Supremum and Infimum: Denote B={|x-y|;x∈A and y∈A}....

Alfredo Dawlabany

New member
Let A be a non-empty bounded set of $$\displaystyle \mathbb{R}$$. Denote B={|x-y|;$$\displaystyle x\in A$$ and $$\displaystyle y\in A$$}.
Prove that Sup B = Sup A - Inf A.

I tried to solve it but didn't know what to do

I said that if A is a bounded set then $$\displaystyle A\subset \left [InfA,SupA\right ]$$
$$\displaystyle \Rightarrow x,y \in \left [ InfA,SupA \right ]$$
so we have $$\displaystyle InfA\leqslant x\leqslant SupA$$ and $$\displaystyle InfA\leqslant y\leqslant SupA$$

Am I going right ? If yes how would I continue ?

tkhunny

Moderator
Staff member
Just hints:

1) Does it matter or help that Sup and Inf are actually members of the sets?
2) I don't see any subtraction in your demonstration, so far. What's up with that?

Last edited:

Alfredo Dawlabany

New member
Just hints:

1) Does it matter or help that Sup and Inf are actually members of the sets?
2) I don't see any subtraction in your demonstration, so far. What's up with that?
What do you mean by subtraction ?

Staff member
|x-y|

Alfredo Dawlabany

New member
OK I understood
But there is the absolute value
-sup A≤-y≤-inf A
So we get
x-y≤sup A - inf A , this implies sup(x-y)=sup A - inf A
Then what?

tkhunny

Moderator
Staff member
|x-y| is just distance. You will lose no generality if you assume x > y. Obviously, if x = y, we have a minimum, but we're not looking for that.

Rephrasing, how far apart can x and y be?

Alfredo Dawlabany

New member
|x-y| is just distance. You will lose no generality if you assume x > y. Obviously, if x = y, we have a minimum, but we're not looking for that.

Rephrasing, how far apart can x and y be?
Okay
I understood everything
Thank you