Answer to Question #3987 in Real Analysis for Junel

Question

let A and B be bounded nonempty subset of R, and let A+B:={a+b:a element A,b element B}.
prove that the Sup(A+B)=Sup A + Sup B and inf (A+B)= inf A + inf B.

Answer 1

Let us show that Sup(A+B)=Sup A + Sup B
By definition x=Sup A
if x>=a for any a from A,
and for any t>0 there exists b from A such that b > x-t
First we show that
(*) Sup(A+B) <= Sup A + Sup B
Let
x=Sup A,
y=Sup B,
z=Sup(A+B)

Then
x>=a for any a from A,
and
y>=b for any b from A,
hence
x+y >= a+b for any a from A and any b from B,
in other words,
x+y >= c for any c from A+B.

This implies (*).

To prove the inverse inequality
(**) Sup(A+B) >= Sup A + Sup B
that is
z >= x+y
suppose that
z<x+y.
Then there exists t>0 such that
z+ 2t < x+y.
In other words,
z < (x-t) + (y-t).

But then there exist a from A and b from B such that
x-t < a and y-t < b.
On the other hand
a+b <= z.

Thus we obtain that
a+b <= z < x-t + y-t < a+b,
which is impossible.
This proves (**), and all the identity
Sup(A+B)=Sup A + Sup B
========================================
The proof of
inf (A+B)= inf A + inf B
is analogous.

Moreover, if we denote by -A the folowing set
-A = { -a | a from A }
consisting of numbers opposite to numbers from A, then
inf A = -sup(-A).

Hence
inf (A+B)= - sup (-A-B) = - sup (-A) -sup(-B) = inf A + inf B.