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Course
18MAT305 (REALANALYSIS)
Institution
Amrita Vishwa Vidyapeetham
The least-upper-bound property is one form of the completeness axiom for the real numbers, and is sometimes referred to as Dedekind completeness.[2] It can be used to prove many of the fundamental results of real analysis, such as the intermediate value theorem, the Bolzano–Weierstrass theorem, t...
Course Material 1.3 Applications of 𝒍𝒖𝒃 property of ℝ (1)
The Least Upper Bound Property of the Real Line
Given a non empty sub set 𝐴 of the real line ℝ which is bounded above then there exists
a real number 𝛼 such that 𝛼 = 𝑙𝑢𝑏𝐴
Remarks.
1. The greatest lower bound property can be stated similarly.
2. The 𝑙𝑢𝑏 property of ℝ ensures that each of its non empty sub set which is bounded
above has a least upper bound and also it states that the least upper bound is
contained in ℝ itself.
3. This is a property of the real line which may not hold for other sets. For example, the
set ℚ of all rational numbers do not satisfy this property as in the above example,
the sub set 𝐴 = 𝑥 ∈ ℚ ∶ 𝑥 2 < 2 of ℚ is bounded above but the 𝑙𝑢𝑏𝐴 = 2 does
not belong to ℚ
Examples
Check whether the following sets satisfy the lub property
1. 𝑃, the set of all irrational numbers
2. The open interval (0, 1)
3. The closed interval [0, 1]
4. The set ℤ of all integers.
, Applications of 𝒍𝒖𝒃 property of ℝ
Theorem (Archimedean Property)
1. The set ℕ of natural numbers is not bounded above
2. Given any two real numbers 𝑥 and 𝑦 with 𝑥 > 0 there exists 𝑛 ∈ ℕ such that 𝑛𝑥 > 𝑦
Proof.
1. Suppose ℕ is bounded above. Then by the least upper bound property of ℝ, the
subset ℕ of ℝ, has a least upper bound say 𝛼 ∈ ℝ. Then 𝑛 ≤ 𝛼 for all 𝑛 ∈ ℕ and 𝛼 is
the least number with this property. Then 𝛼 − 1 is not an upper bound of ℕ. So that,
there exists at least one natural number 𝑚 with 𝛼 − 1 < 𝑚. This implies that
𝛼 < 𝑚 + 1. This shows that there exists an element 𝑚 + 1 of ℕ which is greater
than 𝛼 so that 𝛼 is not an upper bound of ℕ. This is a contradiction and hence
proves that the set of natural numbers ℕ is not bounded above.
2. Suppose the property does not hold for two real numbers 𝑥 and 𝑦 with 𝑥 > 0. That
is, there is no natural number 𝑛 satisfying 𝑛𝑥 > 𝑦 or in other words, for all 𝑛 ∈ ℕ we
𝑦
have 𝑛𝑥 ≤ 𝑦. Then as 𝑥 > 0 we get 𝑛 ≤ 𝑥 . Since this is true for all 𝑛 ∈ ℕ, this means
𝑦
that the number is an upper bound for set ℕ and hence ℕ is bounded above. This
𝑥
contradicts statement 1 and hence the result.
Proposition
The two properties stated in the above theorem are equivalent
Proof.
To prove 1 ⟹ (2)
In the above proof , (2) has been proved using (1) Hence we have 1 ⟹ (2)
To prove 2 ⟹ (1), we assume the property (2) and suppose 𝛼 is any real number then to
show that 𝛼 is not an upper bound for ℕ. Choosing 𝑥 = 1 and 𝑦 = 𝛼 in (2) we have at least
one 𝑛 ∈ ℕ with 𝑛𝑥 > 𝑦 or 𝑛 > 𝛼 and thus proves that 𝛼 is not an upper bound of ℕ.
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