In this case, we say that U is an upper bound for E

Remark: In this sections, all the subsets of R are assumed to be nonempty.

Let E be a subset of R. We say that E is bounded above if there exists a real number U such that x ≤ U for all x ∈ E. In this case, we say that U is an upper bound for E. We say that E is bounded below if there exists a real number L so that x ≥ L for all x ∈ E. In this case, we say that L is a lower bound for E. A subset E of R is said to be bounded if E is both bounded above and bounded below.

Let E be a subset of R.

(1) Suppose E is bounded above. An upper bound U of set E is the least upper bounded of E if for any upper bound U⁰ of E, U⁰ ≥ U. If U is the least upper bound of E, we denote U by sup E. The least upper bound for E is also called supremum of E.

(2) Suppose E is bounded below. A lower bounded L of E is said to be the greatest lower bound of E if for any lower bound L⁰ of E, L ≥ L⁰. If L is the greatest lower bound for E, we denote L by inf E. The greatest lower bound for E is also called the infimum of E.

Example 1.1. Let a, b be real numbers. The set [a, b] = {x ∈ R : a ≤ x ≤ b} is bounded with sup[a, b] = b and inf[a, b] = a.

Example 1.2. Let a be a real numbers. We denote (−∞, a) = {x ∈ R : x < a}. Then (−∞, a) is bounded above but not bounded below.

Example 1.3. Given a sequence (an) of real numbers, let {an∈ R : n ≥ 1} be the image of (a_n), i.e. the set of all values of (a_n). Then (a_n) is bounded (bounded above, bounded below) if and only if the set {an∈ R : n ≥ 1} is bounded (bounded above, bounded below).

Theorem 1.1. (Property of R) In R, the following hold:

(1) Least upper bound property: Let S be a nonempty set in R that has an upper bound. Then S has a least upper bound.

(2) Greatest lower bound property: Let P be a nonempty subset in R that has a lower bound. Then P has a greatest lower bound in R.

Example 1.4. Consider the set S = {x ∈ R : x²+ x < 3}. Find sup S and inf S.

Example 1.5. Let S = {x ∈ Q : x²< 2}. Find sup S and inf S.

Example 1.6. Let S = {x ∈ R : x³ < 1}. Find sup S. Is S bounded below?

Proposition 1.1. Let E be a bounded subset of R and U ∈ R is an upper bound of E.

Then U is the least upper bound of E if and only if for any  > 0, there exists x ∈ E so that x ≥ U − .

Proof. Suppose U = sup E. Then for any  > 0, U −  < U. Hence U −  is not an upper bound of E. Claim: there exists x ∈ E so that x > U − . If there is no x so that x > U − ,

1

then x ≤ U −  for all x ∈ E. This implies that U −  is again an upper bound for E. By the definition of the least upper bound, U ≤ U −  which is absurd since  > 0.

Conversely, let U⁰ = sup E. Since U is an upper bound of E, U ≥ U⁰. Claim U⁰ = U. For any  > 0, choose x ∈ E so that x > U − . Thus U⁰ ≥ x > U − . We know that for any U⁰ > U −  for all  > 0. Since  is arbitrary U⁰ ≥ U. We conclude that U⁰ = U.



Lemma 1.1. Let E be a nonempty subset of R.

(1) If E is bounded above and α < sup E, then there exists x ∈ E so that x > α.

(2) If E is bounded above and β > inf E, then there exists x ∈ E so that x < β.

Proof. The result follows from the definition. 

Corollary 1.1. Let E be a bounded subset of R and L ∈ R is a lower bound of E. Then L is the greatest upper bound of E if and only if for any  > 0, there exists x ∈ E so that x ≤ L − .

Proof. We leave it to the reader as an exercise. 

Let E be a nonempty subset of R. We say that M is a maximum of E if M is an element of E and x ≤ M for all x ∈ E. Using the definition, we immediately know that there is only one maximum of E if the maximum elements of E exists. In this case, we denote M by max E. It also follows from the definition that if the maximum of E exists, it must be bounded above.

We say that m is the minimum of a nonempty subset E of R if m is an element of E and x ≥ m for all x ∈ m. We denote m by min E. It follows from the definition that if a set has minimum, it must be bounded below.

Example 1.7. The following subsets of R are all bounded. Hence their greatest lower bound and their least upper bound exist. Determine whether their maximum or minimum exist.

(1) E₁ = (0, 1).

(2) E2 = (0, 1].

(3) E₃ = [0, 1).

(4) E4 = [0, 1].

Proposition 1.2. Suppose E is a nonempty subset of R. If E is bounded above, then the maximum of E exists if and only if sup E ∈ E. Similarly, if E is bounded below, then the minimum of E exists if and only if inf E ∈ E.

Proof. The proof follows from the definition. 

Proposition 1.3. Let E be a nonempty subset of R. Suppose E is a bounded set. Then inf E ≤ sup E.

Proof. We leave it to the reader as an exercise.



Proposition 1.4. Let E and F be nonempty subsets of R. Suppose that E ⊂ F.

(1) If E and F are both bounded below, then inf F ≤ inf E.

(2) If E and F are both bounded above, then sup E ≤ sup F.

Proof. Let us prove (a). (b) is left to the reader.

For any x ∈ F, x ≥ inf F. Since E is a subset of F, x ≥ inf F holds for all x ∈ E. Therefore inf F is a lower bound for E. Since inf E is the greatest lower bound for E, inf E ≥ inf F.



Theorem 1.2. Every bounded monotone sequence is convergent.

Proof. Without loss of generality, we may assume that (an) is a bounded nondecreasing sequence of real numbers. Let a = sup{a_n : n ≥ 1}. Given  > 0, there exists a_N so that a ≥ a_N > a − . Since (a_n) is nondecreasing, a_n≥ a_N for every n ≥ N. Hence a_n> a −  for every n ≥ N. In this case, |an− a| = a − a_n<  for n ≥ N. We prove that lim

n→∞an= a. 

2. limsup and liminf

Let (a_n) be a bounded sequence of real numbers. Define a new sequence (x_n) by xn= sup{am: m ≥ n}, n ≥ 1,

Since (a_n) is bounded, x_n is a real number for each n ≥ 1. We assume that |a_n| ≤ M for all n ≥ 1. Then |xn| ≤ M for all n ≥ 1. This shows that (x_n) is also a bounded sequence. By Proposition 1.4, (x_n) is nonincreasing. By monotone sequence property, x = lim

n→∞x_nexists.

We denote x by lim sup

n→∞

a_n. Similarly, define a sequence (y_n) by yn= inf{am: m ≥ n}, n ≥ 1.

Then (y_n) is a nondecreasing sequence by Proposition 1.4. By monotone sequence property, y = lim

n→∞yn exists. We denote y by lim inf

n→∞ an. From now on, we will simply denote by

a^∗ = lim sup

n→∞

a_n, a∗= lim inf

n→∞ a_n.

Since (xn) is nonincreasing and bounded below, its limit equals to inf{xn : n ≥ 1}.

Similarly, (y_n) is nondecreasing and bounded above, its limit equals to sup{y_n: n ≥ 1}.

Example 2.1. Find lim sup

n→∞

(−1)ⁿ and lim inf

n→∞ (−1)ⁿ.

Solution: Let us compute these two numbers via definition. Denote (−1)ⁿ by a_n. For each k ≥ 1, we know {an : n ≥ k} = {−1, 1}. For k ≥ 1, x_k = sup{an : n ≥ k} = sup{−1, 1} = 1. Similarly, for k ≥ 1, y_k = sup{a_m : m ≥ k} = inf{−1, 1} = −1. Therefore lim sup

n→∞

an = lim

k→∞xk = 1 while lim inf

n→∞ an = lim

k→∞yk = −1. Notice that the subsequence (a_2n) of (a_n) is convergent to 1 and the subsequence (a_2n−1) of (a_n) is convergent to −1.

Later, we will prove that in general, the limit supremum and the limit infimum of a bounded sequence are always the limits of some subsequences of the given sequence.

Example 2.2. Let (an) be the sequence defined by a_n= 1 − 1

n, n ≥ 1.

Evaluate lim sup

n→∞

a_n and lim inf

n→∞ a_n.

Solution: The sequence (a_n) is increasing and bounded above by 1. Let us prove that sup{an : n ≥ 1} = 1. We have seen that 1 is an upper bound. Now, for each  > 0, choose N_ = [1/] + 1. For n ≥ N_, we see

1 −  < 1 − 1 n.

By Proposition 1.1, we find 1 is indeed the least upper bound for {a_n : n ≥ 1}. Therefore 1 = sup{an: n ≥ 1}. We also know that lim

n→∞an= 1. In other words, we prove

n→∞lim

 1 − 1

n



= sup

 1 − 1

n : n ≥ 1

 .

This gives us an example of Theorem 1.2. Similarly, for each k ≥ 1, we can show that 1 is the least upper bound of the set {1 − 1/n : n ≥ k} and −1 is the greatest lower bound for

{1 − 1/n : n ≥ k}. In other words, we find that x_k= sup

 1 − 1

n : n ≥ k



= 1, y_k= inf

 1 −1

n : n ≥ k



= −1.

This shows that lim

k→∞x_k= 1 and lim

k→∞y_k= −1. In other words, lim sup

n→∞

an= lim inf

n→∞ an= 1.

In this case, (a_n) is convergent to 1 and at the same time, both limit supremum and limit infimum of (an) are also equal to 1. This is not an accident. Later, we will prove that a bounded sequence is convergent if and only if its limit supremum equals to its limit infimum.

Lemma 2.1. Let (a_n) be a bounded sequence and a ∈ R.

(1) If a > a^∗, there exists k ∈ N such that an< a for all n ≥ k.

(2) If a < a^∗, then for all k ∈ N, there exists n ∈ N with n ≥ k such that an> a (3) If a < a∗, there exists k ∈ N such that an> a for all n ≥ k.

(4) If a > a∗, then for all k ∈ N, there exists n ∈ N such that an< a.

Proof. Let (x_k) be the sequence defined as above, i.e. for each k ≥ 1, x_k = sup{a_n: n ≥ k}.

By definition, a^∗= inf{x_k: k ≥ 1}.

(1) If a > a^∗, a is not a lower bound for {xk : k ≥ 1}. Then there exists an element x_k0 ∈ {x_k : k ≥ 1} such that x_k0 < a. Since a > x_k0 and x_k0 is the least upper bound for {a_n: n ≥ k0}, then a is an upper bound for {a_n: n ≥ k0}. Hence a_n< a for all n ≥ k0.

(2) If a < a^∗, then a is a lower bound for {x_k : k ≥ 1}. Then for all k ≥ 1, x_k > a. For each k ≥ 1, x_k is the least upper bound for {an: n ≥ k}. Since a < x_k, a is not an upper bound for {a_n: n ≥ k}. Hence we can choose an element a_n∈ {a_n: n ≥ k} so that a_n> a.

In other words, we can find n ∈ N with n ≥ k such that an> a.

(3) and (4) are left to the reader.

 Definition 2.1. We say that a real number x is a cluster point of a bounded sequence (a_n) if there exists a subsequence (an_k) of (an) whose limit is x.

Corollary 2.1. Let (a_n) be a bounded sequence. Then both lim sup

n→∞

a_n and lim inf

n→∞ a_n are cluster points.

Proof. Since a^∗+ 1 > a^∗, there exists n₁ ∈ N so that an< a^∗+ 1 for all n ≥ n₁ by Lemma 2.1. Since a^∗ − 1 < a^∗, for the given n1, we can find m1 ∈ N with m1 ≥ n₁ such that a_m1 > a^∗− 1. Since m₁ ≥ n₁, a_m1 < a^∗+ 1. We find a^∗− 1 < a_m1 < a^∗+ 1. Inductively, we obtain a subsequence (amk) of (an) so that

a^∗−1

k < a_mk < a^∗+ 1

k, k ≥ 1.

Since lim

k→∞ a^∗−

k = lim

k→∞ a^∗+

k = a^∗, by the Sandwich principle, lim

k→∞amk = a^∗. This shows that a^∗ is a cluster point.

Similarly, we can show that a∗ is a cluster point. 

Lemma 2.2. If (a_n) is a convergent to a, the all the subsequences (a_nk) is also convergent to a.

Proof. Suppose (an) is convergent to a. Then for any  > 0, there exists N∈ N so that

|a_n− a| < , whenever n ≥ N.

Let (a_nk) be any subsequence of (a_n). For any k ≥ N_, n_k ≥ k ≥ N_. Then |a_nk − a| <  whenever k ≥ N. This shows that lim

k→∞an_k = a. 

Corollary 2.2. If (a_n) is convergent to a, then a^∗ = a∗= a.

Proof. Since a^∗ and a∗ are both cluster points, Lemma 2.2 implies that a^∗ = a = a∗.  In fact, the converse is also true:

Theorem 2.1. Let (a_n) be a bounded sequence of real numbers. Then (a_n) is convergent to a real number a if and only if a^∗= a∗ = a.

Proof. We have proved one direction. Conversely, let us assume a∗ = a^∗= a.

For any  > 0, a +  > a^∗. By Lemma 2.1, there exists k_ ∈ N so that for any n ≥ k, an< a^∗+ .

Since a −  < a∗ = a, Lemma 2.1 implies that there exists j_ ∈ N so that an > a −  whenever n ≥ j. Denote N = max{k, j}. Then for all n ≥ N_, we have

a −  < an< a + .

Thus |a_n− a| <  whenever n ≥ N_. This shows that lim

n→∞a_n= a.

 Let (a_n) be a bounded sequence. If E is the set of all cluster points of (a_n), we know that it is nonempty by Bolzano-Weierstrass theorem (since any bounded sequence has a convergent subsequence). For x ∈ E, choose a subsequence (a_nk) of (a_n) so that lim

k→∞a_nk = x. By Theorem 2.1, we know

lim sup

k→∞

ank = lim inf

k→∞ ank = x.

By definition, for any j ≥ 1, the set {a_nk : k ≥ j} is a subset of {a_n: n ≥ j}. This is because for k ≥ j, nk ≥ k ≥ j. For each j ≥ 1,

sup{an: n ≥ j} ≥ sup{ank : k ≥ j}.

Denote the left hand side by xj and the right hand side by αj. Then xj ≥ α_j for all j ≥ 1.

Since both (x_j) and (α_j) are convergent, by Lemma ??, we find a^∗= lim

j→∞xj ≥ lim

j→∞αj = x.

This shows x ≤ a^∗, i.e. E is bounded above by lim sup

n→∞

an. Using a similar argument, we can show that x ≥ a∗, i.e. E is bounded below by a∗. We obtain that E is a nonempty bounded subset of R. Since both a^∗ and a∗ are cluster points, i.e. a^∗, a∗ ∈ E, we obtain that:

Theorem 2.2. Let (an) be a bounded sequence and E be the set of all cluster points of (a_n). Then

lim sup

n→∞

an= max E, lim inf

n→∞ an= min E.

Example 2.3. Construct a sequence with exact three different cluster points.

Solution: We leave it to the reader as an exercise.

Example 2.4. Find the limsup and liminf of the following sequence of real numbers.

(1) a_n= (−1)ⁿ+ 1

n, n ≥ 1.

(2) a_n= cos(nπ +π

6), n ≥ 1.

(3) an=







1

n if n = 3k

1 +_n¹ if n = 3k + 1

−2 + _n¹ if n = 3k + 2.

Theorem 2.3. Let (sn) and (tn) be bounded sequences of real numbers. Suppose that there exists N > 0 so that sn≤ t_n for n ≥ N. Then

(1) lim inf

n→∞ s_n≤ lim inf

n→∞ t_n. (2) lim sup

n→∞

sn≤ lim sup

n→∞

tn.

Proof. We leave it to the reader as an exercise. 

It follows immediately from this theorem that

Corollary 2.3. Let (a_n) be a sequence of real numbers. Then (1) If an≤ M for all n, then lim sup_n→∞an≤ M.

(2) If a_n≥ M for all n, then lim inf_n→∞a_n≥ M.