By David Bressoud @dbressoud

As of 2024, new Launchings columns appear on the third Tuesday of the month.

This month’s column was inspired by an article by Xiaoheng Yan, Ofer Marmur, and Rina Zazkis, “Calculus for Teachers: Perspective and Considerations for Mathematicians.” The authors had interviewed mostly Canadian mathematicians. Unlike Advanced Placement Calculus in the United States, Calculus at the secondary level in Canada is intended as an introduction to the subject rather than an attempt at teaching a university level course to high school students. Nevertheless, as an author of a Calculus textbook (Demana et al, 2020) designed for AP Calculus, it got me thinking about what I would want those teaching from my book to know.

Among the threads identified by Yan et al. were “proof and rigor” and “use/challenge/formalize intuition.” I am not convinced that AP Calculus teachers need a full course of real analysis, but there are some key aspects of 19th century analysis that I believe are important for anyone who would teach Calculus. In particular, while it is enough for students to understand and be able to work with the hypotheses and conclusions of the intermediate value theorem and the mean value theorem, I believe that their teachers should know what sits behind these theorems and how the hypotheses play into their proofs. I will focus on the intermediate value theorem this month and the mean value theorem next month.

Lying at the foundation of both theorems is the completeness of the real number system. While some might argue to include a construction of the reals that implies completeness, I believe that it is sufficient for these students to comprehend the meaning of completeness and the role it plays in mathematical proofs. For Cauchy and others of the early 19th century, this lay in the assumption that any non-empty bounded set of real numbers has a least upper bound, a single least real number that is greater than or equal to every element of the set.

The Intermediate Value Theorem. The only hypothesis needed for the intermediate value theorem is continuity. While most students think of continuity in terms of the flow of the graph of the function, the fact is that continuity is defined point by point. The standard calculus definition of continuity of f at x = a is that the limit as x approaches a of f(x) equals f(a). This is correct given a precise definition of limit, but it gets tangled up in the myriad misunderstandings of limits (see “Beyond the Limit I, II, and III”, Launchings July­­–September, 2014).

I like Hankel’s 1870 definition of continuity in terms of oscillations (see Bressoud, 2008, p. 43). The oscillation of a function over an interval I is the least upper bound of the function over that interval minus the greatest lower bound—think of it as the difference between the maximum and minimum values of the function over that interval. The oscillation at a is the greatest lower bound of the set of all oscillations over all open intervals that contain a. The function f is continuous at a if and only if the oscillation at a is zero. This is completely equivalent to the epsilon-delta definition since the oscillation at a will be zero if and only if we can force f(x) arbitrarily close to f(a) by restricting the distance between x and a. It thus correctly defines continuity in terms of what happens at a single point. This definition avoids both the confusing language of epsilons and deltas and the variety of misconceptions of what is meant by the limit of a function. Continuity at the endpoint of a closed interval simply requires restricting the oscillation to the intersection of each open interval with the closed interval to which we are restricted.

To prove the intermediate value theorem, we assume that f is continuous at every point of [a,b] and that f(a) < f(b). (If f(a) > f(b), just replace f by –f. A function is continuous at a point if and only if its negative is continuous at that point.) Choose any target value T, f(a) < T < f(b). By completeness, there is a least upper bound, call it c, for the non-empty set of values of x in [a,b] for which f(x) < T.  Since f is continuous at c, we can find open intervals containing c on which the oscillation is as small as we wish. If f(c) < T, then we take an open interval containing c on which the oscillation is less than T – f(c), yielding values of x > c at which f(x) < T, contradicting the fact that c was an upper bound of this set. If f(c) > T, then we take an open interval containing c on which the oscillation is less than f(c) – T, yielding values of x < c at which f(x) > T, contradicting the fact that c was the least upper bound. The only remaining possibility is that f(c) = T.

A curious feature of the fact that continuity is defined point by point is that a function can be continuous at one and only one point. Consider the function defined by

G(x) = x if x is rational, = 0 if x is irrational.

This function is continuous at x = 0 since the oscillation over any interval containing 0 is just the distance to the furthest endpoint of the interval. We can make this as small as we wish. On the other hand, if x is not zero, then the oscillation at x will have to be at least |x|.

There is a clever argument that shows that a function can be continuous at every irrational value but not at any rational values other than 0. We define

H(x) = 0 if x is irrational or 0, = 1/q if x = p/q where p and q are integers in lowest terms and q is positive.

At a rational value p/q, the oscillation is at least 1/q, so H is not continuous. It is continuous at 0 since 1/q is always less than or equal to the absolute value of p/q. It takes a bit more insight to see that this function is continuous at all irrational values of x. Choose any positive bound on the oscillation, B. Within one unit of x, there are only finitely many rational numbers with denominators q that are less than 2/B.

For example, let x = pi and B = 0.5. The only rational numbers within one unit of pi with denominators less than 4 are 3/1, 4/1, 5/2, 7/2, 7/3, 8/3, 10/3, and 11/3. The closest of these fractions is 3/1, which is more than 0.14 units away from pi. As long as we stay in the interval (pi – 0.14, pi + 0.14), the oscillation stays less than 0.5.

We can always find an open interval containing x that does not include any of the rational numbers p/q with q < 2/B. Therefore, if p/q is in this interval, then q is greater than or equal to 2/B and therefore H(p/q) = 1/q is less than or equal to 1/(2/B) = B/2 and the oscillation in this interval is at most B.

References

Bressoud, D. (2008). A Radical Approach to Lebesgue’s Theory of Integration. Cambridge University Press. https://www.cambridge.org/us/universitypress/subjects/mathematics/abstract-analysis/radical-approach-lebesgues-theory-integration?format=HB&isbn=9780521884747

Bressoud, D. Beyond the Limit I, Launchings July­­, 2014. https://launchings.blogspot.com/2014/06/beyond-limit-i.html

Bressoud, D. Beyond the Limit II, Launchings August, 2014. https://launchings.blogspot.com/2014_08_01_archive.html

Bressoud, D. Beyond the Limit III, Launchings September, 2014. https://launchings.blogspot.com/2014_09_01_archive.html

Yan, X., Marmur, O. & Zazkis, R. Calculus for Teachers: Perspectives and Considerations of Mathematicians. Can. J. Sci. Math. Techn. Educ. 20, 355–374 (2020). https://doi.org/10.1007/s42330-020-00090-x

David Bressoud is DeWitt Wallace Professor Emeritus at Macalester College and former Director of the Conference Board of the Mathematical Sciences. Information about him and his publications can be found at davidbressoud.org

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