NOTES ON SUP’S, INF’S AND SEQUENCES
LANCE D. DRAGER
The purpose of these notes is to briefly review some material on sup’s, inf’s and
sequences from undergraduate real analysis, to give an introduction to using these
concepts in the extended real numbers, and to give an exposition of lim sup and
lim inf.
1. Sup’s and Inf’s in the Real Numbers
I’ll use the symbol R to denote the set of real numbers.
Let A be a nonempty subset of R. A number u is an upper bound for A if
a ≤ u for all a ∈ A. We say that A is bounded above if it has an upper bound.
Note that saying that a number t is not an upper bound for A is equivalent to
the statement “There is some a ∈ A such that t < a.”
A number s is called the supremum of A (sup) if it is the least upper bound
of A, i.e., s is an upper bound for A and if u is an upper bound for A then s ≤ u.
We denote the supremum of A by sup(A). (A little thought shows there can be at
most one number that satisfies the definition of sup.)
As you recall, the real numbers are constructed by “filling in the holes” in the
rational numbers. One way of saying that the holes have all been filled is the
Completeness Axiom for the Real Numbers, which is stated as follows.
Completeness Axiom for the Real Numbers. If A is a nonempty subset of R
that is bounded above then A has a supremum.
We have similar concepts when working on the other side of A. Again, let A
be a nonempty subset of R. A number ` is a lower bound for A if ` ≤ a for all
a ∈ A. We say A is bounded below if it has a lower bound.
A number i is the infimum of A if it is the greatest lower bound of A, i.e., i is
a lower bound and if ` is a lower bound for A, then ` ≤ i.
It is not necessary to add an axiom about inf’s to the definition of the real
numbers, since the existence of inf’s can be deduced from the existence of sup’s by
what I like to call “The Reflection Trick.” Here we set
−A = {−a | a ∈ A}.
Proposition 1.1. [Reflection Trick] Let A be a nonempty subset of R.
(1) If A is bounded below, −A is bounded above and
inf(A) = − sup(−A).
(2) If A is bounded above, −A is bounded below and
sup(A) = − inf(−A).
Exercise 1.2. Prove the last Proposition.
The following Propositions are frequently used in proofs involving sup and inf.
Proposition 1.3. Let A be a nonempty subset of R.
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2
LANCE D. DRAGER
(1) Suppose that A is bounded above. Then α < sup(A) if and only if α < a
for some a ∈ A.
(2) Suppose that A is bounded below. Then inf(A) < β if and only if a < β for
some a ∈ A.
Proof. To prove the first statement, assume first that α < sup(A). Since α is less
than the least upper bound of A, α is not an upper bound for A. Thus, there is
some a ∈ A so that α < a.
For the second part of the proof, assume that α < a for some a ∈ A. Then
α < a ≤ sup(A) (since sup(A) is an upper bound), so α < sup(A).
The proof of the second statement is similar and is left as an exercise.
Proposition 1.4. Suppose that A is a nonempty subset of R. Then inf(A) ≤
sup(A)
Proof. Since A is nonempty, we can find some a ∈ A. We have inf(A) ≤ a, since
inf(A) is a lower bound and we have a ≤ sup(A) since sup(A) is an upper bound.
Hence inf(A) ≤ a ≤ sup(A).
Proposition 1.5. Suppose that A ⊆ B ⊆ R and A 6= ∅. Then
(1) If B is bounded above, sup(A) ≤ sup(B).
(2) If B is bounded below, inf(A) ≥ inf(B).
Proof. If B is bounded above, sup(B) exists. Since sup(B) is an upper bound for
B, we have b ≤ sup(B) for all b ∈ B. In particular, we have a ≤ sup(B) for all
a ∈ A. Thus, sup(B) is an upper bound for A (so A is bounded above) and we
must have sup(A) ≤ sup(B) since sup(A) is the least upper bound of A.
The proof of the second statement is similar. If B is bounded below, inf(B)
exists. We have inf(B) ≤ b for all b ∈ B. In particular, we have inf(B) ≤ a for all
a ∈ A, so inf(B) is a lower bound for A (so A is bounded below). We must have
inf(B) ≤ inf(A), since inf(A) is the greatest lower bound of A.
2. Sequences of Reals Numbers
In this section, we will give a brief review of the basic material on sequences of real
numbers typically presented in an undergraduate analysis course. The statements
labeled “Fact” should be familiar to you—if the proof does not immediately come
to mind, consider the statement as an exercise.
2.1. Convergence of Sequences. Let N = {1, 2, 3, . . .} be the set of natural
numbers. A sequence of real numbers is a function N → R. It is traditional to
use subscripts (rather than functional notation) to describe the function. Thus the
sequence {an }∞
n=1 is the same thing as the function n 7→ an .
The first and most important concept about sequences is convergence.
Definition 2.1. The sequence {an } ⊆ R converges to a ∈ R if for every ε > 0
there exists an N ∈ N such that |an − a| < ε whenever n ≥ N .
Fact 2.2. A sequence converges to at most one real number.
The statement that {an } converges to a is written in symbols as
lim an = a,
n→∞
or an → a.
Fact 2.3. If an → a and bn → b then an + bn → a + b.
NOTES ON SUP’S, INF’S AND SEQUENCES
3
We will prove the corresponding statement for products in a moment. In order
to do so, we need that following fact.
Fact 2.4. A convergent sequence is bounded, i.e., if {an } converges, there is some
number M so that |an | ≤ M for all n ∈ N .
We also need the following famous statement.
Triangle Inequality. If x and y are real numbers,
|x| − |y| ≤ |x ± y| ≤ |x| + |y|.
We’re now ready to prove the following Proposition.
Proposition 2.5. If an → a and bn → b then an bn → ab.
Proof. The basic idea of the proof is based on the following inequalities.
|an bn − ab| = |an bn − abn + abn − ab|
≤ |an bn − abn | + |abn − ab|
= |bn ||an − a| + |a||bn − b|.
Thus, we have
(2.1)
|an bn − ab| ≤ |bn ||an − a| + |a||bn − b|.
Since {bn } is convergent, there is some number M > 0 so that |bn | ≤ M for all n.
Let ε > 0 be given. Since an → a, there is some N1 ∈ N so that
|an − a| < ε/M
(2.2)
for all n ≥ N1 . Since bn → b, there is some N2 ∈ N such that
ε
(2.3)
|bn − b| <
1 + |a|
for n ≥ N2 .
Set N = max(N1 , N2 ). If n ≥ N both (2.2) and (2.3) hold. Thus, if n ≥ N ,
(2.1) gives us
|an bn − ab| ≤ |bn ||an − a| + |a||bn − b|
≤ M |an − a| + |a||bn − b|
ε
ε
<M
+ |a|
M
1 + |a|
< ε + ε = 2ε,
since 0 ≤ |a|/(1 + |a|) < 1.
Thus, we have
|an bn − ab| < 2ε
for n ≥ N . This completes the proof.
Remark. You might object that we were supposed to get |an bn − ab| < ε instead of
less that 2ε. But this is okay. Look at it this way: you hand me an arbitrary η > 0
and tell me to make the desired quantity |an bn − ab| less than η. I just choose some
number ε so small that 2ε < η. I proceed with my proof to get |an bn −ab| < 2ε < η,
so I’ve met your challenge.
In general it’s okay to show that given ε > 0 you can make the desired quantity
less than Kε, for some constant K. Here “constant” means that K does not depend
on the choice of ε.
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LANCE D. DRAGER
It is convenient to make a few definitions that allow us to restate the definition
of convergence and related concepts concisely.
We say that the sequence {an } is eventually in a set E if there is an N ∈ N so
that an ∈ E for all n ≥ N .
Proposition 2.6. Suppose that {an } is eventually in E1 and eventually in E2 .
Then {an } is eventually in E1 ∩ E2 .
Proof. Since the sequence is eventually in E1 , there is some N1 ∈ N so that an ∈ E1
for n ≥ N1 . Similarly, there is some N2 so that an ∈ E2 for n ≥ N2 . Let
N = max(N1 , N2 ). Then, an ∈ E1 ∩ E2 for n ≥ N .
Let a be a point in R. A set U ⊆ R is a neighborhood of a if there is some
ε > 0 so that (a − ε, a + ε) ⊆ U .
Fact 2.7. Let {an } ⊆ R. Then, the following statements are equivalent.
(1) an → a.
(2) For every ε > 0, the sequence {an } is eventually in the interval (a−ε, a+ε).
(3) The sequence {an } is eventually in every neighborhood of a.
Proposition 2.8. Suppose that an → a and bn → b and that there is some M ∈ N
so that an ≤ bn for n ≥ M . Then a ≤ b.
Proof. The proof is by contradiction. Suppose, for a contradiction, that b < a.
Choose some number ε > 0 so that b + ε < a − ε ( any ε < (a − b)/2 will do).
Since an → a, there is some N1 ∈ N so that an ∈ (a−ε, a+ε) for n ≥ N1 . Similarly,
there is an N2 ∈ N so that bn ∈ (b − ε, b + ε) for n ≥ N2 .
Set N = max(M, N1 , N2 ). On one hand, since N ≥ M we have aN ≤ bN . On
the other hand, we have
bN < b + ε < a − ε < aN ,
so bN < aN . This contradiction shows that our original assumption that b < a
must be wrong.
Fact 2.9. Suppose that an → a and that eventually α ≤ an ≤ β. Then α ≤ a ≤ β.
A sequence {an } is called increasing if an ≤ an+1 for all n, i.e.,
a1 ≤ a2 ≤ a3 ≤ · · · ≤ an ≤ an+1 ≤ · · · .
Similarly, the sequence is called decreasing if an ≥ an+1 for all n.
The following Proposition relates convergence and sup’s.
Proposition 2.10. Let {an } be a sequence of real numbers.
(1) if {an } is increasing and bounded above, then the sequence is convergent
and
lim an = sup{an | n ∈ N}.
n→∞
(2) if {an } is decreasing and bounded below, then the sequence is convergent
and
lim an = inf{an | n ∈ N}.
n→∞
Proof. We’ll prove the first part of the Proposition and leave the second part as an
exercise.
We’re assuming that {an } is bounded above, meaning that the set of values of
the sequence {an | n ∈ N} is bounded above. Thus, s = sup{an | n ∈ N} exists.
NOTES ON SUP’S, INF’S AND SEQUENCES
5
Let ε > 0 be arbitrary. Since s − ε < s, s − ε is not an upper bound for the
values of the sequence, so there is some N so that s − ε < aN . Since the sequence
is increasing an ≥ aN for n ≥ N . Thus, for n ≥ N we have
s − ε < aN ≤ an ≤ s.
Thus, |an − a| < ε for n ≥ N . This shows that an → s.
2.2. Cluster Points and Subsequences. If {an } is a sequence, we say that {an }
is frequently in a set E if, for every N ∈ N, there is an n ≥ N so that an ∈ E.
Another way to say it is that infinitely many terms of the sequence are in E (but
there may also be infinitely many terms that are not in E).
If {an } is a sequence of real numbers, we say p ∈ R is a cluster point of
{an } if the sequence is frequently in every neighborhood of p. Equivalently, every
neighborhood of p contains infinitely many terms of the sequence.
If we have a strictly increasing sequence of natural numbers
n1 < n2 < n3 < · · · ,
{ank }∞
k=1
the sequence
is called a subsequence of {an }.
Fact 2.11. If an → a, then every subsequence of {an } converges to a.
Fact 2.12. A subsequence of a subsequence of {an } is a subsequence of {an }.
The relationship between cluster points and subsequences is given by the following proposition.
Proposition 2.13. Let {an } ⊆ R be a sequence. A point p ∈ R is a cluster point
of {an } if and only if there is a subsequence {ank } of the original sequence so that
ank → p.
Proof. For the first part of the proof, suppose that p is a cluster point of {an }.
Then the sequence is frequently in every neighborhood of p. Since (p − 1, p + 1) is
a neighborhood of p, there is some n1 such that an1 ∈ (p − 1, p + 1).
There are infinitely many terms of the sequence in the neighborhood (p−1/2, p+
1/2) of p, so we can find some n2 > n1 so that an2 ∈ (p − 1/2, p + 1/2). Continuing
in this way (technically an inductive construction), we get a sequence
n1 < n2 < n3 < · · ·
of natural numbers such that ank ∈ (p − 1/k, p + 1/k), equivalent, |ank − p| < 1/k.
It’s fairly clear that the subsequence {ank } converges to p. If k ≥ K, then
(p − 1/k, p + 1/k) ⊆ (p − 1/K, p + 1/K), so |ank − p| < 1/K for k ≥ K. Let ε > 0
be given. Choose K ∈ N so that 1/K < ε. Then |ank − p| < 1/K < ε for k ≥ K.
Thus, ank → p.
For the converse, assume that there is a subsequence {ank } that converges to p.
Suppose that U is a neighborhood of p. Let N be a natural number. Since ank → p,
ank is eventually in U . Thus, there is a K so that ank ∈ U for k ≥ K. Since the
integers nk go to infinity, there is some k ≥ K such that nk > N . Thus, there is
a term ank of the original sequence with index larger than N that is in U . This
shows that {an } is frequently in every neighborhood of p.
Fact 2.14. A cluster point of a subsequence {ank } is a cluster point of the original
sequence {an }.
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LANCE D. DRAGER
3. The Extended Real Numbers
The fact that not all subsets of the real line have a sup leads us to introduce the
extended real numbers, which are essentially just a notational device.
We define the extended real numbers R by R = R ∪ {−∞} ∪ {∞}, where ∞
and −∞ are some objects that are not real numbers. We define the ordering on R
by −∞ < r < ∞ for all r ∈ R. To indicate the ordering, we can use the interval
notation R = [−∞, ∞].
Every subset of R has a sup. If A ⊆ R is bounded above, the sup is the same as
in the reals. If A is not bounded above, sup(A) = ∞. If A ⊆ R contains ∞, then
∞ is the only upper bound for A, so sup(A) = ∞. Similarly, every subset of R has
an inf.
Fact 3.1. [Reflection Trick] If A ⊆ R is a nonempty set, inf(A) = − sup(−A)
and sup(A) = − inf(−A).
We define addition on R by declaring it to be commutative and defining
∞+∞=∞
r + ∞ = ∞,
r ∈ R.
r − ∞ = r + (−∞) = −∞,
−∞ − ∞ = −∞
r∈R
−∞ + ∞ and ∞ − ∞ are undefined.
This addition is associative in the following sense: if a, b, c ∈ R either both of
a+(b+c) and (a+b)+c are undefined or both are defined and a+(b+c) = (a+b)+c.
Proposition 3.2. Suppose that a, A, b, B ∈ R and a ≤ A and b ≤ B. Then
a + b ≤ A + B, provided both sides are defined.
Proof. As usual with the extended reals, the proof is by considering cases. First,
suppose that A and B are real. If a and b are real, we know that a + b ≤ A + B.
The second possibility is that a + b = −∞, in which case a + b ≤ A + B is certainly
true.
Secondly, A + B might be ∞, in which case a + b ≤ A + B is certainly true,
provided that a + b is defined.
The last possibility is that one of A and B is −∞ and the other is −∞ or real.
In this case one of a and b must be −∞ and the other must be −∞ or real. We
must have A + B = −∞ and a + b = −∞, so a + b ≤ A + B is true.
We define multiplication by declaring it to be commutative, and using the principal that multiplication by a negative quantity should reverse ordering. Thus, we
NOTES ON SUP’S, INF’S AND SEQUENCES
7
define
∞·∞=∞
(−∞) · ∞ = −∞
(−∞) · (−∞) = ∞
(
∞,
0<r<∞
r·∞=
−∞, −∞ < r < 0
(
−∞, 0 < r < ∞
r · (−∞) =
∞,
−∞ < r < 0
0 · (±∞) = 0.
The definition 0 · (±∞) = 0 (as opposed to leaving it undefined) is usually
appropriate in integration theory, but requires caution when dealing with limits, as
we will see below. Thus, it’s good to keep in mind that caution is required with
this definition.
Exercise 3.3. The following facts are often used.
(1) Let A ⊆ R be a nonempty set and let c ∈ (0, ∞). Then sup(cA) = c sup(A).
(2) Let X be a set and let f and g be functions X → [0, ∞]. Then
sup{f (x) + g(x) | x ∈ X} ≤ sup{f (x) | x ∈ X} + sup{g(x) | x ∈ X}.
Give an example where strict inequality occurs.
4. Sequences in the Extended Real Numbers
We can easily extend the definition of neighborhood to R. If p ∈ R, a set U ⊆ R
is a neighborhood of p if U contains an interval (p − ε, p + ε) for some ε > 0. A
set U is a neighborhood of ∞ if U contains an interval (α, ∞] for some α ∈ R.
Finally, U is a neighborhood of −∞ if U contains an interval [−∞, β), for some
β ∈ R.
We say that a sequence {an } in R converges to p ∈ R if the sequence is eventually
in every neighborhood of p.
As before, we say that p is a cluster point of {an } if {an } is frequently in every
neighborhood of p.
The following facts are extensions of the facts we know about real sequences to
sequences in R. The proofs are very similar, with a few, easy, added lines to deal
with ±∞.
Fact 4.1. Let {an } be a sequence in R. If an → a, then every subsequence of {an }
converges to a.
Fact 4.2. Let {an } be a sequence in R. Then p is a cluster point of {an } if and
only if there is a subsequence of {an } that converges to p.
Fact 4.3. Let {an } and {bn } be sequences in R. Suppose that an → a, bn → b and
that an ≤ bn (eventually). Then a ≤ b.
As a corollary, if c, d ∈ R and c ≤ an ≤ d (eventually), then c ≤ a ≤ d.
Fact 4.4. Let {an } be a sequence in R. If an is increasing, i.e., an ≤ an+1 , then
an converges to sup{an | n ∈ N}. Similarly, if an is decreasing, it converges to
inf{an | n ∈ N}.
Sums and products of sequences take more care than in the real case.
8
LANCE D. DRAGER
Proposition 4.5. Let {an } and {bn } be sequence in R and suppose that an → a and
bn → b. Then, if a + b is defined, an + bn is eventually defined and an + bn → a + b.
Proof. There are 9 cases for the possible values for a and b, namely
(4.1)
a = ∞,
b=∞
(4.2)
a = ∞,
b∈R
(4.3)
a=∞
b = −∞
(4.4)
a∈R
b=∞
(4.5)
a∈R
b∈R
(4.6)
a∈R
b = −∞
(4.7)
a = −∞
b=∞
(4.8)
a = −∞,
b∈R
(4.9)
a = −∞
b = −∞
In cases (4.3) and (4.7), the sum a + b is undefined, so we don’t need to consider
these cases. We can ignore the case (4.4), since the proof would be the same as in
case (4.2), just interchanging the letters a and b. Similarly, we can ignore (4.8),
since the proof would be the same as in (4.6).
First, consider the case (4.1). Since (−∞, ∞] is a neighborhood of ∞, an is
eventually not equal to −∞. Similarly, bn is eventually not equal to −∞. Thus,
an + bn is eventually defined. Let α ∈ R be arbitrary. Then, an ∈ (α/2, ∞]
eventually. Similarly for bn . Thus, for sufficiently large n, an +bn > α/2 +α/2 = α.
This shows that an + bn → ∞ = a + b.
Next, consider case (4.2). Since R is a neighborhood of b, bn is eventually real.
Thus, an + bn is eventually defined. Since bn → b, {bn } is bounded in the real
numbers, so there is some M ∈ R so that M ≤ bn for all sufficiently large n.
Let α ∈ R be arbitrary. Since an → ∞, an is eventually greater than α − M .
Thus, for sufficiently large n, an + bn > (α − M ) + M = α. This shows that
an + bn → ∞ = a + b.
Consider the case (4.5). As in the last case, both sequences are eventually reals,
so an + bn is eventually defined. The result then follows from the theorem on the
limit of the sum in the real case.
The remaining cases are very similar and are left to the reader.
The proof of the following proposition about products of sequences is left to the
reader.
Proposition 4.6. Let {an } and {bn } be sequences in R, and suppose that an → a
and bn → b. Then an bn → ab provided that ab is not of the form 0 · (±∞).
5. Lim Sup and Lim Inf
Since this material may be new, I will provide more details than in the review
sections.
Let {an } ⊆ R be a sequence. For each n ∈ N, define
An = {ak | k ≥ n} = {an , an+1 , an+2 , . . .},
NOTES ON SUP’S, INF’S AND SEQUENCES
9
the n-th tail of the sequence. Clearly, these form a decreasing sequence of sets, i.e.,
An ⊇ An+1 . Define
A+
n = sup(An ).
Since the An ’s are decreasing, A+
n is a decreasing sequence of extended reals numbers. Thus, this sequence converges and
+
lim A+
n = inf An
n→∞
n∈N
This extended real number is called the limit superior of {an }, and is denoted by
lim sup an
lim an .
or
n→∞
n→∞
Similarly, we define
A−
n = inf(An ).
Since the sets An are decreasing, A−
n is an increasing sequence, so
−
lim A−
n = sup An .
n→∞
n∈N
This extended real number is called the limit inferior of {an } and is denoted by
lim inf an
n→∞
or
lim an .
n→∞
+
We have, of course, A−
n ≤ An , so letting n → ∞ gives
lim inf an ≤ lim sup an .
n→∞
n→∞
There is a straight forward version of the reflection trick for lim sup and lim inf.
The proof is left as an exercise.
Proposition 5.1. [Reflection Trick] Let {an } ⊆ R be a sequence. Then,
(1)
lim sup an = − lim inf (−an ),
n→∞
n→∞
(2)
lim inf an = − lim sup(−an ).
n→∞
n→∞
I will generally do the proofs in this section for lim sup and leave the proofs
for lim inf as an exercise. The can be done by similar arguments, or by using the
reflection trick.
The following is our first basic proposition about lim sup and lim inf.
Proposition 5.2. Let {an } ⊆ R be a sequence.
(1) If
lim sup an < α,
n→∞
then eventually an < α
(2) If, eventually, an ≤ α then
lim sup an ≤ α.
n→∞
(3) If
β < lim inf an
n→∞
then eventually β < an .
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LANCE D. DRAGER
(4) If, eventually, β ≤ an then
β ≤ lim inf an .
n→∞
Proof. Consider the first part of the proposition. If
lim sup an < α,
n→∞
then
inf A+
n < α,
n∈N
A+
N <
so there is some N ∈ N so that
α. But an ≤ A+
N for n ≥ N , so an < α for
n ≥ N.
For the second part of the Proposition, suppose that an ≤ α for n ≥ N . Then,
for any n ≥ N , we have
A+
n = sup{an , an+1 , . . .} ≤ α
and so
lim sup an = inf A+
n ≤ α.
n
n→∞
This completes the proof.
The following Proposition clarifies the meaning of lim sup and lim inf.
Proposition 5.3. Let {an } be a sequence in R. Then, lim supn an is the largest
cluster point of {an } and lim inf n an is the smallest cluster point of {an }, i.e., if p
is a cluster point of {an }, then
lim inf an ≤ p ≤ lim sup an .
n→∞
n→∞
Proof. I’ll do the proof for lim sup. We first want to show that lim sup an is a cluster
point. We distinguish 3 cases.
The first case is lim sup an = ∞. In this case, we have inf n A+
n = ∞, so we must
have A+
=
∞
for
all
n.
Let
α
∈
R
be
given.
Given
any
N
,
we
have α < A+
n
N , so
there must be some element of AN = {aN , aN +1 , . . .} that is bigger than α. This
shows that the sequence {an } is frequently in the neighborhood (α, ∞] of ∞. Since
α was arbitrary, we conclude that ∞ is a cluster point of {an }.
For the second case, assume lim sup an = s ∈ R. Let ε > 0 be arbitrary and let
M ∈ N be given. Since
inf A+
n = lim sup an = s < s + ε
n
n→∞
there is some N ∈ N so that s ≤ A+
N < s + ε. (It follows that ak < s + ε for
+
k ≥ N .) Since A+
n is a decreasing sequence, we have s ≤ An < s + ε for n ≥ N .
+
Let K = max(M, N ), so we have s ≤ AK < s + ε. But then
s − ε < A+
K = sup{ak | k ≥ K}.
Thus, there must be some k ≥ K so that
s − ε < ak ≤ A+
K < s + ε.
Thus, there is some index k ≥ M so that ak is in the interval (s − ε, s + ε). It
follows that the sequence {an } is frequently in (s − ε, s + ε). Since ε was arbitrary,
we conclude that s is a cluster point of an .
NOTES ON SUP’S, INF’S AND SEQUENCES
11
The last case to consider is lim sup an = −∞. Let β ∈ R be arbitrary. Since
inf A+
n < β,
n
there must be some N so that
A+
N
< β. But for any n ≥ N we have
an ≤ A+
N < β.
Thus, the sequence {an } is eventually in [−∞, β). Since β was arbitrary, we conclude that an → −∞, so certainly −∞ is a cluster point.
This completes that proof that lim sup an is a cluster point. Note that we have
proven that an → −∞ if lim sup an = −∞. The corresponding statement for lim inf
is that an → ∞ if lim inf an = ∞. These statements will be used later.
To finish the proof, suppose that p is a cluster point of {an }. Our goal is to
show p ≤ lim sup an . If p = −∞, there is nothing to do, since −∞ ≤ lim sup an no
matter what lim sup an is.
So, suppose that p > −∞. Let α < p be arbitrary. Since p is a cluster point, an
is frequently greater than α (why?). Thus, for any n, the set {ak | k ≥ n} contains
elements that are strictly greater than α, so A+
n = sup{ak | k ≥ n} > α. Thus,
α < A+
for
all
n.
Letting
n
→
∞,
we
conclude
that α ≤ lim sup an . Since α < p
n
was arbitrary, we must have p ≤ lim sup an . This completes the proof.
An easy application of this Proposition is the following.
Proposition 5.4. Let {an } ⊆ R be a sequence, and let {ank } be a subsequence.
Then,
lim inf an ≤ lim inf ank ≤ lim sup ank ≤ lim sup an .
n→∞
k→∞
k→∞
n→∞
Proof. Since lim inf k→∞ ank and lim supk→∞ ank are cluster points of the subsequence {ank }, they are also cluster points of {an }. Thus, the inequality follows
from the last Proposition.
The next Proposition characterizes convergence in terms of lim sup and lim inf.
Proposition 5.5. Let {an } ⊆ R be a sequence. Then
lim an = a
n→∞
if and only if
lim inf an = a = lim sup an
n→∞
n→∞
Proof. If an → a, then a is the only cluster point of the sequence. Since lim inf an
and lim sup an are cluster points, we must have a = lim inf an = lim sup an .
For the second part of the proof, suppose that
lim inf an = a = lim sup an .
n→∞
n→∞
If a = ∞, then lim inf an = ∞. But, as we saw in the proof of Proposition 5.3,
this implies an → ∞ = a. Similarly, if a = −∞, then lim sup an = −∞ and we saw
in the proof of Proposition 5.3 that this implies an → −∞.
This leaves the case where a ∈ R. Let ε > 0 be given. Since lim sup an =
a < a + ε, Proposition 5.2 shows that an is eventually less than a + ε. Since
a − ε < a = lim inf an , Proposition 5.2 shows that an is eventually greater than
a − ε. Thus, an is eventually in the interval (a − ε, a + ε). Since ε > 0 was arbitrary,
we conclude an → a.
To conclude this section, we consider what can be said about the lim sup and
lim inf of the sum of two sequences.
12
LANCE D. DRAGER
Proposition 5.6. Let {an } and {bn } be sequences in R and suppose that an + bn
is (eventually) defined.
(1) We have
(5.1)
lim sup(an + bn ) ≤ lim sup an + lim sup bn ,
n→∞
n→∞
n→∞
provided that the right hand side is defined.
(2) If one of the sequences is convergent then equality holds in (5.1) (still assuming the right hand side is defined).
(3) We have
(5.2)
lim inf an + lim inf bn ≤ lim inf (an + bn ),
n→∞
n→∞
n→∞
provided that the left hand side is defined.
(4) If one of the sequences is convergent, equality holds in (5.2) (still assuming
that the left hand side is defined).
Proof. To prove (5.1), let cn = an + bn and recall the definitions
An = {ak | k ≥ n}
A+
n = sup(An )
lim sup an = lim A+
n
n→∞
n→∞
Bn = {bk | k ≥ n}
Bn+ = sup(Bn )
lim sup bn = lim Bn+
n→∞
n→∞
Cn = {ck | k ≥ n}
Cn+ = sup(Cn )
lim sup cn = lim Cn+ .
n→∞
n→∞
+
Since we are assuming that the right hand side of (5.1) is defined, A+
n + Bn is
eventually defined, so we can consider n’s that are large enough to make this sum
+
defined. For k ≥ n, we have ak ≤ A+
n and bk ≤ Bn . Adding these inequalities
+
+
+
is
an upper bound for Cn , so
.
Thus,
A
+
B
gives ck = ak + bk ≤ A+
+
B
n
n
n
n
+
.
Letting
n
go
to
infinity
gives
(5.1).
+
B
Cn+ ≤ A+
n
n
For the second part, suppose that an → a. Then (5.1) becomes
(5.3)
lim sup(an + bn ) ≤ a + lim sup bn ,
n→∞
n→∞
where we assume that the right hand side is defined. Since lim sup bn is a cluster
point of {bn }, there is a subsequence {bnk } so that bnk → lim sup bn . Since an → a,
we also have ank → a. Thus,
(5.4)
ank + bnk → a + lim sup bn .
n→∞
However, {ank + bnk } is a subsequence of {an + bn }, so the number on the right of
(5.4) is a cluster point of {an + bn }. Thus, we must have
a + lim sup bn ≤ lim sup(an + bn ).
n→∞
n→∞
NOTES ON SUP’S, INF’S AND SEQUENCES
13
by Proposition 5.4. Combining this with (5.3) shows that
lim sup(an + bn ) = a + lim sup bn
n→∞
n→∞
and the proof is complete.
For an example where inequality holds in (5.1) and (5.2), consider the sequences
an = (−1)n and bn = (−1)n+1 . Then lim sup an = lim sup bn = 1, lim inf an =
lim inf bn = −1 and lim sup(an + bn ) = lim inf(an + bn ) = lim(an + bn ) = 0. Thus,
we have
− 2 = lim inf an + lim inf bn < lim inf (an + bn ) = 0
n→∞
n→∞
n→∞
= lim sup(an + bn ) < lim sup an + lim sup bn = 2.
n→∞
n→∞
n→∞
Exercise 5.7. Let {an } be a sequence in R. If c ∈ [0, ∞), show that
lim inf (can ) = c lim inf an
n→∞
n→∞
lim sup(can ) = c lim sup bn .
n→∞
n→∞
What happens if c ∈ (−∞, 0)?
Exercise 5.8. If an ≤ bn eventually, show that
lim inf an ≤ lim inf bn
n→∞
n→∞
lim sup an ≤ lim sup bn .
n→∞
n→∞
Department of Mathematics and Statistics, Texas Tech University, Lubbock TX
79409-1042
E-mail address: drager@math.ttu.edu