Completeness Axiom of Real Numbers: Statement, Example, Application

In this article, we will learn about the completeness property of real numbers with its definition, examples, and applications. Before starting it, let us note down a few important definitions.

ℝ: The set of real numbers.

Let S be a subset of ℝ.

• Upper & Lower Bound: A real number M is called an upper bound of S if x≤M ∀ x∈ S. A real number m is said to be a lower bound of S if x≥m ∀ x∈ S.
• Bounded Above: The set S is called bounded above if S has an upper bound.
• Bounded Below: The set S is said to be bounded below if S has a lower bound.
• Bounded: S is said to be a bounded set if S is bounded above as well as bounded below.

Note:

The least upper bound of S (when exists) is known as the supremum of S and it is denoted by sup S. Similarly, the greatest lower bound of S is known as the infimum of S and it is denoted by inf S. For more details, click on the page “Supremum and Infimum: Definition, Examples, Properties“.

Now, we are going to state the axiom of completeness for real numbers.

Definition of Completeness Axiom of R

Statement of completeness axiom: Every non-empty bounded above subset of real numbers has supremum (i.e., the least upper bound).

The above is also known as the least upper bound (LUB) property of real numbers or the supremum property of real numbers.

Proof:

While constructing the real numbers from the rational numbers, one can able to prove the Completeness axiom of R.

Remark: We also have a similar concept for bounded below sets which we will sate as follows:

Every non-empty bounded below subset of real numbers has infimum (i.e., the greatest lower bound). This is known as the greatest lower bound (GLB) property of real numbers or the infimum property of real numbers.

Note that supremum property ⇔ infimum property.

Examples of Completeness Axiom

The set [0, 1] is bounded above. So by the axiom of completeness, [0, 1] has the supremum. Here, the supremum is 1.

Consider the set S={1/n: n∈ ℕ}. The set S is bounded above and so by the completeness axiom, it has supremum. Note that sup S =1.

Application of Completeness Axiom

As an application of LUB axiom, we can prove the following theorem.

Theorem: If S is a non-empty subset of real numbers, then inf S exists, i.e., S has infimum.

Proof:

Consider the set S’ ={-x : x∈ S}.

As S is non-empty, so is S’.

Given that S is bounded below. So by definition, there exists a real number m such that

x≥m ∀ x∈ S.

⇒ -m ≤ -x ∀ x∈ S

⇒ -m ≤ y ∀ y∈ S’

Thus, S’ is bounded above.

By the axiom of completeness of ℝ, S’ as a supremum.

Assume that M= sup S’. Thus, we have

• x≤M ∀ x∈ S’
• for every ε>0, there is an element y∈ S’ such that M-ε< y.

So we deduce that

-x ≥ -M and -y<-M+ε.

As a result, we can conclude that -M=inf S. ♣

Using the completeness axiom of real numbers, we can prove that ℕ (the set of natural numbers) is unbounded above

Theorem: Prove that the set ℕ of natural numbers is unbounded above.

Proof:

For a contradiction, we assume that the set ℕ of natural numbers is bounded above. Note that 1∈ ℕ. So ℕ is non-empty. Hence, by the completeness axiom of real numbers, ℕ has supremum.

Let M = sup ℕ.

As M-1<M, M-1 cannot be an upper bound of ℕ.

⇒ There is some n ∈ ℕ such that M-1 < n.

⇒ M<n+1.

This contradicts that M = sup ℕ.

Thus, our assumption was wrong. In other words, we have shown that ℕ is unbounded above.

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