Understanding Analysis (2nd Edi on, 2015) –
Solu ons Manual – by Abbo
Axiom of Completeness - answer-Every nonempty set of real numbers that is bounded above
has a least upper bound.
Upper/Lower Bound - answer-A set A⊆R is bounded above (below) if there exists a number b∈R
such that a≤b (a≥b) for all a∈R
least upper (greatest lower) bound - answer-A⊆R has a supremum (infimum) if:
i) s is an upper (lower) bound for A
ii) if b is any upper bound for A then s≤b
wri en as s=supA; s=infA
Maximum (minimum) - answer-A real number a₀ is a maximum (minimum) of the set A if a₀ is
an element of A and a₀≥a (a₀≤a) for all a∈A.
Lemma for Supremum (Infimum) - answer-Assume s∈R is an upper (lower) bound for a set A⊆R.
Then, s=supA (s=infA) iff, for every choice of ε>0, there exists an element a∈A sa sfying s-ε<a.
Nested Interval Property - answer-For each n∈ℵ, assume we are given a closed interval
Iⁿ=[aⁿ,bⁿ]={x∈R:aⁿ≤x≤bⁿ}. Assume also that each Iⁿ contains Iⁿ⁺¹. Then, the resul ng nested
sequence of closed intervals I₁⊇I₂⊇I₃⊇... has a nonempty intersec on; that is ∩∞Iⁿ≠∅
Archimedian Property - answer-(i) Given any number x∈R, there exists an n∈ℵ sa sfying n>x.
(ii) Given any real number y>0, there exists an n∈ℵ sa sfying 1/n<y.
1
,Density of Q in R - answer-For every two real numbers a and b with a<b, there exists a ra onal
number r sa sfying a<r<b.
Corollary of Density of Q in R - answer-Given any two real numbers a<b, there exists an irra onal
number t sa sfying a<t<b
Existence of Square Roots - answer-There exists a real number α∈R sa sfying α²=2.
one-to-one and onto (func ons) - answer-A func on ƒ:A→B is one-to-one if a₁≠a₂ in A implies
ƒ(a₁)≠ƒ(a₂) in B. The func on ƒ is onto if, given any b∈B, it is possible to find an element a∈A for
which ƒ(a)=b
A∼B (same cardinality) - answer-Two sets A and B have the same cardinality if there exists
ƒ:A→B that is one-to-one and onto. We write A∼B.
Countable vs. uncountable - answer-A set A is countable if ℵ∼A. An infinite set that is not
countable is called an uncountable set.
Theorem 1:
(i) The set Q is countable.
(ii) The set R is uncountable.
Theorem 2:
If A⊆B and B is countable, then A is either countable, finite, or empty.
Theorem 3:
The countable union of countable sets is countable
2
, Theorem 4:
The open interval (0,1) = {x∈R:0<x<1} is uncountable.
Sequence - answer-A sequence is a func on whose domain is ℵ.
Convergence of a Sequence - answer-A sequence (aⁿ) converges to a real number a if, for every
posi ve number ε, there exists an N∈ℵ s.t. whenever n≥N it follows that |aⁿ-a|<ε.
or
A sequence (aⁿ) converges to a if, given any ε-neighborhood Vε(a) of a, there exists a point in the
sequence a er which all the terms are in Vε(a).
Boundedness - answer-A sequence (xⁿ) is bounded if there exists a number M>0 such that
|xⁿ|≤M for all n∈ℵ
Theorem:
Every convergent sequence is bounded.
Algebraic Limit Theorem - answer-Let lim aⁿ = a and lim bⁿ = b. Then,
(i) lim (caⁿ) = ca, for all c∈R;
(ii) lim (aⁿ + bⁿ) = a+b;
(iii) lim (aⁿbⁿ) = ab;
(iv) lim (aⁿ/bⁿ) = a/b provided b≠0
Order Limit Theorem
3
Solu ons Manual – by Abbo
Axiom of Completeness - answer-Every nonempty set of real numbers that is bounded above
has a least upper bound.
Upper/Lower Bound - answer-A set A⊆R is bounded above (below) if there exists a number b∈R
such that a≤b (a≥b) for all a∈R
least upper (greatest lower) bound - answer-A⊆R has a supremum (infimum) if:
i) s is an upper (lower) bound for A
ii) if b is any upper bound for A then s≤b
wri en as s=supA; s=infA
Maximum (minimum) - answer-A real number a₀ is a maximum (minimum) of the set A if a₀ is
an element of A and a₀≥a (a₀≤a) for all a∈A.
Lemma for Supremum (Infimum) - answer-Assume s∈R is an upper (lower) bound for a set A⊆R.
Then, s=supA (s=infA) iff, for every choice of ε>0, there exists an element a∈A sa sfying s-ε<a.
Nested Interval Property - answer-For each n∈ℵ, assume we are given a closed interval
Iⁿ=[aⁿ,bⁿ]={x∈R:aⁿ≤x≤bⁿ}. Assume also that each Iⁿ contains Iⁿ⁺¹. Then, the resul ng nested
sequence of closed intervals I₁⊇I₂⊇I₃⊇... has a nonempty intersec on; that is ∩∞Iⁿ≠∅
Archimedian Property - answer-(i) Given any number x∈R, there exists an n∈ℵ sa sfying n>x.
(ii) Given any real number y>0, there exists an n∈ℵ sa sfying 1/n<y.
1
,Density of Q in R - answer-For every two real numbers a and b with a<b, there exists a ra onal
number r sa sfying a<r<b.
Corollary of Density of Q in R - answer-Given any two real numbers a<b, there exists an irra onal
number t sa sfying a<t<b
Existence of Square Roots - answer-There exists a real number α∈R sa sfying α²=2.
one-to-one and onto (func ons) - answer-A func on ƒ:A→B is one-to-one if a₁≠a₂ in A implies
ƒ(a₁)≠ƒ(a₂) in B. The func on ƒ is onto if, given any b∈B, it is possible to find an element a∈A for
which ƒ(a)=b
A∼B (same cardinality) - answer-Two sets A and B have the same cardinality if there exists
ƒ:A→B that is one-to-one and onto. We write A∼B.
Countable vs. uncountable - answer-A set A is countable if ℵ∼A. An infinite set that is not
countable is called an uncountable set.
Theorem 1:
(i) The set Q is countable.
(ii) The set R is uncountable.
Theorem 2:
If A⊆B and B is countable, then A is either countable, finite, or empty.
Theorem 3:
The countable union of countable sets is countable
2
, Theorem 4:
The open interval (0,1) = {x∈R:0<x<1} is uncountable.
Sequence - answer-A sequence is a func on whose domain is ℵ.
Convergence of a Sequence - answer-A sequence (aⁿ) converges to a real number a if, for every
posi ve number ε, there exists an N∈ℵ s.t. whenever n≥N it follows that |aⁿ-a|<ε.
or
A sequence (aⁿ) converges to a if, given any ε-neighborhood Vε(a) of a, there exists a point in the
sequence a er which all the terms are in Vε(a).
Boundedness - answer-A sequence (xⁿ) is bounded if there exists a number M>0 such that
|xⁿ|≤M for all n∈ℵ
Theorem:
Every convergent sequence is bounded.
Algebraic Limit Theorem - answer-Let lim aⁿ = a and lim bⁿ = b. Then,
(i) lim (caⁿ) = ca, for all c∈R;
(ii) lim (aⁿ + bⁿ) = a+b;
(iii) lim (aⁿbⁿ) = ab;
(iv) lim (aⁿ/bⁿ) = a/b provided b≠0
Order Limit Theorem
3
