A Contextual Introduction to Real Analysis: How We Got From There To Here

In deriving the Lagrange and Cauchy forms of the remainder for Taylor series, we made use of the Extreme Value Theorem (EVT) and Intermediate Value Theorem (IVT). In Chapter 8, we produced an analytic definition of continuity that we can use to prove these theorems. Before we can work out the rest of the tools needed for those proofs we need to explore the make–up of the real number system.

To illustrate what we mean, suppose that we have only the rational number system available. We can still use our definition of continuity and can still consider continuous functions such as \(f(x)=x^2\text{.}\)

Notice that \(2\) is a value that lies between \(f(1)=1\) and \(f(2)=4\) so the IVT says that somewhere between \(1\) and \(2\text{,}\) \(f\) must take on the value \(2\text{.}\) That is, there must exist some number \(c\in[1,2]\) such that \(f(c)=2\text{.}\) You might say, “Big deal! Everyone knows \(c=\sqrt{2}\) works.”

However, we are only working with rational numbers and \(\sqrt{2}\) \(\)is not rational. As we saw in Chapter 2 the rational number system has holes in it, whereas the real number system doesn’t. Again, “Big deal! Let’s just say that the real number system contains (square) roots.”

This sounds reasonable and it actually works for square roots, but consider the function \(f(x)=x-\cos x\text{.}\) We know this is a continuous function (Why?). We also know that \(f(0)=-1\) and \(f(\frac{\pi}{2})=\frac{\pi}{2}\text{.}\) Thus according to the IVT, there should be some number \(c\in[0,\frac{\pi}{2}]\text{,}\) where \(f(c)=0\text{.}\) The graph of \(f(x)=x-\cos x\) is shown below.

The situation is not as transparent as before. What is this mysterious \(c\) where the curve crosses the \(x\) axis? Somehow we need to convey the idea that the real number system is a continuum, that it is complete. That is, it has no holes in it.

How about this? Why don’t we just assume that it has no holes in it? Sometimes the simplest answer works best! We could do that. More complicated statements have been taken as axioms in the past. But how are we going to formulate a rigorous proof based on the statement, “There are no holes in the real numbers”?

Also, going back to the Greeks (and possibly further) the thrust of mathematics has always been to keep our axioms as simple and self–evident as possible and to prove more complicated statements from the axioms. If possible. So we’d prefer to make a simpler assumption (axiom) and use it to show that there are no holes in the real numbers.

We will see that there are several different, but equivalent assumptions we can make that will convey the notion of completeness. We will explore some of them in this chapter. For now we adopt the following as our Completeness Axiom for the real number system.

Geometrically, we have the following situation.

Notice that we have two sequences \(\left(x_n\right)\) and \(\left(y_n\right)\text{,}\) one increasing (really non-decreasing) and one decreasing (non-increasing). These sequences do not pass each other. In fact, the following is true:

They are also coming together in the sense that \(\limit{n}{\infty}{\left(y_n-x_n\right)}=0\text{.}\) The NIP says that in this case there is a unique real number \(c\) in the middle of all of this: \(x_n\leq c\leq y_n\) for all \(n\in\NN\text{.}\)

If there was no such \(c\) then there would be a hole where these two sequences come together. The NIP guarantees that there is no such hole. We do not need to prove this since an axiom is, by definition, a self evident truth. We are taking it on faith that the real number system has this property. The next problem shows that the completeness property distinguishes the real number system from the rational number system.

Intuitively, the sequences \(\left(x_n\right)\) and \(\left(y_n\right)\) in the NIP appear to converge to \(c\text{.}\) This is in fact, true and can be proven rigorously. This will prove to be a valuable piece of information for us.

To illustrate the idea that the NIP plugs the holes in the real line, we will prove the existence of square roots of non–negative real numbers.

Notice that we can’t just say, “Let \(c=\sqrt{a}\text{,}\)” since the idea is to show that this square root exists. In fact, throughout this proof, we cannot really use a square root symbol as we haven’t yet proved that square roots exist. The idea behind the proof illustrates how the NIP is used in practice.

The bisection method we employed to prove Theorem 9.1.6 is also pretty typical of how we will use the NIP, as it ensures that we will create a sequence of nested intervals which meet all of the conditions specified in the NIP. We will employ this strategy in the proofs of the IVT and EVT. Deciding how to relabel the endpoints of our intervals will be determined by what we want to do with these two sequences of real numbers. This will typically lead to a fifth property, which will be crucial in proving that the \(c\) guaranteed by the NIP does what we want it to do. Specifically, in the above example, we always wanted our candidate for \(\sqrt{a}\) to be in the interval \([\,x_n,y_n]\text{.}\) This judicious choice led to the extra Condition 5: \(\forall\) \(n,\,x_n^2\leq a\leq y_n^2\text{.}\) In applying the NIP to prove the IVT and EVT, Condition 5 is what will change based on the property we want \(c\) to have.

Before we tackle the IVT and EVT, let’s use the NIP to address an interesting question about the Harmonic Series. Recall that the Harmonic Series, \(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots\text{,}\) grows without bound. That is, \(\sum_{n=1}^\infty\frac{1}{n}=\infty\text{.}\) The question is how slowly does this series grow? For example, how many terms would it take before the series surpasses 100? Or 1000? Or 10000?

We will use the NIP to show that this limit does in fact exist. The details are in the following problem.

We now have all of the tools we need to prove the Intermediate Value Theorem (IVT).

The Intermediate Value Theorem followed pretty readily once we introduced the Nested Interval Property. The proof of Extreme Value Theorem takes a bit more work. First we need the result of Theorem 9.3.1 below.

Sketch of Alleged Proof: Let’s assume, for contradiction, that there is no such bound \(B\text{.}\) This says that for any positive integer \(n\text{,}\) there must exist \(x_n\in[a,b]\) such that \(\abs{f(x_n)}>n\text{.}\) (Otherwise \(n\) would be a bound for \(f\text{.}\))

IF the sequence \(\left(x_n\right)\) converges to, say \(c\in[a,b]\text{,}\) then we would have our contradiction. Indeed, we would have \(\limit{n}{\infty}{x_n}=c\text{,}\) and by the continuity of \(f\) at \(c\) and Theorem 8.2.1 we would have \(\limit{n}{\infty}{f(x_n)}=f(c)\) which implies that the sequence \(\left(f(x_n)\right)\) converges. So by Lemma 6.2.7 the sequence \(\left(f(x_n)\right)\) must be bounded. This would provide our contradiction since, by construction \(\abs{f(x_n)}\gt{}n\text{,}\) for all positive integers \(n\text{.}\) QED?

This would work well except for one little problem. The way it was constructed, there is no reason to expect the sequence \(\left(x_n\right)\) to converge to anything That is why we emphasized the IF above. Fortunately, this idea can be salvaged. While it is true that the sequence \(\left(x_n\right)\) may not converge, part of it will. We will need the following definition.

The idea is that a subsequence of a sequence is a part of the sequence, \((x_n)\text{,}\) which is itself a sequence. However, it is a little more restrictive. We can choose any term in our sequence to be part of the subsequence, but once we choose that term, we can’t go backwards. This is where the condition \(n_1\lt n_2\lt n_3\lt \cdots\) comes in. For example, suppose we started our subsequence with the term \(x_{100}\text{.}\) We could not choose our next term to be \(x_{99}\text{.}\) The subscript of the next term would have to be greater than 100. In fact, the thing about a subsequence is that it is all in the subscripts; we are really choosing a subsequence \(\left(n_k\right)\) of the sequence of subscripts \(\left(n\right)\) in \(\left(x_n\right)\text{.}\)

A very important theorem about subsequences was introduced by Bernhard Bolzano and was later proven independently by Karl Weierstrass. Basically, this theorem says that any bounded sequence of real numbers has a convergent subsequence.

The Bolzano–Weierstrass Theorem says that no matter how the sequence \(\left(x_n\right)\) is produced, as long as it is bounded then it contains a convergent subsequence. This can be very useful. In particular the sequence \((x_n)\) that appears in our Alleged Proof is bounded since \(x_n\in [a,b]\ \forall \ n\in\NN\text{.}\) So it contains a convergent subsequence.

Theorem 9.3.1 says that a continuous function on a closed, bounded interval must be bounded. Boundedness, in and of itself, does not ensure the existence of a maximum or minimum. We must also have a closed, bounded interval. To illustrate this, consider the continuous function \(f(x)=\)tan\(^{-1}x\) defined on the (unbounded) interval \(\left(-\infty,\infty\right)\text{.}\)

This function is bounded between \(-\frac{\pi}{2}\)and \(\frac{\pi}{2}\text{,}\) but it does not attain a maximum or minimum as the lines \(y=\pm\frac{\pi}{2}\) are horizontal asymptotes. Notice that if we restricted the domain to a closed, bounded interval then it would attain its extreme values on that interval (as guaranteed by the EVT).

To find a maximum we need to find the smallest possible upper bound for the range of the function. This prompts the following definitions.

For example, if \(S=(0,1)\text{,}\) then any \(b\) with \(b\geq 1\) would be an upper bound of \(S\text{.}\) Furthermore, the fact that \(b\) is not an element of the set \(S\) is immaterial. Indeed, if \(T=[\,0,1]\text{,}\) then any \(b\) with \(b\geq 1\) would still be an upper bound of \(T\text{.}\) Notice that, in general, if a set has an upper bound, then it has infinitely many since any number larger than that upper bound would also be an upper bound. However, there is something special about the smallest upper bound.

The second condition says that if a number \(c\) is less than \(b\text{,}\) then it can’t be an upper bound, so that \(b\) really is the smallest upper bound.

Also notice that the supremum of the set may or may not be in the set itself. This is illustrated by the examples above as in both cases, \(1=\sup(0,1)\) and \(1=\sup [0,1]\text{.}\) Obviously, a set which is not bounded above such as \(\mathbb{N}=\{1,\,2,\,3,\,\ldots\}\) cannot have a supremum. However, for non-empty sets which are bounded above, we have the following.

Notice that we really used the fact that \(S\) was non-empty and bounded above in the proof of Theorem 9.4.3. This makes sense, since a set which is not bounded above cannot possibly have a least upper bound. In fact, any real number is an upper bound of the empty set so that the empty set would not have a least upper bound.

The following corollary to Theorem 9.4.3 can be very useful.

We now have all the tools we need to tackle the Extreme Value Theorem.

Notice that we used the NIP to prove both the Bolzano–Weierstrass Theorem and the LUBP. This is really unavoidable, as it turns out that all of those statements are equivalent in the sense that any one of them can be taken as the completeness axiom for the real number system and the others proved as theorems. This is not uncommon in mathematics, as people tend to gravitate toward ideas that suit the particular problem they are working on. In this case, people realized at some point that they needed some sort of completeness property for the real number system to prove various theorems. Each individual’s formulation of completeness fit in with his understanding of the problem at hand. Only in hindsight do we see that they were really talking about the same concept: the completeness of the real number system. In point of fact, most modern textbooks use the LUBP as the axiom of completeness and prove all other formulations as theorems. We will finish this section by showing that either the Bolzano–Weierstrass Theorem or the LUBP can be used to prove the NIP. This says that they are all equivalent and that any one of them could be taken as the completeness axiom.

Since the Bolzano–Weierstrass Theorem and the Nested Interval Property are equivalent, it follows that the Bolzano–Weierstrass Theorem is not true in the rational number system.

We now have the machinery in place to prove the Archimedean Property of the real number system as we promised we’d to back in Chapter 2. We mentioned in Chapter 2, that until the invention of Calculus this was taken to be intuitively obvious but we can now prove it as a formal theorem. The proof depends on the completeness of the real number system.

You might reasonably ask if the Archimedean Property is equivalent to the Least Upper Bound Property (and thus could be taken as an axiom). The answer lies in the following problem.