# Proving properties of supremums

If $$E$$ and $$F$$ are bounded subsets of real numbers, show that $$\{x-y:x\in E, y\in F\}= \sup E-\inf F$$

I am stuck trying to prove this. Any pointers on where to start would be great. Even intuitively, I'm not understanding why this is correct.

• I think you missed a "sup" in the left-hand side of your equation – Questioner 2 days ago

For any $$x \in E$$ and $$y \in F$$, we have that $$x \leq \sup E$$ and that $$-y \leq -\inf F$$; thus

$$\forall x \in E \ \forall y \in F \ \big( \ x-y \leq \sup E - \inf F \ \big)$$

which means that $$\sup E - \inf F$$ is an upper bound of the set $$E-F = \{ x-y :\, x\in E \textrm{ and } y\in F \}$$. It follows that (since $$\sup (E-F)$$ is the least upper bound)

$$\sup(E-F) \leq \sup E - \inf F$$

Can you show the reversed inequality? Note that, it is equivalent to showing that $$\sup(E-F) + \inf F$$ is an upper bound for $$E$$.

This is a somewhat immediate application of the definition of "sup": A number $$S$$ is the supremum of a set $$A$$ if - $$S$$ is an upper bound of $$A$$; - if $$s'$$ is another upper bound of $$A$$, then $$S\leq s'$$

and similarly for the "inf", with "lower bounds" and inequalities reversed.

Simply prove that the number $$\sup E-\inf F$$ has these properties wrt the set $$A=\left\{x-y:x\in E, y\in F\right\}$$.