# Answer to Question #4825 in Real Analysis for zainab

Question #4825

let S be a subset of R and let it bounded below , show that infimum of S exists

[ hint: let A= -s :={ -s: s∈S} , show that A is bounded above so sup A exists . now show that inf S= -sup A ]

[ hint: let A= -s :={ -s: s∈S} , show that A is bounded above so sup A exists . now show that inf S= -sup A ]

Expert's answer

Suppose S is bounded below, so there exists K0 such that

K0 < s

for all s from S.

We should show that there exists an infimum of S,

i.e a number C such that

1) C <= s for all s from S

2) for any

eps>0 there exists s from S such that C<= s < C+eps

Let us note

the following statement.

(*) If C belongs to S and C<=s for all s from S,

then C is the infimum of S.

Fix any s0 from S and consider the number

M1 = (K0+s0)/2.

So M1 is the middle point of the segments

[K0,s0]

Consider two cases

a) There exists s1 from S such that s1

<= M1.

Denote K1 = K0, then

K0 = K1 <= s1 <= M1 <=

s0

b) M1 <= s for all S from S.

In this case put s1=s0, and K1=M1,

so

K0 <= M1 = K1 <= s1 = s0

In both cases the segment

[K1,s1] is contained in [K0,s0],

and the length of [K1,s1] <= length of

[K0,s0]/2, i.e.

s1-K1 <= (s0-K0)/2.

Similarly, put

M2

= (K1+s1)/2.

So M2 is the middle point of the segments [K1,s1]

Again

consider two cases

a) There exists s2 from S such that s2 <= M2.

Denote K2 = K1, then

K1 = K2 <= s2 <= M2 <= s1

b) M2

<= s for all S from S.

In this case put s2=s1, and K2=M2, so

K1

<= M2 = K2 <= s2 = s1

Again in both cases the segment [K2,s2] is

contained in [K1,s1],

and the length of [K2,s2] <= length of [K1,s1]/2,

i.e.

s2-K2 <= (s1-K1)/2 <= (s0-K0)/4.

By similar

arguments we will construct a decreasing infinte sequence of segments

[K0,s0] > [K1,s1] > [K2,s2] > ....

such that

a) the

length of [Ki,si] <= length of [K0,s0]/2^i

si-Ki <=

(s0-K0)/2^i.

b) si belongs to S

c) Ki<= s for all s from S and i

Since the lengths of [Ki,si] -> 0, there exists a unique point C

belonging to the intersection of all these segments.

So

c = lim Ki = lim

si,

and

Ki <= C <= si

Let us show that C is the infimum

of S.

1) Take any s from S.

Then

Ki<= s for al i,

whence

C = lim Ki <= s

2) Fix any eps>0.

Then there exists i

such that the length of [Ki,si]<eps, therefore

Ki <= si <= Ki +

eps

on the other hand

Ki <= C <= si

Hence

Ki <= C

<= si <= Ki + eps <= C + eps,

so

C <= si <= C +

eps.

Thus C is the infimum of S.

K0 < s

for all s from S.

We should show that there exists an infimum of S,

i.e a number C such that

1) C <= s for all s from S

2) for any

eps>0 there exists s from S such that C<= s < C+eps

Let us note

the following statement.

(*) If C belongs to S and C<=s for all s from S,

then C is the infimum of S.

Fix any s0 from S and consider the number

M1 = (K0+s0)/2.

So M1 is the middle point of the segments

[K0,s0]

Consider two cases

a) There exists s1 from S such that s1

<= M1.

Denote K1 = K0, then

K0 = K1 <= s1 <= M1 <=

s0

b) M1 <= s for all S from S.

In this case put s1=s0, and K1=M1,

so

K0 <= M1 = K1 <= s1 = s0

In both cases the segment

[K1,s1] is contained in [K0,s0],

and the length of [K1,s1] <= length of

[K0,s0]/2, i.e.

s1-K1 <= (s0-K0)/2.

Similarly, put

M2

= (K1+s1)/2.

So M2 is the middle point of the segments [K1,s1]

Again

consider two cases

a) There exists s2 from S such that s2 <= M2.

Denote K2 = K1, then

K1 = K2 <= s2 <= M2 <= s1

b) M2

<= s for all S from S.

In this case put s2=s1, and K2=M2, so

K1

<= M2 = K2 <= s2 = s1

Again in both cases the segment [K2,s2] is

contained in [K1,s1],

and the length of [K2,s2] <= length of [K1,s1]/2,

i.e.

s2-K2 <= (s1-K1)/2 <= (s0-K0)/4.

By similar

arguments we will construct a decreasing infinte sequence of segments

[K0,s0] > [K1,s1] > [K2,s2] > ....

such that

a) the

length of [Ki,si] <= length of [K0,s0]/2^i

si-Ki <=

(s0-K0)/2^i.

b) si belongs to S

c) Ki<= s for all s from S and i

Since the lengths of [Ki,si] -> 0, there exists a unique point C

belonging to the intersection of all these segments.

So

c = lim Ki = lim

si,

and

Ki <= C <= si

Let us show that C is the infimum

of S.

1) Take any s from S.

Then

Ki<= s for al i,

whence

C = lim Ki <= s

2) Fix any eps>0.

Then there exists i

such that the length of [Ki,si]<eps, therefore

Ki <= si <= Ki +

eps

on the other hand

Ki <= C <= si

Hence

Ki <= C

<= si <= Ki + eps <= C + eps,

so

C <= si <= C +

eps.

Thus C is the infimum of S.

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