Box 3.0c: Proof of the Extreme Value Theorem
© 1997 by Karl Hahn
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 Cauchy sequence definition of real numbers. Remember that a Cauchy sequence is an infinite sequence of numbers that confines itself to as small an interval as you like provided you go far enough into the sequence. Recall that every Cauchy sequence of real numbers converges to some real number in the limit.
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 f(x) = x2 and A is the interval, 2 ≤ x ≤ 3, then the image of A under f(x) is the interval 4 ≤ y ≤ 9.
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 a > c, or equivalently a - c > 0. Let b0, b1, b2, ... be the sequence of upper endpoints of the intervals. All are upper bounds of A. They are all greater than or equal to c. And the sequence of them converges to c, so you can always pick an n large enough to make bn - c as close to zero as you like. That means that you can pick one so that bn - c < a - c. But that means bn < a. If a is a member of A, how can bn, which is an upper bound of A be less than it? It can't. Hence c is an upper bound of A.
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 c - b > 0. Let a0, a1, a2, ... be the sequence of lower endpoints to the intervals. None of them can be upper bounds of A. They also converge to c, and so there is an an that makes c - a n as close to zero as you'd like. In particular, there is one that makes c - an < c - b or equivalently, an > b. But how can something that is not an upper bound of A be greater than something that is? It can't. Hence c is the least upper bound of A.
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, a ≤ x ≤ b (or from any other set), and f(x) is a function, then the set of all f(x)'s that can be formed from x's inside the interval (that is, the image of the interval under f(x)) is a set of real numbers. If that set has an upper bound or a lower bound, then this theorem applies to it.
There are two parts to the theorem: If f(x) is continuous for all x on a closed interval, a ≤ x ≤ b, then
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 δ you can name, f(x) must be unbounded in the interval c - δ ≤ x ≤ c + δ. This would make the delta-epsilon contract for continuity impossible to keep. Since f(x) being continuous was a premise, we must infer that f(x) is bounded over the original 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 δ > 0 be as small as you like. The interval, c - δ ≤ x ≤ c + δ, must contain some of the In's described above. Why? Because we know that In can be made as short as you like by picking n large enough, and In always contains c. That means that In can be chosen short enough so it won't reach from c to c ± δ. And since f(x) is unbounded over In, it must also be unbounded over the interval, c - δ ≤ x ≤ c + δ, which contains In.
But that means that no matter what ε > 0 I choose, you will never be able to find any δ > 0 that guarantees
|f(x) - f(c)| ≤ εwhenever |x - c| ≤ δ.
Why? Because f(x) is unbounded for every interval, c - δ ≤ x ≤ c + δ, and the above inequality is attempting to place bounds of f(c) ± ε on f(x) in such an interval. What all this means is that f(x) cannot be continuous at x = c. If f(x) is unbounded, it always fails the delta-epsilon contract required for continuity at x = c. We must conclude that f(x) cannot be simultaneously continuous on a closed interval and unbounded on that 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, a0, a1, a2, ... , and the upper endpoints, b0, b1, b2, ... , of the intervals each form sequences, and each converges to a limit of c, and since f(x) is continuous, it means the stepping stone theorem applies. It follows from the stepping stone theorem that f(c) exists and is the limit of both f(a0), f(a1), f(a2), ... and f(b0), f(b1), f(b2), ... .
In addition to that, no matter how close to zero you choose a positive δ, there will always be an In contained in the interval c - δ ≤ x ≤ c + δ. This is because every In contains c and you can always find an In short enough so that it won't reach from c to c ± δ.
We now show that f(c) = ylub. For if f(c) > ylub, then ylub would not be an upper bound of f(x). If f(c) < ylub, then pick an ε in the interval 0 < ε < ylub - f(c). According to the delta-epsilon continuity contract, there must be some δ > 0 that guarantees
|f(x) - f(c)| ≤ εwhenever |x - c| ≤ δ.
So whenever c - δ ≤ x ≤ c + δ, it means that f(x) is guaranteed to be within ε of f(c). But ylub is more than ε distant from f(c), and the interval, c - δ ≤ x ≤ c + δ, contains an In over which ylub is the least upper bound of f(x). Yet everything less than ylub and greater than f(c) + ε is also an upper bound of f(x) over In. And that means that ylub is not the least upper bound of f(x) over In. It can't both be and not be f(x)'s least upper bound.
Hence you cannot have f(c) < ylub either. If you can have neither f(c) < ylub nor f(c) > ylub, then they must be equal. So c is a point where where f(x) is maximum over the original closed interval, because f(c) exists and is the least upper bound of f(x) over the original closed interval.
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 perseverance, 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.
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