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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

281 Answers 281

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Teaching introduction to analysis, I had students using the "fact" that if $f: [a,b] \rightarrow \mathbb{R}$ is continuous, then $[a,b]$ can be divided to subintervals $[a,c_1],[c_1,c_2],...,[c_n,b]$ such that $f$ is monotone on every subinterval. For instance you can use this "fact" to "prove" the (true) fact that $f$ must be bounded on $[a,b]$. Also, some students used the same "fact", but with countably many subintervals. I found this mistake hard to explain to students, because constructing a counterexample (such as the Weierstrass function) is impossible at the knowledge level of an introduction course.

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    $\begingroup$ Why not $x \sin(1/x)$ as example? $\endgroup$
    – user9072
    Jan 2, 2014 at 17:33
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    $\begingroup$ It is in the case of finitely many subintervals, but not in the case of countably many subintervals. $\endgroup$ Jan 2, 2014 at 19:17
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    $\begingroup$ You can surely discuss fractal shapes without needing to go into the details of a technical counterexample. The point seems to be that it is hard to imagine that "increasing at a point" and "increasing in a neighborhood of a point" are not the same for continuous functions. You can give easy examples showing that indeed they disagree, locally, and fractals suggest that you can make the disagreement happen everywhere. You can revisit this later, once more technology has been set in place. $\endgroup$ Jan 2, 2014 at 23:44
  • $\begingroup$ While technically it is true one can do it with countably many for the function I gave (if one includes degenerate intervals) I would be surprised if not at least some (or rather most) of the confusion of the students could be addressed by the example (possibly continuing with discussion along the lines suggested by @AndresCaicedo). $\endgroup$
    – user9072
    Jan 5, 2014 at 16:50
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False belief: Any orthonormal basis of a subvectorspace $W\subset V$ of an inner product space $V$ can always be extended to an ONB of $V$.

Counterexample: Let $V$ be $\bigoplus_{i\ge 1} \mathbb{R}$ with the inner product given by $\langle a_*,b_*\rangle =\sum_{i\ge 1} a_ib_i$ and let $W$ be the subvectorspace of $V$ spanned by $e_1+e_i$ for $i\ge 2$. The given set is basis and we can apply Gram-Schmidt to obtain an ONB.

However $W^\perp = 0$ so there is no way to complete it. Related false belief: $(W^\perp)^\perp=W$. These beliefs are all true in finite dimensions, but false in general.

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    $\begingroup$ That's why we like Hilbert spaces (inner product spaces that are complete w.r.t. the inner-product norm) much better than arbitrary inner-product spaces. $\endgroup$ Mar 4, 2016 at 4:08
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If $H$ and $K$ are subgroups of $G$, then $HK$ is a subgroup of $G$.

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    $\begingroup$ Hm, wonder how common that false belief actually is. It seems obviously implausible in the nonabelian case. $\endgroup$
    – Todd Trimble
    Sep 6, 2015 at 15:39
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    $\begingroup$ Its common, specially when undergraduates use product formula : $|HK|=\frac{|H||K|}{| H \cap K | }$ Because all of them are subgroup, except $HK$ probably. $\endgroup$
    – user68208
    Apr 10, 2016 at 17:30
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True: The solution operator of the linear one-dimensional time-dependent ordinary differential equation (ODE) $x' = a(t) x$ is $$ \exp\Bigl(\int\limits_{t_0}^{t} a(s) \, ds\Bigr). $$ True: The solution operator of the linear multi-dimensional time-independent ODE $x' = A x$ is $$ \exp\,(A(t - t_0)). $$ A quite popular misconception, even among research mathematicians:

The solution operator $\Phi(t;t_0)$ of the linear multi-dimensional time-dependent ODE $x' = A(t) x$ is $$ \exp\Bigl(\int\limits_{t_0}^{t} A(s) \, ds \Bigr), $$

perhaps strengthened by Liouville's formula:

$$ \det{\Phi(t;t_0)} = \exp\Bigl(\int\limits_{t_0}^{t} \operatorname{tr}{A(s)} \, ds\Bigr). $$

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  • $\begingroup$ I am holding this belief right now. $\endgroup$
    – Michael
    Apr 27, 2018 at 23:26
  • $\begingroup$ Is there a simple expression that is true? $\endgroup$
    – Hans
    May 3, 2018 at 18:11
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    $\begingroup$ @Hans The Peano-Baker series is, in my opinion, simple, and converges where it should. There is a nice paper by Baake and Schlägel The Peano-Baker series, Proceedings of the Steklov Institute of Mathematics 275 (1) (2011), 155-159 (the paper is behind a paywall on the Publisher's page, but the authors put a copy on ResearchGate (researchgate.net/publication/47702535_The_Peano-Baker_series)). $\endgroup$
    – user539887
    May 4, 2018 at 8:15
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    $\begingroup$ @Hans In contrast, the Magnus expansion is very complicated and may not converge except close to the initial time. $\endgroup$
    – user539887
    May 4, 2018 at 8:20
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    $\begingroup$ @Hans Physicists refer to the time ordered exponential: en.wikipedia.org/wiki/Ordered_exponential $\endgroup$ Aug 27, 2018 at 14:58
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False belief. If a family $(x_n)_{n\geq 1}$ is commutatively convergent (i.e. summable) in a normed space $(V,\|\ \|)$ then $$ \sum_{n\geq 1} \|x_n\|<+\infty\ . $$

This is true in finite dimensions and has counterexamples in infinite dimensions. Details and counterexamples can be found there.

Recall that a family $(x_i)_{i\in I}$ is called summable with sum $S$ iff $$ (\forall \epsilon>0)(\exists F\subset_{finite} I)(\forall F_1\subset_{finite} I)(F\subset F_1\Longrightarrow\\ \|S-\sum_{i\in F_1}x_i\|<\epsilon) $$ This is equivalent with commutative convergence in case $I\subset \mathbb{N}$ is infinite.

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    $\begingroup$ I particularly like Grothendieck's version of this. He proves that unconditional convergence is the same as absolute convergence for a locally convex space if and only if it is nuclear, i.e. every continuous linear map to a normed space is a nuclear map. The falsity of this belief then follows from the fact that a nuclear normed space is finite dimensional. $\endgroup$ Feb 17, 2018 at 7:44
  • $\begingroup$ @RobertFurber Thank you for this learned description (+1) $\endgroup$ Feb 17, 2018 at 8:03
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    $\begingroup$ In fact, a commonly neglected fact is that the representation of an element $x$ of a Hilbert space as sum of its Fourier components in a Hilbert basis $\{u_\lambda\}_{\lambda\in\Lambda}$, namely $x=\sum_{\lambda\in\Lambda} (x\cdot u_\lambda)u_\lambda$, is always a summable family in $H$. Recalling this, the above false belief would just reduce to the very false "$\ell_1(\Lambda)=\ell_2(\Lambda)$" for any set $\Lambda$, that hopefully not many believe! $\endgroup$ Aug 1, 2019 at 10:33
  • $\begingroup$ @ good hint (+1) $\endgroup$ Aug 1, 2019 at 17:06
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When I was studying Banach spaces, I was confused with the following: We know that, in any Banach Space $V$, the closed unit ball is compact in the topology generated by the norm if, and only if, the dimension of $V$ is finite. But thinking about $\mathbb R$ as a vector space over $\mathbb Q$, we have an infinite-dimensional vector space which is complete in the norm (given by the modulus) but the closed unit ball is, of course, compact in topology generated by the norm.

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

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    $\begingroup$ You can define Banach spaces over any complete field. For example, one can define p-adic Banach spaces. But Q isn't complete with respect to any of its norms. $\endgroup$ May 4, 2010 at 22:14
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    $\begingroup$ In fact the Theorem I mentioned in my answer, is based on the Riesz lemma and this lemma is not valid if the scalar fields is not complete. $\endgroup$
    – Leandro
    May 4, 2010 at 22:34
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    $\begingroup$ @QiaochuYuan ...unless you use the trivial norm. $\endgroup$ Oct 20, 2015 at 21:16
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As a student, I thought (for quite a while) that our textbook had stated that tensoring commutes with taking homology groups. It wasn't until calculating the homology groups of the real projective plane over rings Z and Z/2Z that I realized my mistake.

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Two very common errors I see in (bad) statistics textbooks are

(i) zero 3rd moment implies symmetry (though generally stated in terms of "skewness", where skewness has just been defined as a scaled third moment)

(ii) the median lies between the mean and the mode

(I have seen a bunch of related errors as well.)

Another one I often see is some form of claim that the t-statistic goes to the t-distribution (with the usual degrees of freedom) in large samples from non-normal distributions.

Even if we take as given that the samples are drawn under conditions where the central limit theorem holds, this is not the case. I have even seen (flawed) informal arguments given for it.

What does happen is (given some form of the CLT applies) Slutzky's theorem implies that the t-statistic goes to a standard normal as the sample size goes to infinity, and of course the t-distribution also goes to the same thing in the limit - but so, for example, would a t-distribution with only half the degrees of freedom - and countless other things would as well.

The first two errors are readily demonstrated to be false by simple counterexample, and to convince people that they don't have the third usually only requires pointing out that the numerator and denominator of the t-statistic won't be independent if the distribution is non-normal, or any of several other issues, and they usually realize quite quickly that you can't just hand-wave this folk-theorem into existence.

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  • $\begingroup$ In the statistics text at the college where I teach, (ii) is universal among the examples given, so I formulated the conjecture; but when I tried to prove it and thought about what the mode really is, I realised how badly behaved that can be and found immediate counterexamples. (Then this gets me wondering why anybody would bother using the mode as a statistic for anything, since it's pretty much meaningless, but never mind.) $\endgroup$ Apr 4, 2011 at 9:24
  • $\begingroup$ Toby: sure, you use the mode for cases when the domain of the measurement is not an ordered set but just a set without structure and so the median wouldn't make sense. $\endgroup$ Apr 7, 2011 at 12:01
  • $\begingroup$ On (ii) it is easy to find counterexamples, particularly with the mean between the mean and mode for discrete random variables. For example $P(X=0)=0.37, P(X=1)=0.33, P(X=2)=0.3$ with $E[X]=0.93$ between $0$ and $1$ $\endgroup$
    – Henry
    Jan 4 at 14:16
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Draw the graph of a continuous function $f$ (from $\mathbb{R}$ to $\mathbb{R}$). Now draw two dashed curves: one which everywhere a distance $\epsilon$ above the graph of $f$ and one which is everywhere a distance $\epsilon$ below the graph of $f$. Then the open $\epsilon$-ball around $f$ (with respect to the uniform norm) is all functions which fit strictly between the two dashed curves.

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    $\begingroup$ Surely this is true if you are talking about the closed ball, and only just barely false for the open ball (and if we were talking about functions from $[a,b]$ to $\mathbb{R}$ it would be true)? Or else I am one of those with the false belief... $\endgroup$ Oct 10, 2010 at 18:26
  • $\begingroup$ You are right, I should have specified open ball, thanks. I think it is just barely false for the open ball. Honestly, I held this false belief until a couple of days ago, and I haven't thought much about correcting my belief. Probably the real open epsilon ball is the union of all functions that fit between dashed curves a distance strictly less than epsilon away from f? At any rate, I think the above picture is the right way to think about it most of the time. But it gives results such as $tan^{-1}$ being in the open ball of radious pi/2 centered at 0 if you interpret it literally. $\endgroup$
    – user4977
    Oct 10, 2010 at 19:24
  • $\begingroup$ Hmm, very nice (once clarified to the open ball)! Easily dispelled as soon as you question it, but I could easily imagine using it without thinking and missing the alternation of quantifiers that’s going on under the surface. $\endgroup$ Dec 1, 2010 at 15:30
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1- A very common mistake that 1st year students (but not even a single mathematician) think that it is true is "a transitive and symmetric relation on a set is reflexive". But as the empty set is a transitive and symmetric relation but not reflexive on any non-empty set. Of course there lots of non-trivial examples also.

2- Another common mistake is that the expression "countable union of countable sets is again countable" is independent of axiom of choice (AC). Many people make the proof of this statement without mentioning axiom of choice. Indeed, in his holly book Algebra, Lang proves this statement just by taking an ordering from each countable set and continues without the mentioning AC.

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    $\begingroup$ For big-list questions, it's usually best to post independent answers as separate answers. $\endgroup$ Dec 2, 2010 at 15:13
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    $\begingroup$ +1 for #2. Baby Rudin is another offender. And many authors use so-called "diagonalization tricks" for proving compactness theorems like Arzela-Ascoli and Prohorov, which typically reduce to the compactness of $[0,1]^\mathbb{N}$. $\endgroup$ Dec 2, 2010 at 15:21
  • $\begingroup$ Isn't the more right statement that a transitive and irreflexive relation is assymetric? $\endgroup$ Dec 4, 2010 at 22:46
  • $\begingroup$ Similar to 1: a subset of a group that is closed under multiplication and inversion is a subgroup. $\endgroup$ Feb 7, 2021 at 4:17
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False belief: A function being continuous in some open interval implies that it is also differentiable on some point in that interval:

Counterexample:

The Weierstrass function is an example of a function that is continuous everywhere but differentiable nowhere:

$f(x) = \sum_{n=0}^\infty a^n \cos(b^n \pi x)$

Where $a \in (0, 1)$, $b$ is a positive odd integer, and $ab > 1 + \frac{3\pi}{2}$. The function has fractal-like behavior, which leads to it not being differentiable. This notion is rather disheartening to most calculus students, though.

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    $\begingroup$ Related: if f is continuous on the interval I, there must be an interval J in I on which f is monotone. Easily believed by the beginner. $\endgroup$ Aug 31, 2010 at 2:34
  • $\begingroup$ Did you mean "differentiable on some point in that interval:" ? $\endgroup$
    – Rasmus
    Sep 18, 2013 at 19:13
  • $\begingroup$ Haha, figures that I edit a 3 year old post to introduce an even worse typo. Yes, that is what I meant. $\endgroup$ Sep 18, 2013 at 22:53
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I don't know how common this mistake is, but I think it's worth mentioning. I used to think that existence of non-measurable sets is guaranteed by the axiom of choice only.

In the presence of AC, there cannot be a $\sigma$-additive measure on $\mathcal{P}(\mathbb{R})$ that extends the usual Lebesgue measure.

It is true that we cannot extend the Lebesgue measure in a translation-invariant way by various Vitali set constructions. On the other hand, if you do not insist that the extension is translation-invariant, it might be possible to do this relative to a real-valued measurable cardinal assumption.

Theorem (Ulam): If there exists a cardinal $\kappa$ such that there exists an atomless $\kappa$-additive probability measure on $\mathcal{P}(\kappa)$, then there exists a $\sigma$-additive measure on $\mathcal{P}(\mathbb{R})$ extending the Lebesgue measure.

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  • $\begingroup$ I think you need $\kappa\leq\frak c$, no? $\endgroup$
    – Asaf Karagila
    Jan 22, 2015 at 14:44
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    $\begingroup$ @AsafKaragila: I believe the assumption that our measure is atomless already implies that $\kappa \leq 2^{\omega}$. $\endgroup$
    – Burak
    Jan 22, 2015 at 14:45
  • $\begingroup$ Take any measurable cardinal, then there is an atomless probability measure on its power set. It's just that an event is either improbable or its negation is improbable. Unless by probability measure you mean it obtains many values, not just two. $\endgroup$
    – Asaf Karagila
    Jan 22, 2015 at 14:48
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    $\begingroup$ Isn't that measure atomic if you are deriving it from the ultrafilter? (By an atom, I mean any $A$ of positive measure such that for any $B \subseteq A$ either $\mu(B)=0$ or $\mu(B)=\mu(A)$). I will have to catch a course now but the theorem I referred to should be in Kanomori (indeed, I checked the pdf and it is Theorem 2.5) $\endgroup$
    – Burak
    Jan 22, 2015 at 14:53
  • $\begingroup$ Ohhhh, right. I was thinking about atoms in the sense of Boolean algebra, as minimal positive elements. Thanks for the clarification! $\endgroup$
    – Asaf Karagila
    Jan 22, 2015 at 14:55
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Many students believe that every abelian subgroup is a normal subgroup.

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    $\begingroup$ And many other students believe every normal subgroup is an abelian subgroup. $\endgroup$ Jul 3, 2018 at 22:49
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    $\begingroup$ and many believe that a subgroup of an abelian group is normal (which I also believe). $\endgroup$
    – user137767
    Apr 16, 2019 at 21:17
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"A real symmetric matrix is positive-definite iff all the leading principal minors are positive, and positive-semidefinite iff all the leading principal minors are nonnegative."

This paper collects some evidence that this belief is "common", and presents a counterexample (of size $3\times 3$. Exercise: find an example of size $2\times 2$).

(Related to, but not the same as this answer.)

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    $\begingroup$ $$\pmatrix{0&0\cr0&-1\cr}$$ $\endgroup$ Jul 29, 2015 at 3:22
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While this example was already mentioned by Skopenkov in a comment, here it is, explicitly:

If $G\times X\to X$ is a free continuous action of a compact metrizable group on a compact metrizable space $X$, then $X\to X/G$ is a principal $G$-bundle.

This is true if $G$ is a Lie group (Gleason) and, more generally, for proper Lie group actions on locally compact spaces (Palais), but false in general (the example is essentially due to Kolmogorov), see:

R. F. Williams, A useful functor and three famous examples in topology. Trans. Amer. Math. Soc. 106 (1963) 319–329.

In this example, $\dim(X/G)=2, \dim(X)=1$.

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If $A$ and $D$ are $n \times n$ matrices and $D$ is diagonal, then $A \cdot D=D \cdot A$.

Many of my Linear Algebra students believe this, also because it's written in the textbook in some form, but it's only true if $D$ is a multiple of the identity matrix or if both $A$ and $D$ are diagonal.

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    $\begingroup$ Grading their work on similar matrices must be interesting. $\endgroup$
    – nombre
    May 1, 2019 at 21:01
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Some undergraduate common false beliefs that I found

(1) If $H$ is a subgroup of $\mathbb{Z}$ and $H$ and $\mathbb{Z}$ are isomorphic, then $H = \mathbb{Z}$;

(2) In a metric space every two open balls are homeomorphic;

(3) For $ \ p \in [1, \infty] \, $, $ \, L^p(X, \mathfrak{M}, \mu) = \big\{ f \in \mathbb{C}^X : \int_X |f|^p \, d \mu < \infty \big\} \, $ is a $\mathbb{C}$-normed vector space, with norm $ \ \lVert f \rVert_p = \big( \int_X |f|^p \, d \mu \big)^{1/p} \, $.

Belief (1) is very naive, for every nontrivial subgroup of $\mathbb{Z}$ is of the form $n \mathbb{Z}$, all of them isomorphic to $\mathbb{Z}$. For (2) people tend to think of normed vector spaces and forget the discrete metric spaces. For (3) some people just forget that one have to consider the quotient space, where the classes $[f]=[g] \ $ iff $ \ f=g \ $ $\mu$-almost everywhere.

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  • $\begingroup$ Well, in (1) I think I can replace $\mathbb{Z}$ by an arbitrary group $G$, because $\mathbb{Z}$ do not come in mind so quickly. $\endgroup$
    – Gustavo
    Jan 8, 2016 at 4:03
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The following inexact belief can be spotted in many a textbook for undergraduates: the principle of mathematical induction and the well-ordering principle for $\mathbb{N}$ are equivalent.

Lars-Daniel Öhman wrote about this misbelief in his paper Are induction and well-ordering equivalent? (Math. Intelligencer, vol. 41 (2019), no. 3, pp. 33-40.). To make a long story short, one of the (main) points by Öhman in the said article is that if we define the natural numbers à la Peano, i.e. as a set $N$ endowed with a function $S \colon N \to N$ satisfying the following axioms:

  1. $0\in N$,
  2. $\forall n \in N \, (S(n) \neq 0)$,
  3. $S$ is injective, and
  4. (P.M.I.) if $M$ is a subset of $N$ such that $0 \in M$ and $S(m) \in M$ for every $m \in M$, then $M=N$.;

then, the P.M.I. is not equivalent to the well-ordering principle (every nonempty subset of $\mathbb{N}$ has a least element; heretofore, W.O.P.) relative to axioms 1 - 3.

One route that Öhman follows to evince that there are issues with the so-called equivalence of both principles is by illustrating that, in the popular proof of the implication W.O.P. $\Rightarrow$ P.M.I., the existence of an immediate predecessor for every $n \in \mathbb{N}$ is assumed: this is an assumption that can not be obtained as a consequence of 1, 2, 3, and W.B.O., "as evidenced by the existence of a model... in which this property [about immediate predecessors] does not hold [whereas 1, 2, 3, and W.B.O. do]"... In point of fact, the model he provides to exemplify the validity of the assertion between quotation marks is one in which the P.M.I. does not hold either.

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    $\begingroup$ Would we get an equivalence between PMI and WOP, if we replaced axiom 2 with $\forall n\in N\left(n\ne0\Longleftrightarrow\exists m\in N\left(S\!\left(m)=n\right)\right)\right)$ $\endgroup$ Jul 29, 2021 at 19:32
  • $\begingroup$ @VladimirReshetnikov: Good night! I've just corrected the typo that you mentioned in your first comment. Thanks for the links you have shared with me... $\endgroup$ Jul 30, 2021 at 4:38
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    $\begingroup$ @VladimirReshetnikov: Answer to the question in your second comment: Yes. Öhman touches upon this issue on the antepenultimate page of his paper: «One way of actually making the w.o.p. and the p.m.i. equivalent relative to the other axioms is to supplant axiom $\forall n \in N, \, S(n) \neq 0$ with a slightly different version of it, namely: (2') "$\forall n \in N, \, S(n) \neq 0$ and $0$ is the only element that is not a successor". Axiom (2') is obviously stronger than axiom (2). In a sense, this added strength makes up for the weakness of the w.o.p. in relation to the p.m.i.» $\endgroup$ Jul 30, 2021 at 5:05
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Here's one I was reminded recently during lunch in the common room.

A maximal abelian subalgebra of a semisimple Lie algebra is a Cartan subalgebra.

This is true for compact real forms of semisimple Lie algebras, but fails in general. The missing condition is that the subalgebra should equal its normaliser.

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  • $\begingroup$ For example, the usual proof of Schur's theorem on commutative algebras of matrices produces abelian Lie subalgebras of $\mathfrak{sl}_n$ much larger than the rank of $\mathfrak{sl}_n$. $\endgroup$ Aug 4, 2010 at 23:41
  • $\begingroup$ Yes, my favourite example (being a physicist) is the stabiliser in $\mathfrak{so}(1,n)$ of a nonzero zero-norm vector in $\mathbb{R}^{1,n}$ for $n>4$. The stabiliser contains an abelian ideal (infinitesimal null rotations) of dimension $n-1$. $\endgroup$ Aug 5, 2010 at 0:02
  • $\begingroup$ I meant to write $n>3$ above. $\endgroup$ Aug 5, 2010 at 0:03
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    $\begingroup$ ``The missing condition is that the subalgebra should equal its normaliser'', or that the subalgebra consists of semisimple elements, no? (That provides another perspective on why it's true for compact real forms.) $\endgroup$
    – LSpice
    Dec 12, 2013 at 23:26
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An elementary false belief in elementary number theory: for $a, b, c\hspace{.1cm}\varepsilon\hspace{.1cm} \mathbb{N}$

$LCM\left(a,b\right)\times GCF\left(a,b\right) = ab$ .

Thus, $LCM\left(a,b,c\right)\times GCF\left(a,b,c\right) = abc$.

In general, $\left(a_1,a_2,\ldots,a_n\right)[a_1,a_2,\ldots,a_n] = a_1a_2\ldots a_n$.

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  • $\begingroup$ Does GCF mean gcd (greatest common divisor) here? $\endgroup$ Nov 27, 2010 at 19:33
  • $\begingroup$ Yes. GCF means the same: Greatest Common Factor. $\endgroup$
    – Unknown
    Nov 29, 2010 at 6:46
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    $\begingroup$ This kind of stuff is easy to rule out, though; it's dimensionally inconsistent. Replacing a, b, c by ka, kb, kc leads to a quick contradiction. $\endgroup$ Feb 24, 2011 at 21:08
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    $\begingroup$ @Qiaochu, that is a nice quick check. Let the RCF(remnant common factor) be the leftover factor that would make the above second equality true. There seems to be no interesting way of determining $RCF\left(a,b,c\right)$ so that $LCM\left(a,b,c\right)\times RCF\left(a,b,c\right) \times GCF\left(a,b,c\right) = abc$ $\endgroup$
    – Unknown
    Feb 24, 2011 at 22:05
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    $\begingroup$ @Elohemahab: actually, the correct generalization is $\gcd(a, b, c) \text{lcm}(ab, bc, ca) = \text{lcm}(a, b, c) \gcd(ab, bc, ca) = abc$. $\endgroup$ May 8, 2011 at 23:47
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This is more of a false philosophy than a clear mistake, but nevertheless it is very common:

A compact topological space must be "small" in some sense: it should be second countable or separable or have cardinality $ \le 2^{\aleph_0}$, etc.

This is all true for compact metric spaces, but in the general case, Tychonoff's theorem gives plenty of examples of compact spaces which are "huge" in the above sense.

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  • $\begingroup$ More or less along the same lines, one is usually lead to think that "every topological space is Hausdorff". $\endgroup$ May 4, 2011 at 14:43
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A possible false belief is that "a maximal Abelian subgroup of a compact connected Lie group is a maximal torus". Think of the $\mathbf Z_2\times\mathbf Z_2$-subgroup of $SO(3)$ given by diagonal matrices with $\pm1$ entries.

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    $\begingroup$ Fu... I just "proved" that again as an exercise a few days ago. $\endgroup$ Mar 6, 2013 at 0:02
  • $\begingroup$ Or, which is essentially the same example, $\left\langle\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}, \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\right\rangle$ inside $\operatorname U(2)/\operatorname U(1)$. $\endgroup$
    – LSpice
    Aug 5, 2020 at 21:09
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Let $B(r_1) \subset B(r_2)$ be two open balls of radius $r_1$ and $r_2$ respectively. Then $r_1 \leq r_2$.

Bounded metric spaces give trivial counterexamples. Also, $B \left( \frac{1}{6}, \frac{2}{3} \right) \subsetneq B \left( \frac{1}{2}, \frac{1}{2} \right)$ in $(0,+ \infty)$.

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  • $\begingroup$ What is $B(a,b)$ here? $\endgroup$
    – Turbo
    Dec 3, 2013 at 10:09
  • $\begingroup$ Here, $B(a,b)$ is the open ball of radius $b$ centered at $a$. $\endgroup$
    – Seirios
    Dec 3, 2013 at 21:41
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"Let $E$ be a complete locally convex topological vector space (or a complete topological vector space or a complete topological group) and let $F$ be closed vector subspace (or a closed subgroup). Then the quotient $E/F$ is complete."

This just has to be true. One can almost see the proof. And in fact it is true for Banach spaces. So it has to be true for locally convex spaces as well.

Another one with completions:

"Every topological group is a dense subgroup of a complete topological group." True for abelian groups but false in general (take the homeomorphism group of $[0,1]$ with the compact open topology)

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Yet another common false belief is :

"For non constant periodic functions $f: \mathbb{R} \to \mathbb{R}$ and $g: \mathbb{R} \to \mathbb{R}$ with smallest positive periods $p_1 , p_2$ respectively, the sum $f+g$ is periodic if and only if $\frac{p_1}{p_2}$ is rational."

One side of the statement above is true (the latter implies the former, but the former does not necessarily imply the latter. (But it's true for continuous functions.)

As an exercise one may like to prove the following : (Source : Miklos Schweitzer competition)

Given any two positive real numbers $p_1,p_2$, there exists functions $f_1 : \mathbb{R} \to \mathbb{R}$ with smallest positive period $=p_1$, and $f_2 : \mathbb{R} \to \mathbb{R}$ with smallest positive period $=p_2$, such that $f_1+f_2$ is also periodic.

One more interesting false belief is :

"If $f: \mathbb{R} \to \mathbb{R}$ is a continuous function taking unequal values at points $x_1,x_2 \in \mathbb{R}$ with $x_1 < x_2$ then there is some sub-interval of $(x_1,x_2)$ on which $f$ is either strictly increasing or strictly decreasing."

This is wrong. Among the most familiar counterexamples is the Devil's staircase.

I believe this false belief is often due to the habit of seeing continuous functions wearing the glasses provided by the Intermediate Value Property these functions have.

Addendum : Some nice examples from wikipdia, here : https://en.wikipedia.org/wiki/Normally_distributed_and_uncorrelated_does_not_imply_independent

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    $\begingroup$ I think this false belief is mostly due to the underlying "$f$ is continuous" hypothesis that people often carry around $\endgroup$ Jun 12, 2019 at 9:51
  • $\begingroup$ @Max the condition of continuity can be relaxed. In my opinion this false belief often arises due to the negligence of the subtle role played by continuity in proving the statement for continuous functions. $\endgroup$ Jun 12, 2019 at 11:09
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Consider the following well-known result: Let $(E,\leq)$ be an ordered set. Then the following are equivalent: (i) Every nonempty subset of $E$ has a maximal element. (ii) Every increasing sequence in $E$ is stationary.

It is immediate that (i) implies (ii). To prove the converse, one assumes that (i) is false and then "constructs step by step" a strictly increasing sequence.

The common mistake (which I have seen in textbooks) is to describe the latter construction as a proof by induction. In fact, the construction uses the axiom of choice (or at least the dependent choice axiom).

(As a special case, I don't think ZF can prove that every PID is a UFD.)

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    $\begingroup$ It’s not exactly wrong to call it a proof by induction. In ZFC, the proof of dependent choice — or of just about any instance of it, eg the one here — works by combining induction and choice. So I’d agree it’s wrong to sweep the choice under the carpet; but if you’re not explicitly invoking DC, then you will be using induction as well. $\endgroup$ Dec 1, 2010 at 15:34
  • $\begingroup$ Peter, let's state DC as follows: "If $(p_n:X_{n+1}\to X_n)$ is an $\mathbb{N}$-projective system of nonempty sets with all $p_n$ surjective , then projlim($X_n$) is nonempty." Proof from AC: put $X:=\coprod_{n\geq0}X_n$ and $X^+=\coprod_{n>0}X_n$ with obvious map $p:X^+\to X$. Then $p$ is onto, so has a section $s$ (family of sections of all $p_n$'s). Given $x_0\in X_0$, sequence $(s^n(x_0))$ is an element of projlim($X_n$). I agree that we do need induction to define $s^n$. But iteration of a map is such a basic tool that I don't agree to call any proof using it a "proof by induction". $\endgroup$ Dec 7, 2010 at 11:49
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"If a field $K$ has characteristic 0 and $G$ is a group, then all $KG$-modules are completely reducible."

True for finite groups but very false in general.

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I have heard the following a few times :

"If $f$ is holomorphic on a region $\Omega$ and not one-to-one, then $f'$ must vanish somewhere in $\Omega$."

$f(z)=e^z$ of course is a counterexample.

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    $\begingroup$ is it true though if the image is simply-connected? $\endgroup$
    – KotelKanim
    Apr 17, 2011 at 11:33
  • $\begingroup$ thanks. that's great. I never thought about it before, but it just sounded right... $\endgroup$
    – KotelKanim
    Apr 19, 2011 at 7:46
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    $\begingroup$ No true. Take $f(z)=z^3-3z$ and restrict it to the complement of $\lbrace 1,-1\rbrace$ so that $f'(z)$ is never $0$. It maps this domain onto $\mathbb C$. $\endgroup$ May 4, 2011 at 0:16
  • $\begingroup$ @TomGoodwillie: what if both domain and image are simply-connected? $\endgroup$
    – Michael
    Dec 3, 2013 at 0:45
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If $\alpha>0$ is not an integer, the set of functions $f:[a,b]\rightarrow\mathbb R$ such that $$\sup_{y\ne x}\frac{|f(y)-f(x)|}{|y-x|^\alpha}<+\infty$$ is ${\mathcal C}^\alpha([a,b])$.

False for $\alpha>1$, because this set contains only constant functions.

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This is a common error made by mature mathematicians in many books and papers in analysis, especially in differential equations: If $X$ is a closed subspace of a Banach space $Y$, then the $Y^*$ (the dual of $Y$) is isomorphic to a subspace of $X^*$ (the dual of $X$). It is false (of course) since Euclidian space $\mathbb R$ is a subspace of $\mathbb R^2$, yet the dual of $\mathbb R^2=\mathbb R^2$ is not isomorphic to a subspace of the dual of $\mathbb R=\mathbb R$. I guess, sometimes they really, really want it to be true. Cheers Boris

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  • $\begingroup$ I'll take your word for it, but since that statement is false even without introducing norms and topology, it staggers me that people could even believe that. They might say it without thinking, I guess $\endgroup$
    – Yemon Choi
    Oct 20, 2010 at 2:58
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    $\begingroup$ I would also be shocked if this really gets believed often! It seems to be a sort of “mis-dualisation”: they dualise “$X$ is a subobject of $Y$” to “$Y^*$ is a subobject of $X^*$”, where the correct dual is “$X^*$ is a quotient of $Y^*$”. $\endgroup$ Dec 1, 2010 at 15:19
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