PLease prove part b below. Let A and B be sets of real numbers. Define a set A +

Question

PLease prove part b below.

Let A and B be sets of real numbers. Define a set A + B by A + B = {a + b|a A and b B}. Let A = {-1, 0, 3, 5} and B = {-2, 12, 20}. Find A + B. Show that if A and B are bounded sets, then 1.u.b. (A + B) = (1.u.b. A) + (1.u.b. B) and g.1.b. (A + B) = (g.1.b. A) + (g.1.b. B)

Explanation / Answer

Let us assume that lub(A+B) is not equal to lub A+lub B now elements of A+B are defined as {a+b such that a belongs to A and b belongs to B} let x is lub of A+B then x=a+b for some a in A and some b in B now let a1 is lub of A and b1 is lub of B hence three case are possible 1) a>a1 and b>b1 so by simple rules of addition a+b>a1+b1 which is a contradiction so lub of A+B must be a1+b1 hence lub(A+B)=lubA+lubB 2)a>a1 b=b1 then again a+b>a+b1 which is a contradiction so lub of A+B must be a1+b1 hence lub(A+B)=lubA+lubB 3))a=a1 b>b1 then again a+b>a+b1 which is a contradiction so lub of A+B must be a1+b1 hence lub(A+B)=lubA+lubB so in all possible cases we have seen that lub(A+B)=lub(A)+lub(B) by the same logic we can prove the case of gub

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