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From the J. Taylor's book, ''The completeness property is the missing ingredient in most calculus course. It is seldom discussed, but without it, one cannot prove the main theorems of calculus.''

My question is: Why (without it), one cannot prove the main theorems of calculus?

Jason
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4 Answers4

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It is better to state the completeness property which is the topic of the question.

Completeness property of the real number system is the property of real numbers which distinguishes it from the rational numbers.

Apart from this property both the real numbers and rational numbers behave in exactly the same manner. The property can be expressed in many forms (and I am not sure if you can understand all the forms):

  • Dedekind's Theorem: If all the real numbers are grouped into two non-empty sets $L$ and $U$ such that $L \cup U = \mathbb{R}, L \cap U = \emptyset$ and further if every member of $L$ is less than every member of $U$, then there is a unique real number $\alpha$ such that all real numbers less than $\alpha$ belong to $L$ and all real numbers greater than $\alpha$ belong to $U$.

  • Least upper bound property: If $A$ is a non-empty set of real numbers such that no member of $A$ exceeds a constant real number $K$ (say), then there is a real number $M$ with the property that no member of $A$ exceeds $M$, but every real number less than $M$ is exceeded by at least one member of $A$. This number $M$ is called the least upper bound or supremum of $A$.

  • Cauchy Criterion of Completeness: Let $\{a_{n}\} $ be a sequence of real numbers such that for any given real number $\epsilon > 0$ there is a positive integer $n_{0}$ such that $|a_{n} - a_{m}| < \epsilon$ for all positive integers $m, n$ with $m \geq n_{0}, n \geq n_{0}$. Then the sequence $a_{n}$ converges to some real number $L$. In other words, there exists a number $L$ such that for any given $\epsilon > 0$ there is a positive integer $n_{0}$ such that $|a_{n} - L| < \epsilon$ for all $n \geq n_{0}$.

  • Nested Interval Principle: Let $I_{n} = [a_{n}, b_{n}]$ be a sequence of closed intervals such that $I_{n + 1} \subseteq I_{n}$ then there is at least one real number $\alpha$ which lies in all the intervals $I_{n}$ i.e. $\bigcap_{n = 1}^{\infty}I_{n} \neq \emptyset$. This is sometimes also called Cantor's Intersection Theorem.
  • Bolzano-Weierstrass Theorem: If $S$ is an infinite set of real numbers which is bounded then there is a real number $c$ (which may or may not belong to $S$) such that every neighbourhood of $c$ contains a point of $S$ other than $c$. Such a point $c$ is called a limit point or an accumulation point of $S$.
  • Heine Borel Principle: Let $[a, b]$ be a closed interval and let $\mathcal{C}$ be a collection of open intervals $I$ such that each point of $[a, b]$ lies in some interval $I \in \mathcal{C}$. Then it is possible to choose a finite number of intervals from $\mathcal{C}$ say $I_{1}, I_{2}, \ldots, I_{n}$ such that each point of $[a, b]$ lies in some interval $I_{j}$.
  • Absolute Convergence Principle: If $\{a_n\} $ is a sequence then the infinite series $\sum a_n$ is said to be convergent with sum $s$ if the sequence $\{s_n\} $ defined by $$s_n=a_1+a_2+\dots+a_n$$ converges to $s$. If the series $\sum |a_n|$ is convergent then the series $\sum a_n$ is said to be absolutely convergent. Absolute convergence of an infinite series implies its convergence. (This version of completeness was discussed by user @Shahab in comments and has a non-trivial and non-obvious proof.)

None of the above results hold if the numbers concerned are changed into rational numbers. Moreover assuming any one of the above properties, the others can be proved. These are all different notions of completeness which are equivalent in the real number system. Also note that all these properties are sort of existence theorems in the sense that they guarantee the existence of something useful in certain particular contexts.

Out of these the simplest one to understand for a calculus beginner (any student of age 15-16 years) is the first property called Dedekind's Theorem named after Richard Dedekind who developed a rigorous theory of real numbers.

What Taylor's book is saying is very correct and perhaps very very unfortunate for the students who are learning calculus. The subject of calculus builds up on the foundations of real number system and it is a shame that before learning calculus the only machinery available to students is the concept of rational numbers, laws of algebra to manipulate expressions and some knowledge about specific types of irrational numbers like surds or $\pi$.


Any significant theorem of calculus (i.e. theorems which don't deal with algebraic manipulation of various things in calculus) is proven using the completeness property of real numbers. OP wants to know: why is it so?

Well it is difficult to answer your question directly. In my opinion one should say that any theorem of calculus which is based on the completeness property is significant and any theorem of calculus which is not based on the completeness property is insignificant in the sense that it can be obtained by the use of properties of rational numbers and laws of algebra and thereby these are nothing but extensions of algebra in a particular direction.

Thus a theorem like $$\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}$$ is insignificant and is more of an extension of algebra. But the theorem which says that a sequence which is increasing and bounded above converges to a limit is significant. Rolle's Theorem, Mean Value Theorem, Intermediate Value Theorem etc are all significant theorems. It is precisely because the "completeness property" is never introduced in many calculus textbooks that these theorems of calculus are never proved in such textbooks.

For integral calculus too, the story is much worse similar. Guaranteeing the existence of anti-derivative for continuous functions is a difficult result which can't be proved without the completeness property (it uses the Heine Borel Principle in one of the proofs). However you can pretty much invert the differentiation formulas for elementary functions to get anti-derivatives for common functions and this is fine without using completeness.

Finally there is the famous Fundamental Theorem of Algebra (namely that any polynomial of positive degree with complex coefficients has a complex root) which can't be proved without completeness property. This shows that the theorem actually belongs to analysis and is named so because of historical reasons. It does not serve so much of a fundamental purpose in algebra.


Although you did not ask but it is important to throw some light on "why the completeness property is the missing ingredient in most calculus courses". It is because before learning calculus the students are mostly trained in algebraic manipulation of mathematical expressions (along with some basic theorems in Euclidean Geometry) and various textbook authors and instructors think that it will be difficult for students to depart from the algebraic manipulation involving $+,-,\times, /$ and focus on order relations involving $<, >$ (which is true to some extent for many of the students). The completeness property can not even be understood properly (leave its proof) if one does not appreciate the concept of order relations.

At the same time the same textbook authors and instructors feel that it is best if calculus can also be presented as algebraic manipulation of somewhat complicated and new things then the students would sail through a calculus course easily without the need to focus on order relations and completeness property and one champion of this thought is Walter Rudin. On the other hand there are many textbooks like "Principles of Mathematical Analyis" by Rudin which just assume such an important property as completeness as an axiom and justify this approach on the basis of "sound pedagogy".

Such an approach keeps the students limited to their old techniques and attitudes of algebra and they never understand that the fundamental thing to learn in calculus is to appreciate the concept of order relations and thereby study the proper theory of real numbers and the concepts like limit, continuity, derivative and integral based on the theory of real numbers. With this approach calculus seems very mysterious / magical and yes very confusing to students who are new to it. Plus they miss the understanding of all the major and fundamental theorems of calculus.

I now quote my favorite example. In order to learn calculus properly/seriously a student must be able to prove the following two statements:

  • There is no rational number whose square is $2$.
  • If $a$ is a positive rational number such $a^{2} < 2$ then there is a rational number $b > a$ such that $b^{2} < 2$.

The first result belongs to algebra and it is a routine exercise in high school. The second result is the beginning of real-analysis (without mentioning or using anything about real numbers) and if the student can sail through this result, then he can very well learn a proper theory of real numbers via Dedekind's cuts and for him calculus is never going to be mysterious / confusing.

Fortunately such an approach which develops theory of real numbers before developing calculus is available in G. H. Hardy's classic textbook "A Course of Pure Mathematics."

Paramanand Singh
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  • To be fair on the integral point, functions which are familiar to calculus students are $C^1$ or at least continuous and piecewise $C^1$. For such functions one does not need to use the Heine-Cantor theorem (proven using the Heine Borel theorem) to obtain uniform continuity, because one has Lipschitz continuity to work with instead. In this case one can give explicit estimates for convergence of Riemann sums. – Ian May 19 '16 at 13:33
  • Specifically, if your interval has length $I$, you want an error of at most $\varepsilon$, and your Lipschitz constant is $L$, then it is enough to have a partition whose mesh is less than $\frac{\varepsilon}{LI}$. For a piecewise $C^1$ function, $L$ is the bound on the derivative where it exists. – Ian May 19 '16 at 13:38
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    I'm astonished to read that Rudin championed the idea "that _it is best if calculus can also be presented as algebraic manipulation of somewhat complicated and new things then the students would sail through a calculus course easily_ without the need to focus on order relations and completeness property". That doesn't rhyme with what I have read of Rudin. – Daniel Fischer May 19 '16 at 13:53
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    @Ian If you don't have completeness, to what do your Riemann sums converge? – Daniel Fischer May 19 '16 at 13:55
  • @DanielFischer That's a fair point, but that doesn't really support the statement "for integrals the story is much worse", provided you are willing to restrict attention to piecewise $C^1$ functions. It makes it just as bad as everything else: you can see that the Riemann sums are "trying" to converge and only need completeness to make sense of what the limit is. – Ian May 19 '16 at 14:14
  • @DanielFischer Seeing as the Weierstrass function was only invented by *much* later than the beginnings of calculus, was still viewed as a "monster" at the time, and it is not easy to come up with explicit examples of functions which are not piecewise $C^1$ even today, I think that this restriction is reasonable, at least at the introductory level. – Ian May 19 '16 at 14:16
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    @Ian Without completeness, you have fundamental problems way before you think about the Weierstraß function. Can you define $\exp, \sin, \cos$ without using completeness? [Might be possible, but it would be a nightmare.] – Daniel Fischer May 19 '16 at 14:26
  • @DanielFischer Indeed the situation is quite dire, but it is dire pretty much across the board. I was merely contesting the point that it is any worse in integration than anywhere else. – Ian May 19 '16 at 14:35
  • @Ian I'd agree with that. Without completeness, you're basically screwed everywhere, except in things that only need algebraic manipulations. – Daniel Fischer May 19 '16 at 14:43
  • @DanielFischer: Actually Rudin championed the idea that "it is pedagogically unsound to present construction of real numbers" before teaching calculus. But then you get the completeness for free and I seriously don't want to take/give such a big thing for free. I will update my post to reflect this. Sorry if this bothered you a bit. – Paramanand Singh May 19 '16 at 14:43
  • @Ian: Actually there is a proof of integrability for continuous functions without uniform continuity. This I found only in Spivak and no where else. So it seems that it is not well known. But then Spivak uses mean value theorem so it does require completeness. – Paramanand Singh May 19 '16 at 14:45
  • BTW thanks to @DanielFischer for the quote "Without completeness you are screwed everywhere." Too good. – Paramanand Singh May 19 '16 at 14:51
  • @Ian: I think I need to dilute the story for integrals. Both integration and differentiation lack a lot of their power without completeness. Sometimes I go overboard while writing long answers like these but thanks to many people I am learning to restrict myself (that is one hallmark of maturity I need to master). – Paramanand Singh May 19 '16 at 14:55
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    @ParamanandSingh In the preface to the third edition of PMA, he writes "Experience has convinced me that it is pedagogically unsound (though logically correct) **to start off** with the construction of the real numbers from the rational ones." [Emphasis added by me.] So one starts with developing the theory of a complete Archimedean ordered field. Then, when one sees that it's an interesting theory, the students have reached the point where they can appreciate the need to prove that it's not a theory about nothing. I'm not saying it's the only sensible approach, but it's one sensible approach. – Daniel Fischer May 19 '16 at 14:55
  • @DanielFischer: Fully agree. I think it maybe because of the way I learnt calculus from Hardy's Book which presents Dedekind's cuts in the first chapter itself, I am so much so inspired by Dedekind's construction. – Paramanand Singh May 19 '16 at 15:03
  • I have tried to address concerns raised here in comments. I don't know if the downvote was for those concerns or not. If there are any more cocerns (which led to downvote) I would like to know and thereby improve my answer. – Paramanand Singh May 19 '16 at 15:09
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    The statement that the Fundamental Theorem of Algebra "can't be proved without completeness property" is not true. It's true in the field of algebraic numbers, which is not complete. – Robert Israel Jun 15 '16 at 22:48
  • @RobertIsrael: Perhaps you have misunderstood my statement. There are two sort of theorems on existence of roots. 1) Every polynomial of positive degree with complex coefficients has a complex root (this is usually called Fundamental Theorem of Algebra and I mention this theorem in my answer). 2) If $p(x)$ is a polynomial of positive degree over a field $F$ then there is an extension field $K$ of $F$ such that $K$ has a root of $p(x)$. This second result has nothing to do with Fundamental Theorem of algebra or completeness of real numbers. – Paramanand Singh Jun 16 '16 at 04:21
  • @RobertIsrael: When I first read this result about existence of roots of any polynomial over any field $F$ then I was stunned. After some serious reading of its proof I got the essential idea. Here we don't have actually the existence of root in an existing field, rather we create the extension $K$ using the polynomial $p(x)$ and $F$. – Paramanand Singh Jun 16 '16 at 04:22
  • There are various results that might be called the Fundamental Theorem of Algebra. Of course you can't refer to the real or complex numbers without defining them, and that requires completeness. But there's the fact that the algebraic numbers can be expressed as $F[i]$ where $i$ is a square root of $-1$ and $F$ is an ordered field. See e.g. [this paper](http://www.dwc.knaw.nl/DL/publications/PU00018230.pdf) – Robert Israel Jun 16 '16 at 05:53
  • @RobertIsrael: Thanks for the interesting paper. I understand what you have said in comments. I will update my answer to state what I mean by fundamental theorem of algebra. – Paramanand Singh Jun 16 '16 at 07:21
  • Another criterion is that for any series absolute convergence implies convergence. – Shahab Jan 22 '18 at 01:49
  • @Shahab: can you elaborate further? How does this not work in case of rationals? Moreover just considering a series of rationals which converges to a rational is too restrictive. I think it is better to deal with Cauchy sequences of rationals. Do you mean that if $s_n=|u_1|+|u_2|+\dots+|u_n|$ is Cauchy sequence then so is $t_n=u_1+u_2+\dots+u_n$? And how does one show this is equivalent to completeness? – Paramanand Singh Jan 22 '18 at 02:39
  • Yes, the proof uses the Cauchy criterion. The converse is also true. I merely wanted to point out that this is another characterization of completeness. – Shahab Jan 22 '18 at 04:32
  • @Shahab: by the way, thanks for pointing that out! I was not aware of it so it was an opportunity for learning. – Paramanand Singh Jan 22 '18 at 05:00
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    Just want to say this word of encouragement for any student reading this: if you can't solve the two problems posed by Paramanand, it does not automatically mean that you are not ready to learn some Calculus or Analysis. – Blue Apr 28 '21 at 22:07
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    @Blue: thanks for your comment. The problems were meant to show the kind of ideas one usually sees in analysis and not to discourage anyone. But in case someone thinks so, your comment will help them out. +1 for the comment. – Paramanand Singh Apr 28 '21 at 22:21
  • @ParamanandSingh Thank you. I agree that the problems illustrate very well the general ideas. I know you did not mean to be discouraging. – Blue Apr 28 '21 at 22:29
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Suppose we tried to do calculus using only the rational numbers. Define $f$ so that $f(x) = 0$ if $x^2 < 2$, and $f(x) = 1$ otherwise. Then $f'(x) = 0$ for all $x$, but $f$ is not constant. This is the kind of thing that goes wrong without the completeness axiom.

littleO
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  • ''$f'(x)=0$ for all $x$ '' Yes. ''but $f$ is not constant'' I did not understand this. Why? And, You said that '' This is the kind of thing that goes wrong without the completeness axiom'' I did understant this statement. Can you explain again? –  May 16 '16 at 04:02
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    $f$ isn't constant because for example $f(0) = 0$ but $f(2) = 1$. – littleO May 16 '16 at 04:27
  • Thanks. So, what does that have to do with the completness axiom? –  May 16 '16 at 04:44
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    The rationals satisfy all the axioms of the reals except completeness, and this happens without it. The following paper lists and proves lots of properties which are equivalent to completeness. https://arxiv.org/abs/1204.4483 – Patrick Stevens May 16 '16 at 05:08
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    When you look at $f$, you might object that it's not differentiable because it has a discontinuity at $\sqrt{2}$. However, this objection is not valid if we are only working with rational numbers, because then there is no such number as $\sqrt{2}$! The fact that $\sqrt{2}$ is "missing" from the rationals is a symptom of the rationals not being complete. (In order to prove there is a real number $x$ such that $x^2 = 2$, you must use the completeness axiom.) – littleO May 16 '16 at 05:25
  • Are you sure that the definition of derivative make sense out of completeness? Because then Cauchy sequences cant define a value at infinity. – Masacroso May 16 '16 at 05:29
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    BTW using only rational numbers it is bit difficult to prove that $f'(x) = 0$ for all $x$. I am assuming that derivative is defined in usual manner only the limit variable takes rational values. But your example is good. +1 – Paramanand Singh May 16 '16 at 08:34
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Completeness – whether defined in terms of Cauchy sequences, or in terms of existence of $\inf$ and $\sup$ – is a property of the fine structure of ${\mathbb R}$. It allows you to prove the existence of interesting limits without knowing these limits beforehand. Some examples:

We don't need completeness to prove that $$\lim_{x\to1}{x^2-1\over x-1}=2,\qquad{\rm or}\qquad\sum_{k=0}^\infty 2^{-k}=2\ ,$$ because we guessed all along that the limit in these cases is $2$, and it is easy to do some $\epsilon$-$\delta$-chasing in order to provide a proof. But it is another thing with the limit $$\lim_{n\to\infty}\left(1+{1\over n}\right)^n\ .\tag{1}$$ Here completeness does the following for us: If the sequence $a_n:=\left(1+{1\over n}\right)^n$ satisfies certain criteria (monotonicity, boundedness), which can be checked in finite time, i.e., by looking at the given $a_n$ alone, then the limit $(1)$ exists. Similarly for the limit of the sum $$R_n:=\sum_{k=1}^n{1\over n+k}\ ,$$ which is a Riemann sum for the integral $\int_1^2{1\over x}\>dx=\log 2$.

Christian Blatter
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Here is an expansion of littleO's answer. When working in a field such as $\mathbb Q$ that does not satisfy the completeness axiom, the derivative of a function can behave in ways that defy our physical intuition, and it has a number of unsatisfying mathematical properties. For instance, in $\mathbb Q$, it is not true that if $f'(x)=0$ for all $x$ on some interval, then $f$ must be a constant function. This seems absurd on the face of it: if the velocity of a particle is always $0$, then surely it must be standing still! However, the proof of this "obvious" fact in $\mathbb R$ actually requires the mean value theorem, which depends on the extreme value theorem, which in turn depends on the completeness axiom.

Joe
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