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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.

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    $\begingroup$ I think you missed a "sup" in the left-hand side of your equation $\endgroup$ – Questioner 2 days ago
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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 $E$.

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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 $A=\left\{x-y:x\in E, y\in F\right\}$.

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