# Examples of common false beliefs in mathematics

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.

• I have to say this is proving to be one of the more useful CW big-list questions on the site... May 6, 2010 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. May 22, 2010 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. Sep 20, 2010 at 12:39
• It's a thought -- I might consider it. Oct 4, 2010 at 20:13
• Meta created tea.mathoverflow.net/discussion/1165/…
– user9072
Oct 8, 2011 at 14:27

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.

• 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. May 4, 2010 at 21:33
• 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! May 4, 2010 at 22:28
• 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... May 5, 2010 at 10:53
• 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... Jun 5, 2010 at 21:13
• ...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. Jul 14, 2010 at 20:03

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.

• Aug 31, 2010 at 2:22
• 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. Sep 1, 2010 at 9:13
• @ Sune: I think that Dror is suggesting, as the false belief, that people are silly. So he agrees with you. Apr 4, 2011 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. Jul 9, 2011 at 22:57
• This is because one can easily confuse time average with space average at some point in the quick mental calculations. Something along the lines of: 'Oh, the (time) average velocity will be affected in the same way no matter where the stop happens, because no place (wrong, should be time) is better than the other. The space average of the velocity is indeed unaffected, but the time average of the velocity is. The implicitly and incorrectly assumed linearity does not apply to time averaging: an infinite velocity in the moving walkaway would not result in an infinite time-averaged velocity. Aug 28, 2020 at 16:02

Some people believe there is no "formula" for the nth prime number. Of course there are many such formulas, even though not very useful: https://mathworld.wolfram.com/PrimeFormulas.html

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

• 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: math.rutgers.edu/~zeilberg/mamarim/mamarimPDF/enuPCM.pdf Jun 8, 2010 at 14:10
• See my comment to Vania Mascioni's answer, which I made before I saw your post. Jun 10, 2010 at 7:04

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

• Obligatory link: xkcd.com/221 Jul 13, 2010 at 10:20
• And likewise dilbert.com/strips/comic/2001-10-25 Jul 14, 2010 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. Dec 4, 2010 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 $\pi$. Jun 5, 2011 at 15:57
• I find it intriguing that, if one interprets "random" as "Martin-Lof random", then in fact it is.
– usul
Oct 17, 2015 at 19:09

"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.

• 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. Apr 17, 2011 at 0:58
• @Todd Trimble : Very nice! Apr 20, 2011 at 10:34

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 M. Fried: A Note on Automorphism Groups of Algebraic Number Fields; https://www.jstor.org/stable/2043724, https://doi.org/10.2307/2043724. 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.

• Very interesting, I didn't know that result. Apr 12, 2011 at 8:42

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?

• #3 is a great example; I remember being caught off-guard by it indirectly even when I thought I was aware of it. May 5, 2010 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. May 5, 2010 at 20:22
• @Kevin: right, and if -1 is a square it's easy to construct a 3 by 3 nilpotent symmetric matrix May 5, 2010 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. May 5, 2010 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. Sep 23, 2010 at 16:28

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.

• For clarity, the definition of sub-exponential for which this is true is $O(\mathrm{exp}(n^\epsilon))$ for all $\epsilon$. Aug 22, 2010 at 19:02
• What a gem! I always thought this was true. Dec 28, 2017 at 9:36

$\pi$ 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.

• 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. Nov 27, 2010 at 19:36
• Well, doesn't the Bible say that π=3 ?
– ACL
Dec 1, 2010 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... Dec 2, 2010 at 2:00
• This one is depressing... Dec 7, 2010 at 22:26
• @Andres: Well, at least these folks believe something that's true, unlike the ones in Thierry's answer. Dec 7, 2010 at 22:43

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

• "Non-orientable surfaces do not embed in orientable three-manifolds." is also a classic. Apr 10, 2011 at 19:33
• "Covers of planar surfaces are planar." Nov 17, 2011 at 11:16
• Have they never played with a Mobius strip? Aug 23, 2020 at 8:25
• @TobyBartels - The Mobius strip is a surface with boundary. Obviously it is impossible to embed a Klein bottle in an orientable three-manifold... :) Aug 24, 2020 at 11:22
• The projective plane $P^2$ embeds in the projective space $P^3$. So the connect sum $K = P^2 \# P^2$ embeds in the connect sum $P^3 \# P^3$. Sep 1, 2020 at 6:57

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.

• Here "True" means "Holds in some model of ZFC"? Mar 11, 2011 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. Mar 11, 2011 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.) Apr 4, 2011 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." Mar 9, 2015 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. Mar 9, 2015 at 20:59

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).

• 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 $\pm \infty$. Jun 1, 2010 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. Jun 2, 2010 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). Jun 2, 2010 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. Jun 2, 2010 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. Jun 10, 2010 at 6:20

"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.

• @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 :-) May 4, 2010 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". May 4, 2010 at 22:20
• That there are two different notions that are called "locally compact" does not help. May 7, 2010 at 21:21
• @AlfonsoGracia-Saz : More than two! There are four (up to logical equivalence) listed at en.wikipedia.org/wiki/Locally_compact_space#Formal_definition and I've seen all of them in the wild at one time or another. Sep 1, 2020 at 17:39
• This reminds me of another natural wrong belief: "the continuous image of a discrete space is discrete". You could give your way to quash this, e.g., the identity map $\mathbf R_d \to \mathbf R$ where $\mathbf R_d$ is the real line with the discrete topology. But to students that could feel like a dull example. More interesting is the projection map $\mathbf R^2 \to \mathbf R$ to the first coordinate, restricted to the discrete subgroup $\mathbf Z(1,1) + \mathbf Z(\sqrt{2},-\sqrt{2})$: it is discrete in $\mathbf R^2$ but its image $\mathbf Z + \mathbf Z\sqrt{2}$ in $\mathbf R$ is not. Sep 17, 2021 at 18:35

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...

• I used to have the commutator subgroup false belief, but playing around with a Rubik's cube disabused me of that notion really quickly... May 8, 2011 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). Oct 18, 2015 at 12:08
• The smallest such groups are two of order 96, as stated here. Jun 23, 2016 at 9:16
• $SL_2(\mathbb R)$ is a good example. The only element that is not a commutator is $-I$. Jul 13, 2017 at 15:10

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.

• 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. May 5, 2010 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 $x^2=0$. May 5, 2010 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. May 5, 2010 at 2:19
• Well.. It is true when 143 is prime. Apr 12, 2017 at 14:32
• Perhaps solving $x^2 \equiv 1 \pmod8$ is more memorable than solving $x^2 \equiv 1 \pmod{143}$? Jan 5, 2018 at 21:05

Here's one that doesn't seem to be listed here: That ZFC isn't vulnerable to Russell's paradox because it has an axiom which implies that sets cannot be elements of themselves.

This one should be obviously false given even a moment's thought -- you cannot get rid of a paradox by adding axioms! But I've seen it repeated over and over.

The correct statement, of course, is that ZFC isn't vulnerable to Russell's paradox because it doesn't have the axiom of unrestricted comprehension; and that while ZFC does have an axiom which implies that all sets are well-founded, this is irrelevant to Russell's paradox. (It just means that the set of all sets which are not elements of themselves, if it existed, would happen to also be the set of all sets.)

Sometimes this is stated in an ambiguous way that allows for the charitable reading that modern set theory prevents Russell's paradox by preventing one from talking about a set being an element of itself, which is how Russell and Whitehead fixed the problem. But Principia Mathematica isn't exactly "modern set theory"; and I've seen this stated the blatantly wrong way often enough that I'm not too inclined to be charitable about it.

• Why can't you get rid of a paradox by adding axioms? Axioms can certainly constrain as well as construct. Dec 22, 2016 at 5:15
• Because if you establish inconsistency with a proof which uses only axioms $A_1, \ldots, A_n$ in the old system, and add an axiom $A$, then you still have the old proof which used only the axioms $A_1, \ldots, A_n$. Adding axioms increases the potential for proving inconsistency; it can never decrease it. Jul 13, 2017 at 20:10
• As long as "paradox" is defined to mean "contradiction". Mar 4 at 1:51

False belief:

• The category of commutative C*-algebras is opposite equivalent to the category of locally compact Hausdorff spaces.

It's actually not quite that simple! There is some discussion on math.SE.

• Very interesting point. Nov 12, 2014 at 12:42
• Whoops! Thanks for posting this and particularly the link, which shows that the nLab got this one wrong as well. Nov 12, 2014 at 13:03
• Literally just encountered this claim in Borceux and Janelidze's categorical Galois theories book in the portion on Stone duality, much appreciated! Jan 21, 2019 at 23:12

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: $$A=\begin{pmatrix} 1 & 0 \\\\ 0 & 0 \end{pmatrix},\qquad B=\begin{pmatrix} 2 & 1 \\\\ 1 & 1 \end{pmatrix}$$

• I hope it is true when $A,B$ commute. ;) Apr 12, 2011 at 8:49
• I have heard that Ed Witten got this one wrong in a paper. Oct 25, 2015 at 22:13

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 also zero.

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

• The Hopf map is pretty elementary... May 5, 2010 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, 2010 at 2:19
• Yes, contradicting Hurewicz heorem ... May 5, 2010 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. Apr 14, 2011 at 18:51
• This seems to be a constellation of false ideas (some of which I held until corrected by my advisor): $\pi_n(X)$ tends to be nonabelian for all $n$. (In fact $\pi_n(X)$ is abelian for $n \geq 2$. See en.wikipedia.org/wiki/Eckmann%E2%80%93Hilton_argument where it is related that Alexandroff and Hopf were surprised by this.) $H_n(X)$ is (isomorphic to) the abelianization of $\pi_n(X)$ for all $n$. (True for $n \neq 2$. False for $n=2$ unless $X$ is $1$-connected. The Hurewicz theorem gives a homomorphism, $\pi_2 \rightarrow H_2$ when $X$ is not $1$-connected.) Aug 13, 2017 at 2:52

$0^0$ 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 $0^0$.

• 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. May 4, 2010 at 22:17
• 0^0 = 1, because there is one function from the empty set to the empty set May 4, 2010 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? May 5, 2010 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}. May 5, 2010 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. May 5, 2010 at 6:31

I do not think anybody mentioned this example:

If $$M$$ is a $$C^k$$-smooth manifold then the tangent space $$T_pM$$ is isomorphic to the space of derivations of germs (at $$p$$) of $$C^k$$-smooth functions on $$M$$.

This holds for $$k=\infty$$, $$k=\omega$$ (i.e. in the real-analytic category), but not otherwise.

For details and positive results (characterizing derivations which come from tangent vectors even when $$k<\infty$$) see

Newns, W. F.; Walker, A. G., Tangent planes to a differentiable manifold, J. Lond. Math. Soc. 31, 400-406 (1956). ZBL0071.15303.

Abraham, R.; Marsden, J. E.; Ratiu, T., Manifolds, tensor analysis, and applications., Applied Mathematical Sciences. 75. New York: Springer-Verlag. x, 654 p. (1988). ZBL0875.58002.

• +1. Does it isomorphic to the space of derivations of germs of $C^{k-1}$-smooth functions? Aug 27, 2020 at 11:30
• @C.F.G: No, this does not help. Aug 28, 2020 at 21:03
• Do you have any positive result to this? If I am not mistaken, the problem is that Morse's lemma does not hold as is when the smoothness is weakened.
– user20948
Feb 8, 2021 at 19:39
• @Yai0Phah: See the edit. Feb 8, 2021 at 20:27
• Since the first reference is behind a paywall, here is the summary. Let 𝔪 be (the maximal ideal) consisting of functions vanishing at $x$. Then 𝔪² does not contain all functions with vanishing 1-jet at $x$. Indeed any element of 𝔪² has (obviously!) $n+1$ derivatives at $x$. Sep 14, 2022 at 8:51

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.

• 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. Jun 10, 2010 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. Jun 10, 2010 at 8:12

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 terminology 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.
• Is the first common misconception actually common? I have never met anyone who had it (perhaps I hang out with the wrong crowd...) Jan 28, 2011 at 22:51
• Igor, I guess your crowd knows about intersection pairings. Apr 14, 2011 at 18:59
• Could you give a counterexample for the 2nd misconception? Oct 31, 2013 at 8:04

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 $o((x-x_0)^n)$

• +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... Apr 26, 2011 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 $x_0$. Jan 10, 2012 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? Jan 11, 2012 at 21:59
• @Pete: Do you mean the lower derivatives may be wild? Do you have a good counter-example? Jul 30, 2012 at 6:05
• math.stackexchange.com/questions/450743/… Nov 6, 2013 at 4:50

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.)

• 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... Jul 10, 2010 at 3:47
• If that were true, then the class of Abelian groups and nilpotent groups would coincide! Jul 10, 2010 at 3:51
• @Yemon: oops. I wanted "connected." Jul 10, 2010 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). Oct 20, 2010 at 10:51

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

• 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. Apr 14, 2011 at 18:16
• The line with two origins is a simple counterexample. Apr 13, 2022 at 22:17

"Suppose that two features $$[x,y]$$ from a population $$P$$ are positively correlated, and we divide $$P$$ into two subclasses $$P_1$$, $$P_2$$. Then, it cannot happen that the respective features ( $$[x_1,y_1]$$ and $$[x_2,y_2]$$) are negatively correlated in both subclasses

Or more succinctly:

"Mixing preserves the correlation sign."

This seems very plausible - almost obvious. But it's false - see Simpson's paradox

• This seems like much more than a mathematical mistake - really it's a cognitive bias which incidentally can be turned into a mathematically false statement. May 9, 2011 at 22:53
• There were many cases of this error made by reporters and even courts. Dec 3, 2013 at 1:18
• A rather funny real world example: a study of the effect of school kids' smoking on their reading proficiency found a strong positive correlation. The unexpected and very inconvenient discovery caused to take a second look and this time to stratify the pool of school kids by age; that led to a negative correlation in each age group. The positive correlation was due solely to older kids being simultaneously better readers and more likely to smoke. Feb 6, 2020 at 16:45

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.

• 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. May 5, 2010 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. Apr 14, 2011 at 18:45
• @roysmith, I remember how puzzled I was when first learning about algebraic groups to be forced to think of $\mathrm{GL}_n$ as a closed subvariety of an affine space, when it so manifestly wants to be open …. Jan 5, 2018 at 21:07

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.

• 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. May 5, 2010 at 2:25
• Pete: Witt vectors are confusing enough when you're trying to learn about them that I think this is a false belief that is important to clear up if, say, your thesis depends on understanding Witt vectors. May 5, 2010 at 4:13
• @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. May 5, 2010 at 4:57
• P.S.: $W(\overline{\mathbb{F}_p})$ does appear in my thesis...I hope I got it right. :) May 5, 2010 at 6:39
• 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. May 8, 2010 at 16:44

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 $H$.

• What? Is it false? Could you please explain it? Mar 3, 2016 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, 2016 at 10:32