Supremum and Infimum: Definition, Examples, Properties

In this article, let us learn about the supremum and infimum of a set containing real numbers along with examples. We will also provide their properties with proofs.

R stands for the set of real numbers.

At first, we will learn the supremum of a set with examples.

Definition of Supremum

Let S be a bounded above subset of R. A real number M is called the supremum of S, if the following holds:

1. x ≤ M ∀ x∈ S
2. For every ε>0, there is an element y∈ S such that M-ε< y≤ M.

The supremum of a set S is denoted by sup S. From the definition of the supremum of S, we can say that M = sup S.

For example, the set {x: 1≤x≤2} has the supremum 2.

Examples of Supremum

The supremum of a set may or may not exist. See the examples listed below.

1. The set of natural numbers is not bounded above, so it does not contain the supremum.
2. Similarly, Q (the set of rational numbers), R do not have the supremum.
3. If [a, b] is a closed bounded interval, then b is the supremum of [a, b]. For example, 1 is the supremum of the set [0, 1].

Next, we will learn about the infimum of a set with examples.

Read This: List of all Properties of Real Numbers

Definition of Infimum

Let S be a bounded below subset of R. A real number m is called the infimum of S, if the following holds:

1. m ≤ x ∀ x∈ S
2. For every ε>0, there is an element y∈ S such that m≤ y < m+ε.

Notation: The infimum of a set S is denoted by inf S. In this case, we have m = inf S.

Example: The set {x: 1≤x≤2} has the infimum 1.

Remark:

The infimum of a set S may not belong to S. For example, consider the set S={1/n: n∈N}. Then 0 is the infimum of S, that is, 0=inf S but note that 0∉ S.

Examples of Infimum

1. The set N={1, 2, 3, …} of natural numbers has the infimum 1.
2. Q, R do not have the infimum as they are not bounded below.
3. If [a, b] is a closed bounded interval, then a is the infimum of [a, b]. For example, 0 is the infimum of the set [0, 1].

Have You Read These?

• Archimedean Property
• Density Property of Rational Numbers
• Density Property of Real Numbers

Properties of Supremum and Infimum

The theorem below is about the existence and uniqueness of the supremum and infimum of a set. It also provides the relation among them.

Theorem 1: If the supremum or infimum of a set S exists, then it is unique. Moreover, when they exist, we have inf S ≤ sup S.

Proof:

Let M= sup S and m= inf S.

If possible, we assume that there are two suprema M, M’ of S. We have:

• Since M is the least upper bound, M’≤M.
• Again since M’ is the least upper bound, M≤M’.

Thus, we conclude that M=M’. In other words, if sup S exists, then it will be unique.

In a similar way, we can show that the infimum of S is unique whenever it exists.

2nd Part: As sup S and inf S exist, the set S must be non-empty. By the definition of the supremum and infimum, we have that for x∈S,

inf S ≤ x ≤ sup S,

because sup S is an upper bound of S and inf S is a lower bound of S. This completes the proof of the theorem.

Theorem 2: Let S be a bounded below non-empty set of real numbers. Define the set -S by -S={-x : x∈S}. Then inf S = – sup (-S).

Proof:

As S is bounded below, inf A exists. Let m = inf S. So we have

m ≤ x ∀ x∈ S.

⇒ -x ≤ -m ∀ -x∈ -S …(I)

Thus, -S is bounded above and hence sup S exists.

Now, as m = inf S, for each ε>0, there is an element y∈ S such that

y < m+ε

⇒ -m-ε < -y.

So for ε>0, there is an element t∈ -S such that -m-ε < t …(II).

From (I) and (II) and using the definition of the supremum of a set, we conclude that -m=sup (-S), that is, m = – sup (-S).

In other words, inf S = – sup (-S).

Theorem 3: Let S and T be two non-empty bounded sets of real numbers. Show that sup (S+T) = sup S + sup T, where S+T = {x+y : x∈S, y∈T}.

Proof:

As S and T are bounded sets, the suprema of S and T exist. Let M= sup S and N= sup T. Then

x ≤M for all x∈S and y≤N for all y∈T

⇒ x+y≤ M+N for all x∈S, y∈T …(I)

So the set S+T is bounded above and sup(S+T) exists.

Let ε>0 be arbitrary.

By definition, there exist elements (at least one) s∈S and t∈T such that

$M-\dfrac{\epsilon}{2} < s$ and $N- \dfrac{\epsilon}{2} < t$.

⇒ (M+N)-ε < s+t.

So for ε>0, there is an element z∈ S+T such that M+N-ε < z …(II)

From (I) and (II), we can say that M+N is the supremum of S+T, that is, sup(S+T)=M+N. In other words, sup (S+T) = sup S + sup T.

ALSO READ:

Completeness Axiom of Real Numbers

Solved Problems on Supremum and Infimum

Problem 1: Let S and T be two bounded sets of positive real numbers and let A = {y/x : x∈S, y∈T}. Find sup A.

Solution:

Problem 2: Let S ⊂ R be a bounded set. Define a set T = {|x-y| : x, y∈S}. Prove that sup T = sup S – inf S.

Solution:

Let us consider two elements x and y of S. Without loss of generality, we assume that x>y.

Let M=sup S and m=inf S.

We have x≤M and m≤y.

⇒ x≤M and -y≤-m

⇒ x-y ≤M-m

⇒ |x-y| ≤ M-m …(∗)

So the set T is bounded above.

Let ε>0.

As M=sup S, there is an element x₁∈ S such that $M-\dfrac{\epsilon}{2} < x_1$.

Again since m=inf S, there is an element x₂∈ S such that $x_1 < m+\dfrac{\epsilon}{2}$ $\Rightarrow -m-\dfrac{\epsilon}{2} < x_1$.

Combining these two, we get that M-m-ε <x₁-x₂.

So for ε>0, there is an element z∈ T such that M-m-ε < z, where z=|x₁-x₂|…(∗∗)

From (∗) and (∗∗), we deduce that sup T = M-m = sup S – inf S.

The same result holds for y>x in which case |x-y| = y-x.

Problem 3: Find the supremum and infimum of the set S = {1/m +1/n : m,n∈N}.

Solution:

Problem 4: Let S, T ⊂ R be two bounded sets such that x≤y for all x∈S and y∈T. If sup T exists, then sup S also exists. Moreover, sup S ≤ inf T ≤ sup T.

Solution:

Suppose that N is the supremum of T. So by the given condition,

x≤y≤N for all x∈S

⇒ x ≤N for all x∈S

That is, S is bounded above and so sup S exists. Put M=sup S.

Again, by assumption, T is bounded below, so inf T exists. Write n=inf T.

As x≤y for all x∈S and y∈T by assumption, we must have that

M≤n≤N.

Therefore, sup S ≤ inf T ≤ sup T.

FAQs on Supremum and Infimum