Analysis – I 1

LECTURE-2(sec – 1)

Definition-: (Bounded above set) - A set S IR is called bounded above set . If there exist a real number such that

x µ x S

Definition -: (Upper bound of a set) - A real number µ is called an upper bound of a set S IR

Iff x µ x S

EX -: S = {1, } is bounded above by ”1” and hence 1is an upper bound of S.

Ex -: S = {1, …………} is also bounded above by 1

Ex -: S = {-1, ………..} is bounded above by 0.

NOTE -: In this example 0 S

Which means upper bound of a set may or may not belong to the set .

Bounded below set -: A set S IR is said to be bounded below if a real number ‘ ’ such that

µ x x S

Lower bound of a set -: A real number ‘ ’ is called a lower bound of a set S IR

If µ x x S

Bounded set -: A set S is said to be bounded If it is both bounded below and bounded above.

Unbounded set -: A set is said to be unbounded if it is not bounded.

EX -: S = {x : x ≥ 0} is bounded below but not above and hence is not bounded.

Lower bound of S= 0 and has no upper bound

EX -: S = { : n IN} is bounded above by 1.

Prepared by-: Sanjeev Kumar Shukla ALPHA PLUS EDUCATION 9718445143

Analysis – I 2

NOTE -: A real numbers ‘ ’ is not an upper bound of a set S

If y S : y > µ

EX -: S={1,2,3,4,5} µ = 3 is not an upper bound of S as y = 4 S : y = 4 > 3 = µ

Few more example -:

(i)

(ii)

(iii)

(iv)

(v) IR+ is bounded below by “0” but is not bounded above(vi) ]1, [ is bounded below by “1” but is not bounded above.

(vii) { Ø } is not even a subset of real numbers.

Ans -:

(i) Is bounded above by 2 and bounded below by 1.(ii) Is not bounded.

(iii) Is bounded below by 0 and bounded above by 3/2.

(iv) Is bounded below by 2 and is not bounded above.

Supremum of a set -: Let S be a non empty bounded above set then a real number µ is called

supremum [ or least upper bound ] of S If and only if.

(i) x ≤ µ x S (µ is an upper bound of S).(ii) If V is any other upper bound of S then µ ≤ v.

And it is denoted by supS = µ

Theorem -: Supremum of a set if it exists is unique .

Prepared by-: Sanjeev Kumar Shukla ALPHA PLUS EDUCATION 9718445143

Analysis – I 3

Proof -: Do yourself

NOTE-: Supremum of a set may or may not belong to the set

Ex -: S = {1, …………} supS =

Proof -: clearly 1 ≥ x .Also If < 1 then as cant be an upper bound

of S and hence any number smaller then 1 can’t be an upper bound of S 1 is least

upper bound of S and hence is supS.

Remark -: Greatest member of a set, If it exists, is always the supremum of the set .But sup of a

set may not be a greatest member of set.

EX -: S =

0 = supS

Let µ < o we can choose :

As & is not upper bound of S as being arbitrary, we get

no real numbers less than ‘0’ is an upper bound of S. or o ≤ every upper bound of S.

= supS.

Infimum -: Let S be a non empty bounded below set. Then a real number t is said to be Infimum of S, if and only if.

(i) t ≤ s .(ii) If α is any lower bound of S then t ≥ α. Infimum of a set S is denated by Inf(S)

Theorem -: Infimum of a set if it exists is unique.

Proof -: Do yourself.

Lemma -: A number is the supremum of a non-empty set of S Iff satisfies the conditions.

Prepared by-: Sanjeev Kumar Shukla ALPHA PLUS EDUCATION 9718445143

Analysis – I 4

(i)(ii) If , then there exists such that .

Proof -: First assume = supS

If as is supS is l.u.b and hence is not an upper bound

Conversely assume (i) and (ii) hold

then (i) is an upper bound of S also, (ii) no number smaller than can be an upper

bound of S, = l.u.b of S = supS.

Lemma -: An upper bound of a non empty set S in IR is supremum of S Iff

such that .

Proof-: frist let us assume that such that

i.e to show that

(1)

(2) Is least upper bound of S

is an upper bound of S this follows from the given statement.

Now let be any other upper bound of S

Then to show that Let If possible then

Let then by given condition there exists :

which is a contradiction as is upper bounded of S.

Our assumption that is wrong and hence ( )

Prepared by-: Sanjeev Kumar Shukla ALPHA PLUS EDUCATION 9718445143

Analysis – I 5

Conversely assume that is sups then (i) is an upper bound of S and next we need to show

that such that Let be given; as = supS

can’t be an upper bound of S ( sups is least upper bound of S.)

.

Definition -: (Greatest element of a set ) – Let S be any non empty subset of real numbers then the Greatest element of S is an element g of S such that

Definition -: (least element of a set) – Let S be any non empty subset of real numbers then an element is called least element of S

If .

Question -: Every finite set has a greatest element and a least element .

Proof -: Let S = { x1, x2……….xn } be any finite set . Let be any element

If is greatest element then we are done if not then If is

greatest element of S then we are done. If not continuing like this

we’ll get our greatest element. ( this process will stop somewhere as S is finite set ).

IIlr proof can be given for least element of S.

Prepared by-: Sanjeev Kumar Shukla ALPHA PLUS EDUCATION 9718445143