 17, we know that ci can be
written as
ci = .αi1αi2αi3 · · · (ternary)
where αij ∈ {0, 2} for all i, j.
Now suppose that there exists another ternary expansion of ci
ci = .α
′
i1α
′
i2α
′
i3 · · · (ternary)
such that α′ij ∈ {0, 2} for all j. Let k denote the smallest digit such that α′ij 6= αij . Then
α′ik − αik =
∞∑
j=1
αi,k+j − α′i,k+j
3j
But |α′ik −αik| = 2 while
∑∞
j=1
αi,k+j − α′i,k+j
3j
can only take values between −1 and 1. Thus no such k can
exist, so ternary expansions consisting only of 0’s and 2’s are unique.
Consider constructing the sequence
βj =
0 if αjj = 22 if αjj = 0
and define x as
x = .β1β2β3 · · · (ternary)
Then x ∈ C, but its ternary expansion consisting of 0’s and 2’s differs from the unique ternary expansion
consisting of 0’s and 2’s for every element in C. But this implies x cannot be in C, which is a contradiction.
We conclude that C cannot be countable.
Problem 24
Suppose A,B ∈ L0. Fix � > 0. By the approximation theorem, we know that there exists compacts sets K1,
K2 and open set G1, G2 such that
K1 ⊂ A ⊂ G1, λ(G1 ∼ K1) < �/2
K2 ⊂ B ⊂ G2, λ(G2 ∼ K2) < �/2
30
Define K = K1 ∪K2 and G = G1 ∪G2. Then K ⊂ A ∪B ⊂ G and
G ∼ K = (G1 ∩Kc1 ∩Kc2) ∪ (G2 ∩Kc1 ∩Kc2) ⊂ (G1 ∼ K1) ∪ (G2 ∼ K2)
By O5, we have
λ(G ∼ K) ≤ λ(G1 ∼ K1) + λ(G2 ∼ K2) < �
The approximation theorem then implies A ∪B ∈ L0.
Now define K= K1 ∩K2 and G = G1 ∩G2. Then K ⊂ A ∩B ⊂ G and
G ∼ K = (G1 ∩G2 ∩Kc1) ∪ (G1 ∩G2 ∩Kc2) ⊂ (G1 ∼ K1) ∪ (G2 ∼ K2)
Again applying O5, we have
λ(G ∼ K) ≤ λ(G1 ∼ K1) + λ(G2 ∼ K2) < �
so that the approximation theorem implies A ∩B ∈ L0.
Section B
Problem 25
By M4 and M5, we have
λ(A1) = λ
(
A1 ∩
∞⋂
k=2
Ak
)
+ λ
(
A1 ∼
( ∞⋂
k=2
Ak
))
= λ
( ∞⋂
k=1
Ak
)
+ λ
( ∞⋃
k=2
Ack ∩A1
)
= λ
( ∞⋂
k=1
Ak
)
+ lim
k→∞
λ (Ack ∩A1)
If λ(A1) <∞, then we can write
λ
( ∞⋂
k=1
Ak
)
= λ(A1)− lim
k→∞
λ (Ack ∩A1)
= lim
k→∞
λ (A1 ∩Ak)
= lim
k→∞
λ(Ak)
Problem 26
Consider the sequence of sets in R given by
An =
∞⋃
k=1
(
k − 1
2n
, k +
1
2n
)
31
These sets satisfy A1 ⊃ A2 ⊃ · · · and converge to
∞⋂
n=1
Ak =
∞⋃
k=1
{k}
This means that
λ
( ∞⋂
n=1
Ak
)
=
∞∑
k=1
0 = 0
But observe that
λ(An) =
∞∑
k=1
1
n
=∞
for all n, which implies that λ(An) fails to converge to λ(A).
Problem 27
Define α = inf {
∑∞
k=1 λ(Ik) : A ⊂
⋃∞
k=1 Ik}. Suppose A ⊂
⋃∞
k=1 Ik. Fix � > 0 and let I
′
k denote an open
rectangle satisfying
Ik ⊂ I ′k, λ(I ′k) < λ(Ik) + 2−k�
Then
λ∗(A) ≤ λ
( ∞⋃
k=1
I ′k
)
≤
∞∑
k=1
λ(I ′k)
<
∞∑
k=1
(
λ(Ik) + 2
−k�
)
=
∞∑
k=1
λ(Ik) + �
Since � was arbitrary, we must have λ∗(A) ≤
∑∞
k=1 λ(Ik). Therefore λ
∗(A) ≤ α.
Conversely, pick an open set G such that
A ⊂ G, λ∗(A) ≥ λ(G)− �
By Problem 9, G can be written as a countable union of nonoverlapping special rectangles
G =
∞⋃
k=1
Ik
32
Using Problem 10, we have
λ∗(A) ≥ λ
( ∞⋃
k=1
Ik
)
− �
=
∞∑
k=1
λ(Ik)− �
≥ α− �
Since � was arbitrary, we have λ∗(A) ≥ α.
Problem 28
By M11, we have
λ∗(A) + λ∗((A ∪B) ∼ A) = λ(A ∪B) = λ∗(A) + λ∗(B) <∞
This equation implies
λ∗(B ∼ A) = λ∗((A ∪B) ∼ A) = λ∗(B)
Using *1 and *2, we have
λ∗(B) = λ∗(B ∼ A) ≤ λ∗(B) ≤ λ∗(B)
which implies that B is measurable. Since the roles of A and B are symmetric, A can likewise be shown to
be measurable.
Problem 29
By M11, we have
λ(A ∩B) + λ(A ∼ B) = λ(A) (2)
λ(A ∩B) + λ(B ∼ A) = λ(B) (3)
and by M4 we have
λ(A ∪B) = λ(A ∩B) + λ(A ∼ B) + λ(B ∼ A) (4)
Adding equations (2) and (3) and substituting using (4) yields the desired result.
33
Problem 30
Fix � > 0. Pick open sets G1 and G2 satisfying
A ⊂ G1, λ∗(A) ≥ λ(G1)− �/2
B ⊂ G2, λ∗(B) ≥ λ(G2)− �/2
Employing the result in Problem 29, we obtain
λ∗(A) + λ∗(B) ≥ λ(G1) + λ(G2)− �
= λ(G1 ∪G2) + λ(G1 ∩G2)− �
≥ λ∗(A ∪B) + λ∗(A ∩B)− �
Since this expression holds for all �, the result follows.
If λ∗(A) = ∞ or λ∗(B) = ∞, then λ∗(A ∪ B) = ∞ by *2. The result then follows immediately. Suppose
λ∗(A) <∞ and λ∗(B) <∞. Pick compact sets K1 and K2 satisfying
A ⊃ K1, λ∗(A) ≤ λ(K1) + �/2
B ⊃ K2, λ∗(B) ≤ λ(K2) + �/2
Employing the result in Problem 29, we obtain
λ∗(A) + λ∗(B) ≤ λ(K1) + λ(K2) + �
= λ(K1 ∪K2) + λ(K1 ∩K2) + �
≤ λ∗(A ∪B) + λ∗(A ∩B) + �
Since this expression holds for all �, the result follows.
Problem 31
First, we note that any set containing only a single point has measure zero. This is because we can always
construct an open rectangle around a single point with arbitrarily small Lebesgue measure. If A =
⋃∞
k=1{ak},
then by M4 we have λ(A) =
∑∞
k=1 λ({ak}) =
∑∞
k=1 0 = 0.
Problem 32
Consider the sequence of sets
Ak = {a} × [−k, k]n−1
Each Ak is a special rectangle in Rn with λ(Ak) = 0. By M5, we have
λ
(
{a} × Rn−1
)
= λ
( ∞⋃
k=1
Ak
)
= lim
k→∞
λ(Ak) = lim
k→∞
0 = 0
34
Problem 33
Let {qk}∞k=1 denote an enumeration of the rationals. Then we have
A = {(x1, · · · , xn) : xi ∈ Q for some i}
=
n⋃
i=1
{(x1, · · · , xn) : xi ∈ Q}
=
n⋃
i=1
∞⋃
k=1
{(x1, · · · , xn) : xi = qk}
By M4 and Problem 32, we have
λ(A) ≤
n∑
i=1
∞∑
k=1
λ ({(x1, · · · , xn) : xi = qk})
=
n∑
i=1
∞∑
k=1
0
= 0
Problem 34
Fix � ∈ (0, 1). Let {qk}∞k=1 denote an enumeration of Q ∩ [0, 1]. Let rk = �2−k and construct the set
G =
⋃∞
k=1B(qk, rk/2). Observe that
λ(G) ≤
∞∑
k=1
λ(B(qk, rk/2)) =
∞∑
k=1
rk = �
Define A := Gc ∩ [0, 1]. We then have
λ(A) = 1− λ([0, 1] ∩G)
≥ 1− λ(G)
≥ 1− �
> 0
Since G is open, A is closed. Now fix x ∈ A and any δ > 0. Then B(x, δ) must contain a rational number
qk ∈ Q ∩ [0, 1], but qk /∈ A. Thus B(x, δ) 6⊂ A, so x is not an interior point of A.
35
Problem 35
Suppose A ⊂ [0, 1] is closed and λ(A) = 1. Then B := Ac ∩ (0, 1) is open and
λ(B) = λ([0, 1])− λ([0, 1] ∼ B)
= 1− λ(A ∪ {0, 1})
≤ 1− λ(A)
= 0
Thus λ(B) = 0. Problems 6 and 7 then imply B = ∅. But this implies
(0, 1) = (0, 1) ∩Bc
= (0, 1) ∩A
so that (0, 1) ⊂ A. But this means Ao = (0, 1) 6= ∅.
Problem 36
It suffices to prove that the subspace X = [0, 1)n is the union of a countable family of disjoint closed balls
and a set N satisfying λ(N) = 0. Define a box as any set of the form
∏d
i=1[xi, xi + s) where s > 0. Define
a square (round) set as a disjoint union of finitely many boxes (closed balls). By suitably adapting the
argument from Problem 9, one can show that for any � > 0 and any set U ⊂ X open in X, there exists a
square set S ⊂ U with λ(U)− λ(S) < �.
Let a ∈ (0, 1) be a constant (depending on n) such that every box A contains a closed ball B satisfying
λ(B) = aλ(A). Then for any square set S ⊂ X, we can find a round set R ⊂ S such that λ(R) = aλ(S).
Fix � ∈ (0, a). Since X is square, there is a round set R1 ⊂ X with λ(R1) = aλ(X) = a. Thus λ(X ∼ R1) =
1− a ≤ (1− a) + (1− a+ �).
Now suppose that λ(X ∼ Rn) ≤ (1− a)n + (1− a+ �)n holds for a given round set Rn ⊂ X. Since X ∼ Rn
is open in X, there exists a square set S ⊂ X ∼ Rn with λ(X ∼ Rn)− λ(S) ≤ �n+1 and a round set R ⊂ S
with λ(R) = aλ(S). Letting Rn+1 = Rn ∪R we have
λ(X ∼ Rn+1) = λ(X ∼ Rn)− λ(R)
= (λ(X ∼ Rn)− λ(S)) + (1− a)λ(S)
≤ (λ(X ∼ Rn)− λ(S)) + (1− a)λ(X ∼ Rn)
≤ �n+1 + (1− a)[(1− a)n + (1− a+ �)n]
≤ [�n+1 − �(1− a+ �)n] + (1− a)n+1 + (1− a+ �)n+1
≤ (1− a)n+1 + (1− a+ �)n+1
Define A :=
⋃∞
n=1Rn. By construction, A is a disjoint union of closed balls and N := X ∼ A satisfies
λ(N) ≤ λ(X ∼ Rn) ≤ (1− a)n + (1− a+ �)n for all n
36
Since 1− a and 1− a+ � are in (0, 1), this implies λ(N) = 0.
(Taken from Mike F’s solution http://math.stackexchange.com/q/28974)
Problem 37
Choose E ⊂ Rn with λ∗(E) <∞. By definition, there exist open sets Gk ⊃ E such that λ(Gk) < λ∗(E)+k−1
for k = 1, 2, · · · . Define A =
⋂∞
k=1Gk. Then A is measurable by M2 and
λ(A) ≤ λ(Gk) < λ∗(E) + k−1
for all k. This implies λ(A) ≤ λ∗(E). Since A ⊃ E by construction, *2 implies
λ(A) = λ∗(A) ≥ λ∗(E)
Therefore λ(A) = λ∗(E).
Problem 38
By M11, we know that λ∗(E) + λ∗(A ∼ E) = λ(A). Thus if λ∗(A) < ∞, λ(A) = λ∗(E) if and only if
λ∗(A ∼ E) = 0.
Problem 39
Define Ek = B(0, k) ∼ B(0, k − 1). Then λ∗(E ∩ Ek) ≤ λ(Ek) < ∞, so by Problem 37 there exists a
measurable set Ãk ⊃ E ∩Ek with λ(Ãk) = λ∗(E ∩Ek). Define Ak = Ãk ∩Ek. Since Ak ⊃ E ∩Ek, we have
λ∗(E ∩ Ek) ≤ λ∗(Ak) = λ(Ak) ≤ λ(Ãk) = λ∗(E ∩ Ek)
Thus Ak ⊂ Ek is a measurable hull of E ∩ Ek.
Now define A =
⋃∞
k=1Ak and suppose K is an arbitrary compact set such that K ⊂ A ∼ E. Then
K ∩ Ek ⊂ Ak ∼ E = Ak ∼ (E ∩ Ek), so by Problem 38 we have
λ(K ∩ Ek) = λ∗(K ∩ Ek) ≤ λ∗(Ak ∼ (E ∩ Ek)) = 0
Thus λ(K∩Ek) = 0, so that λ(K) =
∑∞
k=1 λ(K∩Ek) = 0. SinceK was arbitrary, this implies λ∗(A ∼ E) = 0.
Problem 40
Let Ak be a measurable hull of Ek and define Bk =
⋂∞
j=k Aj . Then Bk is measurable, Ek ⊂ Bk, and
λ∗(Bk ∼ Ek) ≤ λ∗(Ak ∼ Ek) = 0, so Bk is a measurable hull of Ek. Define B =
⋃∞
k=1Bk and E =
⋃∞
k=1Ek.
Since B1 ⊂ B2 ⊂ · · · , M5 implies
λ (B) = lim
k→∞
λ(Bk) = lim
k→∞
λ∗(Ek)
37
To see that λ (B) = λ∗ (E), define B̃k+1 = Bk+1 ∼ Bk and Ẽk+1 = Ek+1 ∼ Ek with B̃1 = B1 and Ẽ1 = E1.
Then λ∗(B̃k ∼ Ẽk) = λ∗(B̃k ∼ Ek) ≤ λ∗(Bk ∼ Ek) = 0. For any compact set K ⊂ B ∼ E, we have
λ(K) =
∞∑
k=1
λ(B̃k ∩K)
≤
∞∑
k=1
λ∗(B̃k ∼ E)
≤
∞∑
k=1
λ∗(B̃k ∼ Ẽk)
= 0
Since K was arbitrary, λ∗(B ∼ E) = 0. The result then follows from M11.
Problem 41
If
∑∞
k=1 λ(Ak) <∞, then limn→∞
∑∞
k=n λ(Ak) = 0 and λ (
⋃∞
k=1Ak) ≤
∑∞
k=1 λ (Ak) <∞. M6 then implies
λ(lim supAk) = λ
( ∞⋂
n=1
∞⋃
k=n
Ak
)
= lim
n→∞
λ
( ∞⋃
k=n
Ak
)
≤ lim
n→∞
∞∑
k=n
λ(Ak) = 0
Problem 42Let E denote the collection of all nonempty subsets of N which contain no more than d numbers. For any
S ∈ E , define AS as
AS =
(⋂
k∈S
Ak
)
∩
 ⋂
j∈N∼S
Acj

Suppose x ∈ AS . For all S′ ∈ E with S′ 6= S, the set (S ∼ S′) ∪ (S′ ∼ S) contains at least one element j. If
j ∈ S ∼ S′, then x ∈ Aj . But AS′ ⊂ Acj , so x /∈ AS′ . Likewise, if j ∈ S′ ∼ S, then x ∈ Acj . But AS′ ⊂ Aj ,
so we still have x /∈ AS′ . Thus AS and AS′ are disjoint for all S, S′ in E .
Since AS ⊂
⋃∞
k=1Ak for all S ∈ E , we must have
⋃
S∈E AS ⊂
⋃∞
k=1Ak. Now suppose x ∈ Ak for some k.
Then x ∈ AS for S := {j : x ∈ Aj}. By assumption S contains no more than d elements, so we must have
S ∈ E . Thus x ∈
⋃
S∈E AS , so that
⋃∞
k=1Ak ⊂
⋃
S∈E AS .
38
Applying these results, we obtain
∞∑
k=1
λ(Ak) =
∞∑
k=1
(∑
S∈E
1 {k ∈ S}λ(AS)
)
=
∑
S∈E
λ(AS)
∞∑
k=1
1 {k ∈ S}
≤ d
∑
S∈E
λ(AS)
= dλ
( ∞⋃
k=1
Ak
)
where the first equality follows because Ak =
⋃
S∈E:k∈S AS and the AS sets are disjoint, the second equality
follows because it is valid to change the order of a double summation when the terms are non-negative,
the inequality follows by assumption, and the final equality follows because the AS sets are disjoint and⋃
S∈E AS =
⋃∞
k=1Ak.
Problem 43
Define B1 = A1 and Bk = Ak ∼
(⋃k−1
i=1 Ai
)
for k = 2, 3 · · · . Then
⋃∞
k=1Ak =
⋃∞
k=1Bk, the Bk sets are
disjoint, and
λ(Ak) = λ (Bk) + λ
(
Ak ∩
(
k−1⋃
i=1
Ai
))
= λ(Bk) (5)
where the last equality follows from
λ
(
Ak ∩
(
k−1⋃
i=1
Ai
))
≤
k−1∑
i=1
λ(Ak ∩Ai) = 0
But this implies
λ
( ∞⋃
k=1
Ak
)
= λ
( ∞⋃
k=1
Bk
)
=
∞∑
k=1
λ(Bk) =
∞∑
k=1
λ(Ak)
Problem 44
Suppose Bk, k = 1, 2, · · · are defined as in Problem 43. Since
∑∞
k=1 λ(Ak) = λ (
⋃∞
k=1Ak) =
∑∞
k=1 λ(Bk) <
∞, we must have
∑∞
k=1 (λ(Bk)− λ(Ak)) = 0. But since λ(Bk) − λ(Ak) ≤ 0, we must have λ(Bk) = λ(Ak)
for all k. From equation (5), we see that this implies λ
(
Ak ∩
(⋃k−1
i=1 Ai
))
= 0 for all k. But this means
that λ(Ak ∩Ai) = 0 for i = 1, 2, · · · , k − 1, so λ(Aj ∩Ak) = 0 if j 6= k.
Problem 45
First assume that Ai ⊂
[
−k2 ,
k
2
]n
for all i ∈ I. Define I1 = {i ∈ I : λ(Ai) ≥ 1} and Ij = {i ∈ I : 1/(j+ 1) ≤
λ(Ai) < 1/j} for j = 2, 3 · · · . Since λ
([
−k2 ,
k
2
]n)
= kn, Ij can contain at most jkn elements. This implies
Ij is finite, which means that I =
⋃∞
j=1 Ij is countable (Chapter 1, Problem 10).
39
Now define Ek =
[
−k2 ,
k
2
]n ∼ [−k−12 , k−12 ]n for k = 1, 2, · · · and
Ik = {i ∈ I : λ(Ai ∩ Ek) > 0 and λ(Ai ∩ Ej) = 0 for j = 1, 2, · · · , k − 1}
Since Ai ∩Ek ⊂
[
−k2 ,
k
2
]n
, we can apply the above argument to the sets Ai ∩Ek, i ∈ Ik to conclude that Ik
is countable. But this implies I =
⋃∞
k=1 Ik is countable (Chapter 1, Problem 10).
Problem 46
By M6, we have
λ(lim sup
k→∞
Ak) = lim
k→∞
λ
( ∞⋃
n=k
An
)
when λ (
⋃∞
n=1An) <∞. But
λ
( ∞⋃
n=k
An
)
≥ λ(Aj) for all j ≥ k
which means that
λ
( ∞⋃
n=k
An
)
≥ sup
j≥k
λ(Aj)
We therefore have
λ(lim sup
k→∞
Ak) ≥ lim
k→∞
sup
j≥k
λ(Aj) = lim sup
k→∞
λ(Ak)
To see why the condition λ (
⋃∞
n=1An) < ∞ is needed, consider the sequence of sets Ak = (k,∞). Then
lim supk→∞Ak = ∅, so λ(lim supk→∞Ak) = 0. But λ(Ak) =∞ for all k, so lim supk→∞ λ(Ak) =∞.
By M5, we have
λ(lim inf
k→∞
Ak) = lim
k→∞
λ
( ∞⋂
n=k
An
)
But
λ
( ∞⋂
n=k
An
)
≤ λ(Aj) for all j ≥ k
which means that
λ
( ∞⋂
n=k
An
)
≤ inf
j≥k
λ(Aj)
We therefore have
λ(lim inf
k→∞
Ak) ≤ lim
k→∞
inf
j≥k
λ(Aj) = lim inf
k→∞
λ(Ak)
40
Problem 47
Using the result in Problem 46, we have
λ(lim sup
k→∞
Ak) ≥ lim sup
k→∞
λ(Ak) ≥ �
But this implies lim supk→∞Ak is not empty, which means there exists a point which belongs to infinitely
many Ak (see Problem 2, Chapter 1).
Problem 48
First suppose λ(A) <∞. There exists a sequence of compact sets K̃k ⊂ A such that
lim
k→∞
λ(K̃k) = λ∗(A) = λ(A)
Define Kk =
⋃k
n=1 K̃n. Then each Kk is compact (Problem 22(d), Chapter 1) and K1 ⊂ K2 ⊂ · · · ⊂ A. We
also have
λ(A) ≤ lim
k→∞
(
λ(K̃k) + λ
(
Kk ∼ K̃k
))
= lim
k→∞
λ (Kk) ≤ λ(A)
so that
λ(A) = lim
k→∞
(λ (A ∼ Kk) + λ (Kk)) = lim
k→∞
λ (A ∼ Kk) + λ(A)
M6 then implies
0 = lim
k→∞
λ (A ∼ Kk) = lim
k→∞
λ (A ∩Kck) = λ
( ∞⋂
k=1
(A ∩Kck)
)
= λ
(
A ∼
∞⋃
k=1
Kk
)
Now define the sets Ek = B(0, k) ∼ B(0, k − 1) for k = 1, 2, · · · . Then λ(A ∩ Ek) < ∞, so we can find
compact sets Kk1 ⊂ Kk2 ⊂ · · · ⊂ A ∩ Ek such that λ(A ∩ Ek ∼
⋃∞
n=1Kkn) = 0. Define Kn =
⋃n
k=1Kkn.
Each Kn is compact, K1 ⊂ K2 ⊂ · · · ⊂ A, and
λ
(
A ∼
∞⋃
n=1
Kn
)
≤
∞∑
k=1
λ
(
(A ∩ Ek) ∼
∞⋃
n=1
Kn
)
But by M6 we must have
λ
(
(A ∩ Ek) ∼
∞⋃
n=1
Kn
)
= lim
n→∞
λ(A ∩ Ek ∼ Kn)
≤ lim
n→∞
λ(A ∩ Ek ∼ Kkn)
= λ
(
A ∩ Ek ∼
∞⋃
n=1
Kkn
)
= 0
for all k.
41
Chapter 3
Section A
Problem 1
Define the matrix T by
T =
(
F (e1) F (e2) · · ·F (en)
)
where ek is the k-th coordinate vector of Rn. Then for any x = (x1, · · · , xn)′ ∈ Rn we have
F (x) = F (x1e1 + · · ·+ xnen) = x1F (e1) + · · ·+ xnF (en) = Tx
Now suppose there exists some n× n matrix T̃ such that F (x) = T̃ x for all x ∈ Rn. Then
F (ek) = T̃ ek = T̃k = Tk
where T̃k and Tk denote the k-th columns of the matrices T and Tk, respectively. Thus T̃ = T , so the matrix
T is unique.
Problem 2
Define the matrix T by
T =
(
F (e1) F (e2) · · ·F (en)
)
Then for any x = (x1, · · · , xn)′ ∈ Rn, we have
F (x) = F (x1e1 + · · ·+ xnen) = x1F (e1) + · · ·+ xnF (en) = Tx
Now suppose there exists some m× n matrix T̃ such that F (x) = T̃ x for all x ∈ Rn. Then
F (ek) = T̃ ek = T̃k = Tk
Thus T̃ = T , so the matrix T is unique.
Problem 3
Suppose that F (x) = x0 + F0(x) for some fixed x0 and linear function F0. Then for all vectors x and y and
scalars a and b we have
F (ax+ (1− a)y) = x0 + F0(ax+ (1− a)y)
= x0 + aF0(x) + (1− a)F0(y)
= a(x0 + F0(x)) + (1− a)(x0 + F0(y))
= aF (x) + (1− a)F (y)
42
Conversely, suppose F is affine. Define x0 = F (0) and F0(x) as
F0(x) = F (x)− x0
First suppose that a 6= −b. Then we have
F0(ax+ by) = F (ax+ by)− x0
=
a
a+ b
F ((a+ b)x) +
b
a+ b
F ((a+ b)y)− x0
=
a
a+ b
((a+ b)F (x) + (1− (a+ b))x0) +
b
a+ b
((a+ b)F (y) + (1− (a+ b))x0)− x0
= aF (x)− ax0 + bF (y)− bx0
= aF0(x) + bF0(y)
Now suppose that a = −b. Then we have
F0(ax+ by) = F
(
1
2
ax+
1
2
(ax− 2ay)
)
− x0
=
1
2
F (ax) +
1
2
F (ax− 2ay)− x0
=
1
2
F (ax) +
1
2
(−F (−ax) + 2F (−ay))− x0
=
1
2
(aF (x) + (1− a)x0 + aF (x)− (1 + a)x0)− aF (y) + (1 + a)x0 − x0
= aF (x)− ax0 − a(F (y)− x0)
= aF0(x) + bF0(y)
In either case, we see that F0 is linear.
To see that F0 and x0 are uniquely determined, suppose that F (x) = x̃0 + F̃0(x) for some fixed x̃0 and linear
function F̃0. Then
F (0) = x̃0 + F̃0(0)
Since F̃0 is linear, for any x ∈ Rn we must have
F̃0(0) = F̃0(x− x) = F̃0(x)− F̃0(x) = 0
Thus x̃0 = F (0) = x0, which implies F̃0(x) = F0(x).
43
Problem 4
In Problem 1, we showed that the matrix A which uniquely defines the linear transformation F ◦G is given
by
A =
(
(F ◦G)e1 · · · (F ◦G)en
)
=
(
F (G(e1)) · · · F (G(en))
)
=
(
F (Te1) · · · F (Ten)
)
=
(
STe1 · · · STen
)
= ST
Problem 5
Define the matrices T1, · · · , Tk as
T1 =
(
e1 0 · · · 0
)
T2 =
(
0 e2 · · · 0
)
· · · Tn =
(
0 · · · 0 en
)
If STk = TkS, then skk is the only non-zero element in the k-th row and k-th column of S. Thus if STk = TkS
for all k, S must be a diagonal matrix.
Let U(i, j) denote the identity matrix with column i switched with column j. If SU(i, j) = U(i, j)S, then
sii = sjj . Thus if SU(i, j) = U(i, j)S for all i, j, all elements along the diagonal of S must be the same.
We can conclude therefore conclude that if ST = TS for all matrices T , S = c id for some c ∈ R.
Problem 6
Suppose T is an orthogonal matrix. Then
Tx · Ty = xtrT trTy = xtry = x · y
Conversely, suppose Tx · Ty = x · y for all x, y ∈ Rn. Then
Tej · Tek = ej · ek =
1 if j = k0 if j 6= k
But this implies
T trT =

(Te1)
tr
...
(Ten)
tr
(Te1 · · · Ten) = id
so that T tr = T−1.
44
Problem 7
TM is the matrix obtained from T by multiplying row k by the number c.
TAis the matrix obtained from T by adding c times column l to column k.
Problem 8
Let S be defined by
sii = 1 if i 6= k or i 6= l
skl = slk = 1
sij = 0 for all other i, j
Then ST is the matrix obtained by interchanging rows k and l of T .
Problem 9
Let A(k, l, c) denote the adding matrix defined by
aii = 1
akl = c
aij = 0 for all other (i, j)
let M(k, c) denote the multiplying matrix defined by
mii = 1 if i 6= k
mkk = c
mij = 0 if i 6= j
and let S(k, l) denote the switching matrix defined by
sii = 1 if i 6= k or i 6= l
skl = slk = 1
sij = 0 for all other i, j
Then
S(k, l) = M(l,−1)A(l, k, 1)A(k, l,−1)A(k, l, 1)
as can be verified by direct multiplication.
45
Problem 10
The number of steps in the construction of the lemma is given by
n∑
k=1
k +
n∑
k=1
(k − 1) = n(n+ 1)
2
+
n(n− 1)
2
= n2
At least n2 elementary matrices are needed in general because there are n2 elements in a non-singular matrix
that need to be transformed.
Problem 11
Suppose a11 6= 0. Then A can be written as
A =
(
a11
1
)(
1
a21 1
)1 a11a22 − a12a21
a11
1 a12a11
1

Now suppose a11 = 0. Then A can be written as
A =
1 − 1a21
1
( 1
a21 1
)(
1
−a12a21
)1 a12 + a22a21
1

Problem 12
Follow the procedure given in the lemma. At some point, the submatrix to be transformed will have only
0’s in its first column. If all entries in the first row of the submatrix are also zero, move on to the next
submatrix. If not, apply an elementary matrix to the right which transforms the (1, 1) position into a 1.
Then use adding matrices applied to the right to eliminate any remaining non-zero elements in the first row.
This procedure will eventually yield a diagonal matrix diag(a11, · · · , ann) such that aii ∈ {0, 1} for all i. Let
i1, · · · , ik denote the indices satisfying that aijij = 0. Using the notation in Problem 9, this matrix can be
written as
∏n
j=1M(ij , 0). Thus we can write any matrix T as
Er · · ·E1TF1 · · ·Fs = G1 · · ·Gt
where all the E’s, F ’s, and G’s are elementary matrices. Applying inverses, we obtain the formula
T = E−11 · · ·E−1r G1 · · ·GtF−1s · · ·F
−1
1
which shows that T can be written as the product of elementary matrices.
46
Section B
Problem 13
Define Ik as the finite collection of numbers which can be written as
x = .α1α2 · · ·αk (ternary)
Suppose y ∈ A. Let
y = .α1α2 · · · (ternary)
denote a ternary expansion of y. There exists some k such that αn 6= 1 for all n ≥ k. Define x as
x = .α1α2 · · ·αk
Then
y = x+ z
where z is in the Cantor set. But this implies
y ∈ x+ C
We can conclude that
A ⊂
∞⋃
k=1
⋃
x∈Ik
(x+ C)
We therefore have
λ∗(A) ≤
∞∑
k=1
∑
x∈Ik
λ(x+ C) =
∞∑
k=1
∑
x∈Ik
λ(C) = 0
which implies λ(A) = 0 by M8.
Problem 14
Define
B = {x ∈ [0, 1] : The decimal expansions of x contain no 7’s}
Let G1 = (.7, .8) and define
Gk =
10k−1−1⋃
i=0
(
i
10k−1
+
1
10k−1
G1
)
If x ∈ Gk, then the decimal expansions of x have a 7 in the k-th digit. Thus B ⊂ [0, 1] ∼
⋃∞
k=1Gk. But we
also have
λ
(
[0, 1] ∼
n⋃
k=1
Gk
)
= 1− 1
10
n−1∑
k=0
(
9
10
)k
=
(
9
10
)n
which implies that λ∗ (B) ≤ λ ([0, 1] ∼
⋃∞
k=1Gk) = 0.
47
Now define
C = {x ∈ [0, 1] : The decimal expansions of x contain finitely many 7’s}
Using an argument analogous to the one given in Problem 13, it is straightforward to show that λ(C) = 0.
Next define
An = {x ∈ [0, n] : The decimal expansions of x contain finitely many 7’s}
Then An = nC, so λ(An) = nλ(C) = 0. But this implies
λ(A) = λ
( ∞⋃
n=1
An
)
= lim
n→∞
λ(An) = 0
Problem 15
(a) Suppose Φ and Ψ are rigid motions. Then there exist z1, z2 ∈ Rn and orthogonal matrices T1, T2 such
that
Φ(x) = z1 + T1x
Ψ(x) = z2 + T2x
for all x ∈ R. We therefore have
(Φ ◦Ψ)(x) = z1 + T1(z2 + T2x) = z1 + T1z2 + T1T2x = z + Tx (1)
where z = z1 +T1z2 and T = T1T2. But T
trT = T tr2 T
tr
1 T1T2 = T
tr
2 T2 = id, so T is an orthogonal matrix.
(b) There exists z3 ∈ Rn and orthogonal matrix T3 such that
Θ(x) = z3 + T3x
Using equation (1), we see that
((Φ ◦Ψ) ◦Θ)(x) = z1 + T1z2 + T1T2(z3 + T3x)
= z1 + T1(z2 + T2z3 + T2T3x)
= (Φ ◦ (Ψ ◦Θ))(x)
(c) Using equation (1), we see that
Φ ◦ id = z1 + T10 + T1 idx = z1 + T1x = 0 + id z1 + idT1x = id ◦ Φ
(d) Let z2 = −T tr1 z1 and T2 = T tr1 . Then
(Φ ◦Ψ)(x) = z1 + T1(−T tr1 z1) + T1T tr1 x = idx = −T tr1 z1 + T tr1 z1 + T tr1 T1x = (Ψ ◦ Φ)(x)
48
Problem 16
From Problem 13, Chapter 1, |x−y| = |x−z|+ |z−y| if and only if x−z = s(z−y) where s is a nonnegative
scalar. Rearranging this expressions, we obtain z = tx+ (1− t)y where t = 11+s ∈ [0, 1].
Problem 17
Suppose that Φ(0) 6= 0. Define Φ̃ = Φ− Φ(0). Then
|Φ̃(x)− Φ̃(y)| = |Φ(x)− Φ(y)| = |x− y|
so that Φ̃ is an isometry satisfying Φ̃(0) = 0. If Φ̃ is a rigid motion, then Φ = Φ(0) + Φ̃ will also be a rigid
motion. It therefore suffices to work with Φ̃, so we can assume without loss that Φ(0) = 0.
For any s ∈ [0, 1], we have
|Φ(x)− Φ(y)| = |x− y|
= s|x− y|+ (1− s)|x− y|
= |x− ((1− s)x+ sy)|+ |(1− s)x+ sy − y|
= |Φ(x)− Φ((1− s)x+ sy)|+ |Φ((1− s)x+ sy)− Φ(y)|
for all x, y ∈ Rn. Problem 16 then implies
Φ((1− s)x+ sy) = (1− t)Φ(x) + tΦ(y)
for some t ∈ [0, 1]. Rearranging and taking absolute values, we have
|Φ((1− s)x+ sy)− Φ(x)| = t|Φ(y)− Φ(x)|
|(1− s)x+ sy − x| = t|y − x|
s|y − x| = t|y − x|
⇒ s = t
for all x 6= y.
This result implies that for any a > 1, we have
Φ(x) = Φ
(
1
a
ax+
(
1− 1
a
)
0
)
=
1
a
Φ (ax) ⇒ Φ(ax) = aΦ(x)
0 = Φ
(
1
2
(−x) + 1
2
x
)
=
1
2
Φ(−x) + 1
2
Φ(x) ⇒ Φ(−x) = −Φ(x)
49
Now suppose s > 1. Using the above results, we see that
Φ((1− s)x+ sy) = Φ
(
1
s
(1− s)sx+ s− 1
s
s2
s− 1
y
)
=
1
s
Φ ((1− s)sx) + s− 1
s
Φ
(
s2
s− 1
y
)
= Φ((1− s)x) + sΦ(y)
= (s− 1)Φ(−x) + sΦ(y)
= (1− s)Φ(x) + sΦ(y)
Next suppose s < 0. Then we have
Φ((1− s)x+ sy) = Φ
(
1
2
2(1− s)x+ 1
2
2sy
)
=
1
2
Φ (2(1− s)x) + 1
2
Φ (2sy)
= (1− s)Φ(x)− sΦ(−y)
= (1− s)Φ(x) + sΦ(y)
We can conclude that
Φ((1− s)x+ sy) = (1− s)Φ(x) + sΦ(y)
for all scalars s and all vectors x, y ∈ Rn. This implies that Φ is affine, so that Φ(x) = Tx for some matrix
T . T is orthogonal because Φ(x) · Φ(y) = x · y for all x, y ∈ R. This fact is proven in Problem 18.
Problem 18
Since we can assume without loss that Φ(0) = 0, we must have |Φ(x)| = |x| for all x ∈ Rn. We can then see
that
|Φ(x)− Φ(y)|2 = |x− y|2
|Φ(x)|2 − 2 Φ(x) · Φ(y) + |Φ(y)|2 = |x|2 − 2 x · y + |y|2
−2 Φ(x) · Φ(y) = −2 x · y
⇒ Φ(x) · Φ(y) = x · y
This implies
Φ(�i) · Φ(�j) = �i · �j =
1 if i = j0 if i 6= j
which means Φ(�1), · · · ,Φ(�n) are orthonormal.
Define the orthogonal matrix
T =
(
Φ(�1) · · · Φ(�n)
)
50
Then we have
Φ(x) = id Φ(x)
= TT trΦ(x)
=
n∑
k=1
Φ(�k)Φ(�k)
trΦ(x)
=
n∑
k=1
Φ(�k)(�k · x)
=
n∑
k=1
Φ(�k)xk
= Tx
Problem 19
Let B = [0, 1]k and let T = (aij). Then A = TB, so λ(A) = |detT |λ(B) = |detT |.
51
References
Walter Rudin. Principles of Mathematical Analysis. McGraw-Hill, 3rd edition, 1964.
52