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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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    $\begingroup$ I have to say this is proving to be one of the more useful CW big-list questions on the site... $\endgroup$ May 6, 2010 at 0:55
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    $\begingroup$ The answers below are truly informative. Big thanks for your question. I have always loved your post here in MO and wordpress. $\endgroup$
    – Unknown
    May 22, 2010 at 9:04
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    $\begingroup$ 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. $\endgroup$
    – Suvrit
    Sep 20, 2010 at 12:39
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    $\begingroup$ It's a thought -- I might consider it. $\endgroup$
    – gowers
    Oct 4, 2010 at 20:13
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    $\begingroup$ Meta created tea.mathoverflow.net/discussion/1165/… $\endgroup$
    – user9072
    Oct 8, 2011 at 14:27

292 Answers 292

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I have checked the existing answers, but I think this one is not given yet. Sorry, if I missed it.

Although the incompleteness theory of Gödel is generally correctly understood, the consequence of it has multiple false beliefs:

  • Due to the incompleteness theory it is not possible to make an AI. Humans will always be be superior to the AI. This assumes that human thinking is complete and will eventually find the answer on any question.

  • Due to the incompleteness theory, it is not possible to formalize mathematics. This is refuted by many proof systems, which can formalize almost all mathematics.

As side note, I think this is partly fueled how logic is taught. It puts more emphasis on impossibilities (incompleteness theory), than possibilities (a proof system).

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  • $\begingroup$ +1. I always found it wrong that classes in logic put so much emphasis on negative results. (And I wish they had prepared me better for proof assistants... though I guess one semester does not suffice for the ones that exist today.) $\endgroup$ Sep 5, 2015 at 22:17
  • $\begingroup$ There's arguably too much fascination with incompleteness and not enough with completeness, which is more of a cornerstone of model theory. $\endgroup$
    – Todd Trimble
    Sep 6, 2015 at 1:50
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For awhile, I used to think:
If $depth\ M\ge depth\ N$ then $depth\ M_p\ge depth\ N_p$; for any prime ideal $p$ and finite R-modules $M$ and $N$ (Which is not true).

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This example is similar to this earlier answer.

If $k$ is a field, then $k[x] \otimes_k k[y] \cong k[x,y]$. Therefore also $k[[x]] \otimes_k k[[y]] \cong k[[x,y]]$, right?

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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.

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

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    $\begingroup$ For example, ...? $\endgroup$ Nov 5, 2020 at 21:36
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    $\begingroup$ 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.) $\endgroup$ Nov 5, 2020 at 23:23
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    $\begingroup$ 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). $\endgroup$ Nov 6, 2020 at 6:31
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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: \begin{equation} \bigcap\emptyset=\emptyset\label{eq}\end{equation} 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.

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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. There is no simple generalization of the Hodge Theorem to noncompact manifolds.

Belief 2. The most naive statement which would, if true, generalize the Hodge Theorem to noncompact manifolds is this.

The inclusion of the complex of coclosed harmonic forms into the de Rham complex of a riemannian manifold is a quasi-isomorphism.

This statement happens to be true.

Here is a reference:

http://www.iecl.univ-lorraine.fr/~Pierre-Yves.Gaillard/DIVERS/Hodgegaillard/ (Wayback Machine)

The simplest example is that of the real line with its standard metric. In degree zero the complex of coclosed harmonic forms is $\mathbb C\oplus\mathbb Cx$, and in degree one it is $\mathbb Cdx$, which gives the right cohomology.

Here is the (trivial) algebra background.

Let $A$ be a module over some unnamed ring, and let $d,\delta$ be two endomorphisms of $A$ satisfying $d^2=0=\delta^2$. Put $\Delta:=d\delta+\delta d$. Assume $A=\Delta A+A_{d,\delta}$ where $A_{d,\delta}$ stands for $\ker d\cap\ker\delta$. Write $A_{\delta,\Delta}$ for $\ker\Delta\cap\ker\delta$.

We claim that the natural map $$H(A_{\delta,\Delta},d)\to H(A,d)$$ between homology modules is bijective.

Injectivity. Assume $\delta da=0$ form some $a$ in $A$. We must find an $x$ in $A_{\delta,\Delta}$ such that $dx=da$. We have $a=\Delta b+c$ for some $b\in A$ and some $c\in A_{d,\delta}$. One easily checks that $x:=\delta db+c$ does the trick.

Surjectivity. Let $a$ be in $\ker d$. We must find $x\in A$, $y\in A_{d,\delta}$ such that $a=dx+y$. We have $a=\Delta b+c$ for some $b\in A$ and some $c\in A_{d,\delta}$. One easily checks that $x:=\delta b$, $y:=\delta db+c$ works.

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Let $A$ be an $n\times n$-matrix over a noncommutative (but associative and unital) ring. Then, $A$ is invertible if and only if its transpose $A^T$ is invertible, right?

Wrong. A counterexample (which I just learnt from Ulrich Krähmer and Blessing Bisola Oni, Pointed Hopf algebra (co)actions on rational functions, arXiv:2209.06516v3, Example 2.1.1) is the matrix $\begin{pmatrix} a&1 \\ 0&d \end{pmatrix}$ over the ring with two generators $a$ and $d$ and one relation $da=1$ (but not $ad=1$). Its inverse is $\begin{pmatrix} d&-1 \\ 1-ad&a \end{pmatrix}$, and the fact that the two matrices are mutually inverse is straightforward to verify. But its transpose $\begin{pmatrix} a&0 \\ 1&d \end{pmatrix}$ is not invertible, as its inverse would have to contain a right inverse to $a$, but then $a$ would be (two-sidedly) invertible.

Of course, this all stems from the fact that $\left(AB\right)^T$ is not $B^T A^T$ in general when the matrices $A$ and $B$ are over a noncommutative ring.

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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.

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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.
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    $\begingroup$ 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? $\endgroup$
    – Todd Trimble
    Sep 6, 2015 at 1:47
  • $\begingroup$ Yes‌‌‌‌‌‌‌‌‌‌‌‌. $\endgroup$ Sep 6, 2015 at 1:55
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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.

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    $\begingroup$ 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. $\endgroup$
    – KConrad
    Jul 15, 2017 at 3:26
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    $\begingroup$ "are not completely metrizable" is the wording you want. $\endgroup$ Jul 15, 2017 at 3:28
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    $\begingroup$ @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!) $\endgroup$ Jul 15, 2017 at 4:15
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    $\begingroup$ 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 $\mathbb{N}^\mathbb{N}$). $\endgroup$
    – Todd Trimble
    Jul 23, 2017 at 21:07
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    $\begingroup$ @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 $\endgroup$ Aug 11, 2017 at 14:41
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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.

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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).

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    $\begingroup$ 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. $\endgroup$
    – Todd Trimble
    Oct 8, 2017 at 3:14
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    $\begingroup$ @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? $\endgroup$ Oct 10, 2017 at 11:56
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    $\begingroup$ 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 $f'(0) = 1$. $\endgroup$
    – Todd Trimble
    Oct 10, 2017 at 13:22
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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

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    $\begingroup$ Can you give a concrete example why it is a "mistake"? $\endgroup$ Apr 25, 2020 at 13:15
  • $\begingroup$ 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 ... $\endgroup$ Sep 16, 2020 at 13:00
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    $\begingroup$ 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. $\endgroup$ Oct 16, 2020 at 11:39
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I already thought that the following two sets of matrices are one. $$M(\color{blue}{\Bbb R},2n)\qquad \text{and}\qquad M(\color{red}{\Bbb C},n).$$

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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.

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  • $\begingroup$ 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? $\endgroup$ Jan 12, 2018 at 21:53
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    $\begingroup$ @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 $\endgroup$ Jan 12, 2018 at 22:23
  • $\begingroup$ @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? $\endgroup$ Mar 16, 2018 at 23:24
  • $\begingroup$ @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. $\endgroup$ Mar 17, 2018 at 8:27
  • $\begingroup$ @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. $\endgroup$ Mar 17, 2018 at 14:35
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"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.

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  • $\begingroup$ 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. $\endgroup$ Feb 7, 2022 at 10:34
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    $\begingroup$ 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 $n$. $\endgroup$
    – gowers
    May 23, 2022 at 11:10
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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 divisible elements doesn't help.)

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    $\begingroup$ 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? $\endgroup$ Feb 7, 2021 at 6:37
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I searched through the answers and hope that I haven't missed this one:

The Minkowski sum of two closed and convex sets is closed and convex.

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  • $\begingroup$ Can you explain what goes wrong? $\endgroup$ Mar 3, 2023 at 22:17
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    $\begingroup$ Closedness is not necessarily true: Add the real line (in 2d) and the epigraph of exp (both closed and convex) and you get the open upper half plane. $\endgroup$
    – Dirk
    Mar 3, 2023 at 22:23
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    $\begingroup$ At least this is true if the sets are compact. $\endgroup$ Aug 24, 2023 at 12:28
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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 $4$.

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

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    $\begingroup$ 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). $\endgroup$ Oct 22, 2010 at 15:29
  • $\begingroup$ 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. $\endgroup$
    – Karol
    Nov 16, 2010 at 21:06
  • $\begingroup$ @AndreasBlass's reference, clickably: Hodges - 6 impossible rings. $\endgroup$
    – LSpice
    Feb 5, 2019 at 1:09
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If every collection of disjoint open sets in a topological space is at most countable, then the space is separable

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A set is compact iff it is closed and bounded.

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    $\begingroup$ This is perhaps a common false belief among undergraduates, but one that is dispelled by just a superficial acquaintance with functional analysis. $\endgroup$
    – Todd Trimble
    Dec 9, 2013 at 2:45
  • $\begingroup$ @ 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. $\endgroup$ Dec 13, 2013 at 8:34
  • $\begingroup$ I thought "bounded" is only defined on metric spaces, and this is true on metric spaces. Is that wrong? $\endgroup$ Sep 1, 2015 at 2:48
  • $\begingroup$ 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". $\endgroup$ Oct 20, 2015 at 21:27
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    $\begingroup$ @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. $\endgroup$
    – ACL
    Apr 21, 2016 at 13:43
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The "conditional Vitali convergence theorem": Let $X_n$ be a uniformly integrable sequence of random variables with $X_n \to X$ almost surely, and $\mathcal{G}$ a sub-$\sigma$-field. Then $\mathbb{E}[X_n \mid \mathcal{G}] \to \mathbb{E}[X \mid \mathcal{G}]$ almost surely (FALSE).

I believed this one until I read Uniformly integrable sequence such that a.s. limit and conditional expectation do not commute. It is particularly startling because the conditional versions of the monotone convergence theorem, the dominated convergence theorem, and Fatou's lemma are all true!

What is true is that $\mathbb{E}[X_n \mid \mathcal{G}] \to \mathbb{E}[X \mid \mathcal{G}]$ in $L^1$, so you do have a subsequence converging almost surely.

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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 $2$.

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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."

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    $\begingroup$ 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 $\mathbb{RP}^\infty = B(\mathbb{Z}/(2))$. $\endgroup$ Mar 3, 2020 at 15:55
  • $\begingroup$ @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. $\endgroup$ Mar 4, 2020 at 19:46
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(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 ?" .

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People seem to believe that conventional computation (for example, running a chaotic irreversible cellular automaton) can be as efficient as one wants simply with good engineering, but this is not the case. Landauer's principle states that erasing a bit of information always takes $\ln(2)\cdot k\cdot T$ energy where $k$ is Boltzmann's constant ($k=1.38065\cdot 10^{-23}$ Joules/Kelvin) and $T$ is the temperature. Landauer's principle is a consequence of the second law of thermodynamics since if Landauer's principle were violated, then entropy would decrease. Landauer's principle means that conventional irreversible computation always must take $\ln(2)\cdot k\cdot T$ energy per bit erased (and one can erase data just by running it through AND and OR gates, so every irreversible gate must take a minimum amount of energy by Landauer's principle). However, Landauer's principle does not apply to reversible computation since reversible computers are not allowed to erase data.

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    $\begingroup$ Okay, I see what you are aiming at with this last edit. The idea that ordinary computation can be made arbitrarily efficient is a reasonably common false belief about our physical world (and may even have a somewhat solid mathematical interpretation). I withdraw my initial objection. $\endgroup$
    – S. Carnahan
    Aug 11, 2017 at 14:04
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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]]).

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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 $P^A=NP^A$.

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    $\begingroup$ 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. $\endgroup$ Jan 13, 2018 at 5:16
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