a Let S be a nonempty bounded subset of R Prove that sup S i
Solution
a). z = sup S
1. z s for all s S;
2. If z\' < z, then there exists s\' S such that s\' > z
Suppose x = sup S and y = sup S
It is not the case that x < y
Suppose x < y. Letting x take the role of z\' and y take the role of z in Property 2 reveals that there exists s\' S such that s\'> x. But then x is not equal to sup S because x does not satisfy Property 1
b). Suppose m = max S, and n = max S. We will show m = n. To do so we recall the defining
characteristics for the general statement k = max S
1. k S;
2. s k for all s S.
Now, n = max S, so n S. Further, because m = max S we can say (letting m take the role of k
in Property 2) that n m. Similarly, because m S, we can say (letting n take the role of k in
Property 2) that m n. Thus, we must have m = n
