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

I've seen a lot of definitions of things like boundary points, accumulation points, continuity, etc, and axioms 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 area of a circle and I know the definition of $\pi$ and an integral, I can figure it out.

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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
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, ...). –  ABC 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
@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. –  ABC Dec 23 '13 at 0:39
@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 1:13

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.

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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). –  user72694 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. –  user72694 Dec 24 '13 at 17:46

The rigorous formulations of the $\delta,\varepsilon$ definitions of the limit and continuity were developed in their today's form by Karl Theodor Wilhelm Weierstrass.

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.

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+1 for Bolzano. We don't give that guy enough credit. –  Eric Stucky Dec 22 '13 at 23:03
@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. –  user72694 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. –  user72694 Jan 22 at 17:04
@user72694: See updated version. –  Yiorgos S. Smyrlis Jan 22 at 17:13
@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... –  user72694 Jan 22 at 17:29

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.)

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And the question arises as to who came up with these: most likely the point set topology Russian school in the 20s? –  WetSavannaAnimal aka Rod Vance 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 $p$) –  Hurkyl Feb 15 at 9:44