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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 ]
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.

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