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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