Prove that $\sup(A-B) = \sup(A) - \inf(B)$ $A-B = \{a-b: a\in A, b\in B\}$. 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 $x=\sup \left( A\right), y=\inf \left( B\right)$
$a\in A\implies a\leq x$
$b\in B \implies b\geq y$
$a+b\geq x+y$
 A: 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.
A: 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)
