SOLUTIONS MANUAL
, Solutions 3
Solutions Manuals For Metric Space.
Chapter 1: Metrics
Q 1.1 Certainly, if d is a metric on X then it has these properties. For the
converse, suppose that the conditions are satisfied. Then, putting a = b in
the given inequality, we have 0 = d(a, a) ≤ d(z, a) + d(z, a) for all a, z ∈
X, so that d satisfies the positive condition of 1.1.1. Also, putting b = z in
the given inequality, we have d(a, b) ≤ d(b, a) + d(z, z) = d(b, a), so that d(a, b)
≤ d(b, a). Reversing the roles of a and b, we have also d(b, a) ≤ d(a, b), so
that d satisfies the symmetry condition of 1.1.1. It then also satisfies the
triangle inequality because, for all a, b, z ∈ X, we have d(a, b) ≤ d(z, a)+d(z, b)
= d(a, z)+d(z, b).
Q 1.2 Using the triangle inequality twice, d(x, y) ≤ d(x, z)+d(z, w)+d(w, y);
so d(x, y)−d(z, w) ≤ d(x, z)+d(w, y). Similarly, d(z, w) ≤ d(z, x)+d(x, y)+d(y, w)
yields d(z, w) − d(x, y) ≤ d(z, x) + d(y, w). These two inequalities and the
symmetry of d together gives the result.
Q 1.3 d clearly satisfies the positive and symmetric criteria. For the
triangle inequality, we have d(a, b) = 0 ≤ d(a, c) + d(c, b) when a = b; and
otherwise d(a, b) = 1 ≤ d(a, c) + d(b,c) because either a /= c or b /= c.
Q 1.4 When n = 1, µ2 is simply (a, b) '→ |b − a|, the usual metric on R. For
2 2 2 2 2
all real numbers u,v,w,z, we have (wv − uz) ≥ 0, so that 2 u v w z ≤ w v + u z ,
whence (uv +wz)2 ≤ (u2 +w2)(v2 +z 2 ) and (uv +wz) ≤ √(u2 + w2)(v2 + z2),
yielding
√ √ 2
(u + v)2 + (w + z)2 ≤ u2 + w 2 + v2 + z 2 .
Suppose inductively that k ∈ N and that the function µ2 is a metric on Rk ×Rk.
Let a, b, c ∈ Rk+1. Then, by the inductive hypothesis,
, 2
,,
u u
k+1 k k
Σ uΣ uΣ
(bi − ai)2 ≤ , (bi −ci)2 + , ( c i −ai)2 + (bk+1 − ak+1)2,
i=1 i=1 i=1
which, since certainly bk+1 − ak+1 = (bk+1 − ck+1) + (ck+1 − ak+1), does
qΣ qΣ 2
k+1 k+1
not exceed i=1 (bi − ci)2 + i=1(ci − ai)2 , by the inequality pre-
viously displayed; taking square roots gives the triangle inequality for
µ2 on Rk+1 × Rk+1. The other metric properties are obvious. So the
,Principle of Induction ensures that the appropriate µ2 is a metric on
Rn for each n ∈ N .
Q 1.5 The symmetric and positive criteria are clearly satisfied by µ1
and µ∞. For the triangle inequalities, we have, for a, b, c in R2,
2 2
Σ Σ
µ1(a, b) = |bi − ai| ≤ (|bi − ci| + |ci − ai|)
i=1 i=1
, Solutions 5
2 2
Σ Σ
= |bi − ci| + |ci − ai|
i=1 i=1
= µ1(a,c) + µ1(c,b)
and µ∞(a,b) = max{|bi − ai| i = 1,2}
≤ max{|bi − ci| + |ci − ai| i = 1,2}
≤ max{|bi − ci| i = 1,2} + max{|ci − bi| i = 1,2}
= µ∞(a,c) + µ∞(c,b).
Q 1.6 Consider X = {a, b, c}, where a, b and c are distinct. Define d and e
symmetrically on X × X with d(a, b) = 2, d(b, c) = 4, d(a, c) = 6, e(a, b) =
3, e(b, c) = 3, e(a, c) = 6 and d(x, x) = e(x, x) = 0 for x = a, b, c. It is
easy to check that both d and e are metrics on X. However, if g is
defined by g(x, y) = min{d(x, y), e(x, y)} for all x, y ∈ X, we get g(a, c) =
6, whereas g(a, b) = 2 and g(b, c) = 3, yielding g(a, c) > g(a, b) + g(b, c);
so g is not a metric on X.
Now suppose that X is an arbitrary non-empty set, that d is the dis-
crete metric on X, that e is any metric on X and that g is as
defined in the question. Then g certainly satisfies both the positive
and symmetric cri- teria. Moreover, for all x, y, z ∈ X, we have g(x, z) =
min{1, e(x, z)}, whereas g(x, y) + g(y, z) = min{1, e(x, y)} + min{1, e(y, z)} ≥
min{1, e(x, y) + e(y, z)}, which, by the triangle inequality for e, cannot be
less than min{1, e(x, z)}, which is g(x, z).
Q 1.7 One might add the difference in lengths of the words to the
number of letters that occur in the same position in the words but are
different. For example, d(kiss, curse) = |4 − 5| + 3 = 4 because the words
have four and five letters, respectively, and they differ in their first,
second and third letters. It is easy to show this is a metric. Moreover,
d(complement, compliment) = 1. This method will not always produce
good results because words like ‘torture’ and ‘pleasure’ , despite their
obvious similarity, are further apart than might be wished; in fact,
d(torture, pleasure) = |7 − 8| + 7 = 8.
Q 1.8 Certainly d is non-negative symmetric. If A, B ∈ F(S) and d(A, B)
= 0, then A\B = ∅ = B\A, which implies that B ⊆ A and A ⊆ B,
whence A = B. For the triangle inequality, we note that, for A, B, C ∈
F(S),
∆(A,B) = (A\B) ∪ (B\A)
= (A ∪ B)\(A ∩ B)
⊆ ((A ∪ B)\C) ∪ (C\(A ∩ B))
= (A\C) ∪ (B\C) ∪ (C\A) ∪ (C\B)
, Solutions 3
Solutions Manuals For Metric Space.
Chapter 1: Metrics
Q 1.1 Certainly, if d is a metric on X then it has these properties. For the
converse, suppose that the conditions are satisfied. Then, putting a = b in
the given inequality, we have 0 = d(a, a) ≤ d(z, a) + d(z, a) for all a, z ∈
X, so that d satisfies the positive condition of 1.1.1. Also, putting b = z in
the given inequality, we have d(a, b) ≤ d(b, a) + d(z, z) = d(b, a), so that d(a, b)
≤ d(b, a). Reversing the roles of a and b, we have also d(b, a) ≤ d(a, b), so
that d satisfies the symmetry condition of 1.1.1. It then also satisfies the
triangle inequality because, for all a, b, z ∈ X, we have d(a, b) ≤ d(z, a)+d(z, b)
= d(a, z)+d(z, b).
Q 1.2 Using the triangle inequality twice, d(x, y) ≤ d(x, z)+d(z, w)+d(w, y);
so d(x, y)−d(z, w) ≤ d(x, z)+d(w, y). Similarly, d(z, w) ≤ d(z, x)+d(x, y)+d(y, w)
yields d(z, w) − d(x, y) ≤ d(z, x) + d(y, w). These two inequalities and the
symmetry of d together gives the result.
Q 1.3 d clearly satisfies the positive and symmetric criteria. For the
triangle inequality, we have d(a, b) = 0 ≤ d(a, c) + d(c, b) when a = b; and
otherwise d(a, b) = 1 ≤ d(a, c) + d(b,c) because either a /= c or b /= c.
Q 1.4 When n = 1, µ2 is simply (a, b) '→ |b − a|, the usual metric on R. For
2 2 2 2 2
all real numbers u,v,w,z, we have (wv − uz) ≥ 0, so that 2 u v w z ≤ w v + u z ,
whence (uv +wz)2 ≤ (u2 +w2)(v2 +z 2 ) and (uv +wz) ≤ √(u2 + w2)(v2 + z2),
yielding
√ √ 2
(u + v)2 + (w + z)2 ≤ u2 + w 2 + v2 + z 2 .
Suppose inductively that k ∈ N and that the function µ2 is a metric on Rk ×Rk.
Let a, b, c ∈ Rk+1. Then, by the inductive hypothesis,
, 2
,,
u u
k+1 k k
Σ uΣ uΣ
(bi − ai)2 ≤ , (bi −ci)2 + , ( c i −ai)2 + (bk+1 − ak+1)2,
i=1 i=1 i=1
which, since certainly bk+1 − ak+1 = (bk+1 − ck+1) + (ck+1 − ak+1), does
qΣ qΣ 2
k+1 k+1
not exceed i=1 (bi − ci)2 + i=1(ci − ai)2 , by the inequality pre-
viously displayed; taking square roots gives the triangle inequality for
µ2 on Rk+1 × Rk+1. The other metric properties are obvious. So the
,Principle of Induction ensures that the appropriate µ2 is a metric on
Rn for each n ∈ N .
Q 1.5 The symmetric and positive criteria are clearly satisfied by µ1
and µ∞. For the triangle inequalities, we have, for a, b, c in R2,
2 2
Σ Σ
µ1(a, b) = |bi − ai| ≤ (|bi − ci| + |ci − ai|)
i=1 i=1
, Solutions 5
2 2
Σ Σ
= |bi − ci| + |ci − ai|
i=1 i=1
= µ1(a,c) + µ1(c,b)
and µ∞(a,b) = max{|bi − ai| i = 1,2}
≤ max{|bi − ci| + |ci − ai| i = 1,2}
≤ max{|bi − ci| i = 1,2} + max{|ci − bi| i = 1,2}
= µ∞(a,c) + µ∞(c,b).
Q 1.6 Consider X = {a, b, c}, where a, b and c are distinct. Define d and e
symmetrically on X × X with d(a, b) = 2, d(b, c) = 4, d(a, c) = 6, e(a, b) =
3, e(b, c) = 3, e(a, c) = 6 and d(x, x) = e(x, x) = 0 for x = a, b, c. It is
easy to check that both d and e are metrics on X. However, if g is
defined by g(x, y) = min{d(x, y), e(x, y)} for all x, y ∈ X, we get g(a, c) =
6, whereas g(a, b) = 2 and g(b, c) = 3, yielding g(a, c) > g(a, b) + g(b, c);
so g is not a metric on X.
Now suppose that X is an arbitrary non-empty set, that d is the dis-
crete metric on X, that e is any metric on X and that g is as
defined in the question. Then g certainly satisfies both the positive
and symmetric cri- teria. Moreover, for all x, y, z ∈ X, we have g(x, z) =
min{1, e(x, z)}, whereas g(x, y) + g(y, z) = min{1, e(x, y)} + min{1, e(y, z)} ≥
min{1, e(x, y) + e(y, z)}, which, by the triangle inequality for e, cannot be
less than min{1, e(x, z)}, which is g(x, z).
Q 1.7 One might add the difference in lengths of the words to the
number of letters that occur in the same position in the words but are
different. For example, d(kiss, curse) = |4 − 5| + 3 = 4 because the words
have four and five letters, respectively, and they differ in their first,
second and third letters. It is easy to show this is a metric. Moreover,
d(complement, compliment) = 1. This method will not always produce
good results because words like ‘torture’ and ‘pleasure’ , despite their
obvious similarity, are further apart than might be wished; in fact,
d(torture, pleasure) = |7 − 8| + 7 = 8.
Q 1.8 Certainly d is non-negative symmetric. If A, B ∈ F(S) and d(A, B)
= 0, then A\B = ∅ = B\A, which implies that B ⊆ A and A ⊆ B,
whence A = B. For the triangle inequality, we note that, for A, B, C ∈
F(S),
∆(A,B) = (A\B) ∪ (B\A)
= (A ∪ B)\(A ∩ B)
⊆ ((A ∪ B)\C) ∪ (C\(A ∩ B))
= (A\C) ∪ (B\C) ∪ (C\A) ∪ (C\B)
