# Can a mathematical definition be wrong?

This question originates from a bit of history. In the first paper on quantum Turing machines, the authors left a key uniformity condition out of their definition. Three mathematicians subsequently published a paper proving that quantum Turing machines could compute uncomputable functions. In subsequent papers the definition of quantum Turing machine was revised to include the uniformity condition, fixing what was clearly a mathematical error the original authors made.

It seems to me that in the idealized prescription for doing mathematics, the original definition would have been fixed, and subsequent papers would have needed to use a different term (say uniform quantum Turing machine) for the class of objects under study. I can think of a number of cases where this has happened; even in cases where, in retrospect, the original definition should have been different.

My question is: are there other cases where a definition has been revised after it was realized that the first formulation was "wrong"?

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You only have to read e.g. James' History of Topology to learn of countless reformulations of the definition of manifold, fundamental group, homology... –  Qiaochu Yuan Jul 11 '10 at 4:53
I think whether or not 0 is a natural number and whether or not all "rings" have 1 are still not settled issues! –  Greg Friedman Jul 11 '10 at 6:14
@Greg: I will settle the latter controversy right now: Bourbaki says all rings have a multiplicative identity. This is a plain and simple argument from authority. –  Harry Gindi Jul 11 '10 at 6:59
The real question is why anyone thinks the phrase "natural numbers" is preferable to "nonnegative integers" or "positive integers." –  JBL Jul 11 '10 at 19:34
I want to praise the author of this question for the philosophical neutrality of the phrasing. –  Alexander Woo Jul 11 '10 at 23:45

Here's my favorite example. Imre Lakatos' book Proofs and Refutations contains a very long dialogue between a teacher and pupils who debate what are good definitions of polyhedra, with respect to a claimed proof that $V-E+F=2$ is true for polyhedra. It's common that a good definition (or reformulation) of a concept can help yield proofs of theorems, and this book promotes the "dual" view that a proof of a theorem can lead to a good definition in hindsight.

The footnotes of this dialogue show that Lakatos is actually tracing the history of the Euler characteristic in the mathematical literature. In short, both the definition(s) and the proof(s) went through substantial revisions over time.

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This issue is also masterfully discussed in Euler's Gem by Richeson: amazon.com/Eulers-Gem-Polyhedron-Formula-Topology/dp/0691126771 –  Qiaochu Yuan Jul 11 '10 at 5:45
The first appendix to Proofs and Refutations contains another example: definitions of continuity. –  Seamus Jul 15 '10 at 14:50
The problem is that Lakatos discusses a mathematical praxis before the "modern" revolution and the definition of polyhedron is not even stated. –  Leo Alonso Dec 1 '10 at 10:07

I was involved in such a case. My thesis advisor (S. Husseini) and his coauthor (E. Fadell) defined the "category weight" of a cohomology class in a way that allowed them to prove the theorem they wanted. But when I started looking at it, I noticed that it was not homotopy invariant, and all the applications were homotopy invariant. So I introduced the homotopy invariant version and called it "essential category weight." At roughly the same time Y. Rudyak made the same observation, and also defined a homotopy invariant version, calling it "strict category weight." After a few years of using competing terminology for the same concept, and not using the original "category weight" at all, everybody agreed to call the homotopy invariant version "category weight."

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Thanks. This is exactly the kind of case I was looking for when I asked the question. –  Peter Shor Mar 16 '13 at 15:51

This is not quite an answer to the question you are asking, but the definition of a function is an obvious example of a concept that underwent considerable change, though perhaps one might argue that the eighteenth-century notion of a function was never formally defined. Amusingly, there are many textbooks that attempt to give a formal definition of function but get it wrong. (This is not my observation but something spotted by a colleague of mine who studies mathematical language.)

What the books do is say, "A function from A to B is a subset $F$ of AxB such that for every x in A there is a unique y in B such that $(x,y)\in F$." But if that is all you say, then two functions can be equal even if they have different codomains, which the authors of these same textbooks clearly don't intend if they ever mention surjections, bijections or inverse functions. There's an easy fix, which is to define a function to be an ordered triple consisting of A, B and the subset of AxB, but almost no books do this. (I'm talking here about introductions to undergraduate-level mathematics rather than books about axiomatic set theory.)

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There was a discussion of this point here very recently: mathoverflow.net/questions/30381/definition-of-function –  JBL Jul 12 '10 at 12:19
In one undergraduate text on algebra the domain of a function is required to be nonempty. I think the author wrote this in order to be able to assert later on in the book that every one to one function has a left inverse. –  Tom Goodwillie Jul 12 '10 at 15:36
@gowers: Many set theorists actually define a function this way. –  Harry Gindi Jul 14 '10 at 6:22
It seems inaccurate to portray the logician's standard definition of function as an amusing failed attempt at precision. Although other definitions are also common, this particular definition is used with consistency and precision throughout logic and set theory and other areas, including introductory undergraduate texts. (There are no problematic issues with surjections, bijections or inverse functions.) The link in JBL's comment has further explanation. –  Joel David Hamkins Dec 2 '10 at 1:00
@Stefan, there is no function from a non-empty set to the empty set. (Not even "the empty function".) –  gowers Dec 2 '10 at 12:18

I think there are many examples, spread out across a continuum of how "wrong" the definition really was. Of course, strictly speaking a definition cannot be "wrong", or can only be wrong in the logical sense of not umambiguously denoting a class of examples. E.g., Newton's and Leibniz's definition of the derivative was wrong -- or better, not well-defined! -- because it used infinitesimals in a way that was not formalized and could not be formalized in the context of known mathematics.

There are a lot of definitions that in retrospect look too limited or pedestrian: e.g., defining a manifold to be a certain kind of subset of Euclidean space. (Some people would say that the definition of a Riemann integrable function is "wrong" in this sense. I disagree -- the notion of Riemann integrability is a natural one that comes up e.g. in characterizations of uniform distribution of sequences.)

It seems like you are looking for examples of the following kind: the definition is given and then, in the same paper (or book, or whatever) a theorem is given using the definition. But contemporary mathematicians who look back at the theorem agree that the conclusion is not the desired one.

I can think of one instance of this, although it is of relatively minor importance. R.G. Bartle's 1955 paper Nets and Filters in Topology was one of the first to try to explicitly work out the folkloric "equivalence" between nets and filters when studying convergence on topological spaces. The way to do this is to, given a net on a topological space, associate a filter, and conversely, and then prove theorems about these associated nets and filters having the same convergence properties. But the definition Bartle gives of how to associate a net to a filter is "wrong", in the sense that certainly you want that when you in turn associate a filter to that net you get the filter that you started with, but his definition does not have this property (and the right definition does!). See for instance the last page of

http://math.uga.edu/~pete/convergence.pdf

for some more discussion of this.

In general, I would think that one has to be rather well-read in a subject area to come up with such examples, because -- thankfully! -- a truly "wrong" definition is usually swiftly drowned out by the correct definition.

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Wow,Pete-that's one of my favorite papers and I never noticed that!!! But it's a very subtle point and something even an analyst of Bartle's stature-and he was a VERY good one,apparently-missed. –  The Mathemagician Jul 11 '10 at 6:47
Dear Pete L. Clark, there is a typo in the first reference in your above-mentioned paper. –  Rasmus Bentmann Jul 11 '10 at 13:18
@Rasmus: Thanks for pointing it out. I have reformatted the references. –  Pete L. Clark Jul 11 '10 at 17:25
"Of course, strictly speaking a definition cannot be "wrong", or can only be wrong in the logical sense of not umambiguously denoting a class of examples." I am not inclined to agree with this. One ought instead to say that the standard codified canons of logic that we are all taught do not tell us what right and wrong definitions are (beyond saying they must be unambiguous; they must not be circular, etc.). We should not rule out the possibility of future codification of such things based on understanding that has not yet been accomplished. –  Michael Hardy Aug 7 '10 at 22:36

In higher category theory, there have been examples of "wrong definitions". An example of such a definition is the definition of a strict 3-category. It is "wrong" because it doesn't include the fundamental 3-groupoid of the 2-sphere, see http://lanl.arxiv.org/abs/math/9810059.

Among mathematicians (unlike physicists!), it is common practice to accompany one's claims by proofs.
But we are not used to accompanying our definitions by proofs.

In higher category theory, things are different:
We should probably prove that our definitions are correct before being allowed to go on.

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You know,this is why I never feel comfortable with category theory no matter how hard I try to-insane things like THIS happen in it......... –  The Mathemagician Aug 6 '10 at 16:27
This is not insane. –  Kevin H. Lin Aug 7 '10 at 23:37
No, it's not insane. In fact I don't agree with Andre that this notion is "wrong" (although I understand why he says that) -- it's just a natural extension of defining a 2-category as a Cat-enriched category: a strict n-category is an (n-1)-Cat-enriched category. Strict n-categories are still a technically useful notion; for example, Batanin's notion of weak oo-category starts with the monad on globular sets whose algebras are strict oo-categories, and develops an associated notion of operad. Weak oo-categories are then algebras over certain contractible such operads. –  Todd Trimble Sep 19 '10 at 12:53
In this question, "definition" is really intended to mean something like "definition, together with some intuition about the thing thus defined". Neither can really be wrong on their own, but they can be mismatched. I guess Andre is claiming there's a standard mismatch here. –  James Cranch Oct 1 '14 at 18:47

An important historical example is the difficult evolution of the correct definition of "integer" in algebraic extensions, i.e. defining algebraic integers. It was only with great difficulty that Dedekind discovered the necessity of passing to integrally closed extensions in order to obtain nice factorization theories. Similar struggles were encountered while distilling the correct notion of integral elements for quaternion rings.

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Here's a rather mundane example: a basis of a vector space. A basis is usually defined to be "a set of vectors such that...." The problem with this is the following: $$\begin{bmatrix}0 & 0 & 1 \\\\ 1 & 1 & 0\end{bmatrix}$$Do the columns form a basis for $\mathbb{R}^2$? The answer is "yes" if a basis is a set of vectors...but this is obviously false.

The same applies to the usual definition of linear independence. Are the columns of the following matrix linearly independent? $$\begin{bmatrix}1 & 1 \\\\ 1 & 1\end{bmatrix}$$The answer is apparently "yes," since in my experience linear [in]dependence is usually defined only for a set of vectors, and the set of the columns of that matrix consists of a single nonzero vector...again, obviously false: the columns of a square matrix shouldn't be linearly independent unless the matrix is invertible.

In my mind, the collections of vectors to which pretty much all linear algebra concepts apply are tuples of vectors, not sets. This stems from the fact that the fundamental operation on a vector space is the linear combination, which operates on a tuple of vectors and a tuple of scalars (indexed by the same set). If I want to figure out whether a collection of vectors spans a space or is linearly independent, the next thing I'm going to do is consider a linear combination of those vectors. Therefore, it only makes sense for that collection to be something to which "linear combination" readily applies, namely a tuple! Still, I've seen reputable textbooks define everything only in terms of sets.

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I agree that tuples are better than sets in this context, but I think indexed families are better yet. If the indexing set happens to be the integers from 1 to $n$, then you have tuples. But if you're dealing with (the underlying vector space of) the group algebra of a finite group $G$, then an obvious basis is indexed by $G$, and it seems unnecessary and unnatural to impose an (arbitrary) ordering on $G$ just to get tuples. (For similar reasons, I like to think of matrices as having their rows and columns labeled by some indices but not necessarily by natural numbers.) –  Andreas Blass May 17 '11 at 13:56
@Andreas - thanks for clarifying. That is in fact exactly what I meant! I understand "tuple" in the generalized sense, as in an "X-tuple" is nothing but a function with domain X. X doesn't need to be {1,...,n} or {0,...,n-1}; it could be any index set. (I'm not making this up: tuples are treated this way in Bergman's An Invitation to General Algebra and Universal Constructions, and the Wikipedia article mentions this perspective as well.) –  Darsh Ranjan May 17 '11 at 17:31

This happened with model categories, where Quillen's original definition only required the existence of finite limits and colimits as well as a factorization of all maps that wasn't necessarily functorial. Almost all books today take model category to mean what Quillen called a closed model category (and in fact, they go one step further. They add functorial factorization as well), since it makes the proofs easier and the conclusions much more far-reaching.

This is at the cost of losing some categories of some kinds of finite chain complexes (I have never run into one of these in practice, but I suppose that some people do) as model categories, but this stronger definition includes almost every usage of model-category theory in homotopy theory.

I will note, however, that while model categories are almost always taken to be closed (outside of Quillen's original paper), functorial factorization is not nearly as standard (if I remember correctly, Jacob Lurie doesn't require functorial factorization in his definition in Higher Topos Theory).

For more, you can read the introductions to:

Model Categories by Mark Hovey

Model Categories and their Localizations by Phil Hirschhorn

Homotopy Limit Functors on Model Categories and Homotopical Categories by Dwyer, Hirschhorn, Kan, and Smith

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@Andrew L : I think you miss the point here. The opposite of what you described is true : Quillen's original definition was actually too general! –  Andy Putman Jul 11 '10 at 4:57
Andrew L, Quillen has even adopted the newer definition (he was one of the first to do so, in fact). (cf. Rational Homotopy Theory (1969) by Dan Quillen). –  Harry Gindi Jul 11 '10 at 5:00
As I say in my response below, whether and to what extent a definition is "wrong" is certainly up for debate. But I think this definition is not "wrong" in the sense of Shor's example or mine -- in Harry's example it's just that a different definition (which the author also gave!) turned out to be more useful. For Shor's example and mine, it seems that no one should want to use the "wrong" definition for anything: it doesn't have the intended consequences. –  Pete L. Clark Jul 11 '10 at 5:06
An example in the opposite direction is the change in terminology from "pre-scheme" to "scheme" in the second edition of EGA. –  Victor Protsak Jul 11 '10 at 7:09
For the record, I am interested in changes of definition even in the case where in cannot be argued that the original definition was "wrong" (whatever that means for a definition). –  Peter Shor Jul 11 '10 at 23:58

If a definition can be tentative, it can also be wrong. Lakatos has been mentioned already. This is actually a fairly basic issue in understanding how "formal" mathematics advances. Something as fundamental as integration has seen inadequate definitions such as the Riemann integral pushed out by the Lebesgue integral. In this case the "wrong" definition has not simply been supplanted, though.

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Charles,this is a bad example since the Riemann integral is perfectly fine if the functions are defined on compact support and the number of discontinuities is at most countable. It wasn't that the Riemann integral was "wrong",it was simply discovered to be too restrictive for the purposes of modern analysis. –  The Mathemagician Jul 11 '10 at 20:06
No, it's not perfectly fine if you want a theory of Fourier transforms. –  Charles Matthews Jul 11 '10 at 20:44
@Charles If the same restrictions are imposed,I believe the theory of Fourier transforms can be established just fine with the Riemann integral. But of course,this is too restrictive for a general theory,which is why the Lebesgue integral supplanted it. –  The Mathemagician Jul 11 '10 at 23:55
In Rudin's Real and Complex Analysis he introduces the Lebesgue measure as the measure (say on the closed unit interval) such that integration with respect to this measure gives you the positive linear functional on C([0,1]) that is called "Riemann integral". So he actually defines the Lebesgue measure in terms of the Riemann integral. I agree that this approach is sort of odd, but very elegant from a certain point of view. My students didn't like this approach, though. I think they would have preferred one of the more common approaches to the Lebesgue integral. –  Stefan Geschke Dec 2 '10 at 9:10

An example from algebraic geometry :

At some point during the redaction of the EGA by Grothendieck and Dieudonne, Grothendieck discovered how to make parts of the theory work without finiteness (noetherian) hypotheses on schemes, by strenghtening finiteness for morphisms (finite presentation instead of finite type). The study of morphisms of finite presentation was carried out in EGAIV.

Unfortunately, some definitions of properties of morphisms were made before this discovery. In particular, the definition of a proper morphism in EGA only includes finite type and not finite presentation, mostly because it was first used for noetherian schemes. This can lead (and lead some fine french mathematicians) to spend entire classes repeating the words "morphismes propres de présentation finie"...

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Dear Simon, It's not actually clear that the definition of proper is "wrong" because it has finite type rather than finite presentation. Somewhere in the literature surrounding the stacks project there is a discussion stating that finite type is the natural condition for certain properties, including quasi-finiteness, proper, and unramified (which should require only locally finite type, not locally finite presentation). One relevant remark is that locally of finite type implies that the diagonal is actually locally of finite presentation. I would like to say more to justify this, but ... –  Emerton Jun 4 '13 at 23:02
... unfortunately can't, since I haven't completely thought through all the issues here. Anyway, I thought you might be interested to hear about a different view-point. Regards, –  Emerton Jun 4 '13 at 23:02
Thank you for this addendum, I will have a look if I can track down the discussion in the stacks project and put the reference here. –  Simon Pepin Lehalleur Jun 5 '13 at 9:14

As someone who regularly posts questions on MO asking things like "what is the correct definition for XYZ", I strongly believe that the answer to the question is "yes". For me, a definition is correct if it both matches and helps refine intuition.

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The word "chaos" entered the mathematical literature in the paper by Li and Yorke, Period three implies chaos, Amer Math Monthly 82 (1975) 985-992, MR0385028 (52 #5898). Li and Yorke proved that a continuous function on an interval with a point of period 3 has points of period $k$ for all positive integers $k$.

A lively discussion of chaos ensued, with various definitions being proposed. A definition due to Devaney (see his text, An Introduction to Chaotic Dynamical Systems, Benjamin/Cummings 1986, MR0811850 (87e:58142)) seems to be the one most commonly accepted today. By the Devaney definition, period three does not imply chaos; for example, a function with a stable 3-cycle is not Devaney-chaotic.

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I'm not terribly familiar with this material, but since there's a good chance nobody will say something about it, I'll chip in with an example (I think?):

The Hardy spaces $H^p$ were originally defined in terms of complex functions on the unit disk. Namely, an element of $H^p$ is a holomorphic function $f$ on the unit disk such that $\sup_{r} \int_{0}^{2 \pi} |f(re^{i \theta})|^p d \theta$ is finite. This quantity (in analog with the $L^p$ norms) is used to define the norm on $H^p$, so this is a Banach space for $p \geq 1$ (and a Hilbert space for $p=2$). There is a very rich and interesting theory of these complex Hardy spaces. For instance, radial limits exist almost everywhere (though this is also true for the broader case of $f$ in the Nevanlinna class), and have vanishing Fourier coefficients at negative indices. The function $f$ can be reconstructed from the boundary data via a Poisson integral. More interestingly, the corona theorem is a statement about the spectrum of the Banach algebra $H^\infty$; it states that the ideals $M_z := \{f: f(z)=0\}$ for $z$ in the unit disk are dense. All this is based upon the complex-variable theory, which came first.

However, this original definition via complex-variable theory was "wrong" in the sense in that it had to be modified to allow for the real-variable theory in higher dimensions. The "real-variable" definition of an element of $H^p$ is defined in terms of distributions with "maximal functions" (defined with respect to a normalized Schwarz function) in $L^p$. Much of the modern theory of Hardy spaces (e.g. duality of $H^1$ and $BMO$, stability under singular integrals) was developed, I think, in this more general setting (which, according to Stein, took hold in the 1960s).

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This is a slightly different kind of example, namely, one where the original definition had to be revised when it later was realized that it was useless in certain contexts. Probably most readers have encountered the Dorroh extension [1] as a way to adjoin 1 to a rng (ring without unit). While various arguments can be made for the naturality of this construction, it turns out that this is the wrong definition in many contexts because it doesn't preserve crucial properties of the source rng and/or doesn't satisfy various minimality properties. For much further discussion see [2]. I mention it primarily as another perspective on the way that definitions may evolve. A word of warning: I've mentioned this many times over the years and almost always someone argues tooth-and-nail for the naturality of the Dorroh unital extension without first appreciating the issues that arise in contexts outside their expertise, e.g. see the AAA thread [3]. To avoid that here I highly recommend first perusing [2] before commenting.

Added later: remarks in the comments below lead me to believe that posting the the following excerpt from the introduction of [2] may help serve to alleviate any further confusion:

"It was observed long ago as 1932 (Dorroh's Theorem) that any non-unital ring $R$ may be embedded in a ring with unity. This is done by adjoining a copy of $\mathbb{Z}$, the ring of integers, to $R$. This does not preserve all the nice properties which $R$ might have, nor is it minimal in any of various senses; and so over the decades many embeddings have been invented to serve diverse purposes. For example, if $R$ is regular (or some generalization of regular such as $\pi$-regular) one would like to embed $R$ into a regular ring (or the generalization). There are other sorts of properties (semiprime, artinian domain, Ore domain) which one may wish to preserve in going from $R$ to some ring with 1, say $R^1$, all the while without adjoining anything more than necessary. It turns out that there is one construction [...] which will give all the main results as well as some new ones, although there is not yet one proof by which to do it. In the case of the generalized sorts of regularity, the ring formed [...] satisfies a universal property with respect to the adjunction of 1."

[2] W.D. Burgess; P.N. Stewart.
The characteristic ring and the "best" way to adjoin a one.
J. Austral. Math. Soc. 47 (1989) 483-496 http://anziamj.austms.org.au/JAMSA/V47/Part3/Burgess/p0483.html

[3] rings, ideals and correspondence theorem -- clarification requested. Ask an Algebraist, 8/4/2008

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I would like to request that Mr. Dubuque remove reference [3] from his response. It's not an example of mathematics but rather of several people squabbling in a rather unprofessional way (ostensibly) about mathematics. In the interest of full disclosure, let me add that one of those people is me, and this thread contains the things that I most regret having posted to the internet, ever. –  Pete L. Clark Jul 11 '10 at 20:00
I vote out of respect for Pete's wishes,Mr.Dubuque comply.That being said-and with the disclaimer I haven't carefully read it-that Pete shouldn't be afraid to act human in front of his collegues and they should be understanding if such incidents occur when passions flare. Scientists are human and there should be no shame in that. –  The Mathemagician Jul 11 '10 at 20:03
PS I just noticed that PLC revealed his anonymous participaton in the linked thread. Frankly I had forgotten about that since I have seen so many such discussions before that I didn't recall the particpants. I suggest instead of censoring my link that PLC simply remove his comment - thus alleviating any potential embarassment on his part. –  Bill Dubuque Jul 11 '10 at 21:27
I didn't downvote this because of what Pete said. Rather, I downvoted because I don't think this answers the question. Other mathematicians have not adopted the alternate definition despite the fact that it has been available for many years. Given that the notion of a "correct definition" is in many ways a question of aesthetics, the long-standing community consensus seems to trump your private campaign. –  Andy Putman Jul 11 '10 at 22:57
At the risk of commenting on an old discussion: this being a CW question, all answers are also CW, and so anyone, e.g. Pete, is doing nothing wrong by removing the reference themself. –  Theo Johnson-Freyd Aug 6 '10 at 21:10

Yes, in the sense that a careless definition can actually introduce unsoundness into a formal framework. See Norm Megill's explanation [1] as well as Raph Levien's Ghilbert [2] which is (apparently) a solution to the problem.

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The second link seems to be gone. Is ghilbert-app.appspot.com a good replacemnt? –  Gerry Myerson Oct 2 '14 at 1:06
@GerryMyerson: Yes, thanks. –  Charles Oct 2 '14 at 13:09

The Euler Characteristic in the statement of Riemann-Roch was redefined by Grothendieck to be in the K-group, which he presumably(ie, I imagine so) constructed for this purpose.

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Here is a bad definition of integrability that my analysis instructor taught me about and explained why it was bad. Consider a real-valued function f with the interval [a, b] as its domain. Partition [a, b] evenly into n intervals of length (b-a)/n. If the upper Darboux sum and lower Darboux sum of f in these partitions converge and equal each other as n approaches infinity, then call f integrable on [a,b]. To see why this definition fails, consider the following function:

f(x) = n if x = 1/n for some natural number n

f(x) = 0 otherwise

On [0,1], the upper Darboux sums of f on partitions from this definition must be greater than or equal to 1 and the lower Darboux sums converge to 0. So f is not integrable with respect to this definition. But f is Riemann integrable.

Edit: Okay, I don't remember the exact example he used.

"Actually, I'm pretty sure the condition you wrote down is equivalent to Darboux integrability. The "bad" definition we spoke about is an exercise in Pugh's book: looking at convergence of Riemann sums with equally-spaced subintervals as you do, but only allowing the midpoints of these subintervals as sample points. Then according to this definition, the characteristic function of the irrationals on [0,1] is integrable (which we know is not, at least if you're talking Riemann/Darboux) since all sample points would be rational so all Riemann sums would be 0, and hence they converge. This shows why we need to consider arbitrary sample points in the definition of Riemann integrability."

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Riemann-integrable functions should be bounded by the usual definition, so f is not Riemann-integrable. But nonetheless the limit $R-\int_a^1 f(x)\,dx$ for $a\to 0$ exists. –  Johannes Hahn Dec 1 '10 at 16:12
Agreed, this function $f$ is not Riemann integrable according to the usual definition. But is the above is a bad definition? Can we "simplify" the usual definition (allowing arbitrary finite partitions) by requiring only "equally spaced" partitions? Isn't this an exercise in "advanced" presentations such as Rudin? –  Gerald Edgar Dec 1 '10 at 16:31
Ah, I still think there is a Riemann integrable function that messes up this definition. I just can't think of it. –  Patrick Tam Dec 2 '10 at 8:21

There's an easy way to make the mistake. To define something, it must exist and be unique. Therefore "define the square root of a real number to be the real number so that when you square it you get the original number." This is wrong because "the" is broken in two ways: it implicitly asserts something exists (it doesn't for negative numbers), and that it is unique (it doesn't for positive numbers).

See also assuming unique antiderivatives, assuming unique coset representations of groups, etc. In general confusing a representation with a definition does the trick. See whether .999... equals 1. The anti-intuition comes from the fact that most people associate numbers with their decimal representation very strongly (and, from a mathematical theory standpoint, incorrectly) so that if two numbers "look" different in their representation on paper, they must also be different.

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