Let S be a nonempty subset of R In class we said that S is b

Let S be a nonempty subset of R. In class we said that S is bounded if it has both an upper bound and a lower bound. Show that S is bounded if and only if there is some M E R such that |s| <= M for every sES. (The proof each way should be very short.)

Solution

Let l be a lower bound of S: x S, l x. So x S, x l, in other words,

y S, y l.

Thus S is bounded above because l is an upper bound of S.

Since S is nonempty, it follows that x S, and so we obtain that x S. Hence S is nonempty.

As S is nonempty and bounded above, it follows that sup(S) exists, by the least upper bound property of R.

Since sup(S) is an upper bound of S, we have:

y S, y sup(S),

that is, x S, x sup(S),

that is, x S, sup(S) x.

So sup(S) is a lower bound of S.

Next we prove that sup(S) is the greatest lower bound of S.

Suppose that l 0 is a lower bound of S such that sup(S) < l 0 . Then we have

x S, sup(S) < l 0 x,

that is, x S, x l 0 < sup(S),

that is, y S, y l 0 < sup(S).

So l 0 is an upper bound of S, and l 0 < sup(S), which contradicts the fact that sup(S) is the least upper bound of S.

Hence l 0 sup(S).

Consequently, inf S exists and inf S = sup(S).