Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

The first thing to say is that this is not the same as the question about interesting mathematical mistakes. I am interested about the type of false beliefs that many intelligent people have while they are learning mathematics, but quickly abandon when their mistake is pointed out -- and also in why they have these beliefs. So in a sense I am interested in commonplace mathematical mistakes.

Let me give a couple of examples to show the kind of thing I mean. When teaching complex analysis, I often come across people who do not realize that they have four incompatible beliefs in their heads simultaneously. These are

(i) a bounded entire function is constant;
(ii) sin(z) is a bounded function;
(iii) sin(z) is defined and analytic everywhere on C;
(iv) sin(z) is not a constant function.

Obviously, it is (ii) that is false. I think probably many people visualize the extension of sin(z) to the complex plane as a doubly periodic function, until someone points out that that is complete nonsense.

A second example is the statement that an open dense subset U of R must be the whole of R. The "proof" of this statement is that every point x is arbitrarily close to a point u in U, so when you put a small neighbourhood about u it must contain x.

Since I'm asking for a good list of examples, and since it's more like a psychological question than a mathematical one, I think I'd better make it community wiki. The properties I'd most like from examples are that they are from reasonably advanced mathematics (so I'm less interested in very elementary false statements like $(x+y)^2=x^2+y^2$, even if they are widely believed) and that the reasons they are found plausible are quite varied.

share|improve this question
66  
I have to say this is proving to be one of the more useful CW big-list questions on the site... –  Qiaochu Yuan May 6 '10 at 0:55
18  
The answers below are truly informative. Big thanks for your question. I have always loved your post here in MO and wordpress. –  Unknown May 22 '10 at 9:04
15  
wouldn't it be great to compile all the nice examples (and some of the most relevant discussion / comments) presented below into a little writeup? that would make for a highly educative and entertaining read. –  Suvrit Sep 20 '10 at 12:39
16  
It's a thought -- I might consider it. –  gowers Oct 4 '10 at 20:13
22  
Meta created tea.mathoverflow.net/discussion/1165/… –  quid Oct 8 '11 at 14:27

186 Answers 186

Everyone knows that for any two square matrices $A$ and $B$ (with coefficients in a commutative ring) that $$\operatorname{tr}(AB) = \operatorname{tr}(BA).$$

I once thought that this implied (via induction) that the trace of a product of any finite number of matrices was independent of the order they are multiplied.

share|improve this answer
35  
In fact Tr$(AB)=$Tr$(BA)$ holds also for non-square matrices $A,B$ for which both $AB$ and $BA$ are defined. Now for determinants, det$(AB)$=det$(BA)$ holds for square matrices, but of course not for non-square matrices (consider the case where $A$ is a column vector and $B$ a row vector). –  user2734 May 4 '10 at 21:46
14  
@Nate: If you want high-powered generalities, the most general situation I know where one can prove this statement is in a ribbon category. These have a graphical calculus where tr(ABC...) corresponds to a closed loop on which A, B, C... sit as labels in order, which clearly shows that the only invariance one should expect is under cyclic permutation. See, for example, the beginning of Turaev's "Quantum Invariants of Knots and 3-Manifolds." –  Qiaochu Yuan May 4 '10 at 22:32
5  
@Harry, if you think about what happens when you split a product $abcdefgh$ in the middle and interchange the two halfs, you'll see where Nate is going... –  Mariano Suárez-Alvarez May 5 '10 at 3:30
16  
@unknown: nonetheless, the characteristic polynomials of AB and BA are the same up to a power of $\lambda$ (A is m by n and B is n by m), which generalizes both properties –  Victor Protsak May 5 '10 at 6:54
6  
@Victor Protsak: Nice! BTW, one way to get what you say is from det$(I_m+AB)=$det$(I_n+BA)$, which funnily doesn't hold for the trace in case of non-square matrices (there is a difference of $m-n$). –  user2734 May 5 '10 at 7:44

"It is impossible in principle to well-order the reals in a definable manner."

To be more precise, the belief I am talking about is the belief that well-orderings of the reals are provably chaotic in some sense and certainly not definable. For example, the belief would be that we can prove in ZFC that no well-ordering of the reals arises in the projective hierarchy (that is, definable in the real field, using a definition quantifying over reals and integers).

This belief is relatively common, but false, if the axioms of set theory are themselves consistent, since Goedel proved that in the constructible universe $L$, there is a definable well-ordering of the reals having complexity $\Delta^1_2$, which means it can be obtained from a Borel subset of $R^3$ by a few projections and complements. See this answer for a sketch of the definition of the well-order.

The idea nevertheless has a truth at its core, which is that although it is consistent that there is a definable well-ordering of the reals (or the universe), it is also consistent that there is no such definable well-ordering. Thus, there is no definable relation that we can prove is a well-ordering of the reals (although we also cannot prove that none is).

share|improve this answer
16  
I think that you are taking an imprecise statement and making it precise in such a way that it is wrong. –  Marcos Cossarini May 5 '10 at 1:00
3  
Marcos, the false belief I was aiming at was the belief that it is impossible in principle for a well ordering of the reals to be definable. That is indeed a false belief, since it IS possible in principle for there to be such a definable well-ordering, as there is one in Goedel's universe L. (Note: the constructible universe L has nothing at all to do with constructivism or constructive proofs, in the sense of your comment; Goedel used only classical logic.) And my remark was not aimed particularly at undergrads or even grads, but rather at research mathematicians holding that false view. –  Joel David Hamkins May 5 '10 at 1:29
4  
The best part of this answer is how the constructible universe subverts all the intuition you learn about AC from doing non-model theory. Experience would lead one to think that "AC = nonconstructive" in "the usual model of the real numbers", not realizing that there is no usual model. Your (my, everyone's) mental image of the reals is a sort of "lazy evaluation" (to use a programming term) of the model we would really like but haven't even specified fully. As you show in your answer, once given the facts we wouldn't even know which model that would be. –  Ryan Reich Oct 20 '10 at 11:21

It's easy when you're an amateur to topology to assume any continuous bijection has a continuous inverse.The inverse of an arbitrary continuous bijection in a topological space is open,but it's NOT necessarily continuous. Continuity turns out to be a stronger condition.

share|improve this answer
23  
Comments like this lead me to point the finger at some people's undergraduate education ;-) It was very common in my UG years to have "definition/basic properties/interesting counterexamples to plausible-sounding statements/example sheet with harder examples". For example, although it's not continuity but differentiability, I vividly remember being shown the map x-->x^3 on the reals very shortly after being told the definition of a diffeomorphism. –  Kevin Buzzard May 4 '10 at 21:33
21  
Kevin's right. It's just as important to ask students for counterexamples as it is to tell them theorems. But I think this is hard for undergraduates in a way we tend to forget because we have slowly but surely rewired our brains to think in this way. When I was 20 I took a reading course from R. Narasimhan, and towards the end he told me to "Ask yourself the natural questions". Now, [harrumph] years later, this is second nature to me, but I remember at the time thinking that it was not so easy! –  Pete L. Clark May 4 '10 at 22:28
10  
I think one source of this problem are definitions (in the first lectures) like: a bijective morphism of groups is called an isomorphism. Introducing categories (very roughly!), defining the general notion of an isomorphism in a category and mentioning that it's awesome that for groups we just have to check bijectivity could really prevent this... –  user717 May 5 '10 at 10:53
18  
There is at least two important cases where this intuition does hold, though: (a) when the continuous bijection is a linear transformation between Banach spaces (the open mapping theorem), and (b) when the bijection is from a compact space to a Hausdorff space. Of course, one only really appreciates these facts when one knows that the claim is false in general... –  Terry Tao Jun 5 '10 at 21:13
7  
...Some mathematicians have perfected this technique -- or possibly its converse? -- to the point where they become "venus fly traps": if an open problem flies too close to the domain of things that the mathematician understands ridiculously well, then...SNAP! That's the end of the problem. I have long admired Yuri Zarhin in this way, for instance. –  Pete L. Clark Jul 14 '10 at 20:03

I don't know if this is common or not, but I spent a very long time believing that a group $G$ with a normal subgroup $N$ is always a semidirect product of $N$ and $G/N$. I don't think I was ever shown an example in a class where this isn't true.

share|improve this answer
2  
Argh! Me too! What is a good example? –  Nate Eldredge May 4 '10 at 21:28
95  
umm Z/4Z contains Z/2Z? –  Kevin Buzzard May 4 '10 at 21:30
42  
This suggests that we do a terrible job of talking about semi-direct products no? –  Kevin McGerty May 5 '10 at 0:27
31  
Schur--Zassenhaus says that this is true if $N$ and $G/N$ have coprime orders, so there is some intrinsic pressure in the subject towards this. Coupled with the fact that it is true for the first non-trivial non-abelian example ($A_3$ inside $S_3$), it's easy to see how this misconception arises. –  Emerton May 5 '10 at 1:49
12  
Remember being confused by this too. It became much clearer when I formally was taught about short exact sequences. Then you can see exactly the obstruction to such a decomposition. –  Fabrizio Polo May 10 '10 at 11:14

In measure-theoretic probability, I think there is sometimes an idea among beginners that independent random variables $X,Y$ should be thought of as having "disjoint support" as measurable functions on the underlying probability space $\Omega$. Of course this is the opposite of the truth.

I think this may come from thinking of measure theory as generalizing freshman calculus, so that one's favorite measure space is something like $[0,1]$ with Lebesgue measure. This is technically a probability space, but a really inconvenient one for actually doing probability (where you want to have lots of random variables with some amount of independence).

share|improve this answer
1  
+1 nice example! –  Gil Kalai May 5 '10 at 11:51
1  
A student this last semester made precisely this mistake, and it was a labor of three people to convince him otherwise. –  Andres Caicedo May 17 '10 at 0:28
1  
This disjoint support misconception reinforces the incorrect idea that pairwise independent implies independent. –  Douglas Zare Oct 20 '10 at 18:47

I think, there are different types of false beliefs. The first kind are statements which are quite natural to believe, but a moment of thought shows the contradiction. Of this type is the sin-example in the opening post or a favorite of mine (also occured to me):

  • The underlying additive group of the field with $p^n$ elements is $\mathbb{Z}/p^n\mathbb{Z}$.

The other type is also quite natural to believe, but one has really to think to construct a counter example:

  • Every contractible manifold is homeomorphic to $\mathbb{R}^n$.
  • Every manifold is homotopy equivalent to a compact one.
  • Quotients commute with products in topological spaces.
  • Every connected component of a topological space is open and closed. Or related to this:
  • To give a continuous action of a topological group $G$ on a discrete space $X$ is the same as to give an action of the group of connected components of $G$ on $X$.
share|improve this answer
27  
Along the same vein as the first example: the field with $p^2$ elements is a subfield of the field with $p^3$ elements (etc...) –  Sean Kelly May 5 '10 at 17:06
2  
Profinite groups are not the topological space, most topologists are most familiar with. –  Lennart Meier Dec 3 '10 at 14:42
1  
2  
@Lennart: True, but topologists know the rationals. They are (usually) not the ones with this mistaken belief. –  Gerald Edgar May 4 '11 at 13:14

$0^0$ is undefined.

EDIT: People write things like $\sum_{k=0}^\infty x^k$ all day, but somehow $x=k=0$ is still scary when written as $0^0$.

share|improve this answer
6  
Are you saying that's a false belief?? –  Kevin Buzzard May 4 '10 at 22:11
10  
This isn't a false belief so much as a matter of convention. I think the OP is after statements which are believable and provably false. –  Qiaochu Yuan May 4 '10 at 22:17
32  
0^0 = 1, because there is one function from the empty set to the empty set –  Steven Gubkin May 4 '10 at 23:04
21  
Some would say ... $0^0=1$ where the exponent is the integer $0$, but $0^0$ is undefined where the exponent is the real number $0$. Does this fit all of the comments so far? –  Gerald Edgar May 5 '10 at 0:15
16  
@Dmitri Pavlov: I disagree. A polynomial is a purely algebraic object. The notation $t^0$ is just shorthand for a certain polynomial, otherwise called $1$. (If you identify $R[t]$ with the set of finitely nonzero functions from $\mathbb{N}$ to $R$, it is the function which maps $0$ to $1 \in R$ and everything else to zero.) Evaluation of a polynomial at an element of $R$ is a certain ring homomorphism $R[t] \rightarrow R$ which in particular sends the polynomial $1$ to the element $1$. We never "evaluate" $0^0$ in any sense, because indeed it has no independent algebraic definition. –  Pete L. Clark May 5 '10 at 6:31

"Any subspace of a separable topological space is separable, too." Sounds natural.

share|improve this answer
10  
(and it is true of metric spaces, and natural generalizations...) –  Mariano Suárez-Alvarez May 4 '10 at 23:31
11  
This seems to be the fault of the "divorce" of second countability and separability in general topological spaces: they coincide for metrizable spaces, but for general spaces the favorable properties were split up in a custody hearing: see en.wikipedia.org/wiki/…. (I think I wrote this part of the article.) –  Pete L. Clark May 5 '10 at 2:14
11  
I think I kind of believed that until today. –  Olivier May 5 '10 at 8:17
25  
By an amusing coincidence, I came across this for the first time a couple of days ago. There was a Cambridge exam question in 2008 where you had to show that products and subspaces of separable metric spaces were separable, and then you were given a topological space and asked to show that its square was separable, and that a certain subspace of it was not separable. I had to stare at it for about a minute before I understood why I had not just proved a contradiction. –  gowers May 5 '10 at 9:18
19  
Perhaps some discontinuities between metric space theory and topology arise because when studying metric spaces, distances are mysteriously required to be symmetric, and the requirement is dropped when switching to topological spaces. So you can have a point x that has y as a limit (i.e. it is in all neighbourhoods of y) but y doesn't have x as a limit. With this idea in mind, you can make any topological space separable by adding a single point, and making it belong to all open sets. –  Marcos Cossarini May 5 '10 at 14:53

"If any two of the $3$ random variables $X,Y,Z$ are independent, all three are mutually independent." In fact, they may be dependent; the simplest example is probably $(X, Y, Z)$ chosen uniformly from $\{(0, 0, 0), (1, 1, 0), (1, 0, 1), (0, 1, 1)\}$.

share|improve this answer
35  
or in words: take two fair $0-1$ coins and the random variables ""value first coin", "value second coin" and "sum modulo 2". –  Alekk May 5 '10 at 8:25
11  
A physics professor assigned proving this false statement as homework in a statistical mechanics course. –  Douglas Zare May 5 '10 at 17:43
1  
A related thread: mathoverflow.net/questions/7998/… –  Yoo Jun 15 '10 at 18:21

"A continuous image of a locally compact space is locally compact."

This is tempting because it is true without the "locally"s and it is often the case that topological properties and statements can be "localized". This came up in a problem session for my [number theory!] course this semester, and although the students were too experienced to accept it without proof, they had no alarm bells in their heads to prevent them from entertaining the possibility.

The way to quash this (as well as Andrew L.'s answer, which reminded me of this) is to realize that if it were true, every space $X$ would be locally compact, since the identity map from $X$ endowed with the discrete topology to $X$ is a continuous bijection.

share|improve this answer
5  
@Pete: related story. Is it true that a continuous image of a compact rigid space is compact? Recall that rigid spaces (in Tate's sense) only have a Grothendieck topology on them, not a topology, so "compact" means that every admissible cover has an admissible finite subcover. The problem with the usual proof is that you admissibly cover the target, pull it back to the source, write down a finite admissible subcover, push it forward, and it might not be admissible! Ofer Gabber once presented me with a counterexample to the assertion written on the back of a napkin :-) –  Kevin Buzzard May 4 '10 at 22:13
3  
"Recall" -- what, from your course in 2002(ish)? No problem, this is one of the few things I remember lo these many years later. I particularly recall your intuition for Tate's Grothendieck topology as requiring the open sets to be "big and chunky". –  Pete L. Clark May 4 '10 at 22:20
2  
That there are two different notions that are called "locally compact" does not help. –  Alfonso Gracia-Saz May 7 '10 at 21:21

When I was studying Banach spaces, I was confused with the following: We know that, in any Banach Space $V$, the closed unit ball is compact in the topology generated by the norm if, and only if, the dimension of $V$ is finite. But thinking about $\mathbb R$ as a vector space over $\mathbb Q$, we have an infinite-dimensional vector space which is complete in the norm (given by the modulus) but the closed unit ball is, of course, compact in topology generated by the norm.

I took some time to discover that my mistake was that I thought about $\mathbb R$ over $\mathbb Q$ as a Banach space. In fact, this vector space is a complete metric space (in the sense of Cauchy sequences), but I realized later that the word Banach space is reserved only for vector spaces defined over the fields $\mathbb R$ or $\mathbb C$.

share|improve this answer
10  
You can define Banach spaces over any complete field. For example, one can define p-adic Banach spaces. But Q isn't complete with respect to any of its norms. –  Qiaochu Yuan May 4 '10 at 22:14
2  
In fact the Theorem I mentioned in my answer, is based on the Riesz lemma and this lemma is not valid if the scalar fields is not complete. –  Leandro May 4 '10 at 22:34

In geometric combinatorics, there is a widespread belief that polytopes of equal volume are not scissor congruent (as in Hilbert's third problem) only because their dihedral angles are incomparable. The standard example is a cube and a regular tetrahedron, where dihedral angles are in $\Bbb Q\cdot \pi$ for the cube, and $\notin \Bbb Q\cdot \pi\ $ for the regular tetrahedron. In fact, things are rather more complicated, and having similar dihedral angles doesn't always help. For example, the regular tetrahedron is never scissor congruent to a union of several smaller regular tetrahedra (even though the dihedral angles are obviously identical). This is a very special case of a general result due to Sydler (1944).

share|improve this answer

If a topological space has an open cover by Hausdorff spaces, it is Hausdorff.

share|improve this answer

For vector spaces, $\dim (U + V) = \dim U + \dim V - \dim (U \cap V)$, so $$ \dim(U +V + W) = \dim U + \dim V + \dim W - \dim (U \cap V) - \dim (U \cap W) - \dim (V \cap W) + \dim(U \cap V \cap W), $$ right?

share|improve this answer
63  
Wait, that isn't true? –  Simon Rose May 4 '10 at 23:19
92  
Take three distinct lines in R^2 as U, V, W. All intersections have 0 dimensions. The LHS is 2, the RHS is 3. The problem is that $(U+V)\cap W \neq U\cap W + V\cap W$. –  Willie Wong May 4 '10 at 23:38
18  
Take 3 lines in $\mathbb{R}^2$... –  Tom Smith May 4 '10 at 23:38
28  
This is perhaps a shameful comment for math overflow, but: ROFL (in the best possible sense) :-) excellent answer! –  Kevin McGerty May 5 '10 at 0:26
8  
100 upvotes! The first "Great Answer" badge! (Besides Anton's fluke from the moderator election.) –  Kevin H. Lin Jul 20 '10 at 18:14

An incredibly common false belief is:

For a (say smooth, projective) algebraic variety $X$ the $K_X$-negative part of the cone $NE(X)$ is locally polyhedral.

A right statement of the theorem of the cone is

$\overline{NE(X)} = \overline{NE(X)}_{K_X \geq 0} + \sum_{i} \mathbb{R}[C_i]$ for a denumerable set $\{ C_i \}$ of rational curves, which accumulate at most on the hyperplane $K_X = 0$.

At a first glance this seems to imply that $\overline{NE(X)}_{K_X < 0}$ is locally poyhedral, but this is not true. It depends on the shape of the intersection $\overline{NE(X)} \cap \{ K_X = 0 \}$.

For instance if this latter intersection is round, and there is only one curve $C_i$, the half-cone $\overline{NE(X)}_{K_X < 0}$ is actually a circular cone! Definitely not polyhedral in any sense. I believe this behaviour can happen even with varieties birational to abelian varieties.

The strange thing about this false belief is that it is held true by many competent mathematicians (and indeed I don't believe that many undergraduates meet the theorem of the cone!).

share|improve this answer
68  
Incredibly common? The number of people who can even understand the statement, let alone believe it, isn't all that large... –  Victor Protsak May 5 '10 at 6:57
5  
Yes, but among those, almost all believe that the wrong version is true. –  Andrea Ferretti May 5 '10 at 10:13
7  
And about 50% of the large community who cannot understand the point will believe that the right version is true! Rather high percentage... –  Wadim Zudilin May 5 '10 at 11:41

I just finished quadratic congruences in my number theory class. I am not any more surprised to see how strong is students' belief in the fact that $x^2\equiv a\pmod{m}$ has at most 2 solutions. Even after you discuss an example $x^2\equiv1\pmod{143}$ (with solutions $x\equiv\pm1,\pm12\pmod{143}$) in details.

And, of course, a lot of wrong beliefs in real analysis. Like an infinitely differentiable function, say in a neighbourhood of origin, must be analytic at the origin.

share|improve this answer
1  
I think you mean that they think $x^2 \cong a \pmod m$ can only have 2 solutions. This is the case over the reals, which is probably where the students get it from. –  Michael Lugo May 5 '10 at 0:33
1  
@Michael: It might have 0 and 1 solutions as well (for example, when $m$ is prime and $a$ is a quadratic nonresidue or 0 respectively). The "real" analogues are $x^2=-1$ and $x^2=0$. –  Wadim Zudilin May 5 '10 at 1:05
1  
Michael's comment is because you have said that students believe that $x^2 \equiv a \pmod m$ may have more than $2$ solutions. But as you go on to indicate, this is a correct belief. So there must be a typo in your answer. –  Pete L. Clark May 5 '10 at 2:19
1  
Thanks, Pete! I corrected the point. –  Wadim Zudilin May 5 '10 at 2:43

The standard projection map in a first course in topology is open. How could it not be closed? I always forget the standard homework exercise in which people first try to use this non-fact.

share|improve this answer
13  
The standard counter-example is $\mathbb{R} \times \mathbb{R} \to \mathbb{R}$, with the closed hyperbola $xy = 1$ mapping to the open set $x \neq 0$. I remember making this mistake, but not what problem prompted me to do so. –  David Speyer May 5 '10 at 12:44

Some things from pseudo-Riemannian geometry are a bit hard to swallow for students who have had previous exposure to Riemannian geometry. Aside from the usual ones arising from sign issues (like, in a two dimensional Lorentzian manifold with positive scalar curvature, time-like geodesics will not have conjugate points), an example is that in Riemannian manifolds, connectedness + geodesic completeness implies geodesic connectedness (every two points is connected by a geodesic). This is not true for Lorentzian manifolds, and the usual example is the pseudo-sphere.

share|improve this answer

All higher homotopy groups of spheres are zero.

Proof: The higher homology groups of spheres are zero, and the higher homotopy groups are abelian, and since homology groups are abelianizations of homotopy groups the higher homotopy groups are alse zero.

This misconception is also made more difficult by the fact that even the simple counterexamples can't be drawn easily.

share|improve this answer
3  
The Hopf map is pretty elementary... –  Andy Putman May 5 '10 at 0:47
17  
I agree that it's elementary, but it's not a particularly easy example to come up with on your own. –  Inna May 5 '10 at 2:19
1  
For anyone who knows the definition of the projective line, the equivalence classes on the 3-sphere in C^2 defined by complex line sections through the origin, has equivalence space CP^1 ≈ S^2, i.e. the Hopf map. I believe I thought of this as a student, but probably did not show it is non trivial. –  roy smith Apr 14 '11 at 18:51

Many students believe that 1 plus the product of the first $n$ primes is always a prime number. They have misunderstood the contradiction in Euclid's proof that there are infinitely many primes. (By the way, 2 * 3 * 5 * 7 * 11 * 13 + 1 is not prime.)

Much later edit: As pointed out elsewhere in this thread, Euclid's proof is not by contradiction; that is another widespread false belief.

Much much later edit: Euclid's proof is not not by contradiction. This is another very widespread false belief. It depends on personal opinion and interpretation what a proof by contradiction is and whether Euclid's proof belongs to this category. In fact, if the derivation of an absurdity or the contradiction of an assumption is a proof by contradiction, then Euclid's proof is a proof by contradiction. Euclid says (Elements Book 9 Proposition 20): The very thing (is) absurd. Thus, G is not the same as one of A, B, C. And it was assumed (to be) prime.


Nb. The above edits were not added by the OP of this answer.

share|improve this answer
112  
When I was 11 y.o. I was screamed at by a teacher and thrown out of class for pointing this out when he claimed the false belief stated (it wasn't class material, but the teacher wanted to show he was smart). I found the counterexample later at home. I didn't let the matter drop either... I knew I was right and he was wrong, and really had a major fallout with that math teacher and the school; and flunked math that year. –  Daniel Moskovich May 5 '10 at 1:19
71  
@Daniel: Sorry to hear that. When my daughter Meena was the same age (11), her teacher asserted that 0.999... was not equal to 1. Meena supplied one or two proofs that they were equal, but her teacher would not budge. Maybe this is another example of a common false belief. –  Ravi Boppana May 5 '10 at 2:59
57  
@Daniel: I've heard a worse story. A college instructor claimed in Number Theory class that there are only finitely many primes. When confronted by a student, her reply was: "If you think there are infinitely many, write them all down". She was on tenure track, but need I add, didn't get tenure. –  Victor Protsak May 5 '10 at 5:38
79  
This false belief leads to a proof of the Twin Prime conjecture: For every $n$, $(p_1 p_2 \cdots p_n -1, p_1 p_2 \cdots p_n +1)$ are twin primes, right? –  David Speyer May 6 '10 at 15:50
113  
Daniel, about the same age, I was asked to leave class for claiming that pi is not 22/7. The math teacher said that 3.14 is an approximation and while some people falsly believe that pi=3.14 but the true answer is 22/7. Years later an Israeli newspaper published a story about a person who can memorize the first 2000 digits of pi and the article contained the first 200 digits. A week later the newspaper published a correction: "Some of our readers pointed out that pi=22/7". Then the "corrected" (periodic) 200 digits were included. Memorizing digits of pi is a whole different matter if pi=22/7. –  Gil Kalai May 11 '10 at 5:45

Here are two group theory errors I've seen professionals make in public.

1) Believing that if $G_1 \subset G_2 \subset \cdots$ is an ascending union of groups such that $G_i$ is free, then $\bigcup_{i=1}^{\infty} G_i$ is free. Probably the vague idea they have is that any relation has to live in some $G_i$, so there are no nontrivial relations.

2) Consider a group $G$ acting on a vector space $V$ (over $\mathbb{C}$, say). Assume that $G$ acts as the identity on a subspace $W$ and that the induced action of $G$ on $V/W$ is trivial. Then I've seen people conclude that the action of $G$ on $V$ is trivial. Of course, this is true if $G$ is finite since then all short exact sequences of $G$-modules split, but it is trivial to construct counterexamples for infinite $G$.

share|improve this answer
19  
So people think $\mathbb{Q}$ is a free group? Curious. –  Pete L. Clark May 5 '10 at 23:39
23  
Sadly enough, I suspect that many people who care about geometric/combinatorial group theory do not think of Q as a group... –  Andy Putman May 6 '10 at 0:28
11  
Pete, they don't think... they BELIEVE! –  Victor Protsak May 14 '10 at 6:56
13  
Come on Steve -- aren't all abelian groups finitely generated? =) –  Andy Putman May 15 '10 at 15:04
7  
@yatima You probably mean $G_i=\{n/i!\}$. –  The User May 11 '13 at 23:58

After learning that the Witt vectors of a finite field of size $p^n$ is the ring of integers of the unramified extension of ${\mathbf Q}_p$ of degree $n$, I think lots of people then think that the Witt vectors of $\overline{\mathbf F}_p$ (the algebraic closure of ${\mathbf F}_p$) is the ring of integers of the maximal unramified extension of ${\mathbf Q}_p$. It isn't: the integers of the maximal unramified extension is the union of the Witt vectors of the finite fields of $p$-power size whereas the Witt vectors of $\overline{\mathbf F}_p$ is the $p$-adic completion of the integers of the maximal unramified extension; the distinction turns on being able to write Witt vectors over $\overline{\mathbf F}_p$ as series with coefficients that are prime-to-$p$ roots of unity of increasingly large degree instead of having bounded degree.

I was at a conference last fall where a famous mathematician was confused by this point, although to be fair he really never worked seriously with Witt vectors before.

share|improve this answer
1  
If you're going to hold a false belief, this is a good one to have. While the maximal unramified extension of $\mathbb{Q}_p$ is not the fraction field of $W(\overline{\mathbb{F}_p})$ -- the latter is complete, the former isn't -- one is a subfield of the other and the two fields have very similar properties. One way to express/justify this is by saying that the inclusion map is an elementary embedding in the sense of model theory. For arithmetic geometers, this is related to work of Greenberg on rational points over Henselian fields. –  Pete L. Clark May 5 '10 at 2:25
1  
@K: Well, sure, believing false things is never desirable. All I'm saying is that there are some things working underneath the surface to prevent this particular false belief from really screwing you up. I seem to remember once writing something like "where $\mathbb{Q}_{p^{\infty}}$ is, according to the reader's preference, either the maximal unramified extension of $\mathbb{Q}_p$ or its completion." Not that this is especially good mathematical writing, but the point was that it manifestly didn't matter which. –  Pete L. Clark May 5 '10 at 4:57
5  
Pete: It can screw you up very badly. For example, F-isocrystals are very different over the maximal unramified extension of Q_p and its completion. –  JS Milne May 8 '10 at 16:44

This is a bit specialized, but a common misconception in low-dimensional topology (particularly in knot theory) is that any change of basis in homology is realized by a diffeomorphism, hence (for a surface) by an action of a mapping class. I think this is exactly the type of false belief being described (I falsely believed it for a long time myself).

Common misconception: Let F be a genus 2g surface, and let $b_1,\ldots,b_{2g}$ be a primitive basis for $H_1(F)$, represented as embedded curves in F. Any change of basis for $H_1(F)$ is realized by an action of the mapping class group of F on the embedded curves.

This is rubbish- the action of the mapping class group on homology is by $Sp_{2g}(\mathbb{Z}))$, which for $g>1$ is a proper subgroup of $GL_{2g}(\mathbb{Z})$, the group of base-changes of $H_1(F)$.
As an example of what you can't do with a diffeomorphism of a surface, consider a disc with 4 bands A,B,C,D attached, so the order of the end sements is $A^+B^-A^-B^+C^+D^-C^-D^+$, together forming a surface. A basis a,b,c,d for $H_1(F)$ is given by this picture as 4 loops going through the cores of the bands A,B,C,D correspondingly. You can add a to b, b to c, c to d, or d to a by diffeomorphism of F (sliding adjacent bands over one another). However, although you can add a to c algebraically, because bands A and C are "not adjacent in F", there is no corresponding diffeomorphism of $F$.
One place this mistake manifests itself (cranking up the level of terminilogy for a second) is in thinking that unimodular congruence of a Seifert matrix corresponds to ambient isotopy of a Seifert surface.
A related common mistake (closely related to this question):

Common misconception: Any homology class is represented as a submanifold. Maybe even as an embedded submanifold.
share|improve this answer
1  
Is the first common misconception actually common? I have never met anyone who had it (perhaps I hang out with the wrong crowd...) –  Igor Rivin Jan 28 '11 at 22:51

Sequence $\{a_n\}$ has a limit $A$ in $\mathbb{R}$ and a limit $B$ in $\mathbb{Q}_p$. Then $A$ is rational iff $B$ is rational.

share|improve this answer
3  
Or: if a sequence has a rational limit in Q_p and in Q_r, then they're the same. –  Qiaochu Yuan May 5 '10 at 4:16
1  
But if a rational sequence has a limit in all Q_p, including Q_\infty ... –  Gerald Edgar May 5 '10 at 12:17

I don't know if this is what you are looking for, but I keep hearing that "a differentiable function is one that is locally linear", not one whose local variation can be approximated linearly. No one stops to think about e.g, x2, and the fact that its graph does not look like a line at any value of x.

share|improve this answer
5  
I would say this is more a heuristic than a false statement; as such, it would be more appropriate as an answer to mathoverflow.net/questions/2358/most-harmful-heuristic (although I do not think anyone interprets it the way you apparently do). –  Qiaochu Yuan May 5 '10 at 4:53

Linear algebra: 1. If V is a vector space spanned by {ei} and W is a subspace of V then W is spanned by ek's contained in it. Actually, this is widely believed with bases in place of spanning sets. Or

2.   (U+V)∩W = U∩W + V∩W.

Both these "properties" are closely related to the current leader (by Tilman).

3. Every element of V⊗W is v⊗w with v∈ V, w∈ W.

All three are probably due to interpolating our intuition about sets to vector spaces.

4. Every symmetric matrix is diagonalizable.

Wait, didn't we prove this? ("True for the real matrices, so must be true in general").

Algebraic groups: if G is a linear algebraic group acting on a vector space V then the (Krull) dimension of the invariant ring satisfies the inequality

    dim k[V]G ≥ dim V-dim G,

or even a more precise belief that dim k[V]G=dim V-dim Gx for a generic x. This is true in the differentiable situation for the dimension of the quotient, when a compact Lie group acts smoothly on a manifold, and algebraic actions are "nicer", right?

share|improve this answer
5  
#3 is a great example; I remember being caught off-guard by it indirectly even when I thought I was aware of it. –  Qiaochu Yuan May 5 '10 at 15:40
3  
(0 1;1 0) isn't diagonalisable over the field with 2 elements. It has 1 as an eigenvalue twice, but isn't the identity matrix. –  Kevin Buzzard May 5 '10 at 20:22
1  
@Kevin: right, and if -1 is a square it's easy to construct a 3 by 3 nilpotent symmetric matrix –  Victor Protsak May 5 '10 at 21:07
13  
#3 gets to play in physics-land, too --- the fact that there are non-simple tensors is the same as the fact that particles can become entangled in quantum mechanics. –  Matt Noonan May 5 '10 at 23:56
2  
@unknown. Just take ${\mathbb C}$. Every square matrix is similar to a symmetric one. Since there are non-diagonalizable complex matrices, some complex symmetric matrices are not diagonalizable. –  Denis Serre Sep 23 '10 at 16:28

These are actually metamathematical (false) beliefs that many intelligent people have while they are learning mathematics, but usually abandon when their mistake is pointed out, and I am almost certain to draw fire for saying it from those who haven't, together with the reasons for them:

The results must be stated in complete and utter generality.

Easy examples are left as an exercise to the reader.

It is more important to be correct than to be understood.

(Applicable to talks as well as papers.)

Reasons: 1. Von Neumann is in the audience. 2. This is just a generalization of Lemma 1.2.3 in volume X of Bourbaki. 3. The results are impressive and speak for themselves.

share|improve this answer
3  
IMO "It is more important to be correct than to be understood" is not a false belief. –  Michael Oct 16 at 18:32

Here are two things that I have mistakenly believed at various points in my "adult mathematical life":

For a field $k$, we have an equality of formal Laurent series fields $k((x,y)) = k((x))((y))$.

Note that the first one is the fraction field of the formal power series ring $k[[x,y]]$. For instance, for a sequence $\{a_n\}$ of elements of $k$, $\sum_{n=1}^{\infty} a_n x^{-n} y^n$ lies in the second field but not necessarily in the first. [Originally I had $a_n = 1$ for all $n$; quite a while after my original post, AS pointed out that that this actually does lie in the smaller field!]

I think this is a plausible mistaken belief, since e.g. the analogous statements for polynomial rings, fields of rational functions and rings of formal power series are true and very frequently used. No one ever warned me that formal Laurent series behave differently!

[Added later: I just found the following passage on p. 149 of Lam's Introduction to Quadratic Forms over Fields: "...bigger field $\mathbb{R}((x))((y))$. (This is an iterated Laurent series field, not to be confused with $\mathbb{R}((x,y))$, which is usually taken to mean the quotient field of the power series ring $\mathbb{R}[[x,y]]$.)" If only all math books were written by T.-Y. Lam...]

Note that, even more than KConrad's example of $\mathbb{Q}_p^{\operatorname{unr}}$ versus the fraction field of the Witt vector ring $W(\overline{\mathbb{F}_p})$, conflating these two fields is very likely to screw you up, since they are in fact very different (and, in particular, not elementarily equivalent). For instance, the field $\mathbb{C}((x))((y))$ has absolute Galois group isomorphic to $\hat{\mathbb{Z}}^2$ -- hence every finite extension is abelian -- whereas the field $\mathbb{C}((x,y))$ is Hilbertian so has e.g. finite Galois extensions with Galois group $S_n$ for all $n$ (and conjecturally provably every finite group arises as a Galois group!). In my early work on the period-index problem I actually reached a contradiction via this mistake and remained there for several days until Cathy O'Neil set me straight.

Every finite index subgroup of a profinite group is open.

This I believed as a postdoc, even while explicitly contemplating what is probably the easiest counterexample, the "Bernoulli group" $\mathbb{B} = \prod_{i=1}^{\infty} \mathbb{Z}/2\mathbb{Z}$. Indeed, note that there are uncountably many index $2$ subgroups -- because they correspond to elements of the dual space of $\mathbb{B}$ viewed as a $\mathbb{F}_2$-vector space, whereas an open subgroup has to project surjectively onto all but finitely many factors, so there are certainly only countably many such (of any and all indices). Thanks to Hugo Chapdelaine for setting me straight, patiently and persistently. It took me a while to get it.

Again, I blame the standard expositions for not being more explicit about this. If you are a serious student of profinite groups, you will know that the property that every finite index subgroup is open is a very important one, called strongly complete and that recently it was proven that each topologically finitely generated profinite group is strongly complete. (This also comes up as a distinction between the two different kinds of "profinite completion": in the category of groups, or in the category of topological groups.)

Moreover, this point is usually sloughed over in discussions of local class field theory, in which they make a point of the theorem that every finite index open subgroup of $K^{\times}$ is the image of the norm of a finite abelian extension, but the obvious question of whether this includes every finite index subgroup is typically not addressed. In fact the answer is "yes" in characteristic zero (indeed $p$-adic fields have topologically finitely generated absolute Galois groups) and "no" in positive characteristic (indeed Laurent series fields do not, not that they usually tell you that either). I want to single out J. Milne's class field theory notes for being very clear and informative on this point. It is certainly the exception here.

share|improve this answer
6  
Milne also, in his notes on field and Galois theory, takes the time to point out (and prove using Zorn's lemma and the group $\mathbb{B}$ above) that the absolute Galois group of $\mathbb{Q}$ has non-open subgroups of index $2^n$ for all $n>1$. He adds as a footnote a quote of Swinnerton-Dyer where he mentions the "unsolved [problem]" of determining whether every finite index subgroup of $G_\mathbb{Q}$ is open or not, observing that this problem seems "very difficult." –  Keenan Kidwell May 5 '10 at 12:45
3  
Nice examples! Actually it is known that any finite groups arises as a Galois group over K=C((x,y)). Since K is Hilbertian, it is enough to prove it for K(t). Now, we know that if L is a large field (ie any smooth L-curve has infinitely many L-points as soon as it has one), then any finite groups arises as a Galois group over L(t) (see F.Pop, Embedding problems over large fields, Ann. of Math., 1996). And F. Pop recently proved that if R is a domain which is complete wrt a non-zero ideal (Henselian's enough), then its fraction field is large (see Henselian implies Large on his webpage). –  Jérôme Poineau May 5 '10 at 13:52
21  
@Pete: I remember once reading a paper of Katz and being bewildered by what he was saying until I realised that Q_p[[x]] was much bigger than Z_p[[x]] tensor_{Z_p} Q_p. –  Kevin Buzzard May 5 '10 at 20:19
9  
It's funny that you are illustrating yourself how tricky the distinction between $k((x))((y))$ and $k((x,y))$ can be, by giving a wrong example: in fact $\sum_{i \geq 0}x^{−i}y^i \in k((x,y))$. (Isn't it just $x/(x - y)$? Think a bit about convergence issues.) But I believe that $\sum_{i \geq 0} x^{-i^2} y^i \not\in k((x,y))$ - and I think I can prove this using the Weierstrass preparation theorem for Laurent series over complete DVRs, or something like that. –  Wanderer Jul 8 '10 at 18:07

There are cases that people know that a certain naive mathematical thought is incorrect but largely overestimate the amount by which it is incorrect. I remember hearing on the radio somebody explaining: "We make five experiments where the probability for success in every experiment is 10%. Now, a naive person will think that the probability that at least one of the experiment succeed is five times ten, 50%. But this is incorrect! the probability for success is not much larger than the 10% we started with."

Of course, the truth is much closer to 50% than to 10%.

(Let me also mention that there are various common false beliefs about mathematical terms: NP stands for "not polynomial" [in fact it stands for "Nondeterministic Polynomial" time]; the word "Killing" in Killing form is an adjective [in fact it is based on the name of the mathematician "Wilhelm Killing"] etc.)

share|improve this answer
10  
And the Killing field has nothing to do with Pol Pot. –  Nate Eldredge May 5 '10 at 14:40
2  
Unfortunately I often slip up in class and say that the Killing vector field $T$ kills the metric term (well, I use the verb kills when a differential operator hits something and makes it zero, because, you know, bad terms are always "the enemy"). I'm not sure how much damage I did to the students' impressions... –  Willie Wong May 5 '10 at 17:19
3  
"Kills" is one of those terms I hear mathematicians use surprisingly often. The other one is "this guy." I never really understood the prevalence of either. –  Qiaochu Yuan May 6 '10 at 7:38
15  
"Guy" is a pretty standard English colloquialism for "person"; combine this with humans' tendency to anthropomorphize and this usage is understandable. (Though we shouldn't anthropomorphize mathematical objects, because they hate that.) –  Nate Eldredge May 6 '10 at 14:51
10  
In the only lecture I saw by David Goss he started with "guy", quickly went to something like "uncanny fellow" and then stayed with "sucker" for most of the talk. I don't know what those poor Drinfeld modules had done to him the day before :-) –  Peter Arndt May 19 '10 at 12:24

a student, this afternoon: "this set is open, hence it is not closed: this is why [...]"

share|improve this answer
63  
The terminology is rather unfortunate. –  Nate Eldredge May 5 '10 at 14:39
37  
Either that or topologists need a sit-down about the facts of life in life, where they are told how unfortunate their notation is... –  Kevin Buzzard May 5 '10 at 20:25
97  
Munkres is fond of saying "sets are not doors." –  Qiaochu Yuan May 5 '10 at 20:53
31  
On my office door I once put "clopen the door" –  hypercube May 26 '10 at 22:52
52  
Actually, topologists have studied spaces where every set is open or closed (or both, of course), and they're called "Door spaces".... –  Henno Brandsma Jun 6 '10 at 11:46

protected by François G. Dorais Oct 15 '13 at 2:34

Thank you for your interest in this question. Because it has attracted low-quality answers, posting an answer now requires 10 reputation on this site.

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.