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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 $\mathbb{C}$;
(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 $\mathbb{R}$ must be the whole of $\mathbb{R}$. 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.

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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
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
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
It's a thought -- I might consider it. – gowers Oct 4 '10 at 20:13
Meta created… – user9072 Oct 8 '11 at 14:27

218 Answers 218

Occasionally seen on this site: if a polynomial $P:\mathbb{Q}\rightarrow\mathbb{Q}$ is injective, so must be its extension to $\mathbb{R}$.

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+1 Yaakov. $P(x) = x^3 - 5x$ is a counterexample. – Todd Trimble Mar 31 '11 at 21:47
That's another funny one, because you would be very unlikely to see the same mistake with $\mathbb{C}$. So why are people liklely to make it with $\mathbb{R}$? – Thierry Zell Apr 7 '11 at 0:30
@Tierry: continuity, of course. $\mathbb{R}$ is far from dense in $\mathbb{C}$ but $\mathbb{Q}$ is dense in $\mathbb{R}$. – Ryan Reich Apr 22 '11 at 18:06

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?

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#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
(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
@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
#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
@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

"The image of a category under a functor is a category."

This is a small one, but it lasted for six months when I was starting in category theory.

A finite counterexample exists, with just 3 objects. Even under a connectivity requirement, a small finite counterexample still exists. In fact, it is dead wrong to think anything like this holds.

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You only need two objects 0, 1 and a single non-identity arrow 0->1. Consider the functor from this to Set which takes 0 and 1 to the natural numbers and the arrow to the successor function. – Todd Trimble Apr 17 '11 at 0:58
@Todd Trimble : Very nice! – Jérôme JEAN-CHARLES Apr 20 '11 at 10:34

False belief: "There are no known sub-exponential time algorithms for NP-complete problems."

This one is tricky for a couple of reasons. The first is that the term "sub-exponential" is sometimes defined in different ways. With a sufficiently strong definition of "sub-exponential" the above statement is true, in the sense that there is no known separation of the complexity classes NP and EXPTIME (EXPTIME being the class of languages decidable in time $2^{p(n)}$ where $p(n)$ is a polynomial). However, it is quite common to refer to an $O(2^{\sqrt{n}})$ algorithm as "sub-exponential." It is trivial to construct an NP-complete problem that can be solved in sub-exponential time in this sense, because the standard definition of a reduction allows you to expand the size of the input from $n$ to $n^2$ say (e.g., by padding with zeros). But less artificial examples also exist, such as the planar traveling salesman problem, which was shown by Smith to be solvable in $2^{O(\sqrt{n})}$ time without any artificial padding. What is true is that there are many NP-complete problems, such as 3SAT, for which no subexponential algorithms are known if you do not artificially pad the representation of the instances. (Reducing 3SAT to planar TSP does not work because the instance size blows up during the reduction.)

Often this false belief shows up in the following form: "Factoring cannot be NP-complete because there are subexponential algorithms for factoring." It is true that factoring is not known to be NP-complete but the reasoning is wrong. Showing that factoring is NP-complete would not automatically yield subexponential algorithms for all other NP-complete problems.

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For clarity, the definition of sub-exponential for which this is true is $O(\mathrm{exp}(n^\epsilon))$ for all $\epsilon$. – Peter Shor Aug 22 '10 at 19:02

People are silly. Did you ever notice how at airports, say, people happily walk around but when they come to a moving walkway, they tend to stop and take a break? Walking on a moving walkway is not any harder than walking on an ordinary walkway, and resting for 10 seconds here or there will get you to your destination 10 seconds later. So why do people stop for a rest in one place but hardly ever in the other?

The embarrassing bit is that I believed this logic myself for some time, and thought that people were indeed silly, until my son corrected me. I'm not sure if this falls under "a common false belief in mathematics", but it's certainly an amusing and confusing mistake to make.

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Amusing - thanks for the reference. --- Dror. – Dror Bar-Natan Aug 31 '10 at 10:35
That is not a mistake: If you take a 10 sec break at the moving walkway you will arrive faster than if take a 10 sec break on the ordinary walkway. Imagine two people walking from A to B. One person take a 10 sec break just before the moving walkway the other take a 10 sec break as soon as he is on the walkway. They will begin to walk again at (almost) the same time, and the second person will have moved the distance the moving walkway moves in 10 sec. – Sune Jakobsen Sep 1 '10 at 9:13
@ Sune: I think that Dror is suggesting, as the false belief, that people are silly. So he agrees with you. – Toby Bartels Apr 4 '11 at 7:48
@Sune: Nice explanation. Another possible way to think about it: Suppose there are several different moving walkways, some faster than others. Suppose further that you're required to rest for 10 seconds somewhere. On which moving walkway should you take your 10 second rest? Clearly on whichever one moves the fastest. – idmercer Jul 9 '11 at 22:57

That the notion of "picking a random number" is well-defined without providing any further information.

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Obligatory link: – Charles Stewart Jul 13 '10 at 10:20
And likewise – Nate Eldredge Jul 14 '10 at 19:59
In fact Knuth makes a similar joke in TAOCP chapter 3.1 (though he proves a different point with it). He asks whether 2 is a random number. – Zsbán Ambrus Dec 4 '10 at 22:54
I've long suspected that the Dilbert comic alludes to the "Feynman point" where six 9's occur surpisingly early (before digit 1000) in the decimal expansion of $\pi$. – Noam D. Elkies Jun 5 '11 at 15:57
I find it intriguing that, if one interprets "random" as "Martin-Lof random", then in fact it is. – usul Oct 17 '15 at 19:09

It is a difficult open problem whether every finite group is isomorphic to the group of automorphisms of a finite extension $K$ of $\mathbb{Q}$. In fact, this result is true! See for instance The actual "inverse Galois problem" also requires $K$ to be a Galois extension of $\mathbb{Q}$. The true theorem is equivalent to the statement that every finite group is isomorphic to a quotient group of a Galois group over $\mathbb{Q}$. I once observed a famous expert on algebraic number theory being confused on this issue.

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Thanks for the reference! – Andrés E. Caicedo Mar 5 '11 at 23:46
Very interesting, I didn't know that result. – Martin Brandenburg Apr 12 '11 at 8:42

If $f$ is (Lebesgue) integrable on $\mathbb{R}$, then $f(x) \to 0$, as $x \to \infty$. False: there exists a continuous integrable function on $\mathbb{R}$ such that $\limsup_{\infty} f = \infty$ (an exercise in Stein and Shakarchi's Real Analysis).

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It's easy to prove the true statement, but I was startled at first by having it pointed out, because I recognize your false statement as something I've heard many physicists claim. Presumably they are assuming (knowingly or not) that $f$ is reasonably regular at $\pm \infty$. – Mark Meckes Jun 1 '10 at 16:12
I think some hold this belief by analogy with the behavior of infinite series. At least, that's the common thread I've noticed with some confused students. – Andrés E. Caicedo Jun 2 '10 at 1:49
On the other hand, this is true for uniformly continuous functions (which is what most students would have in their head in the first place). – Peter Humphries Jun 2 '10 at 2:03
@Peter: that probably explains the physicists as well, since I expect they are thinking of smooth functions such that $f'(x)\to 0$ as $x\to \pm \infty$, which in particular implies uniform continuity. – Mark Meckes Jun 2 '10 at 12:07
On the other hand, physicists have no problem with saying that $e^{ikx}$ belongs to the $L^2$ Hilbert space. Actually, they are right: (our) Lebesgue integral isn't the right instrument to formalize (their) physical intuition. – Victor Protsak Jun 10 '10 at 6:20

False belief: Saying that ZFC is consistent is the same as saying that if ZFC proves "there are infinitely many twin primes" (for example) then there really are infinitely many twin primes.

Everybody realizes that if ZFC is inconsistent then a formal ZFC proof of "there are infinitely many twin primes" tells us nothing about whether there really are infinitely many twin primes. A lot of people, without necessarily realizing it, turn this around and assume that the consistency of ZFC is the only condition needed to ensure that its theorems are "trustworthy." But this is not the case, even if we restrict our attention to first-order statements about the natural numbers. We say that ZFC is arithmetically sound if all first-order sentences about the natural numbers that are provable in ZFC are true. The arithmetical soundness of ZFC is a stronger condition than the consistency of ZFC. For example, Goedel's 2nd incompleteness theorem says that if ZFC is consistent, then ZFC doesn't prove "ZFC is consistent." So it's conceivable that ZFC is consistent but that "ZFC is inconsistent" is a theorem of ZFC. Then we would have an example of a theorem of ZFC that asserts something false, even though ZFC is consistent.

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Here "True" means "Holds in some model of ZFC"? – Louigi Addario-Berry Mar 11 '11 at 15:39
@Louigi: No, "true" means "holds of the natural numbers." For example, we say that "there are infinitely many twin primes" is true if and only if there are infinitely many twin primes. Some people will claim not to know what it means to say that there are infinitely many twin primes. But if we step next door into the number-theory classroom, then suddenly their alleged confusion disappears and they know exactly what it means to say that there are infinitely many twin primes. So all you need to do is to avoid deleting your memory as you walk back from the number theory classroom. – Timothy Chow Mar 11 '11 at 16:17
If you don't know what it means to say that there are infinitely many twin primes, then you don't know what it means to say that ZFC is consistent. (Both are of the form $\forall\, n, P(n)$ for decidable $P$.) It's a respectable attitude to take that we don't really know what this means, that its truth may depend on the model, etc; but then you have to apply this across the board. (Incidentally, this is the same trap that postmodernist philosophers fall into when they try to analyse science that they don't understand.) – Toby Bartels Apr 4 '11 at 7:24
"Both are of the form ∀n,P(n) for decidable P." Can you elaborate further? It seems to me that there is no way to disprove "there are infinitely many twin primes" by a counterexample, whereas there is for "ZFC is consistent." – Ibrahim Mar 9 '15 at 16:06
@Ibrahim : You're correct. However, there are plausible $\Pi^0_1$ strengthenings of the twin prime conjecture that give computable bounds on the spacing between twin primes. – Timothy Chow Mar 9 '15 at 20:59

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.

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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
@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
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
Thanks, Pete! I corrected the point. – Wadim Zudilin May 5 '10 at 2:43

The commutator [H,K] of two subgroups H,K is the set of commutators [h,k] with h in H and k in K. (Instead, it is the group generated by those commutators. Confusingly, the convention with products HK usually goes the other way.)

In a similar vein: a 2-vector $\omega \in \bigwedge^2 V$ is always the wedge product of two 1-vectors. (Instead, this is merely an important special case of a 2-vector.) Part of the difficulty here is that the statement is true in the important three-dimensional case. Once one is aware of the Plucker embedding, this confusion goes away, but that can take a while...

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I used to have the commutator subgroup false belief, but playing around with a Rubik's cube disabused me of that notion really quickly... – Harrison Brown May 8 '11 at 11:18
I remember teaching in a second year algebra exercise session and telling my students that "of course" the set of commutators is not necessarily a subgroup. Then they asked me for an example... And I was unable to provide an elementary one at once, the only example I could come up with in a minute was free groups, which they had never seen so I could not tell them, and it is not quite easy anyway. As it turns out, the smallest finite group for which this occurs has order > 100 (I cannot recall right now). – Arnaud Chéritat Oct 18 '15 at 12:08
The smallest such groups are two of order 96, as stated here. – Rosie F Jun 23 at 9:16

Some people believe there is no "formula" for the nth prime number. Of course there are many such formulas, even though not very useful:

The reason given for disbelieving the existence of a "prime number formula" is also curious: "because the primes are unpredictable". This believe is in contradiction with the simple fact that anyone can come up with an easy algorihtm which gives the nth prime number. There is something mystical associate with this ill-defined term "formula".

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Doron Zeilberger's article in the PCM has an illuminating discussion in the very beginning about what constitutes an "answer" to a question like "what is the nth prime," albeit in the context of combinatorics: – Qiaochu Yuan Jun 8 '10 at 14:10
See my comment to Vania Mascioni's answer, which I made before I saw your post. – Victor Protsak Jun 10 '10 at 7:04

Real projective space ${\mathbb{RP}}^3 = (\mathbb R^4 - 0)/\mathbb R^*$ is non-orientable.

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"Non-orientable surfaces do not embed in orientable three-manifolds." is also a classic. – Sam Nead Apr 10 '11 at 19:33
"Covers of planar surfaces are planar." – Sam Nead Nov 17 '11 at 11:16

"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.

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@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
"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
That there are two different notions that are called "locally compact" does not help. – Alfonso Gracia-Saz May 7 '10 at 21:21

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.
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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
Igor, I guess your crowd knows about intersection pairings. – roy smith Apr 14 '11 at 18:59
Could you give a counterexample for the 2nd misconception? – Michael Oct 31 '13 at 8:04

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.

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The Hopf map is pretty elementary... – Andy Putman May 5 '10 at 0:47
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
Yes, contradicting Hurewicz heorem ... – Qfwfq May 5 '10 at 12:13
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

$\pi$ is equal to 22/7.

This was touched upon in the comments to a totally unrelated answer but I think this false belief is important enough to warrant its own answer (and as far as I could tell it does not have one yet, my apologies if I overlooked one.)

Of course, it's unlikely anyone on this site believes this, or ever believed it, which is why I think it's important to insist on this: it does not really resonate with us, we are unlikely to warn students against it, yet we probably see in front of us many students who have that false belief and then will move on to spread it around.

A Piece of Evidence

Let me offer as evidence this gem taken off the comments section of an unrelated (but quite thought-provoking) article on Psychology Today, of all places! When Less is More: The Case for Teaching Less Math in Schools (The title is a misnomer, it's a case for starting math later, but I think that with such a scheme you should be able to teach more math overall; anyway, read it for yourselves.)

Some years ago, my (now ex-) wife was involved in a "trivia night" fundraiser at her elementary school, and they wanted me on their "teacher team" to round out their knowledge. They had almost everything covered except some technology-related topics and I was an IT guy. In round four, my moment to shine arrived, as the category was "Math & Science" and one of the questions was, "give the first five digits of pi." I quickly said, "3.1415." The 9 teachers at the table ignored me and wrote down "22/7" on scrap paper and began to divide it out. I observed this quietly at first, assuming that 22/7ths gave the right answer for the first 5 digits, but it doesn't. It gives something like 3.1427. I said, "Whoops, that won't work." They ignored me and consulted among themselves, concluding that they had all done the division properly on 22/7ths out to five digits. I said, "That's not right, it's 3.1415." [...]

I'm cutting it off here because it's a long story: hilarity ensues when the non-teacher at the table stands up for the truth (when he finds out that the decimals of 22/7 were the expected answer!) The final decision of the judges:

"We've got a correction on the 'pi' question, apparently there's been confusion, but we will now be accepting 3.1415 as a correct answer as well" [as 3.1427].

The Moral of the Story

I used to dismiss out of hand this kind of confusion: who could be dumb enough to believe that $\pi$ is 22/7? (Many people apparently: in the portion of the story I cut was another gem - "I'm sorry, but I'm a civil engineer, and math is my job. Pi is 22/7ths.")

Now, I treat this very seriously, and depending on where you live, you should too. Damage wrought during the influential early years is very hard to undo, so that the contradictory facts "$\pi$ is irrational" and "$\pi$=22/7" can coexist in an undergraduate's mind. And when that person leaves school, guess which of the two beliefs will get discarded: the one implanted since childhood, or the one involving a notion (rational numbers) which is already getting fuzzy in the person's brain? I'm afraid it's no contest there, unless this confusion has been specifically addressed.

So if you have any future teachers in your classes (and even if you don't, cf. the civil engineer above), consider addressing this false belief at some point.

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You know what the most funny part is? They can't divide, because actually 22/7 is between 3.1428 and 3.1429, but this way they got something closer to the real answer. – Zsbán Ambrus Nov 27 '10 at 19:36
Well, doesn't the Bible say that π=3 ? – ACL Dec 1 '10 at 22:52
ACL: I don't have the passage handy now, but as I recall, it does not outright say perimeter over diameter equals three; you can make the case that it's a matter of overenthusiastic rounding down of the value of the perimeter. It is a bit sloppy though, even for that time period... – Thierry Zell Dec 2 '10 at 2:00
This one is depressing... – Andrés E. Caicedo Dec 7 '10 at 22:26
This is from the AFP report on the recent computation of the first $5\times 10^{12}$ (decimal) digits of $\pi$: "Pi, the ratio of a circle's circumference to its diameter, starts with 3.14159 in a string whose digits are believed to never repeat or end."… – Andrés E. Caicedo Dec 7 '10 at 22:29

A false belief which I meet not infrequently this time of year while marking exams is the following:

The exponential map is surjective for a connected Lie group.

This is true for compact Lie groups, but certainly false in general. A (finite-dimensional) connected Lie group is generated by the image of the exponential map, but already $SL(2,\mathbb{R})$ shows that there are elements which are not in the image of the exponential map.

Interestingly, for a connected real Lie group, every element can be written as the product of at most two exponentials.

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Well, exponential function isn't surjective either, is it? More seriously, not only does the exp map fail to be surjective, it also fails to be a local diffeomorphism! But I wonder what beliefs like that say about us. Certainly, unexpected things happen when you push familiar notions into an unfamiliar terrain. Isn't it our job as teachers to make the students perfectly aware of what properties do and do not continue to hold? Where is that intuition supposed to come from? I once taught a Lie groups course, and the book I used (Rossmann) made a big deal about matrix exponentials being unusual. – Victor Protsak Jun 10 '10 at 6:36
But the exponential function, thought of as the exponential map from the one-dimensional Lie algebra $\mathbb{R}$ to the Lie group $\mathbb{R}^+$ (under multiplication) is surjective. I too teach this course based on Wulf Rossmann's book and I think that I do emphasise, if not perhaps all the unusualness of the exponential map, certainly the fact that it may fail to be surjective on the identity component. We even do the case of $SL(2,\mathbb{R})$ is some detail as an application of the formula for the exponential of a $2\times 2$ matrix. – José Figueroa-O'Farrill Jun 10 '10 at 8:12

Let $A,B$ be Hermitian matrices. If $0_n\le A\le B$, then $A^2\le B^2$.

False, but subtle! Loewner's theory characterizes those numerical functions $f:[0,\infty)\rightarrow {\mathbb R}$ such that $0_n\le A\le B$ implies $f(A)\le f(B)$ (operator monotone functions). These are the traces over $[0,\infty)$ of holomorphic functions mapping the Poincaré half-plane ${\mathcal H}$ into itself, and of course real on $[0,\infty)$. Thus the square root is operator monotone: $$(0_n\le A\le B)\Longrightarrow(\sqrt A\le\sqrt B),$$ but the square map is not. Counter-example: $$A=\begin{pmatrix} 1 & 0 \\\\ 0 & 0 \end{pmatrix},\qquad B=\begin{pmatrix} 2 & 1 \\\\ 1 & 1 \end{pmatrix}$$

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I hope it is true when $A,B$ commute. ;) – Martin Brandenburg Apr 12 '11 at 8:49
I have heard that Ed Witten got this one wrong in a paper. – J Tyson Oct 25 '15 at 22:13

The quotient $G/Z(G)$ of a group by its center is centerless. I definitely thought this until it was pointed out to me in a Lie theory textbook that this wasn't true in general, but is true for (edit: connected) Lie groups with discrete center. (It is also true if $G$ is perfect by Grun's lemma.)

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Indeed. Though in your second sentence I guess your Lie group has finitely may components? For a concrete example, Heisenberg group (over ${\mathbb R}$ or ${\mathbb Z}$) will be abelian modulo its centre... – Yemon Choi Jul 10 '10 at 3:47
If that were true, then the class of Abelian groups and nilpotent groups would coincide! – Abhishek Parab Jul 10 '10 at 3:51
@Yemon: oops. I wanted "connected." – Qiaochu Yuan Jul 10 '10 at 4:07
For a counter-example, just consider a non-abelian $p$-group $G$. Because $G/Z(G)$ is still a $p$-group, its center is non trivial (classical). – Denis Serre Oct 20 '10 at 10:51

$0^0$ 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 $0^0$.

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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
0^0 = 1, because there is one function from the empty set to the empty set – Steven Gubkin May 4 '10 at 23:04
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
@Qiaochu: More precisely, the notation a^b denotes two completely different notions. The first notion requires that a is an element of some monoid and b∈N={0,1,2,…}. The second notion requires that a∈C and b belongs to the universal cover of C \ {0}. Then a^b = exp(b * Log(a)), where Log denotes the natural logarithm defined as a function on the universal cover of C \ {0}. Accidentally, these two functions coincide when a∈C and b∈N \ {0}. – Dmitri Pavlov May 5 '10 at 0:53
@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

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

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of course all algebraic geometers have some picture such as that on pages 34 of Eisenbud-Harris, or that on p. 37 of Mumford's (new) red book, tattooed on their eyeballs as a reminder of the difference between a scheme and a separated scheme. – roy smith Apr 14 '11 at 18:16

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.

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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
this is another of the standard algebraic geometry examples from the "red book", whereby one shows the affine open set f(x) ≠ 0 is isomorphic to the closed set yf(x) = 1, in one dimension higher ambient space. – roy smith Apr 14 '11 at 18:45

a function $f:\mathbb{R}\rightarrow \mathbb{R}$ is $n$ times differentiable if and only if for each real $x_0$ it may be approximated near $x_0$ by a polynomial of degree at most $n$ with remainder $o((x-x_0)^n)$

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+1: I posted an answer to a question on math.SE using this "characterization" of differentiability. Didier Piau pointed out that it was completely wrong. It is true for $n = 1$, so even now I find it somewhat weird that it doesn't hold for higher derivatives... – Pete L. Clark Apr 26 '11 at 15:03
It becomes true, and a deep fact, in the $C^n$ version: the function $f$ is $C^n$ iff it has a n-order polynomial expansion at any $x_0$, with coefficients depending continuously on $x_0$. – Pietro Majer Jan 10 '12 at 13:48
Pietro, what do you mean exactly? "There exist continuous in $x_0$ functions $h_0(x_0)$, $h_1(x_0)$, $\dots$, $h_n(x_0)$ such that for any fixed $x_0$ one has $f(x_0+t)=h_0(x_0)+h_1(x_0)t+\dots+t^n h_n(x_0)+o(t^n)$ when $t$ goes to 0." Or something different? – Fedor Petrov Jan 11 '12 at 21:59
@Pete: Do you mean the lower derivatives may be wild? Do you have a good counter-example? – Kerry Jul 30 '12 at 6:05

Any subgroup of the direct product $G \times H$ of two groups is of the form $A \times B$, where $A$ is a subgroup of $G$ and $B$ is a subgroup of $H$.

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What? Is it false? Could you please explain it? – Nil Mal Mar 3 at 9:19
@Nilan: For $G=H$, consider the diagonal subgroup $\{ (g,g) : g \in G\}$. It's not of the form above unless $G$ is the trivial group. – Mark Mar 3 at 10:32

As a teaching assistant in an elementary number theory course, I've seen the following quite often :

If $a$ divides $bc$ and $a$ does not divide $b$, then $a$ divides $c$.

That's of course true if $a$ is prime, but people seem to forget that hypothesis.

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the correct assumption is $gcd(a,b)=1$. – Martin Brandenburg Oct 5 '10 at 8:23
i was never able to eradicate this mistake, even by my best students, in spite of naming it the prime divisibility property. – roy smith May 9 '11 at 2:15
@MartinBrandenburg : What does it mean for an assumption to be incorrect or correct? – Kalim Feb 7 '14 at 18:40

I'm not sure how common this is, but it confused me for years. Let $f : \mathbb{C} \to \mathbb{C}$ be an analytic function and $\gamma$ a path in $\mathbb{C}$. In your first class in complex analysis, you define the integral $\int_{\gamma} f(z) dz$.

Now let $a(x,y) dx + b(x,y) dy$ be a $1$-form on $\mathbb{R}^2$ and let $\gamma$ be a path in $\mathbb{R}^2$. In your first class on differential geometry, you define the integral $\int_{\gamma} a(x,y) dx + b(x,y) dy$.

It took me at least three years after I had taken both classes to realize that these notations are consistent. Until then, I thought there was a "path integral in the sense of complex analysis", and I wasn't sure if it obeyed the same rules as the path integral from differential geometry. (By way of analogy, although I wasn't thinking this clearly, the integral $\int \sqrt{dx^2 + dy^2}$, which computes arc length, is NOT the integral of a $1$-form, and I thought complex integrals were something like this.)

For the record, I'll spell out the relation between these notions. Let $f(x+iy) = u(x,y) + i v(x,y)$. Then $$\int_{\gamma} f(z) dz = \int_{\gamma} \left( u(x,y) dx - v(x,y) dy \right) + i \int_{\gamma} \left( u(x,y) dy + v(x,y) dx \right)$$ The right hand side should be thought of as multiplying out $\int_{\gamma} (u(x,y) + i v(x,y)) (dx + i dy)$, a notion which can be made rigorous.

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I think it is customary to say "contour integral" for the complex analysis gadget, "line integral" for the multivariable calculus gadget, and "path integral" for a (not necessarily rigorously defined) integral over a space of fields. – S. Carnahan Jan 13 '11 at 2:47

I'm not sure that anyone holds this as a conscious belief but I have seen a number of students, asked to check that a linear map $\mathbb{R}^k \to \mathbb{R}^{\ell}$ is injective, just check that each of the $k$ basis elements has nonzero image.

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Higher-level version: $n$ vectors are linearly independent iff no two are proportional. I've seen applied mathematicians do that. – darij grinberg Apr 10 '11 at 18:45

In descriptive set theory, we study properties of Polish spaces, typically not considered as topological spaces but rather we equip them with their "Borel structure", i.e., the collection of their Borel sets. Any two uncountable standard Borel Polish spaces are isomorphic, and the isomorphism map can be taken to be Borel. In practice, this means that for most properties we study it is irrelevant what specific Polish space we use as underlying "ambient space", it may be ${\mathbb R}$, or ${\mathbb N}^{\mathbb N}$, or ${\mathcal l}^2$, etc, and we tend to think of all of them as "the reals".

In Lebesgue Sur les fonctions representables analytiquement, J. de math. pures et appl. (1905), Lebesgue makes the mistake of thinking that projections of Borel subsets of the plane ${\mathbb R}^2$ are Borel. In a sense, this mistake created descriptive set theory.

Now we know, for example, that in ${\mathbb N}^{\mathbb N}$, projections of closed sets need not be Borel. Since we usually call reals the members of ${\mathbb N}^{\mathbb N}$,

it is not uncommon to think that projections of closed subsets of ${\mathbb R}^2$ are not necessarily Borel.

This is false. Note that closed sets are countable union of compact sets, so their projections are $F_\sigma$. The actual results in ${\mathbb R}$ are as follows: Recall that the analytic sets are (the empty set and) the sets that are images of Borel subsets of $\mathbb R$ by Borel measurable functions $f:\mathbb R\to\mathbb R$.

  • A set is Borel iff it and its complement are analytic.

  • A set is analytic iff it is the projection of the complement of the projection of a closed subset of ${\mathbb R}^3$.

  • A set is analytic iff it is the projection of a $G_\delta$ subset of $\mathbb R^2$.

  • There is a continuous $g:\mathbb R\to\mathbb R$ such that a set is analytic iff it is $g(A)$ for some $G_\delta$ set $A$.

  • A set if analytic iff it is $f(\mathbb R\setminus\mathbb Q)$ for some continuous $f:\mathbb R\setminus\mathbb Q\to\mathbb R$. (Note that if $f$ is actually continuous on $\mathbb R$, then $f(\mathbb R\setminus\mathbb Q)$ is Borel.)

(See also here.)

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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).

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

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

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