Who came up with the $\varepsilon$-$\delta$ definitions and the axioms in Real Analysis?

Question

I've seen a lot of definitions of notions like boundary points, accumulation points, continuity, etc, and axioms for the set of the real numbers. But I have a hard time accepting these as "true" definitions or acceptable axioms and because of this it's awfully hard to believe that I can "prove" anything from them. It feels like I can create a close approximation to things found in calculus, but it feels like I'm constructing a forgery rather than proving.

What I'm looking for is a way to discover these things on my own rather than have someone tell them to me. For instance, if I want to derive the area of a circle and I know the definition of $\pi$ and an integral, I can figure it out.

If $\epsilon - \delta $ is where it's at, how did other people get there so I can get there too? Did they make a giant plane out of ad hoc and tiny pockets large enough to hide their infinities away in to fly there? — Kainui, Dec 22 '13 at 22:38
Historically, Leibniz/Newton started this, though rigor (in the form of $\epsilon\delta$) was added only later by Weierstrass. For an alternative (possibly matching better what Leibniz/Newton had in mind) see Robinson's hyperreal. — Hagen von Eitzen, Dec 22 '13 at 22:44
What exactly do you have problems with accepting when it comes to $\varepsilon$-$\delta$ definitions? — Brian, Dec 22 '13 at 22:44
The way you can 'discover' those definitions by yourself is by using the following dictionary: $|a-b|$ is the distance between $a$ and $b$. Saying that $|a-b|<\epsilon$ means that the distance between $a$ and $b$ is small when compared to $\epsilon$. This $\epsilon$ becomes here as the measuring to which yo compare to call something small. When you replace these in any of those definitions you get directly definitions that are pretty much natural language definitions of the notions you expect (limit, boundary, interior, ...). — OR., Dec 22 '13 at 22:52
The text of the question is almost entirely disconnected from its title. — Did, Dec 22 '13 at 23:02
The definition of continuity bothers me, suppose we have f(x)=x^2 on $[\Bbb R \rightarrow \Bbb R]$ or $[(\Bbb R / \Bbb Q) \rightarrow \Bbb R]$. It seems to me that both are considered continuous functions, however we've removed an infinite number of points. — Kainui, Dec 22 '13 at 23:40
@Kainui Indeed, they are both continuous functions. What is the problem? — Potato, Dec 22 '13 at 23:46
How can R\Q possibly continuous since we've taken out all the rationals? I have a function mapping to the reals, but I can't get to 4 since I can't plug in 2. This feels discontinuous to me.
A quick contrast of a discontinuous: f(x)=(x-2)^3/(x-2) — Kainui, Dec 22 '13 at 23:52
@Potato: Because $\mathbb{R} \setminus \mathbb{Q}$ is totally disconnected? The intuitive picture of continuity has a lot to do with connectedness (when I talk about a continuous function in freshman calc I say mostly that it is, roughly, a function whose graph is a "nice, unbroken curve"; maybe once or twice in the entire course will I mention the "true" $\epsilon$-$\delta$ definition), but the formal definition has -- on the surface! -- nothing to do with it. I think this is something that beginners find confusing. The OP has done a better than average job verbalizing the confusion. — Pete L. Clark, Dec 22 '13 at 23:52
@Kainui Being continuous is a property of a function, not the space on which the function is defined. It is true that $\mathbb R\backslash \mathbb Q$ is not connected, but this is a different property. — Potato, Dec 22 '13 at 23:59
@Potato The space in which a function is defined is part of the definition of the function. What may not be related to the continuity of the function is the disconnectedness of its domain. — OR., Dec 23 '13 at 00:08
I have never heard of connected, I've only just gotten through Real Analysis MATH 3000 at my university after taking cal 1,2,and 3 along with differential equations along with several chemistry and physics courses that use calculus. I love calculus, but I feel like I'm forced to sort of take claims of Real Analysis on faith and it disturbs me when I have already used calculus a great deal with actual, physical results.
I'm just not seeing how set theory has any kind of precedence as being somehow more fundamental than calculus; it feels like these absurd definitions work only coincidentally. — Kainui, Dec 23 '13 at 00:08
@PeteL.Clark The fact that "nice, unbroken curve" has nothing to do with the definition of continuity is because it has nothing to do with it. You are simply telling the students the wrong common language notion. The common language definition of continuity is "values at nearby point approach the value at the point". — OR., Dec 23 '13 at 00:11
@ABC So what is continuous really, if not anything to do with being broken or unbroken? Obviously the definition involves saying if you can make a spot in the range, you can find an entire spot in the domain that maps inside there. But that's not satisfying. — Kainui, Dec 23 '13 at 00:14
@Kainui I already said it in the post immediately above yours: "values at nearby points stay as close as we want to the value at the point". Of course, as always, common language is a bit rough. Another thing to take into account is that the word itself "continuous" shouldn't be the indicator of your intuition. Some names of concepts change over time. The meaning of the common language word can also change over time. — OR., Dec 23 '13 at 00:17
So to phrase my question another way: if the functions f,g:[R\Q->R], $f(x)=(x-2)^3/(x-2)$ is equal to $g(x)=(x-2)^2$ right? — Kainui, Dec 23 '13 at 00:23
@Kainui For example, the very same word continuous in Latin means uninterrupted, but also means hanging together. The former meaning may lead us to thing that people calling a function continuous tried to formalize the former meaning, but it is really the latter the one that really fits the definition from Bolzano. It is very likely that when the concept of continuous was invented, those who did it were more familiar with Latin, than we are now. — OR., Dec 23 '13 at 00:23
@ABC I definitely agree with you, that I shouldn't rely on my intuition here. I guess the problem is I have nothing to rely on at all except that whomever is telling me these definitions isn't lying to me. Suppose a better definition of continuous is slightly more refined than this with one little extra stipulation? How do I know. — Kainui, Dec 23 '13 at 00:28
@Kainui Definitions are whatever you want them to be. Just like your name is whatever you want it to be. The only problem might be if people normally call you some other thing, or if people normally use a different definition under the same name. To your question above. Yes, your $f$ and $g$ are the same function from $\mathbb{R}\setminus\mathbb{Q}$ to $\mathbb{R}$. But notice they are not the same function from $\mathbb{R}$ to $\mathbb{R}$. The domain is part of the definition of the function, part of the function itself. — OR., Dec 23 '13 at 00:31
@ABC I don't believe that we can make mathematical definitions whatever we want like this. I hate to try to go down a slippery slope but I feel like that's equivalent to saying that I can define the derivative of x^2 to be 3x instead of 2x. But upon looking we can definitely see that the slope will be 2x despite our attempt at redefining it. — Kainui, Dec 23 '13 at 00:36
@Kainui If derivative has already been defined the fact that the derivative of $x^2$ is $2x$ is not a definition, it is a deduction. This is very different. Now, when you are defining derivative you can put many different definitions. Actually, there are definitions of derivatives that do not agree with each other (for more complicated object than $x^2$). The question about definitions doesn't begin form their names. It begins from the notion you want to give a name. — OR., Dec 23 '13 at 00:39
You do not like the fact that the function $f\colon\mathbb R\setminus\mathbb Q\to\mathbb R$, $f(x)=x^2$ is called continuous. Fair enough. Introductory classes may indeed discuss continuity only for functions which are defined on a real interval. For these you surely find that the $\epsilon$-$\delta$-definition yields the right concept? Now, later you want to extend the concept for example to functions between metric spaces. And $\mathbb R\setminus\mathbb Q$ is one, and if one applies the definition then our $f$ is continuous. — Carsten S, Dec 23 '13 at 00:41
@ABC: I disagree with "The fact that "nice, unbroken curve" has nothing to do with the definition of continuity is because it has nothing to do with it. You are simply telling the students the wrong common language notion." I don't want to start an argument, and your dismissive attitude to my teaching doesn't make me want to continue the conversation. So I will suffice with the remark that there are certainly sound mathematical grounds to think about continuous real functions in terms of connectedness of the graph: see e.g. "Continuous Functions and Connected Graphs" by C.E. Burgess. — Pete L. Clark, Dec 23 '13 at 00:51
@ABC Thanks, you're helping me really understand and I appreciate it a lot, but I'm not quite where I'd like to be yet. So now you've got me on board and I see how definitions are different from deductions. Now, is there a way of proving that our definitions aren't somehow contradictory? A kind of meta-real-analysis? My intuition is telling me any time you have more than one definition, you have two possibilities: that they have a possibility of being either contradictory in some unanticipated way or they could both be stated better as one single definition. — Kainui, Dec 23 '13 at 00:52
@Kainui: As you go on in mathematics you will meet many concepts which can be given several different-looking but ultimately equivalent definitions. Really one mark of the naturalness of a mathematical concept is that there are several ways to capture it via a formal definition. Comparing different definitions to each other and discovering that they are equivalent in one context but not another is a big part of mathematics (and especially, learning mathematics).... — Pete L. Clark, Dec 23 '13 at 00:57
...It is natural to want to "synthesize" different definitions into a common definition and sometimes good things come of this, but it is also very often the case that the various definitions have merits that do not completely subsume the other. Perhaps the first good example of this in mathematics is the determinant. When you study it you can hardly get away with just seeing just one definition: really one needs at least three or so in order to be able to use it properly. — Pete L. Clark, Dec 23 '13 at 00:59
@PeteL.Clark I just said that you are saying to your students is wrong (which you also said in your own post), and telling you which should be one way of talking about continuity using common language. Read also my post about the meaning of continuus in Latin, and how the name could have been adopted because it meant hanging together and not uninterrupted. The fact that continuous functions map connected to connected is as relevant as continuous functions preserving any other property. — OR., Dec 23 '13 at 01:00
@Kainui Definitions cannot lead to contradictions (below we discuss when it can happen). Defining something is like name-calling. If you give names to people in any way there is not problem. The only thing that can happen is if someone else also gives names and uses names different from yours. Or if you got confused and gave the same person two different names. In the same way, as long as your definition is not calling the same function continuous and not-continuous at the same time. There is no problem. — OR., Dec 23 '13 at 01:05
@ABC: A function $f: \mathbb{R} \rightarrow \mathbb{R}$ is continuous if and only if its graph is connected and closed. This is a formalization of the pre-formal idea that a function is continuous when its graph is a "nice, unbroken curve". So why do you insist on saying that I am telling students something which is wrong? I have also spent years honing ways to talk about $\epsilon$-$\delta$ in ways which will make some sense to students. In my experience if you just say "values at nearby points approach the value at the point", then most students take little or nothing from this.... — Pete L. Clark, Dec 23 '13 at 01:07
If you want to give your actual name and talk about your own teaching experiences and techniques, so be it: it is very likely that I could learn something useful from that. But criticism from an anonymous internet entity is not appealing to me. — Pete L. Clark, Dec 23 '13 at 01:08
@PeteL.Clark I can definitely agree. However, these definitions may take you to the right answer, but is there a better way? Is this way "right" or only coincidentally, since we knew the correct outcome? I feel like Real Analysis approaches math as if it's created, not discovered. I mean, where does logic come from in the first place? I don't quite believe there's a platonic realm out there, but by just seeing how well math models physical reality, it seems that math is something we're not just defining and inventing, but discovering. On top of that, not ever using complex numbers disturbs me. — Kainui, Dec 23 '13 at 01:10
@Kainui: All of the questions in your last comment are great. They are however well beyond the scope of a site like this to answer. I recommend that you find a math professor that you like to talk to you and try to explore these issues with her. — Pete L. Clark, Dec 23 '13 at 01:13
Well, thanks, and damn. I just get furiously angry whenever I think of Real Analysis and I just visualize Real Analysis using all these words such as "real" and "proof" when in reality all numbers are imaginary and the things they put forth as proof is really just passing the buck by tucking infinity away into an epsilon or whatever and hoping you're not clever enough to figure out what new assumptions have been made... At any rate, thanks. — Kainui, Dec 23 '13 at 01:16
@kainui : people prove (no quotation marks needed) things using the epsilon-delta definition of the limit all the time, as well as the definitions of the other things you mentioned at the beginning of your questions. Just crack open a real analysis book and read a proof. People computed things like areas under a curve without rigorous definitions long ago, but mathematics advanced a lot farther after rigorous definitions were found for the ideas people were using. Intuition and hand-waving have their place, but are no substitute for rigor. — Stefan Smith, Dec 23 '13 at 03:58
@StefanSmith, the idea that the infinitesimal approach to continuity and other concepts of the calculus amounts to "hand-waiving" was blown out of the water by Abraham Robinson fifty years ago or in 1961 to be more precise, but not everybody noticed. — Mikhail Katz, Dec 24 '13 at 17:48
@StefanSmith I'm not denying the value of rigor. I'm denying that Real Analysis is capable of supplying a foundation sturdy enough to produce anything of rigor. It seems made up, rather than discovered. For instance, suppose I have prime numbers and you start defining something that only uses numbers that end in 1,3,7, and 9. Sure they will be coincidentally similar and with a few extra modifications to this list we can "construct" the prime numbers. This is how I view set theory's relation to calculus. — Kainui, Dec 24 '13 at 18:35
@Kainui: The key insight is that if you can come up with approximations to something, then there is a thing that you are approximating. If you come up with approximations in such a way that there is only one thing they could possibly be approximating, then voila: you have performed an exact computation. This was known at least as far back as the ancient Greeks, e.g. in the form of the principle of exhaustion. For example, if the area of a circle is smaller than any number greater than $\pi$ and larger than any number less than $\pi$, then it must be exactly $\pi$. — , Mar 22 '14 at 13:14

Yiorgos S. Smyrlis · Answer 1 · 2014-10-01T22:23:30.030

23

The rigorous formulations of the $\delta,\varepsilon$ definitions of the limit and continuity, as well as the $\varepsilon$ definition of the convergence of sequences, in their today's form, were developed by Karl Theodor Wilhelm Weierstrass (1815-1897).

Weierstrass presented this rigorous formulation of Mathematical Analysis for the first time in the lectures of a course named Differential Rechnung during the academic year 1859-60 in the Königliche Gewerbeinstitut in Berlin (now Technical University of Berlin). (See also Wikipedia's article.)

Nevertheless, sufficiently rigorous (with today standards) definition of the limit was given by Bernard Bolzano in 1817.

The first major step, however, towards a $\delta,\varepsilon$ definition appears in the work of Augustin Louis Cauchy Cours d'Analyse (1821), where he wrote:

The function $f(x)$ is continuous with respect to $x$ between the given limits if, between these limits, an infinitely small increment in the variable always produces an infinitely small increment in the function itself.

Although Cauchy never used $\delta,\varepsilon$ definitions, he occasionally used $\delta,\varepsilon$ arguments in his proofs.

edited Oct 01 '14 at 22:23

answered Dec 22 '13 at 23:01

Yiorgos S. Smyrlis

83,933

2

@Yiorgos, you will not find an epsilon, delta definition of continuity in Cauchy even if you look with a microscope. On the other hand, you will find his definition of continuity in terms of infinitesimals: every infinitesimal increment $\alpha$ necessarily produces an infinitesimal change $f(x+\alpha)-f(x)$ in the function. – Mikhail Katz Dec 23 '13 at 13:57
@Yiorgos, both of the wiki pages you connected clearly state that Cauchy did not give an epsilon delta definition of continuity but rather an infinitesimal one. – Mikhail Katz Jan 22 '14 at 17:04
@user72694: See updated version. – Yiorgos S. Smyrlis Jan 22 '14 at 17:13
2

@YiorgosS.Smyrlis, thanks for your openmindedness regarding this issue. However, there are two separate tracks for the development of analysis: the Archimedean, or A-track, in the context of the real continuum, where the epsilon, delta definition belongs; and the Bernoullian, or B-track, in the context of an infinitesimal-enriched continuum. The definition Cauchy gave was a B-track definition. In fact, the definition of continuity over the hyperreal contiuum is identical with Cauchy's. This point was already made by Robinson in his 1966 book. Thus, Cauchy's definition was not a step... – Mikhail Katz Jan 22 '14 at 17:29
...toward a Weierstrassian epsilon, delta definition of continuity, but rather a step toward a rigorous infinitesimal definition as ultimately formalized by Hewitt, Los, and Robinson. – Mikhail Katz Jan 22 '14 at 17:30
@user18921: Thanx! – Yiorgos S. Smyrlis Mar 02 '14 at 00:25
You're welcome! – goblin GONE Mar 02 '14 at 10:57
@YiorgosS.Smyrlis : I find it alleged on meta that you will not receive any notification of my comments at the URL below. Is that true? http://math.stackexchange.com/questions/1258577/error-function-etymology-why-the-name – Michael Hardy Apr 30 '15 at 22:28
@MichaelHardy: I don't understand the question! – Yiorgos S. Smyrlis May 01 '15 at 14:58
@YiorgosS.Smyrlis : All of the comments below that question have been deleted including mine. I don't know how that was done or who did it. I was addressing everyone who participated in closing that question, explaining why closing it was obviously a mistake. – Michael Hardy May 01 '15 at 17:41
@MichaelHardy: I do not have such privileges, and I did not mind any of comments. It sounds quite strange though! – Yiorgos S. Smyrlis May 02 '15 at 08:10

score 4 · Accepted Answer · answered Dec 23 '13 at 14:02

Contrary to a common misconception, one will not find an epsilon, delta definition of continuity in Cauchy even if you look with a microscope. On the other hand, you will find his definition of continuity in terms of infinitesimals: every infinitesimal increment $\alpha$ necessarily produces an infinitesimal change $f(x+\alpha)-f(x)$ in the function. More specifically, the recent translation

Bradley, Robert E.; Sandifer, C. Edward Cauchy's Cours d'analyse. An annotated translation. Sources and Studies in the History of Mathematics and Physical Sciences. Springer, New York, 2009

contains the following material on Cauchy's definition. Cauchy's Section 2.2 is entitled Continuity of functions. Cauchy writes: "If, beginning with a value of $x$ contained between these limits, we add to the variable $x$ an infinitely small increment $\alpha$, the function itself is incremented by the difference $f(x+\alpha)-f(x)$", and states that "the function $f(x)$ is a continuous function of $x$ between the assigned limits if, for each value of $x$ between these limits, the numerical value of the difference $f(x+\alpha)-f(x)$ decreases indefinitely with the numerical value of $\alpha$." Cauchy goes on to provide an italicized definition of continuity in the following terms:

The function $f(x)$ is continuous with respect to $x$ between the given limits if, between these limits, an infinitely small increment in the variable always produces an infinitely small increment in the function itself.

Thanks! However this seems to imply that f(x)=5 is not a continuous function with respect to x since no matter how much we increment x, f(x) will never increment. Of course this is "defined" this way so it doesn't necessarily mean what my intuition tells me it should. But it still sort of leaves a bad taste in my mouth. — Kainui, Dec 24 '13 at 17:29
@Kainui, "will never increment" means that the increment is zero. Now zero is an infinitesimal, so constant functions are certainly continuous in Cauchy's definition (or any other definition). — Mikhail Katz, Dec 24 '13 at 17:43
@Kainui, I am very sympathetic with your comment above concerning the common tendency of "just passing the buck by tucking infinity away into an epsilon or whatever". More specifically, Cauchy's lucid definition of continuity in terms of infinitesimals is typically replaced by the epsilon, delta definition using real numbers to the exclusion of infinitesimals. What's worse, Cauchy is actually "blamed" for the epsilontic definition, which is a rather brazen approach to historical accuracy. — Mikhail Katz, Dec 24 '13 at 17:46

score 0 · Answer 3 · answered Dec 22 '13 at 23:53

Well, originally, definition of differential and integral used infintesimal numbers, but they were unknown and seemed nonsense, so they came up with the $\varepsilon$-$\delta$ definitions of limit, continuity and others, to make it precise. (Later those 'infinitesimal numbers' were found, in the sense that the set of real numbers can be nicely extended with them, staying logically consequent.)

But, this was just a middle step in the story, at least for limit and continuity. E.g., in general topology, it turned out that the most effective definition of continuity of a function $f:A\to B$ is that

The preimage $f^{-1}(V)$ of any open subset $V$ of $B$ is an open subset of $A$. $\quad\quad \quad\quad(1)$

In any metric space $(X,d)$, (i.e. where $X$ is a set and $d$ stands for 'distance' defined for pair of elements of $X$) the open subsets are defined to be the unions of (arbitrarily many) open balls $B_x(r):=\{y\in X\mid d(x,y)<r\}$. (In particular, in $\Bbb R$ the distance is given by $d(x,y):=|x-y|$ and $B_x(r)$ is the open interval $(x-r,x+r)$.)

Try to prove that the definition $(1)$ coincides with the $\varepsilon$-$\delta$ definition for $\Bbb R\to\Bbb R$ functions.
(Hints: taking preimage preserves arbitrary unions, so we can reduce to the case when $V$ itself is an open ball.)

And the question arises as to who came up with these: most likely the point set topology Russian school in the 20s? — Selene Routley, Dec 23 '13 at 14:07
FYI, there is a non-standard version of the general definition too. I can't find a reference, but I think it's that a standard function $f$ is continuous if and only if "For all standard $x$: $f(\mu(x)) \subseteq \mu(f(x))$", where $\mu(p)$ means the monad of $p$. (the set of non-standard points contained in every standard open neighborhood of $p$) — , Feb 15 '14 at 09:44

Who came up with the $\varepsilon$-$\delta$ definitions and the axioms in Real Analysis?

3 Answers3