# Answer to Question #3987 in Real Analysis for Junel

Question #3987

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.

prove that the Sup(A+B)=Sup A + Sup B and inf (A+B)= inf A + inf B.

Expert's answer

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.

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.

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