# 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

I didn't notice this in the long list. A student beginning to learn group theory may believe that the converse of Lagrange's Theorem is true, because it is true for subgroups of prime power. They may also believe that a Sylow subgroup is normal because it has a special name. A counter example to both is $A_{4}$ of order $12$ which has no subgroup of order $6$ and whose four different Sylow $3$-subgroups are all conjugates of one another.

• "Normal because it has a special name" belief looks pretty weird. All people have special names, do they consider all of them normal? Mar 21, 2021 at 20:49

I always thought that the GCD of two elements existed if and only if the LCM of those two elements existed, because of my experience with GCDs and LCMs in the ring $$\mathbb{Z}$$. Only recently did I discover that this isn't true in other commutative rings.

• For example, ...? Nov 5, 2020 at 21:36
• See link.springer.com/article/10.1007/BF02837870 There are two elements of the integral domain $\mathbb{Z}[\sqrt{-3}]$ that have an LCM but not a GCD. (Theorem 2 gives a nice relationship between LCM's and GCD's in general.) Nov 5, 2020 at 23:23
• Thanks – but, at 35 euros, I think I'll pass. For free, there are examples at math.stackexchange.com/questions/1449521/… (where it clarifies that if there's an LCM, then there's a GCD; only the converse fails). Nov 6, 2020 at 6:31

This might not be common, but it gave me a headache once. I'll delete if it gets heavily downvoted.

I once had to think really hard about a contradiction in the great scheme of things that followed from my unwitting assumption that if $f$ was a function from a semigroup to a semigroup, then if its kernel was a congruence, $f$ had to be a homomorphism. I encountered a function whose kernel clearly was a congruence but which clearly wasn't a homomorphism, and it took about an hour's walk in a park for my vague notions and incoherent thought to produce the necessary realization.

A few mistakes I remember:

• The quotient groups $\frac{G}{N}$ and $\frac{H}{K}$ are isomorphic if $G \thicksim H$ and $N\thicksim K$.
• A closed interval of a complete lattice is a complete sublattice.
• Two homeomorphic topologies on a set are the same.
• The set of all compatible uniformities of a topological group forms a complete lattice.
• The trace of the identity matrix is 1.
• A closed interval of a complete lattice does form a lattice that is complete, right? So that the mistake is that sups and infs in the interval (particularly the sup and inf over the empty set) are not necessarily computed as they would be in the ambient complete lattice; is that what you have in mind? Sep 6, 2015 at 1:47
• Yes‌‌‌‌‌‌‌‌‌‌‌‌. Sep 6, 2015 at 1:55

I'm seven years late to the game, but here is mine:

False belief: The irrational numbers, in their usual topology as a subset of $\mathbb{R}$, are not a complete metric space.

• Could you write that more carefully? You mean the false belief to be that the irrationals, as a topological space, can't be complete for some metric. What you wrote can be easily confused with saying the irrational numbers are not complete for the usual metric coming from $\mathbf R$, which is true rather than false. Consider the "false belief" that $(-1,1)$ with its topology from $\mathbf R$ can't be made into a complete metric space for some metric. Certainly it's not complete for the usual metric, but it is if we use $\tan(\pi x/2)$ to identify $(-1,1)$ with $\mathbf R$ topologically. Jul 15, 2017 at 3:26
• "are not completely metrizable" is the wording you want. Jul 15, 2017 at 3:28
• @KConrad and Andres: The wording is part of what made that false belief so believable! At the time I didn't think about the fact that there could be multiple metrics, much less that the completeness of those metrics wasn't a topological property. I only realized my mistake when I was introduced to the ideas contained in your two comments. (That, and picturing the irrationals as complete is HARD!) Jul 15, 2017 at 4:15
• Pace, you probably realize by now that continued fractions make the picture a lot easier (whereby the space of irrationals between $0$ and $1$ can be identified with a product space $\mathbb{N}^\mathbb{N}$). Jul 23, 2017 at 21:07
• @DuchampGérardH.E. A topological subspace of a completely metrizable topological space is completely metrizable if and only if it is a $G_\delta$, that is a countable intersection of open sets. One can use Baire's category theorem to show that $\mathbb{Q}$ is not a $G_\delta$. All this can be found at: en.wikipedia.org/wiki/G%CE%B4_set Aug 11, 2017 at 14:41

Given a bundle $E \to X$, let $\mathcal{E}$ denote its sheaf of sections.

False belief: Given a map $f: Y \to X$, the inverse image sheaf $f^{-1}\mathcal{E}$ is the sheaf of sections of the pullback bundle $f^* E \to X$.

This is true if $E \to X$ is a local homeomorphism (e.g. a covering space), or if $f: U \hookrightarrow X$ is the inclusion of an open subset, but not for general maps and bundles.

For instance, taking $x^{-1}\mathcal{E}= \mathcal{E}_x$ for $x: 1 \to X$ the inclusion of a point and $\mathcal{E}$ the sheaf of smooth functions on a manifold will demonstrate that it is false.

For vector bundles (or sheaves of modules over the structure sheaf of a ringed space in general), the correct statement is obtained by using the pullback functor $$f^*\mathcal{V} = \mathcal{O}_Y \otimes_{f^{-1}\mathcal{O}_X} f^{-1} \mathcal{V}$$ which is the inverse image followed by extension of scalars.

One issue which leads to this false belief is that texts on sheaves often use $f^*$ in place of $f^{-1}$ for the inverse image functor, rather than reserving the former for sheaves of modules over ringed spaces.

False belief: It is obvious how to prove that $\sin'=\cos$.

Not so much... if $\cos$ and $\sin$ are defined geometrically. You need to prove geometrically that $$\lim\limits_{x\to 0}\frac{\sin x}x=1$$ and a (non-circular) proof of that is not obvious (see here).

Personally I'm aware of that just today! (thanks to a remedial course given to my niece).

• The string of comments below the Math.SE link shows that really pursuing this gets one down a rabbit hole of rigorous discussions of arc length and such. On the other hand, one can prove that if $S$ (think "sine") and $C$ (think "cosine") are continuous functions which satisfy the standard addition formulas and the Pythagorean theorem $C^2 + S^2 \equiv 1$, then $S'(0)$ exists and $S'(x) = S'(0)C(x)$. There is a whole family of sine-like functions $S_a: x \mapsto S(ax)$; adjusting the parameter $a$ so that $S_a^\prime(0) = 1$, you can define the standard sine to be this $S_a$: a neat finesse. Oct 8, 2017 at 3:14
• @ToddTrimble: Should we interpret your comment as for or against this false belief? Or else, is it just a neutral complement? For a full proof, is your way really easier? Oct 10, 2017 at 11:56
• It's mainly in agreement with you, with a shade of neutral. I don't claim that my suggestion really makes it easier, but only that one can prove things rigorously without getting into considerations of arc length, with (theoretically anyway) no prerequisites past introductory differential calculus. Mainly it's based on convexity arguments; I have a write-up here: ncatlab.org/toddtrimble/published/…; see theorem 3.1 and the crucial lemma 3.4. The "finesse" is akin to how we adjust parameter $a$ in $f: x \mapsto a^x$ to $a = e$ to get $f'(0) = 1$. Oct 10, 2017 at 13:22

I already thought that the following two sets of matrices are one. $$M(\color{blue}{\Bbb R},2n)\qquad \text{and}\qquad M(\color{red}{\Bbb C},n).$$

• ... matrices, presumably? Nov 5, 2020 at 4:17

This is more sort of a convention issue than an outright false belief (connected to the usual $$\emptyset$$ vs $$\{\emptyset\}$$ stuff), but I find it funny. I guess a fair share of mathematicians believe that: $$$$\bigcap\emptyset=\emptyset\label{eq}$$$$ while retaining the standard definition for intersection: $$\bigcap S:=\{x\ \text{such that}\ \forall Y(Y\in S\implies x\in Y)\}$$ according to which in fact: $$\bigcap\emptyset=V$$ where $$V$$ is the universal class. The condition in round brackets is of course vacuously true. So in a way - this is what I find funny - the former is the worst possible tentative solution of an equation ever.

False belief: << Let $$M$$ be the von Neumann algebra generated by a $$\rm{C}^{\star}$$-algebra $$\mathcal{A}$$. >>

The false belief is to think that the above sentence makes sense. In fact, a von Neumann algebras and a $$\rm{C}^{\star}$$-algebra don't have the same status. A von Neumann algebra is an operator algebra by definition, i.e. it is defined inside $$B(H)$$ for some separable Hilbert space $$H$$. Now, some subalgebras of $$B(H)$$ are (separable) $$\rm{C}^{\star}$$-algebras, but a $$\rm{C}^{\star}$$-algebra can also be defined abstractly. It can next be represented and a given representation $$H$$ (defined for example by GNS construction for a given state), if it is faithful, induces an embedding in $$B(H)$$.
So to make sense, the sentence above should be modified as:

<< Let $$M$$ be the von Neumann algebra generated by $$(\mathcal{A},\rho)$$, a couple of $$\rm{C}^{\star}$$-algebra and state. >>
or
<< Let $$M$$ be the von Neumann algebra generated by a $$\rm{C}^{\star}$$-algebra $$\mathcal{A}$$ represented on $$H$$. >>

Then, $$M = \pi_H(\mathcal{A})''$$. We can use $$M$$ to characterize the representation $$H$$, for example, we can talk about a representation of type $${\rm I}$$, $${\rm II}$$ or $${\rm III}$$ if $$M$$ is a von Neumann algebra of type $${\rm I}$$, $${\rm II}$$ or $${\rm III}$$. There is a $$\rm{C}^{\star}$$-algebra with representations of every type, for example the Cuntz algebra.

Finally, there exists a universal representation for every $$\rm{C}^{\star}$$-algebra (i.e. the direct sum of the corresponding GNS representations of all states; it is faithful). The associated von Neumann algebra is called the enveloping von Neumann algebra (it can also be defined as the double dual); it contains all the operator-algebraic information about the given $$\rm{C}^{\star}$$-algebra.

• So there is no abstract version of the notion of a von Neumann algebra? Like, say, isomorphism classes of "usual" von Neumann algebras, or something like that? Jan 12, 2018 at 21:53
• @მამუკაჯიბლაძე: A von Neumann algebra can be defined abstractly as a (non-necessarily separable) $\rm{C}^{\star}$-algebra that have a predual; but it is not the usual definition, some authors call this abstract version a $\rm{W}^{\star}$-algebra, see the last paragraph of en.wikipedia.org/wiki/Von_Neumann_algebra#Definitions Jan 12, 2018 at 22:23
• @SebastienPalcoux If one takes this abstract definition (or something equivalent to it), how does one recognise the concrete von Neumann algebras, i.e. what condition on a continuous *-homomorphism $\rho: A\to B(H)$ is equivalent to $\rho(A)$ being a von-Neumann-algebra? I'd guess it is something like "$\rho$ is still continuous if one chooses certain other natural topologies on $A$ and $B(H)$ instead of the norm topologies". Is that the case? Mar 16, 2018 at 23:24
• @JohannesHahn: A $\rm{C}^{\star}$-algebra (resp. von Neumann algebra) can be defined concretely as a $\star$-subalgebra of $B(H)$ closed by the operator norm topology (resp. the weak operator topology). The problem in your question is that these topologies are operator topologies, and $A$ is abstract. You could be satisfied by the following paragraph on the predual. Mar 17, 2018 at 8:27
• @SebastienPalcoux That paragraph is part of the reason why I'm asking. I checked wikipedia first of course. I don't see if or how it answers my question. Since the predual is intrinsic, the ultraweak topology can be defined intrinsically as well. So it makes sense to say "$\rho$ is continuous w.r.t. the ultraweak topologies on $A$ and $B(H)$". That's what makes me think that a characterisation like what I'm asking is even possible. But I don't see if it's true. Mar 17, 2018 at 14:35

"It is less risky to place 100 one-dollar bets on 100 independent coin flips than to place a single one-dollar bet on a single coin flip".

This is quite evidently false for any reasonable definition of "risky"; for example, in the former case the probability of losing more than a dollar is about 38%; in the latter case the probability is zero.

Nevertheless I am prepared to attest that at least two excellent mathematicians have told me that they believed this all their lives until it was pointed out to them that nothing like it can be true.

What is true is that it is less risky to place 100 one-dollar bets on a 100 independent coin flips than to place a single hundred dollar bet on a single coin flip.

A related fallacy (or the same fallacy in different language), which I have heard many times from both students and professional colleagues, is that "insurance works because the insurance company takes on many independent risks, thereby reducing their overall risk to nearly zero". This is of course false. The more independent risks the company takes on, the more risk it faces.

The correct statement is that insurance works because each risk is apportioned among a great many shareholders, each of whom is now effectively betting a very small amount on each of a great many independent coin flips. If there were only one shareholder, there would be no point in insurance unless either a) the single shareholder was pathologically risk-preferring or b) the premiums were sufficiently actuarially unfair for the company to earn supernormal profits, which for some reason did not get competed away.

• I've seen this phenomenon in other instances where something grows sublinearly with size, so that it's proportionally shrinking. An example I remember catching me out is thinking about how the speed of a centrifuge depends on its size if you want to maintain a constant $1g$ acceleration. Does it get faster or slower as you make it larger? Answer: It rotates fewer times per second, but the outer edge moves faster. Feb 7 at 10:34
• The statement about insurance companies becomes true if the premiums are set in such a way that the expected gain for the insurance company is positive. If, for example, their gain from each transaction is normal with mean $\mu>0$ and standard deviation $\sigma$, then their gain from $n$ independent transactions is normal with mean $n\mu$ and standard deviation $\sigma\sqrt n$, so the probability of making a loss decreases rapidly with $n$. May 23 at 11:10

"A closed unit-ball (in a Banach space) is compact!"

It only true in finite-dimensional spaces.

• Is this belief common? Alaoglu's theorem is pretty famous, and I would expect it to alert most people, even students, to the fact that the compactness of a closed unit ball is a subtle matter. Apr 19 at 22:05
• It would be common only for complete noobs to functional analysis. Apr 20 at 21:18

These are 2 instances which i have seen to happen with my friends. If $A$ and $B$ are 2 matrices, then they believe that $(A+B)^{2}=A^{2}+ 2 \cdot A \cdot B +B^{2}$.

Another mistake is if one i asked to solve this equation, $\displaystyle\frac{\sqrt{x}}{2}=-1$, people generally square both the sides and do get $x$ as $4$.

• What "people"? Non-mathematicians? May 4, 2011 at 0:03
• @Todd: No i was talking of high school students.
– C.S.
May 4, 2011 at 4:08
• @S.C.:if squarring both sides will not give the solution then how can second problem be solved? Oct 25, 2017 at 16:08

"It cannot be shown without some form of AC that the union (or disjoint union) of countably many countable sets is countable. I have a countably infinite set X of countably infinite sets. Therefore, the union of X cannot be shown to be countable without Choice."

The fallacy is that in many cases of interest, it is possible to exhibit an explicit counting of every element of X. In such a case a counting of X by antidiagonals is easily constructed. The usual counting of the rationals is an example of this.

I think this may even be an example of a more general phenomenon of "people think AC is necessary for a certain construction, but in fact it turns out not to be necessary for the example they have in mind". For example, AC is necessary to find a maximal ideal in an arbitrary ring ... but it isn't if you're prepared to assume the ring is Noetherian.

• If "Noetherian" is defined by the ascending chain condition or by requiring all ideals to be finitely generated, then in order to deduce the existence of maximal ideals, you still need a weak form of the axiom of choice. The usual argument uses the axiom of dependent choice. (Of course, if you define "Noetherian" to mean that every set of ideals has a maximal element, then deducing the existence of maximal ideals is a choiceless triviality.) A good reference is "Six impossible rings" by Wilfrid Hodges (J. Algebra 31 (1974) 218-244). Oct 22, 2010 at 15:29
• Thanks Andreas! I had a feeling there was a technicality somewhere there, but couldn't remember what it was. As a philosophical point I personally think that of course in the absence of AC you want to define Noetherian so that my original statement is true, but admittedly that's a harder sell than my countable-sets example. Nov 16, 2010 at 21:06
• @AndreasBlass's reference, clickably: Hodges - 6 impossible rings. Feb 5, 2019 at 1:09

A set is compact iff it is closed and bounded.

• This is perhaps a common false belief among undergraduates, but one that is dispelled by just a superficial acquaintance with functional analysis. Dec 9, 2013 at 2:45
• @ Todd Trimble: true, but then also the belief about $sin$ suggested by the OP is only common among people who have not completed a course in complex analysis. Dec 13, 2013 at 8:34
• I thought "bounded" is only defined on metric spaces, and this is true on metric spaces. Is that wrong? Sep 1, 2015 at 2:48
• I have seen analysis textbooks take this as a definition. I hope they realize that they are contributing to future confusion in their readers once they move on to topology or even metric spaces. @AkivaWeinberger, The Heine-Borel theorem stated in this way makes sense for arbitrary metric spaces, but it is only true for complete metric spaces for which balls are totally bounded. The correct statement of H-B for general metric spaces is "a metric space is compact iff it is complete and totally bounded". Oct 20, 2015 at 21:27
• @AkivaWeinberger: Yes, it is wrong. The closed unit ball of an normed vector space is compact if and only if the space is finite dimensional.
– ACL
Apr 21, 2016 at 13:43

In ${\mathbb F}_p^\times$, the non-squares are the opposite of the squares. In other words, $a$ is square iff $-a$ is not a square.

This is a confusion with the facts that the kernel of $x\mapsto x^2$ is $\{1,-1\}$ and the subgroup of squares has index $2$.

Most people that study Riemannian geometry for their first time make the following assumption at some point: "Let $(e_1,\dots,e_n)$ be a local orthonormal frame of $TM$ such that all Lie brackets $[e_i,e_j]$ vanish..."

This one is not so common (maybe special to me), but here we go: "$\mathbb{RP}^\infty$ and $\mathbb{CP}^\infty$ are Eilenberg-Mac Lane spaces, so $\mathbb{HP}^\infty$ is one, too."

• Although it's not an Eilenberg-MacLane space, $\mathbb{HP}^\infty$ is a classifying space (specifically $B(SU(2))$), just as $\mathbb{CP}^\infty = B(U(1))$ and $\mathbb{RP}^\infty = B(\mathbb{Z}/(2))$. Mar 3, 2020 at 15:55
• @RobertFurber That's right. And because $BO(1)$ and $BU(1)$ are Eilenberg-Mac Lane spaces, you can immediately classify real and complex line bundles by a single cohomology class. But not quaternionic line bundles. Mar 4, 2020 at 19:46

Anytime I wanted to write an answer to this question, I doubted maybe it is not as common as worthy of mentioning here. In fact, I am also not sure how common is the false belief that I observed today in a PDE class. I didn't observe that in many years of teaching calculus, but today four or five students in a small PDE class when calculating a definite integral by parts only applied the limits of the integral to the "second" integral, that is:

$$\int_{a}^b{f(x) g'(x) dx}=f(x) g(x) - \int_{a}^b{f'(x) g(x) dx}$$

Haven't I observed well enough in my calculus classes?

• Definitely integrals are numbers and $f(x)g(x)$ is a function of variable $x$. Formula as written is something very strange. Apr 20, 2016 at 18:58
• @FedorPetrov More strange is that most students don't see such a very strange something :) Apr 20, 2016 at 19:50

(1) All Lebesgue-null sets are countable, or are strongly measure zero. (2) The following,verbatim, was a Q in American Mathematical Monthly : " A student asserted that any uncountable real set has a closed uncountable subset. Is this true ?" .

I once misunderstood the definition of monads, and thought that for a monad $(T,\eta,\mu)$, we have $T\eta_X = \eta_{TX}$ (or fmap return == return in Haskell). Of course this is not the case (in case of $T=$[], fmap return [1,2] is [[1],[2]], whereas return [1,2] is [[1,2]]).

This might not be common, but I once believed the following.

Let $A, B$ be integers, and define a sequence by the linear recurrence $s_n = A s_{n-1} + B s_{n-2}$ with the base case $s_0 = 0$, $s_1 = 1$. Two important special cases are the Fibonacci sequence ($A = B = 1$) and the sequence $s_n = 2^n - 1$ (where $A = 3$, $B = -2$). Then, for any integers $n$ and $k$, $\gcd(s_n, s_k) = s_{\gcd(n,k)}$.

This is true in the two mentioned special cases, so it's tempting to believe it's true in general. But there's a counterexample: $A = B = k = 2$, $n = 3$.

Update: corrected the powers of two minus one example from $B = 2$ to $B = -2$. Thanks to Harry Altman.

• Quick correction, that should be A=3, B=-2 for 2^n-1. Apr 7, 2011 at 21:23
• Hmm, this raises an obvious question of whether it is true whenever (A,B)=1. Apr 22, 2020 at 23:09

Here are two beliefs. I think everybody will agree that one of them, at least, is false. I adhere to the second one.

Belief 1. The simplest way to compute the exponential $e^A$ of a complex square matrix $A$ is to use the Jordan decomposition.

Belief 2. It's simpler and more efficient to use the following fact.

Let $f(z)$ be the minimal polynomial of $A$, let $g(z)$ be $f(z)$ times the singular part of $e^z/f(z)$, and observe $e^A=g(A)$.

(By abuse of notation $z$ is at the same time an indeterminate and a complex variable.) (The problems of computing the exponential of $A$ and that of computing the Jordan decomposition of $A$ have the same difficulty level. But, to solve one of them, there is no need to refer to the other.) Here are two references

http://www.iecl.univ-lorraine.fr/~Pierre-Yves.Gaillard/DIVERS/Constant_coefficients/

Jordan decomposition is often mentioned in relation with matrix exponentials. I'm convinced (rightly or wrongly) that the association of these notions in this context is purely irrational. I think somebody once made this association by accident, and then many people repeated it mechanically.

Here is another attempt to describe the situation.

Put $B:=\mathbb C[A]$. This is a Banach algebra, and also a $\mathbb C[X]$-algebra ($X$ being an indeterminate). Let $$\mu=\prod_{s\in S}\ (X-s)^{m(s)}$$ be the minimal polynomial of $A$, and identify $B$ to $\mathbb C[X]/(\mu)$. The Chinese Remainder Theorem says that the canonical $\mathbb C[X]$-algebra morphism $$\Phi:B\to C:=\prod_{s\in S}\ \mathbb C[X]/(X-s)^{m(s)}$$ is bijective. Computing exponentials in $C$ is trivial, so the only missing piece in our puzzle is the explicit inversion of $\Phi$. Fix $s$ in $S$ and let $e_s$ be the element of $C$ which has a one at the $s$ place and zeros elsewhere. It suffices to compute $\Phi^{-1}(e_s)$. This element will be of the form $$f=g\ \frac{\mu}{(X-s)^{m(s)}}\mbox{ mod }\mu$$ with $f,g\in\mathbb C[X]$, the only requirement being $$g\equiv\frac{(X-s)^{m(s)}}{\mu}\mbox{ mod }(X-s)^{m(s)}$$ (the congruence taking place in the ring of rational fractions defined at $s$). So $g$ is given by Taylor's Formula.

This can be summarized as follows:

There is a unique polynomial $E$ such that $\deg E<\deg\mu$ and $e^A=E(A)$. Moreover $E$ can be uniquely written as $$E=\sum_{s\in S}\\ E_s\\ \frac{\mu}{(X-s)^{m(s)}}$$ with (for all $s$) $\deg E_s < m(s)$ and $$E_s\equiv e^s\ e^{X-s}\\ \frac{(X-s)^{m(s)}}{\mu}\mbox{ mod }(X-s)^{m(s)},$$ the congruence taking place in $\mathbb C[[X-s]]$.

• Dear Johannes, please reread my post. May 12, 2010 at 15:18
• Even a cursory examination of Nick Higham's book amazon.co.uk/Functions-Matrices-Computation-Nicholas-Higham/dp/… will show that both these opinions on the evaluation of matrix exponentials are hopelessly naive. May 15, 2010 at 9:17
• Dear Robin, Thanks for your answer. I don't have Nick Higham's book. I was wondering if you could be more precise. Your comment is very surprising to me: I thought I was stating a triviality. Here are two references en.wikipedia.org/wiki/Matrix_exponential#Alternative iecn.u-nancy.fr/~gaillard/DIVERS/Constant_coefficients Looking forward to hearing from you. May 15, 2010 at 10:51
• Your opinions are normative statements: "one should" and "it is better". It is naive to suppose that there is one best method that one should use to compute the matrix exponential. May 15, 2010 at 14:07
• I don't think the OP wants examples of normative statements. As I read it, the question is about conceptual errors regarding non-normative mathematical statements. May 17, 2010 at 6:19

False belief: relativization is well-defined and the corresponding notation $C^A$ is unambiguous. Which is not quite true because $P=NP$ would not imply $P^A=NP^A$.

• Maybe some more explanation would be useful. If decision problems and oracles are subsets of $\mathbb{N}$, and complexity classes are subsets of $P(\mathbb{N})$, then there is in general no such operation as relativization. I'm not sure how common of a false belief this is, but once I settled on my preference for the set point of view and saw what was going on here I lost some interest in the idea of relativization. Jan 13, 2018 at 5:16

I used to think that the subset of even norm vectors in an integral lattice is a sub-lattice. This is true for the "classically integral" lattice defined by $<u,v> \in \mathbb{Z}$ for $u,v$ in the lattice because the even vectors is the kernel of the group homomorphism $v \rightarrow <v,v>$ mod 2. However this fails for the more general notion of "integer norm" lattice where we only require the quadratic form is integer valued (ie. the coefficients are all integral or that the off diagonal entries in the Gram matrix may be half integral eg $x^2+xy+2y^2$). For the hexagonal lattice $x^2+xy+y^2$ which is not classically integral, the even vectors is a sub-lattice but for a different reason that it is the lattice scaled by 2.

If $t = 3$ mod 4, the lattice $L_t$ with quadratic form $x^2+ty^2$ is classically integral. Its even sublattice $L_{t0}$ has quadratic form $4(x^2+xy+(t+1)y^2/4)$ which clearly equals 2$W_t$ where $W_t$ is the lattice with quadratic form $(x^2+xy+(t+1)y^2/4)$. If $t=7$ mod 8, the coefficients of the form is [odd,odd,even], the even vectors is not a subgroup since for example $[0,1]$ and $[1,1]$ has even norm but $[0,1]+[1,1]=[1,2]$ has odd norm. If $t$ is 3 mod 8, the form $(x^2+xy+(t+1)y^2/4)$ can only be even only if both $x,y$ are even since all coefficients are odd. So the even vectors in $W_t$ turn out to be $2W_t$. It is a sub-lattice and it is the subset of even vectors but it is index 4 in $W_t$. It is the 2-scaled sub-lattice.

If $v \in L_t$, $2v \in 2L_t \subset L_{t0}=2W_t$, so $L_t \subset W_t$. So the picture is $2W_t=L_{t0} \subset L_t \subset W_t$ with each containment is index 2.

Just today, I realised that I had been mis-interpreting the FTFGAG. One can speak of the $2^\infty$-torsion subgroup or the $2'$-torsion subgroup, the $3^\infty$-torsion or the $3'$-torsion subgroup, or even just the torsion subgroup of a FGAG or the maximal torsion-free quotient … so surely one can speak of the torsion-free subgroup, right? In fact, when I was corrected on this, my first thought was to reply: "just take the subgroup consisting of all infinite-order elements", and it only occured to me as I was saying it to wonder how the identity element would squeeze its way into this so-called 'subgroup'.

• A good way to think of this is as a natural, exact sequence of abelian groups $1\to tor(G)\to G\to G/tor(G)\to 1$ (and similar sequences). If $G$ is f.g., the sequence happens to split, but not in a natural way. Thus there is not such thing as "the" complement of the torsion subgroup. Mar 16, 2018 at 23:08
• Consider the unit group $\mathbf Z[\sqrt{2}]^\times = \{\pm 1\} \times (1+\sqrt{2})^{\mathbf Z}$. Applying conjugation turns $(1+\sqrt{2})^{\mathbf Z}$ into $(1-\sqrt{2})^{\mathbf Z}$ and these are not the same subgroup complementing the torsion subgroup. Mar 16, 2018 at 23:39
• To be clear, I am aware that this is false! Mar 17, 2018 at 0:46
• @JohannesHahn, what was misleading me was the fact that, if $\pi$ is any set of primes and $G$ is torsion (as well as finitely generated Abelian), then $1 \to G[\pi^{\mathbb Z}] \to G \to G/G[\pi^{\mathbb Z}] \to 1$ does split canonically; so it seems so tempting to think that a similar result should hold without the torsion hypothesis …. Mar 17, 2018 at 1:52

It is a common mistake to believe that epimorphisms are either identical to surjections or that they are a better concept. Unfortunately this is rarely the case; epimorphisms can be very mysterious and have unexpected behavior

• Can you give a concrete example why it is a "mistake"? Apr 25, 2020 at 13:15
• If you mean set theoretic surjective then in every concrete category a morphism is surjective then the morphism is epi and in all the category that I know epi are something very reasonable for example in Hausdorff spaces epi are exactly the map with dense image ... Sep 16, 2020 at 13:00
• I think the best example where this constitutes a mistake is the category of rings. For example the inclusion $\mathbb{Z}\to \mathbb{Q}$ is an epimorphism. Oct 16, 2020 at 11:39

The following analogue of this result is clearly false:

Falsehood. If $$M$$ is a module over a commutative ring $$R$$, then $$M^\vee = \operatorname{Hom}_R(M,R)$$ is at least as big as $$M$$ (e.g. in terms of cardinality or rank).

For example, if $$R$$ is a domain and $$M$$ is torsion, then $$M^\vee = 0$$. But what's much more surprising is that the following is still false:

False belief: If $$M$$ is a torsion-free module over a principal ideal domain $$R$$ (even $$R = \mathbf Z$$), then $$|M^\vee| \geq |M|$$ and/or $$\operatorname{rk}(M^\vee) \geq \operatorname{rk}(M)$$.

(Even assuming $$M$$ has no divisble elements doesn't help.)

• One might add that what is being overlooked here is that the "correct dualizer" is an injective cogenerator, rather than a projective generator such as $R$. What makes it work for vector spaces is that a field is an injective cogenerator over itself. By the way, what are the rings for which injective hull of $R$ works? It seems to always work for torsion frees but not for all - even over integers. Presumably $R$ must be local for that? Feb 7, 2021 at 6:37

The "curse of dimensionality" means that in a hypercube the volume is increasingly concentrated in the corners as the number of dimensions increase.

In fact half the volume of a hypercube is closer to the centre than to the nearest vertex, with any number of dimensions.

The real curse is that the vast majority of the points of a unit hypercube of dimension $$n$$ are a distance less than $$\frac{5}{n}$$ from the outside of the hypercube, distances $$\sqrt{\frac{n}{12}} \pm \frac12$$ both from the centre and from the nearest vertex, and a distance $$\sqrt{\frac{n}{6}} \pm 1$$ from the vast majority of other points, which for large $$n$$ are narrow bands.

I don't think I've seen it in here:

Every vector space has a non-trivial dual space ($L^p$ for $0 < p < 1$ was a counter-example only mentioned during one of the classes in measure theory)

And of course there's the common false belief of people outside of mathematics that "mathematicians work with numbers and formulae all day long" :)

• Well, it is true that every vector space has a dual space, even $L^{1/2}$... and it is even true that every topological vector space has a continuous dual space... What you mean is that it is not true that every topological vector space has a non-trivial continuous dual space (or, that the continuous dual of a topological vector space does not necessarily separate points) Jul 7, 2010 at 18:54
• You are indeed correct. I'll do better not to dismiss the trivial case the next time. Jul 7, 2010 at 19:31
• The existence of a nonzero functional on a locally convex space is guarenteed by the Hahn-Banach theorem. $p$-Banach spaces are in general not locally convex if $p<1$.
– user20948
Feb 8, 2021 at 22:45

False Belief: "The suspension spectrum map from spaces to (edit: symmetric) spectra preserves smash-products"

The facts that one denotes the smash product of spectra and the smash product of a space with a spectrum (levelwise) with the same $\wedge$ and tends to leave away the $\Sigma^\infty$ when one embeds a space into spectra are also not helpful in getting used to the harsh reality that the above is wrong.

• Yay! 100th answer! Oct 4, 2010 at 21:33
• I don't see that this qualifies as a false belief. In order for the question of whether it is true or false to even be meaningful, you have to first commit yourself to one of the many different notions of spectrum, not to mention smash product of spectra. Oct 5, 2010 at 0:35
• True. I meant symmetric spectra with the smash product coming from their description as modules over the symmetric sequence of spheres. Oct 5, 2010 at 10:52