Supremum or Least Upper Bound

If there exists a number X such that ∀ x ∈ S : x < X then S is said to be

bounded above and X is called an upper bound.

A set bounded above may or maynot have a greatest member.

Theorem : For every set S bounded above ∃ a number Y such that (i) ∀ x ∈ S ≤
Y and (ii) For ε > 0 however small, there is at least one y ∈ S such that
y > Y − ε.

Proof : Divide all real numbers into two classes L and R a number ∈ L if it is
smaller than at least one member of S , otherwise it is in R . This division then
defines a section.
Thus there exists a number Y such that any x < Y ∈ L and any x > Y ∈ R .
Any number Y − ε < Y therefore belongs to L , thus smaller than at least one
member of S .
Let x ∈ S > Y , then all numbers Y < y < x ∈ R as each of them is greater
than Y . But each of them is less than x ∈ S , so also belong to L . This is
impossible. So all members of S < Y .
Thus the set of upper bounds of a set bounded above is bounded below and has
a least member. This is called the least upper bound or supremum of S
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 Limiting points
A member x ∈ S is called a limiting pont if every open neighbourhood of x
contain an infinite number of members of S .

Finite sets can have no limiting points.

Infinite sets may also have no limiting points, e.g. Z .

A limiting point may not belong to the set, e.g. a real number for Q .

Bolzano Weierstrass’ Theorem : Every infinite bounded set has at least one limiting
point.
Proof : Since S is bounded, there exists an upper bound X and a lower bound Y
and ∀ x ∈ S Y < x < X. Divide all real numbers into two classes L and R
such that we put a real number x ∈ L if it is greater than a finite number of
members of S , otherwise we put it in R . This division is a section.
So there exists a g such that all members of L < g. Thus g − ε ∈ L and can
exceed only a finite number of members of S . Similarly g + ε ∈ R : so must be
greater than an infinite number of members of S .
Thus the open interval ]g − ε, g + ε[ contains an infinite number of members of S .
It is a limiting point.

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 Derivative Sets
The set of limiting points of a bounded set S is called the first derivative of the set
and written as S 0

Theorem : The first derivative of a bounded set S is itself bounded and attains its
bounds

Proof : Let S be bounded, ∃ a supremum and an infremum X, Y such that

X ≤ x ≤ Y ∀ x ∈ S . Thus no limiting point of S , i.e. any member g ∈ S 0 can lie
outside this interval. Thus S 0 is also bounded. Let g, G be its infremum and
supremum.
Since G is a supremum of S 0 , ∃ ξ ∈ S 0 such that G − ε < ξ ≤ G. This
neighbourhood contains an infinite number of members of S as ξ is a limiting
point of S . This shows that G is also a limiting point of S so G ∈ S 0 .
Similarly g also belongs to S 0 .
G is called the upper limit and g the lower limit of S

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 Sequences
A sequence is a function F : N −→ <. We represent this F as

{an }∞
n=1 where F(n) = an n∈N and an ∈ <

A sequence is said to converge if :

Given an ε > 0quad (however small), ∃ n0 ∈ N (however large) and
∃ a ∈ < such that :
|an − a| < ε ∀ n > n0

If above is true we write :

lim {an } = a
n→∞

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 Examples
Let 1 1
Given ε > 0 choose
, , . . . n1 }
ˆ1˜
{an }∞
1= {1, 2 3
n0 = ε
+1
such that :
˛ ˛
˛1
˛ − 0˛ = 1 < 1 < ε
˛
˛n ∀ n > n0
˛ n n0

lim {an } = 0
n→∞

Let {an }∞
1 = {1, 3 7 2n −1
, , . . . 2n−1
2 4
. . .}

˛ ˛
1 1
∀ n > max {1, [1/ε] + 1}
˛ ˛
|an − 2| = ˛2 − n−1 − 2˛˛ = n−1 < n < ε
˛
2 2

lim {an } = 2
n→∞

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 Examples
Let {an }∞
1 = {n1/n }∞
1 = {1, 21/2 , 31/3 , . . .}

Suppose n1/n = 1 + δn for n > 1 δn > 0

n(n−1) 2 n(n−1) 2
n = (1 + δn )n = 1 + nδn + 2
δn ... ≥ 1+ 2
δn
q
So, n − 1 ≥ n(n−1) 2
2
δn , for n > 1 n 2
δ
2 n
≤ 1 ⇒ δn ≤ 2
n
q q
ε2
Thus if n0 2
> 1 for n > n0 0 ≤ δn ≤ 2
n
≤ 2
n0
<ε

Thus ,

» –
2
˛ ˛
˛ 1/n
˛n − 1˛ < ε ∀n > n0 = +1
˛
ε2

lim {n1/n } = 1
n→∞

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 Examples
Let {an }∞
n
1 = { 2n! }∞
1

2 2 2 2 4
an = · · ··· < ∀n>3
1 2 3 n n

Given ε > 0, choose

n0 = max {[4/ε] + 1, 3} ⇒ |an | < ε ∀ n > n0

lim {an } = 0
n→∞

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 Examples
Theorem : If {an }∞
1 is a convergent sequence, then for every ε > 0
(however small) ∃ n0 such that

|an − am | <  ∀ n, m > n0

Proof: Since the sequence converges, given ε > 0 (however small),

∃ n0 and a ∈ < such that

|an − a| < ε/2

So,
|am − an | ≤ |am − a| + |an − a| < ε ∀ n, m > n0

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