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

[pisrationalhahaha]

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]

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?

 

[pisrationalhahaha]

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 ?

 

[tkhunny]

|x-y|

 

[pisrationalhahaha]

|x-y|

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]

|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?

 

[pisrationalhahaha]

|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