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

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

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

$a+b\geq x+y$

Marc
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  • How do the last four lines here finish the proof?? You haven't proved that they are equal. – bob Aug 07 '19 at 03:50

2 Answers2

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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.

egreg
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    I think that we need that sets $X$ and $Y$ are bounded in some way for first hint. – Cortizol Oct 22 '14 at 18:20
  • @Cortizol: It's enough for them to be nonempty. But it's a good point nonetheless. – tomasz Oct 22 '14 at 19:06
  • @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 $t>-\infty$. – egreg Oct 22 '14 at 19:35
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Since A and B are bounded nonempty, both set must be bonded above and bounded below as well as. Consider that A is bounded below and B is bounded above. Since, A is bounded below, by infimum property, A must consists an infimum say w. Therefore w ≤ x for all x that belongs to A.....(i) Similarly in set B, it is bounded above. By supremum property, it also consist a supremum say u. Therefore u ≥ y for all y belongs in B........(ii) Adding i & ii we get---- w+y ≤ x+u => w-u ≤ x-y for all x-y belongs to A-B So clearly w-u is a lower bound for A-B. By infimum property, A-B must consist an infimum say g. So, g= Inf(A-B). Therefore we can infer that g ≥ w-u...iii We also have g ≤ x-y for A-B => g+y ≤ x for x belongs to A Hence it is clear that g+y is a lower bound of A and from very beginning we have w is Inf A, the below relation will be hold g+y ≤ w => g-w ≤ -y => w-g ≥ y for y belongs to B Since w-g is an upper bound of B and u= Sup B Then, w-g ≥ u => -g ≥ u-w. => g≤ w-u....iv Combining iii and iv we get g = w-u which is equally to say Inf(A-B) = Inf A- Sup B Similarly we can proof Sup(A-B) = SupA - Inf B.

---By Ronit Debnath . (Student of class Eleven)

Ronit Debnath
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