Why don’t we teach a topological view of continuity instead of epsilon-delta?

Question

I would like a critique of this approach to teaching continuity to calculus 1 students.

1. Show them that for an increasing function on (a,b) we have that (a,b) is contained in the set of solutions to $f(a) < f(x) < f(b)$.

2. For decreasing function we get (a, b) is contained in the set of solutions to $f(b) < f(x) < f(a)$.

3. Define pre-image of a point and then of an open interval. The solutions to the equation $f(x) =a$ is the pre-image of a. The solutions to $a<f(x)<b$ are the pre-image of $(a,b)$.

4. A function is continuous at x if the pre-image of every open interval containing f(x) contains an open interval containing x. In English this is hard to understand/state but it’s clear geometrically by looking at graphs of functions.

It is easy to show that functions like $\sin{x}$, $x^2$, $\frac{1}{x}$, $\sqrt{x}$ are increasing or decreasing on certain intervals and thus it’s relatively simple to show continuity of these functions at specific points.

My main problem with epsilon-delta proofs is that we don’t need to find delta. It’s clear that if c is in (a, b) then there is an open interval centered at c contained in (a,b). Given epsilon I just need to show that the set of solutions to $L-\epsilon < f(x) < L+\epsilon$ contains an open interval containing c.

After teaching continuity I then plan to introduce limits. Limit at x existing is weaker than continuity because we just need the pre-image of an open interval to contain an open interval with x removed. A punctured interval.

Am I crazy for doing this approach?

EDIT: What’s easier to do: epsilon-delta proof for continuity of $x^2$ at $x=2$ or show that $4-\epsilon< x^2<4+\epsilon$ can be solved easily since $x^2$ is increasing for positive numbers and the solutions are $\sqrt{4-\epsilon} < x < \sqrt{4+\epsilon}$. This latter interval obviously contains 2.

Answer 1

Yes. You are crazy to be spending time on another effort to bring real analysis style rigor into a calc 1 course. In this case, even off brand real analysis.

Instead of a "different" way to deal with continuity, don't obsess on it at all. Do the bare minimum and move on. Teach them the chain rule and how to integrate by parts.

I'm serious...you need to prioritize. You are constrained in time, student interest, and student intellect. Get the basics done, that they need for physics class. Even that can be hard! Make a limited objective and get a victory. Not a trainwreck.

https://scholarship.claremont.edu/cgi/viewcontent.cgi?article=1234&context=hmnj

Answer 2

You aren't crazy because Leonard Gillman wasn't crazy. See Emphasizing order relations rather than metric concepts in real analysis

The family of all open intervals (A, B) about a point L assuredly constitutes a base for the neighborhood system at L. There is no need to restrict oneself to intervals (L - e , L + e), whose very notation suggests computation. Indeed the order relations A < L < B are usually sufficient, the distances of A and B to L being a distraction. One advantage is to simplify the notation and, in notable cases, to drastically simplify a proof. Another is to obtain a definition of limit that is less intimidating than epsilon-delta. Examples are given from elementary and semi-advanced calculus. Other, somewhat related topics are a characterization of limit that is very easy to understand, yet is completely rigorous; the composition of limits; the exceptional case in the proof of the chain rule; and a simple proof of L'Hôpital's rule.

My personal view is that Howard Jerome Keisler is even less crazy than Gillman. Engineers, scientists, and basically everyone except mathematicians and students of history should have the epsilon-delta definition of the limit entirely removed from their curriculum and rely on the infinitesimal quantities of Leibniz brought up to full rigor by the hyperreal number system of Abraham Robinson in the unfortunately-named nonstandard analysis.

Answer 3

If you want the comment on the approach yourself rather than on whether you should try to implement it, the key words in your post are

Given epsilon I just need to show that the set of solutions to $L−\varepsilon<f(x)<L+\varepsilon$ contains an open interval containing $c$.

That is absolutely correct, but even for strictly increasing functions, your (1) is of little help in that task unless I misunderstand your logic entirely, so I just do not see how you are going to manage for any function that doesn't have an obvious formula for its inverse, say a third degree polynomial.

As for "it’s clear geometrically by looking at graphs of functions", do not forget that those graphs are drawn from the idea that we have continuity. Besides, I have never seen a perfectly accurate graph of $x^2$ in my life: even the computer generated images cannot go beyond the resolution of the screen and the hand-drawn ones hardly ever go beyond the 5% accuracy, so even for some relatively large $\varepsilon$ like $10^{-239}$ you cannot see much on any of them, forget about really small values. The whole point (if you want to introduce rigor at all) is to teach the students to develop healthy doubts about what they "clearly see" and to convert "obvious" pictures into impeccable arguments. Replacing one hand-waving style with another hand-waving style will accomplish very little, if anything.

One personal story in the end. When I was in the 8-th grade, I was on some math. olympiad where there was a problem about a few quadratic equations (something in the style of if two quadratic polynomials with some coefficients depending on a couple of free parameters attain some values somewhere, then so does the third). I figured out that the third quadratic polynomial had two points where it was above and below the target and happily said to the examiner "since a quadratic polynomial is a continuous function, it must take the intermediate value as well". The response was "and what is a continuous function?". I said "Well, it is a function whose graph is a curve", to which the response was "and what is a curve?". After several more attempts, with a sweaty back, I managed to get away with some explicit algebraic monster using square roots about which I proved that their arguments were positive.

Four years later, when I saw the Weierstrass $\varepsilon-\delta$ definition and the proof of the IVT in the first analysis university course, my reaction was "OMG! How simple and elegant!". What's the moral? IMHO, it is "One cannot appreciate the beauty before he is thrown face down into the ugliness". Whether you want to make your students sweat a bit before revealing what you call the "$\varepsilon-\delta$ monstrosity" is up to you, but I'm ready to bet that without that sweat they'll never appreciate how amazingly simple and clear-cut it really is...

Answer 4

If you want to teach a topological view of continuity, then you have to teach topology first. The set of open intervals is a basis of a topology, but it's not itself a topology, and the topology generated by it is not the only topology possible on the real line. So you'll have to either introduce a bunch of stuff about abstract topology in general, and then proceed to say "But never mind that, we're only going to concern ourselves with one particular topology", or you have to pretend that open intervals are a topology, leaving any students who do go on to study topology to have to re-learn what "topology" means.

Moreover, open intervals are equivalent to the $\delta$ definition. If there is some open interval about $x$, it can be expressed as $(x-d_1, x+d_2)$, and if we take $\delta$ as the minimum of $\{d_1,d_2\}$, then $(x-\delta, x+\delta)$ is a subset of that interval.

Also, you used $a$ to mean two different things, which you should avoid when teaching.

It is easy to show that functions like $\sin{x}$, $x^2$, $\frac{1}{x}$ are increasing or decreasing on certain intervals and thus it’s relatively simple to show continuity of these functions at specific points.

Increasing functions aren't necessarily continuous. Your implicit argument seems to be that given an open interval $(A, B)$ containing $f(x)$, we can express $(A, B)$ as $(f(a), f(b))$ for some $a,b$. But that isn't necessarily true. Given two values that $f$ takes, we can use the IMV to show that it takes all values in between, but the IMV requires continuity, which is what we're trying to prove.

Answer 5

A positive attribute of the standard $\epsilon$-$\delta$ definition continuity is that the same formalism can be mutated to define similar things such as limits at infinity or the limit of a sequence $\displaystyle{\lim_{n\to\infty}a_n}$. How would your formalism be adaptable to these?

Answer 6

One approach that has been advocated several times is to significantly decrease the focus the focus on “pointwise continuity” and “pointwise differentiability” notions (or entirely omit them) and instead work with their uniform versions. Two people I can think of who have advocated this approach in the last 20-30 years are Peter D. Lax (see his letter on p. 6 of the Jan 1998 Notices of the AMS, which I believed generated several follow-up letters in subsequent months) and Mark Bridger (see his 1999 paper co-authored with Stolzenberg and his 2007 real analysis textbook which was reprinted in 2019).

Regarding Bridger, in the late 1990s (beginning I think around May 1996, and maybe continuing into the first few years of the 2000s) he made a lot of posts in Math Forum groups (especially CALC-REFORM) in which he strongly advocated using uniform notions of continuity and differentiability in calculus, and even in elementary real analysis. Unfortunately, NCTM took over Math Forum in the mid 2010s and within a few years they took down the tens of thousands (hundreds of thousands?) of Math Forum posts, some of which were of extremely high quality, especially back when Bridger’s posts were made. Perhaps someone better than I am at finding such posts at the Internet Archive can locate some of them, at least if they exist there. (The calc-reform URLs in the late 1990s began with http://forum.swarthmore.edu/epigone/calc-reform and at some point in the early 2000s this initial portion was changed to http://mathforum.org/epigone/calc-reform)

I've also seen this advocacy of “uniform over pointwise” notions argued in some non-research journal papers (going back to at least the early 1930s, I think), but I don't know any specific references at the moment. I think I have photocopies of some of these papers in the notebook binder I used for many of the references in this answer, but I can’t find that binder now.

Answer 7

Introduction

I take it that the OP, now departed from the site, was exploring a pedagogical way to introduce the concept of continuity. I do not think they were trying to put continuity on a new rigorous basis. I once gave a talk, "Limits without delta," in which I described a middle ground between the "intuitive notion of a limit" and the "rigorous definition of a limit," as they are often called in US textbooks. The problem with learning a rigorous approach to limits is the algebraic handling of inequalities involving distance between numerical quantities. Students are comparatively okay at determining the sign of factored expressions. But distances involving the absolute value are an algebraic mess in their hands. They might manage the nested quantifiers — who knows? Others say no, but I claim the students' difficulties at coordinating the $\delta$ and $\epsilon$ inequalities (not their fault, due to it not being taught at all) makes testing that hypothesis impossible. I intended the middle ground to give the students some practice at thinking with inequalities, a skill that benefits from recurrent practice over a few semesters. The idea started from Hardy's approach to sequences in A Course of Pure Mathematics and later got a boost while I was rereading Marsden and Tromba's Vector Calculus. In Marsden and Tromba, I noticed they kept using the squeeze theorem to manage $\epsilon$ without any need to solve for $\delta$. Another attack on $\epsilon-\delta$ is that it is more complicated than how mathematicians actually prove limits, and I wondered why we don't teach the squeeze theorem more effectively. In most textbooks, it resembles an appendix on the verge of appendicitis.

The definition below introduces continuity without reference to limits, in accordance with OP's course structure. It is nothing more than $\epsilon$-$\delta$ in disguise. An advantage is obtained by quickly introducing some squeeze theorems that the students can use in practical work. Intuitive language based on the common notion of distance allow the students to avoid much of the algebra of inequalities that usually accompanies $\epsilon$-$\delta$. The work is not devoid of inequalities, but it is simplified. Again, the main goal is to understand the concept of continuity, and only in simplified contexts. In my mind the importance of continuity has its root source in our experience of motion and change in the world. Most of what we see or feel are continuous changes. Sudden changes — collisions, for instance — are imagined to be occur over a very small interval of time and not be "quantum leaps." In class, I place continuous change centered in the analysis of the change $f(x)-f(c)$ and emphasize that continuity is not just a kind of exam question but a model of change in the world.

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Definition

The setup to be assumed:

• $c$ is a number;
• $f$ is a function;
• Some notion that $c$ is a "limit point" of the domain of $f$: For first-time calculus students, let's say we will assume the domain of $f$ contains an open interval containing $c$. This is stronger than necessary but it keeps things simple.

$f$ is continuous at $c$ if for every positive distance, $f(x)$ lies within that distance of $f(c)$ for all $x$ that are sufficiently close to $c$ and belong to the domain of $f$.

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Theorems and examples

Theorem 1. $f(x) = x$ is continuous at every real number.

Corollary 1.1. If you can find a positive constant $M$ such that $|f(x)-f(c)| \le M\cdot|x-c|$ for all $x$ sufficiently close to $c$, then $f$ is continuous at $c$.

Proofs: Theorem 1 is a tautology, more or less, and your best approach is not to try to prove what no one doubts. One might say, however close we would like $f(x)$ and $f(c)$ to be, well, that's the same as how close $x$ and $c$ would have to be, since $f(x)=x$ and $f(c)=c$. -- [Perhaps you would like to point out the meaning of being within some distance. Since $f(x)-f(c)$ is $x-c$, then what we need to show to prove continuity is that $|x-c|$ is small provided $|x-c|$ is small, a tautology. Even in formal $\epsilon$-$\delta$ proof, we quickly derive a tautology after letting $delta=\epsilon$.]
     For the corollary, we have to divide by $M$ as follows: However close we would like $f(x)$ and $f(c)$ to be, say, less than some desired distance, such closeness is guaranteed whenever $x$ and $c$ are within $1/M$ times the desired distance. -- [If you want to introduce $\epsilon$-$\delta$, this is a good place to do it: Let $\epsilon$ be the distance that we desire $f(x)$ and $f(c)$ to be within. Then we would have $M\cdot|x-c| < \epsilon$ whenever $|x-c| < \epsilon/M$; etc.]

Discussion. For the corollary, warm-up proofs of special cases can be helpful. Take $f(x)=10x$. Then $f(x)-f(c)=10(x-c)$. So, if we want the distance between $f(x)$ and $f(c)$ to be less than $1/10$, how close to $c$ should $x$ be to be close enough? If the desired distance were $1/100$? Whatever distance I pick, how close would be close enough?

Likewise, we can show:

Corollary 1.2. If $g$ is continuous at $c$ and $|f(x)-f(c)| \le |g(x)-g(c)|$ for all $x$ sufficiently close to $c$, then $f$ is continuous at $c$.

With these tools it is as easy as calculating derivatives to show $\sqrt{x}$ and $x^n$ are continuous at $c \ne 0$. (The special case $c=0$ is easier but uses a different approach for roots.) The basic idea for $c\ne 0$ is to factor out $x-c$ from $f(x)-f(c)$ and bound the remaining factor. For well-chosen examples, finding the bound $M$ is a simple matter. For complicated examples, it is not. I avoid the complicated examples, since my goal is an understanding of the concept and not to train the students to be consummate provers of continuity. As for which examples are worth understanding, I would say the basic elements that make up functions expressed by formulas: powers, trig., log., and exp., as well as the ways functions are combined, $+$, $-$, $\times$, $\div$, and composition. One should end by remarking that a function given by a single formula in terms of elementary functions is continuous at each point in its domain.

Example from the OP. $x^2$ is continuous at $2$.

Proof: $x^2-2^2=(x+2)(x-2)$. If $x$ is closer to $2$ than one unit, then $1<x+2<3$. In that case, $|x^2-2^2| \le 5\cdot|x-2|$. By Corollary 1.1, $x^2$ is continuous at $2$.

Extension 1 of the example. $x^2$ is continuous at all numbers $c$ in the interval $(1,3)$.

Proof: Changing $5 = 3+2$ into $3+c$, it's still the case that $|x^2-c^2| \le (3+c)\cdot|x-c| \le 6\cdot|x-c|$ for all $x$ in $(1,3)$.

Extension 2 of the example. $x^2$ is continuous at all numbers $c$.

Proof: Changing $5 = (2+1)+2$ into $(|c|+1)+|c|$, it's still the case that $|x^2-c^2| \le (2|c|+1)\cdot|x-c|$ for all $x$ within $1$ unit of $c$.

Example 2. The unit step function $u(x)$ is discontinuous at $0$.

Proof: Over any open interval containing, the values of $u(x)$ are $0$ and $1$. So we cannot make $u(x)$ be closer to $u(0)$ than one unit for all $x$, no matter how close we require $x$ and $0$ to be.

Theorem 2. If $f$ and $g$ are continuous at $c$, then so is $f+g$.

Proof: $(f+g)(x)-(f+g)(c)=[f(x)-f(c)]+[g(x)-g(c)]$, so $|(f+g)(x)-(f+g)(c)|\le|f(x)-f(c)|+|g(x)-g(c)|$. By the continuity of $f$ and $g$ at $c$, each of $|f(x)-f(c)|$, $|g(x)-g(c)|$ may be made as small as we please. Thus their sum may be made as small as we please, provided $x$ is sufficiently close to $c$. -- [If you feel the above is not pedantic enough: Suppose we'd like $(f+g)(x)$ and $(f+g)(c)$ to be closer than some desired distance. Then in the first case, if $x$ is sufficiently close to $c$, $|f(x)-f(c)|$ will be less than some part of it, say, half the deired distance. Likewise in the second case, if $x$ is sufficiently close to $c$, $|g(x)-g(c)|$ will also be less than half the deired distance. In the first case, $x$ has to be within one distance of $c$; in the second, $x$ has be within a second distance of $c$. Thus if $x$ is closer to $c$ than each distance, then $|f(x)-f(c)|$ and $|g(x)-g(c)|$ will each be less than half the "desired distance." Therefore their sum will be less than the desired distance, and $|(f+g)(x)-(f+g)(c)|$ is at most their sum.]

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Conclusion

One can then develop derivatives à la Stephen Kuhn (The Derivative à la Carathéodory. Am. Math. Monthly, 1991), which has conceptual advantages over the standard limit of the difference quotient; see this answer. But the OP indicated that they would do limits next.

One warning: I find it easier to characterize various kinds of singularities with limits: discontinuities, nondifferentiable points, asymptotes. But getting across the main gist of calculus, which deals with smooth functions, seems simpler without limits. Furthermore, students need to understand limits in subsequent courses, so one should not dispense with limits. They can be included without much trouble, including the characterizations of continuity and differentiability in terms of limits.

The upshot is that $\delta$ need only be found for $f(x)=x$, by the teacher, and for $f(x)=x^{1/n}$ in the case that $|x| < \delta \land \delta = \epsilon^n$ implies $|x^{1/n}| < \epsilon$, which the teacher can do. (For $n$ even, we would have to discuss one-sided continuity, which I've omitted from this answer.) The teacher might also elect to omit the $\epsilon$-$\delta$ proofs. The approach frames continuity in terms of change in a (often physical) quantity, analyzes $f(x)-f(c)$ which analysis will be echoed in the subsequent discussion of differentiability, introduces inequalities related to the notion of "arbitrarily small," and cultivates "number sense." The squeeze theorems teach the student to think in terms of bounds and estimate them. It feels satisfying when they realize that if $M=2$ is good enough to apply Corollary 1.1, then so is $M=100$. Of course, there is something satisfying to the students when they can actually think with the concepts and not just push symbols around and calculate.