# Thread: supremum if bounded subset

1. ## supremum if bounded subset

Let A be a non-empty bounded subset of $\displaystyle \mathbb{R}$ Define:

$\displaystyle D=\{|x-y|:x,y\in A\}$

show that sup D = sup A - inf A.

I can intuitively understand this however I am not too sure how to go about a rigorous proof.

Thanks for any help

2. ## Re: supremum if bounded subset

Originally Posted by hmmmm
Let A be a non-empty bounded subset of $\displaystyle \mathbb{R}$ Define:

$\displaystyle D=\{|x-y|:x,y\in A\}$

show that sup D = sup A - inf A.
Let $\displaystyle U=\sup( A)~\&~L=\inf( A)$ and $\displaystyle \{x,y\}\subset A$.

Then $\displaystyle L \leqslant x \leqslant U\;\& \, - U \leqslant - y \leqslant - L$ which implies
$\displaystyle - (U - L) \leqslant x - y \leqslant U - L$ or $\displaystyle |x-y|\le U-L$.

Can you finish?

3. ## Re: supremum if bounded subset

errm no sorry, I get that you have shown that it is an upper bound, but i don't know how to show that it is the least upper bound, sorry.

thanks for any help

4. ## Re: supremum if bounded subset

Originally Posted by hmmmm
errm no sorry, I get that you have shown that it is an upper bound, but i don't know how to show that it is the least upper bound, sorry.
Let $\displaystyle \alpha=\sup(D)$ suppose that $\displaystyle \alpha<U-L$.
Use $\displaystyle \beta=\frac{U-L-\alpha}{4}>0$.
There must be $\displaystyle y'\in A$ such that $\displaystyle L\le y'<L+\beta.$
There must be $\displaystyle x'\in A$ such that $\displaystyle U-\beta< x'\le U.$

Now you show that $\displaystyle \alpha< |x'-y'|$ which is a contradiction.
So what does that prove?

5. ## Re: supremum if bounded subset

It shows that if there is any number less than U-L then you can find a member of the set that is bigger than it and so it is not the supermum.

Thanks for the help