Definition. A set
of real numbers is bounded above if there is a number
such that
Such a number
is called an upper bound for
.
Definition. A number is a least upper bound of if
(1) is an upper bound of .
and
(2) if is an upper bound of , then
If is a set of real numbers, and if is bounded above, then has a least upper bound.
Theorem If is continuous on and then there is some number in such that . |
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Proof. Define
Clearly , since is in ; in fact,.there is some such that contains all points satisfying ; this is so since is continuous on and Similarly, is an upper bound for and, in fact, there is a such that all points satisfying are upper bounds for since From the least upper bound axiom it follows that has a least upper bound and that We now wish to show that , by eliminating the possibilities and .
Suppose first that . Then there is a such that for . Now there is some number in which satisfies (because otherwise would not be the least upper bound of This means that is negative on the whole interval . But if is a number between and , then is also negative on the whole interval . Therefore is negative on the interval , so is in . But this contradicts the fact that is an upper bound for ; our original assumption that must be false.
Suppose, on the other hand, that . Then there is a number such that for . Once again we know that there is an in satisfying ; but this means that is negative on , which is impossible, since . Thus the assumption also leads to a contradiction, leaving as the only alternative.
Theorem. If is continuous on and then there is some number in such that
Theorem. If is continuous on and then there is some number in such that
Proof. Consider the function
Theorem. If is continuous on and then there is some number in such that
Proof. Consider the function
Theorem If is continuous at , then there is a number such that is bounded above on the interval
Proof. Since , there is, for every a such that, for all ,
It is only necessary to apply this statement to some particular
,
for example,
.
We conclude that there is a
such that, for all x,
It follows, in particular, that if
then
This completes the proof: on the interval
the function
is bounded above by
Theorem. If is continuous on , then is bounded above on .
Proof. Let
Clearly (since is in ),and is bounded above (by ), so has a least upper bound . Notice that we are here applying the term ''bounded above'' both to the set , which can be visualized as lying on the horizontal axis, and to , i.e., to the sets which can be visualized as lying on the vertical axis. Our first step is to prove that we actually have Suppose, instead, that By the previous theorem there is such that is bounded on Since is the least upper bound of there is some in satisfying This means that is bounded on . But if is any number with then is also bounded on . Therefore is bounded on so is in contradicting the fact that is an upper bound for . This contradiction shows that . One detail should be mentioned: this demonstration implicitly assumed that (so that would be defined on some interval ; the possibility can be ruled out similarly: using the existence of a such that is bounded on
The proof is not quite complete--we only know that is bounded on for every not necessarily that is bounded on However, only one small argument needs to be added.
There is a 0 such that is bounded on There is in such that Thus is bounded on and also on , so is bounded on
Theorem If is continuous on then there is a number in such that for all in
Proof. We already know that
is bounded on
which
means that the set
is bounded. This set is obviously not
,
so it has a least upper bound
.
Since
for
in
it suffices to show that
for some
in
.
Suppose instead that
for all
in
Then the function
defined by
is continuous on
,
since the denominator of the right side is never
.
On the other hand,
is the least upper bound of
;
this means that
This, in turn, means that
But this means that
is not bounded on
,
contradicting the previous theorem.