An Introduction to Metric
Spaces 1st Edition
ST
UV
SOLUTIONS
IA
MANUAL
_A
PP
Dhananjay Gopal
RO
Aniruddha Deshmukh
VE
Comprehensive Solutions Manual for
Instructors and Students
D?
© Dhananjay Gopal & Aniruddha Deshmukh. All rights reserved. Reproduction or
distribution without permission is prohibited.
©MedConnoisseur
, Solutions Manual for An Introduction to Metric
Spaces, 1e Dhananjay Gopal, Aniruddha
Deshmukh (All Chapters)
Chapter 1
Set Theory
ST
Ex 1.1. Collection of lines with constant slope and changing intercepts (or, vice-versa) does the job.
That is,
F = {Lr |r ∈ R} ,
UV
where, Lr = {(x, y) ∈ R2 |x + y = r} are lines with slope −1 and intercept r. Other similar sets can be
constructed which can be indexed by R.
Ex 1.2. A few subsethood relations are
IA
N⊆Z⊆Q⊆R
N ⊆ Q+ ⊆ R+ ⊆ R
N ⊆ Q∗ ⊆ R∗ ⊆ R
_A
Similar relations can be obtained by comparing elements.
Ex 1.3. The results for N are given. Other results can be constructed in a similar manner.
N∪Z=Z
PP
N∩Z=N
N∪Q=Q
N∩Q=N
N ∪ Q+ = Q+
RO
N ∩ Q+ = N
N ∪ Q− = {x ∈ Q|x < 0 or x ∈ N}
N ∩ Q− = ∅
N ∪ Q∗ = Q∗
N ∩ Q∗ = N
VE
N ∪ R + = R+
N ∩ R+ = N
N ∪ R− = {x ∈ R|x < 0 or x ∈ N}
N ∩ R− = ∅
D?
N ∪ R∗ = R∗
N ∩ R∗ = N
A few observations that can be made are as follows. The reader can make as many observations as they
want to.
1. A ⊂ B implies A ∪ B = B and A ∩ B = A.
2. Some of the (newly formed) sets are merely a result of union/intersection of known sets.
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1
, Ex 1.4. If F is a pair-wise disjoint family indexed by Λ, then
T
Aλ 6= ∅ would imply the existence
λ∈Λ
of some element x in all of Aλ ’s contradicting the pair-wise disjointness. Thus, any pair-wise disjoint
family is disjoint.
The family {{1, 2} , {2, 3} , {1, 3}} is disjoint but not pair-wise disjoint, since {1, 2} ∩ {2, 3} = {2} =
6 ∅.
Ex 1.5. Example in Solution to exercise 1.4 does the job.
Ex 1.6. Prove the two-way subsethood for each of the equalities using the basic principles of logic and
properties of conjunction and disjunction operators.
Ex 1.7. Prove the two-way subsethood for each of the equalities using the basic principles of logic and
ST
properties of conjunction and disjunction operators.
Ex 1.8. P (∅) = {∅}.
Ex 1.9. The three sets cannot be equal since the elements in all three sets are different. R3 contains
UV
3-tuples, while (R × R) × R and R × (R × R) contains 2-tuple. Also, the first entry in a 2-tuple of
(R × R) × R is again a 2-tuple, while for elements in R × (R × R), the second entry is a 2-tuple.
Q S
Ex 1.10. Aλ = ∅, because if not so, the choice function f : ∅ → Aλ will be such that f (λ) ∈ Aλ .
λ∈∅ λ∈∅
This is possible only when we have such λ’s. Many also argue that since ∅ does not have any elements,
IA
this is vacously true. However, even in such an argument, the empty product is not defined. If there is
a difficulty to the reader in understanding this, the reader may take it as a convention.
Ex 1.11. (x1 , x2 , · · · , xn ) L (y1 , y2 , · · · , yn ) if and only if
_A
6 y1 ⇒ x1 R1 y1 ) and (x1 = y1 and x2 6= y2 ⇒ x2 R2 y2 ) and · · · and
(x1 =
(x1 = y1 , x2 = y2 , · · · , xn−1 = yn−1 ⇒ xn Rn yn )
PP
Ex 1.12.
≤ : Refexive, Anti-symmetric, Transitive
⊆ : Refexive, Anti-symmetric, Transitive
= : Refexive, Symmetric, Anti-symmetric, Transitive
RO
| : Refexive, Anti-symmetric, Transitive
≡ : Refexive, Symmetric, Transitive
Ex 1.13. Generalising modulo relation on Z for n ∈ N, we have m1 ≡ m2 (mod n) if and only if
n|m1 − m2 .
VE
The equivalence classes for this relation are {[0] , [1] , [2] , · · · , [n − 1]}, and an integer m is in an equiv-
alence class j if and only if the remainder after dividing m by n is r.
Ex 1.14. The equivalence classes for = relation on X for each x ∈ X are [x] = {x}.
The equivalence classes for ≡ (modulo relation) are as in solution to exercise 1.13.
D?
Ex 1.15. In the set X = {{1} , {2} , {3} , {1, 2} , {2, 3} , {1, 3}} equipped with ⊆ relation, there is no
least element.
Ex 1.16. In the example of solution to exercise 1.15, the sets {1} , {2} , {3} are minimal elements and
the sets {1, 2} , {2, 3} , {1, 3} are maximal elements..
Ex 1.17. In general, it is not true. For X = {1, 2, 3}, the POSET (P (X) , ⊆) has a least element,
namely ∅, but the subset {{1} , {2} , {3}} of P (X) has no least element.
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Spaces 1st Edition
ST
UV
SOLUTIONS
IA
MANUAL
_A
PP
Dhananjay Gopal
RO
Aniruddha Deshmukh
VE
Comprehensive Solutions Manual for
Instructors and Students
D?
© Dhananjay Gopal & Aniruddha Deshmukh. All rights reserved. Reproduction or
distribution without permission is prohibited.
©MedConnoisseur
, Solutions Manual for An Introduction to Metric
Spaces, 1e Dhananjay Gopal, Aniruddha
Deshmukh (All Chapters)
Chapter 1
Set Theory
ST
Ex 1.1. Collection of lines with constant slope and changing intercepts (or, vice-versa) does the job.
That is,
F = {Lr |r ∈ R} ,
UV
where, Lr = {(x, y) ∈ R2 |x + y = r} are lines with slope −1 and intercept r. Other similar sets can be
constructed which can be indexed by R.
Ex 1.2. A few subsethood relations are
IA
N⊆Z⊆Q⊆R
N ⊆ Q+ ⊆ R+ ⊆ R
N ⊆ Q∗ ⊆ R∗ ⊆ R
_A
Similar relations can be obtained by comparing elements.
Ex 1.3. The results for N are given. Other results can be constructed in a similar manner.
N∪Z=Z
PP
N∩Z=N
N∪Q=Q
N∩Q=N
N ∪ Q+ = Q+
RO
N ∩ Q+ = N
N ∪ Q− = {x ∈ Q|x < 0 or x ∈ N}
N ∩ Q− = ∅
N ∪ Q∗ = Q∗
N ∩ Q∗ = N
VE
N ∪ R + = R+
N ∩ R+ = N
N ∪ R− = {x ∈ R|x < 0 or x ∈ N}
N ∩ R− = ∅
D?
N ∪ R∗ = R∗
N ∩ R∗ = N
A few observations that can be made are as follows. The reader can make as many observations as they
want to.
1. A ⊂ B implies A ∪ B = B and A ∩ B = A.
2. Some of the (newly formed) sets are merely a result of union/intersection of known sets.
Downloaded by: tutorsection | Want to earn $1.236
Distribution of this document is illegal extra per year?
1
, Ex 1.4. If F is a pair-wise disjoint family indexed by Λ, then
T
Aλ 6= ∅ would imply the existence
λ∈Λ
of some element x in all of Aλ ’s contradicting the pair-wise disjointness. Thus, any pair-wise disjoint
family is disjoint.
The family {{1, 2} , {2, 3} , {1, 3}} is disjoint but not pair-wise disjoint, since {1, 2} ∩ {2, 3} = {2} =
6 ∅.
Ex 1.5. Example in Solution to exercise 1.4 does the job.
Ex 1.6. Prove the two-way subsethood for each of the equalities using the basic principles of logic and
properties of conjunction and disjunction operators.
Ex 1.7. Prove the two-way subsethood for each of the equalities using the basic principles of logic and
ST
properties of conjunction and disjunction operators.
Ex 1.8. P (∅) = {∅}.
Ex 1.9. The three sets cannot be equal since the elements in all three sets are different. R3 contains
UV
3-tuples, while (R × R) × R and R × (R × R) contains 2-tuple. Also, the first entry in a 2-tuple of
(R × R) × R is again a 2-tuple, while for elements in R × (R × R), the second entry is a 2-tuple.
Q S
Ex 1.10. Aλ = ∅, because if not so, the choice function f : ∅ → Aλ will be such that f (λ) ∈ Aλ .
λ∈∅ λ∈∅
This is possible only when we have such λ’s. Many also argue that since ∅ does not have any elements,
IA
this is vacously true. However, even in such an argument, the empty product is not defined. If there is
a difficulty to the reader in understanding this, the reader may take it as a convention.
Ex 1.11. (x1 , x2 , · · · , xn ) L (y1 , y2 , · · · , yn ) if and only if
_A
6 y1 ⇒ x1 R1 y1 ) and (x1 = y1 and x2 6= y2 ⇒ x2 R2 y2 ) and · · · and
(x1 =
(x1 = y1 , x2 = y2 , · · · , xn−1 = yn−1 ⇒ xn Rn yn )
PP
Ex 1.12.
≤ : Refexive, Anti-symmetric, Transitive
⊆ : Refexive, Anti-symmetric, Transitive
= : Refexive, Symmetric, Anti-symmetric, Transitive
RO
| : Refexive, Anti-symmetric, Transitive
≡ : Refexive, Symmetric, Transitive
Ex 1.13. Generalising modulo relation on Z for n ∈ N, we have m1 ≡ m2 (mod n) if and only if
n|m1 − m2 .
VE
The equivalence classes for this relation are {[0] , [1] , [2] , · · · , [n − 1]}, and an integer m is in an equiv-
alence class j if and only if the remainder after dividing m by n is r.
Ex 1.14. The equivalence classes for = relation on X for each x ∈ X are [x] = {x}.
The equivalence classes for ≡ (modulo relation) are as in solution to exercise 1.13.
D?
Ex 1.15. In the set X = {{1} , {2} , {3} , {1, 2} , {2, 3} , {1, 3}} equipped with ⊆ relation, there is no
least element.
Ex 1.16. In the example of solution to exercise 1.15, the sets {1} , {2} , {3} are minimal elements and
the sets {1, 2} , {2, 3} , {1, 3} are maximal elements..
Ex 1.17. In general, it is not true. For X = {1, 2, 3}, the POSET (P (X) , ⊆) has a least element,
namely ∅, but the subset {{1} , {2} , {3}} of P (X) has no least element.
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Distribution of this document is illegal extra per year?
