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$A-B = \{a-b: a\in A, b\in B\}$. Prove that $\sup(A-B) = \sup(A) - \inf(B)$

OK, let $x=\sup(A), y=\sup(B)$:

$a\in A \implies a\leq x$

$b\in B \implies b\leq y$

$a+b\leq x+y$ is a upper bound

Take $\varepsilon > 0$ and find $a,b$ s.t.:

$a>x-\dfrac {\varepsilon}{2}, b>y-\dfrac {\varepsilon}{2}$

$a+b>x+y-\varepsilon$

It means that every potential smaller upper bound $x+y-\varepsilon$ is not really an upper bound.

Therefore $\sup(A+B) = \sup(A) + \sup(B)$

$A-B = \{a-b: a\in A, b\in B\}$.

Proving that $\sup(A-B) = \sup(A) - \inf(B)$

Let $x=\sup \left( A\right), y=\inf \left( B\right)$

$a\in A\implies a\leq x$

$b\in B \implies b\geq y$

$a+b\geq x+y$

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  • $\begingroup$ How do the last four lines here finish the proof?? You haven't proved that they are equal. $\endgroup$
    – bob
    Aug 7 '19 at 3:50
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Define similarly

  • $X+Y=\{x+y:x\in X, y\in Y\}$
  • $-X=\{-x:x\in X\}$

Hint 1: $\sup(X+Y)=\sup X+\sup Y$

Hint 2: $\sup(-Y)=-\inf Y$

Proving that $\sup(X+Y)=\sup X+\sup Y$ is psychologically simpler than the statement you have, but they're actually the same, in view of hint 2.

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    $\begingroup$ I think that we need that sets $X$ and $Y$ are bounded in some way for first hint. $\endgroup$
    – Cortizol
    Oct 22 '14 at 18:20
  • $\begingroup$ @Cortizol: It's enough for them to be nonempty. But it's a good point nonetheless. $\endgroup$
    – tomasz
    Oct 22 '14 at 19:06
  • $\begingroup$ @Cortizol Yes, at least one should be non empty, or we might have a $\infty-\infty$ situation. Otherwise one has $\infty+t=\infty$ for any $t>-\infty$. $\endgroup$
    – egreg
    Oct 22 '14 at 19:35

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