If is a maximizing set for on , and
, then is
also a maximizing set for on .

If
is any function (with
), then is a
maximizing set for on , so every function with non-empty domain has a
maximizing set.

Let

Then every disc is a maximizing set for , since if we can find with ; then , so , so and . This argument shows that is also a maximizing set for .

Proof: Suppose is a maximizing set for on , but is not a maximizing set for on . Then there is some such that for all , . Since is a maximizing set for on , there is an element in such that , so for all , so , so . Now, for every there is an element in with . If , then the element satisfies so there is some element with (if , take ; if , take .) Hence is a maximizing set for on .

Proof: We will construct a binary search sequence with such that each interval is a maximizing set for on . We put

By the preceding lemma (and induction), we see that each interval is a maximizing set for on . Let be the number such that and let . Since is a maximizing set for on , there is a number with . Since

we have , so . By continuity of , . Since , it follows by the inequality theorem for limits that

Hence is a maximum point for on . This shows that has a maximum. Since is also a continuous function on , has a maximum on ; i.e., there is a point such that for all . Then for all , so has a minimum at .

Conversely, if a subset of has an upper bound and a lower bound ,then is bounded, and is a bound for , since

Proof: By the extreme value theorem, there are points such that

Hence has an upper bound and a lower bound, so is bounded.

- a)
- , is not bounded.
- b)
- , is continuous, is not bounded.
- c)
- , is continuous, is not bounded.
- d)
- , is strictly increasing, is continuous, is bounded.
- e)
- , is continuous, is not bounded.