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.
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.
