You are to be commended for your pluck and curiosity, coming here. This proof is hard, and if you get stuck, try working on something else for a while, then coming back here in a few hours or even on another day. You never know when something has clicked in your mind while you were asleep. And do keep in mind that this is optional material. Don't lose the time you need to understand the required material by struggling with this.
We shall be relying heavily on the method of trapping a real number inside a sequence of nested intervals -- the method used in the proof of the intermediate value theorem. We shall use this method several times.
In addition, we shall have to rely upon two other theorems, which I have named The Camel and the Tent Theorem and Stepping Stone Theorem. The latter is given as an exercise in section 3.0. Although at first it may not seem clear where I'm going with these two, in the end, you will see that they form the foundation of the proof of the main theorem -- that is, every function that is continuous over a closed interval has a maximum and a minimum on that interval.
Everthing rests ultimately on the
If you get through this whole development and understand it, go and treat yourself to something yummie. You deserve it. You will understand more about the real numbers than the average student does after passing two semesters of calculus. So stand tall and march boldly on...
If A is any nonempty set or collection of real numbers, then a real number, x, is said to be an upper bound of A if and only if x is greater than or equal to every real number that is a member of A. So, for example, if A is the set of all real numbers between 1 and 2, then 3 is an upper bound of A, and so is 1000.
A lower bound is defined the same way, except that x is a lower bound of A if and only if it is less than or equal to every real number that is a member of A. Using the same example of A being all the real numbers between 1 and 2, 0 would be a lower bound of A, and so would -1000.
We say that a set is bounded if it has both an upper bound and a lower bound.
If f(x) is a function and A is some nonempty set of real numbers,
then the image of A under f(x) is the set of
all real numbers that can be produced by applying f(x) to a
member of A. So if a is a member of A, then
f(a) is a member of the image if A under f(x).
The function, f(x), is said to be bounded over A
if the image of A under f(x) is bounded. In this
proof we will consider A to be a closed interval (that is, it
includes its endpoints) and f(x)
to be a function that is continuous over that interval. For example,
if
If A again is a nonempty set of real numbers, then we say that a real number, x, is the least upper bound of A if and only if it is an upper bound of A and it is less than or equal to all other upper bounds of A. If A is the set of real numbers between 1 and 2, then 2 is the least upper bound of A.
Likewise, a real number, x, is the greatest lower bound of A if and only if it is greater than or equal to all other lower bounds of A. In our example of the numbers between 1 and 2, 1 is the greatest lower bound of A.
If riding around the desert on a camel and sleeping in a tent at night were your life, you probably wouldn't want the camel to sleep in the tent with you. The proverbial camel, though, always tries to get his nose into the tent, hoping that he can gradually scooch the rest of himself in unnoticed.
So you lay down the law to the camel. The camel's nose can go this far toward the door of the tent, and no farther. When his nose reaches that point, the camel is still ok. But if he moves his nose even the tiniest fraction farther into the tent, you're going to slap it.
The point here is that of all the places the camel can put his nose, which is to say anywhere outside the tent, there is a spot that he can put it that is as close to being inside the tent as it can go without being slapped.
Imagine that all the places inside the tent are represented by the set, A. And imagine that all the places the camel can put his nose are upper bounds of A. Then that spot he can put his nose that is closest to being inside the tent is the least upper bound of A. This theorem guarantees that that spot always exists.
If A is a nonempty set of real numbers that has an upper bound, then A also has a least upper bound.
Between any point inside the tent (where the camel's nose is not allowed) to any point outside the tent (where the camel's nose is allowed) you can draw a straight line to connect them. The midpoint of that line must either be inside or outside the tent. So by proper choice from the two fragments you can make a line of half the length that still goes from inside the tent to outside the tent. And you can keep doing that indefinitely, getting shorter and shorter lines that always go from inside the tent to outside the tent. We shall show that those shorter and shorter lines collapse onto a single point, and that it is the point where the camel's nose is allowed no farther.
We know that there are real numbers in A because it is nonempty. If you pick any real number that is a member of A, and then pick a to be less than that member, then a is guaranteed not to be an upper bound of A. So let that a be the lower endpoint of a closed interval. We have presupposed that A has an upper bound. So let b be some upper bound of A, and let it also be the upper endpoint of a closed interval. So we have an interval that goes from not-an-upper-bound to is-an-upper-bound.
Divide that interval precisely in half and make both fragments closed intervals. One of two situations exist:
Here is the crux of this proof. c must be the least upper
bound of A. Why? First we show that c is indeed
an upper bound. If it weren't, then there would have to be some
real number, a, that was a member of A such that
You can use a similar argument to show that no upper bound of
A can be less than c. If there were such an upper
bound, b, then we have
So the camel always has a most favorable spot where he is allowed to rest his nose. The corollary to this is that every set that has a lower bound also has a greatest lower bound. The proof is identical, except you turn all the inequalities around.
If x is chosen from an interval,
There are two parts to the theorem:
If f(x) is continuous for all x on a closed interval,
Since the theorem is in two parts, we shall prove them separately. To
prove the first part, we shall observe that if f(x) is
unbounded over the closed interval, then it must be unbounded over
at least one half of that closed interval, and that half can be
made to be a closed interval as well. So we can apply the operation
again to get an even smaller interval. And we can continue to do
so indefinitely, trapping a region in which f(x) is unbounded inside
a smaller and smaller closed interval. We can demonstrate that both
sequences of endpoints converge to the same real number, c.
We then show that for every d
you can name, f(x)
must be unbounded in the interval
To prove the second part, again we divide the interval in half repeatedly to trap an x where f(x) has its least upper bound. We then use the camel and the tent theorem and the stepping stone theorem to show that continuity forces the existence of an x where f(x) is maximum and another x where it is minimum.
Suppose that f(x) is continuous over some closed interval but is not bounded. Divide that interval precisely in half, making each fragment a closed interval. One of the following two situations must be the case:
Let
But that means that no matter what
|f(x) - f(c)| £ ewhenever
Why? Because f(x) is unbounded for every interval,
We have already shown that if f(x) is continuous over a closed interval, then it is bounded over that interval as well. That means that the image of the interval under f(x) has an upper bound and a lower bound. According to the camel and the tent theorem, it must also have a least upper bound and a greatest lower bound. We shall use the properties of the least upper bound to show that f(x) has a maximum on the closed interval. Nearly identical logic on the greatest lower bound proves that f(x) also has a minimum on the interval.
We know from the above that f(x) has a least upper bound over the interval. Call the real number that is f(x)'s least upper bound, ylub. If you divide the interval precisely in half, making each fragment a closed interval, then ylub is an upper bound of f(x) over each fragment, and it is the least upper bound of of f(x) over at least one of the fragments. So one of two situations must be the case:
Since both the lower end points,
In addition to that, no matter how close to zero you choose a positive
d,
there will always be an In contained in the interval
We now show that
|f(x) - f(c)| £ ewhenever
So whenever
Hence you cannot have
Again, the proof for the existence of a minimum for continuous f(x) over a closed interval is nearly identical, except you use the greatest lower bound instead of the least upper bound, and some of the inequalities are turned around the other way.
It follows almost directly from this theorem and from the intermediate value theorem that if f(x) is continuous over a closed interval, then the image of that closed interval under f(x) is also a closed interval.
Well, was getting through this proof a glorious victory or what? Thanks to your perseverence, you have taken your first steps into a field of mathematics with the impressive name of Real Analysis. What's more important, though, is that you have begun to think like a mathematician. And that will take you a long way in your calculus training.
email me at hahn@netsrq.com