Can a mathematical definition be wrong? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T05:09:01Z http://mathoverflow.net/feeds/question/31358 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/31358/can-a-mathematical-definition-be-wrong Can a mathematical definition be wrong? Peter Shor 2010-07-11T04:18:27Z 2011-05-17T06:02:53Z <p>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. </p> <p>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 <em>uniform quantum Turing machine</em>) 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. </p> <p>My question is: are there other cases where a definition has been revised after it was realized that the first formulation was "wrong"? </p> http://mathoverflow.net/questions/31358/can-a-mathematical-definition-be-wrong/31359#31359 Answer by Harry Gindi for Can a mathematical definition be wrong? Harry Gindi 2010-07-11T04:39:37Z 2010-07-11T04:39:37Z <p>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 <em>model category</em> to mean what Quillen called a <em>closed model category</em> (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. </p> <p>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.</p> <p>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). </p> <p>For more, you can read the introductions to:</p> <p><em>Model Categories</em> by Mark Hovey</p> <p><em>Model Categories and their Localizations</em> by Phil Hirschhorn</p> <p><em>Homotopy Limit Functors on Model Categories and Homotopical Categories</em> by Dwyer, Hirschhorn, Kan, and Smith</p> http://mathoverflow.net/questions/31358/can-a-mathematical-definition-be-wrong/31360#31360 Answer by Pete L. Clark for Can a mathematical definition be wrong? Pete L. Clark 2010-07-11T04:56:17Z 2010-07-11T04:56:17Z <p>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.</p> <p>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.) </p> <p>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.</p> <p>I can think of one instance of this, although it is of relatively minor importance. R.G. Bartle's 1955 paper <em>Nets and Filters in Topology</em> 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 </p> <p><a href="http://math.uga.edu/~pete/convergence.pdf" rel="nofollow">http://math.uga.edu/~pete/convergence.pdf</a></p> <p>for some more discussion of this. </p> <p>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. </p> http://mathoverflow.net/questions/31358/can-a-mathematical-definition-be-wrong/31361#31361 Answer by Ryan Williams for Can a mathematical definition be wrong? Ryan Williams 2010-07-11T05:30:17Z 2010-07-11T05:54:51Z <p>Here's my favorite example. Imre Lakatos' book <em>Proofs and Refutations</em> contains a very long dialogue between a teacher and pupils who debate what are good <em>definitions</em> of polyhedra, with respect to a claimed <em>proof</em> 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.</p> <p>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.</p> http://mathoverflow.net/questions/31358/can-a-mathematical-definition-be-wrong/31381#31381 Answer by Charles Matthews for Can a mathematical definition be wrong? Charles Matthews 2010-07-11T09:09:38Z 2010-07-11T09:09:38Z <p>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.</p> http://mathoverflow.net/questions/31358/can-a-mathematical-definition-be-wrong/31431#31431 Answer by Bill Dubuque for Can a mathematical definition be wrong? Bill Dubuque 2010-07-11T17:06:00Z 2010-07-11T19:51:42Z <p>An important historical example is the difficult evolution of the correct definition of "integer" in algebraic extensions, i.e. defining <strong>algebraic integers</strong>. 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.</p> http://mathoverflow.net/questions/31358/can-a-mathematical-definition-be-wrong/31452#31452 Answer by Akhil Mathew for Can a mathematical definition be wrong? Akhil Mathew 2010-07-11T19:13:38Z 2010-07-11T19:13:38Z <p>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?):</p> <p>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.</p> <p>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).</p> http://mathoverflow.net/questions/31358/can-a-mathematical-definition-be-wrong/31457#31457 Answer by Bill Dubuque for Can a mathematical definition be wrong? Bill Dubuque 2010-07-11T19:47:05Z 2010-07-12T15:32:02Z <p>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 <strong>Dorroh extension</strong> [1] as a way to <strong>adjoin 1 to a rng</strong> (ring without unit). While various arguments can be made for the naturality of this construction, it turns out that this is the <strong>wrong definition</strong> 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 <em>highly</em> recommend first perusing [2] before commenting.</p> <p>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:</p> <p>"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."</p> <p>[1] <a href="http://en.wikipedia.org/wiki/Pseudo-ring" rel="nofollow">http://en.wikipedia.org/wiki/Pseudo-ring</a></p> <p>[2] W.D. Burgess; P.N. Stewart.<br> The characteristic ring and the "best" way to adjoin a one.<br> J. Austral. Math. Soc. 47 (1989) 483-496 <a href="http://anziamj.austms.org.au/JAMSA/V47/Part3/Burgess/p0483.html" rel="nofollow">http://anziamj.austms.org.au/JAMSA/V47/Part3/Burgess/p0483.html</a> </p> <p>[3] rings, ideals and correspondence theorem -- clarification requested. Ask an Algebraist, 8/4/2008<br> <a href="http://at.yorku.ca/cgi-bin/bbqa?forum=ask_an_algebraist_2008;task=show_msg;msg=2138.0002.0001.0001.0001.0002.0001" rel="nofollow">http://at.yorku.ca/cgi-bin/bbqa?forum=ask_an_algebraist_2008;task=show_msg;msg=2138.0002.0001.0001.0001.0002.0001</a></p> http://mathoverflow.net/questions/31358/can-a-mathematical-definition-be-wrong/31473#31473 Answer by Charles for Can a mathematical definition be wrong? Charles 2010-07-11T22:15:28Z 2010-07-11T22:15:28Z <p>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.</p> <p>[1] <a href="http://us.metamath.org/mpeuni/mmset.html#definitions" rel="nofollow">http://us.metamath.org/mpeuni/mmset.html#definitions</a></p> <p>[2] <a href="http://wiki.planetmath.org/AsteroidMeta/Ghilbert" rel="nofollow">http://wiki.planetmath.org/AsteroidMeta/Ghilbert</a></p> http://mathoverflow.net/questions/31358/can-a-mathematical-definition-be-wrong/31508#31508 Answer by gowers for Can a mathematical definition be wrong? gowers 2010-07-12T07:09:37Z 2010-07-12T22:58:10Z <p>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.)</p> <p>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.)</p> http://mathoverflow.net/questions/31358/can-a-mathematical-definition-be-wrong/34687#34687 Answer by Anweshi for Can a mathematical definition be wrong? Anweshi 2010-08-05T21:16:38Z 2010-08-05T21:16:38Z <p>The <a href="http://en.wikipedia.org/wiki/Euler_characteristic" rel="nofollow">Euler Characteristic</a> in the statement of <a href="http://en.wikipedia.org/wiki/Riemann-Roch" rel="nofollow">Riemann-Roch</a> was redefined by Grothendieck to be in the <a href="http://en.wikipedia.org/wiki/K-group" rel="nofollow">K-group</a>, which he presumably(ie, I imagine so) constructed for this purpose.</p> http://mathoverflow.net/questions/31358/can-a-mathematical-definition-be-wrong/34737#34737 Answer by Simon Pepin Lehalleur for Can a mathematical definition be wrong? Simon Pepin Lehalleur 2010-08-06T06:51:22Z 2010-08-06T06:51:22Z <p>An example from algebraic geometry :</p> <p>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.</p> <p>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"...</p> http://mathoverflow.net/questions/31358/can-a-mathematical-definition-be-wrong/34759#34759 Answer by AndrĂ© Henriques for Can a mathematical definition be wrong? AndrĂ© Henriques 2010-08-06T12:33:30Z 2010-08-06T12:33:30Z <p>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 <a href="http://lanl.arxiv.org/abs/math/9810059" rel="nofollow">http://lanl.arxiv.org/abs/math/9810059</a>.<br><br></p> <p>Among mathematicians (unlike physicists!), it is common practice to accompany one's claims by proofs.<br> But we are not used to accompanying our <em>definitions</em> by proofs.</p> <p>In higher category theory, things are different:<br> We should probably <strong>prove that our definitions are correct</strong> before being allowed to go on.</p> http://mathoverflow.net/questions/31358/can-a-mathematical-definition-be-wrong/34797#34797 Answer by Theo Johnson-Freyd for Can a mathematical definition be wrong? Theo Johnson-Freyd 2010-08-06T21:12:19Z 2010-08-06T21:12:19Z <p>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 <em>correct</em> if it both matches and helps refine intuition.</p> http://mathoverflow.net/questions/31358/can-a-mathematical-definition-be-wrong/34804#34804 Answer by Jeff Strom for Can a mathematical definition be wrong? Jeff Strom 2010-08-06T23:36:51Z 2010-08-06T23:36:51Z <p>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."</p> http://mathoverflow.net/questions/31358/can-a-mathematical-definition-be-wrong/47877#47877 Answer by Patrick Tam for Can a mathematical definition be wrong? Patrick Tam 2010-12-01T06:59:14Z 2010-12-02T21:20:07Z <p>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:</p> <p><STRIKE>f(x) = n if x = 1/n for some natural number n</p> <p>f(x) = 0 otherwise</p> <p>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.</STRIKE></p> <p>Edit: Okay, I don't remember the exact example he used.</p> <p>Edit 2: I asked my instructor about this and this was his response:</p> <p>"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."</p> http://mathoverflow.net/questions/31358/can-a-mathematical-definition-be-wrong/65204#65204 Answer by Darsh Ranjan for Can a mathematical definition be wrong? Darsh Ranjan 2011-05-17T06:02:53Z 2011-05-17T06:02:53Z <p>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: $$\begin{bmatrix}0 &amp; 0 &amp; 1 \\ 1 &amp; 1 &amp; 0\end{bmatrix}$$Do the columns form a basis for $\mathbb{R}^2$? The answer is "yes" if a basis is a <em>set</em> of vectors...but this is obviously false.</p> <p>The same applies to the usual definition of linear independence. Are the columns of the following matrix linearly independent? $$\begin{bmatrix}1 &amp; 1 \\ 1 &amp; 1\end{bmatrix}$$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. </p> <p>In my mind, the collections of vectors to which pretty much all linear algebra concepts apply are <em>tuples</em> 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. </p>