Proof related to Supremum and Infimum

Question

S is set

x belongs to real number R

Show that

Sup (S+x)= Sup S+ x

Answer 1

To make the notation easier, let m = Sup(S). We need to show two things:

1. m + x is an upper bound of the set (S + x)
2. If M is any upper bound of of the set (S + x), then m + x ≤ M.

For (1), let y be any element of (S + x). Then y = a + x for some element a in S. Thus a ≤ m and therefore y = a + x ≤ m + x.

For (2), let M be an upper bound of (S + x). Let a be any element of S. Thus a + x is an element of (S+x), therefore a + x ≤ M and thus a ≤ M - x. This shows M - x is an upper bound for S; which means m ≤ M - x (since m is the least upper bound of S) and thus m + x ≤ M.