Let A and B be non-empty bounded subsets of R. DefineA − B := {a − b : a ∈ A, b ∈ B}.Prove thatsup(A − B) = sup A − inf B

Step 1: Define the sets

Let A and B be non-empty bounded subsets of R. We define A - B as the set of all differences a - b, where a is an element of A and b is an element of B.

Step 2: Show that sup(A - B) is an upper bound for A - B

We know that for any a in A and b in B, a - b is less than or equal to sup A - inf B. This is because a is less than or equal to sup A and -b is greater than or equal to -inf B. Therefore, sup A - inf B is an upper bound for A - B.

Step 3: Show that sup(A - B) is the least upper bound

Now we need to show that sup(A - B) is the least upper bound for A - B. Suppose there is another upper bound U for A - B that is less than sup A - inf B. Then there exists an a in A such that a > U + inf B. Similarly, there exists a b in B such that b < sup A - U. Subtracting these inequalities gives a - b > sup A - inf B, which contradicts the assumption that U is an upper bound for A - B. Therefore, sup(A - B) is the least upper bound for A - B.

Step 4: Conclusion

We have shown that sup(A - B) is an upper bound for A - B and that it is the least upper bound. Therefore, sup(A - B) = sup A - inf B.