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.

  • 3
    $\begingroup$ I think you missed a "sup" in the left-hand side of your equation $\endgroup$ – Questioner Feb 14 at 4:11

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.