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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 '10 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 '10 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 '10 at 12:39
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    $\begingroup$ It's a thought -- I might consider it. $\endgroup$
    – gowers
    Oct 4 '10 at 20:13
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    $\begingroup$ Meta created tea.mathoverflow.net/discussion/1165/… $\endgroup$
    – user9072
    Oct 8 '11 at 14:27

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An incredibly common false belief is:

For a (say smooth, projective) algebraic variety $X$ the $K_X$-negative part of the cone $NE(X)$ is locally polyhedral.

A right statement of the theorem of the cone is

$\overline{NE(X)} = \overline{NE(X)}_{K_X \geq 0} + \sum_{i} \mathbb{R}[C_i]$ for a denumerable set $\{ C_i \}$ of rational curves, which accumulate at most on the hyperplane $K_X = 0$.

At a first glance this seems to imply that $\overline{NE(X)}_{K_X < 0}$ is locally poyhedral, but this is not true. It depends on the shape of the intersection $\overline{NE(X)} \cap \{ K_X = 0 \}$.

For instance if this latter intersection is round, and there is only one curve $C_i$, the half-cone $\overline{NE(X)}_{K_X < 0}$ is actually a circular cone! Definitely not polyhedral in any sense. I believe this behaviour can happen even with varieties birational to abelian varieties.

The strange thing about this false belief is that it is held true by many competent mathematicians (and indeed I don't believe that many undergraduates meet the theorem of the cone!).

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  • $\begingroup$ You meant: I believe this behaviour can happen even with (varieties birationally isomorphic to) abelian varieties. Nice example although perhaps too technical for MO. $\endgroup$
    – VA.
    May 5 '10 at 3:27
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    $\begingroup$ Incredibly common? The number of people who can even understand the statement, let alone believe it, isn't all that large... $\endgroup$ May 5 '10 at 6:57
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    $\begingroup$ Yes, but among those, almost all believe that the wrong version is true. $\endgroup$ May 5 '10 at 10:13
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    $\begingroup$ And about 50% of the large community who cannot understand the point will believe that the right version is true! Rather high percentage... $\endgroup$ May 5 '10 at 11:41
  • $\begingroup$ I'm not sure to what extent this is a "false belief", and to what extent people are just being sloppy with the terminology "locally polyhedral". But I agree, it's disturbing to hear experts happily making this false statement, without any further comment. <i>Mea culpa:</i> An old version of the wikipedia article entitled "Cone of curves" contained this false statement. If one looks through the article history, it's not hard to see who is to blame... $\endgroup$
    – user5117
    May 6 '10 at 7:24
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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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The fundamental group of the Klein bottle is $D_\infty$, the infinite dihedral group (which is $\mathbb Z \rtimes \mathbb Z_2$).

I believed this for some time, and I seem to recall some others having the same confusion.

The group that has been mistaken for $D_\infty$ is in fact $\mathbb Z \rtimes\mathbb Z$, which can also be written with the presentation $x^2y^2=1$. The former abelianizes to $\mathbb Z_2\oplus \mathbb Z_2$, the latter to $\mathbb Z\oplus \mathbb Z_2$.

A 2-dimensional Lie group is a product of circles and lines, in particular it is abelian.

I don't know if anyone else suffered this one. The mistake is (a) in forgetting that the classification of surfaces doesn't apply since homeomorphic Lie groups are not necessarily isomorphic (e.g., the (bijective, orientation preserving) affine transformations $x\mapsto ax+b$, where $a>0, b\in \mathbb R$ are homeomorphic to $\mathbb R^2$, though not isomorphic) and (b) that Lie groups aren't necessarily connected, in particular $\mathbb R^2$ cross any finite non-abelian group is non-abelian.

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  • $\begingroup$ Count me in for the 2nd fallacy. $\endgroup$
    – Michael
    Dec 3 '13 at 0:41
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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 '10 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 '10 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 '10 at 15:30
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A degree $k$ map $S^n\to S^n$ induces multiplication by $k$ on all the homotopy groups $\pi_m(S^n)$.

(Not sure if this is a common error, but I believed it implicitly for a while and it confused me about some things. If you unravel what degree $k$ means and what multiplication by $k$ in $\pi_m$ means, there's no reason at all to expect this to be true, and indeed it is false in general. It is true in the stable range, since $S^n$ looks like $\Omega S^{n+1}$ in the stable range, "degree k" can be defined in terms of the H-space structure on $\Omega S^{n+1}$, and an Eckmann-Hilton argument applies.)

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    $\begingroup$ If $n$ is even and $x \in \pi_{2n-1}(S^n)$ and $f$ a degree $k$ map and $H$ the Hopf invariant, then $H(f_* (x)) = k^2 H(x)$. A related misbelief: if $M$ is a framed manifold and $N\to $M a finite cover, of degree $d$. Then the framed bordism classes satisfy $[N]=d [M]$. Completely wrong. $\endgroup$ Apr 14 '11 at 9:04
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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 '10 at 2:34
  • $\begingroup$ Did you mean "differentiable on some point in that interval:" ? $\endgroup$
    – Rasmus
    Sep 18 '13 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 '13 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 '15 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 '15 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 '15 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 '15 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 '15 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 '18 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 '19 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 '15 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 '19 at 21:01
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If $X$ is uncountable, then $X^{\mathbb{N}}$ is in bijection with $X$.

König's theorem implies that $|X^{\mathbb{N}}|>|X|$ whenever the cardinality of $X$ has countable cofinality.

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    $\begingroup$ Of course, this is well known to people who are used to this sort of things, but I have found that most of my non-set-theorist friends (and me) believed this. $\endgroup$ Oct 16 '20 at 5:38
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    $\begingroup$ It is especially tricky since it's not that easy to come up with cardinals having countable cofinality, and most familiar uncountable sets do have this property. $\endgroup$ Oct 16 '20 at 5:40
  • $\begingroup$ Yes. The cofinality of $\mathbb{R}$ with its usual ordering is $ℵ_0$, since $\mathbb{N}$ is cofinal in $\mathbb{R}$. But the cofinality of its cardinality $c$ has cofinality strictly greater than $ℵ_0$ (the usual ordering of $\mathbb{R}$ is not order isomorphic to $c$, so that the cofinality depends on the order). Question: Is there really an uncountable cardinal with countable cofinality? reference? $\endgroup$ Oct 16 '20 at 11:12
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    $\begingroup$ But $|X^\mathbb{N}|\geqslant 2^{\mathbb{N}}$ that may be more than $|X|$ if continuum hypothesis is not true. $\endgroup$ Oct 16 '20 at 12:26
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    $\begingroup$ @SebastienPalcoux I believe the standard example is $\aleph_\omega$ $\endgroup$ Oct 16 '20 at 12:51
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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 '11 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 '11 at 12:01
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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 '10 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 '10 at 0:02
  • $\begingroup$ I meant to write $n>3$ above. $\endgroup$ Aug 5 '10 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 '13 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 '10 at 19:33
  • $\begingroup$ Yes. GCF means the same: Greatest Common Factor. $\endgroup$
    – Unknown
    Nov 29 '10 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 '11 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 '11 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 '11 at 23:47
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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 '10 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 '10 at 15:21
  • $\begingroup$ Isn't the more right statement that a transitive and irreflexive relation is assymetric? $\endgroup$ Dec 4 '10 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 at 4:17
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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 '11 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 '13 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 '20 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? $\endgroup$
    – Mr.
    Dec 3 '13 at 10:09
  • $\begingroup$ Here, $B(a,b)$ is the open ball of radius $b$ centered at $a$. $\endgroup$
    – Seirios
    Dec 3 '13 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 '19 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 '19 at 11:09
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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 '10 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 '10 at 22:34
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    $\begingroup$ @QiaochuYuan ...unless you use the trivial norm. $\endgroup$ Oct 20 '15 at 21:16
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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 '10 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 '10 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 '11 at 11:33
  • $\begingroup$ thanks. that's great. I never thought about it before, but it just sounded right... $\endgroup$
    – KotelKanim
    Apr 19 '11 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 '11 at 0:16
  • $\begingroup$ @TomGoodwillie: what if both domain and image are simply-connected? $\endgroup$
    – Michael
    Dec 3 '13 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 '10 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 '10 at 15:19
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Multiplication of differential forms is inherently anti-commutative. Thus, if $x$ and $y$ are coordinates on a surface, then $dx \wedge dy$ makes sense but $(dx)^2+(dy)^2$ is either nonsense or, if it means anything, is $0$.

I'm not sure why I believed this, but I did for several years. I tried my best to avoid creating this impression in my students, but I think it still happened in some of them, simply because the curriculum spends a lot of time on integration and Stokes theorem and very little time on metrics, curvature, etc.

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  • $\begingroup$ This is about notations. You do the multiplication of differential forms in exterior algebra, but for metrics, you do it in tensor algebra. By the way, the textbooks that I read don't write dx², but write a symmetric bilinear form explicitly. $\endgroup$
    – user20948
    Feb 8 at 22:05
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A polynomial $p(x)$ of degree $n$, with coefficients in a commutative ring $R$, has at most $n$ roots, counting multiplicity. This is true if $R$ is an integral domain, but it can fail in the presence of zero divisors.

For instance, $p(x) = x^2+5x$ has four solutions when $R = \mathbb{Z}/6\mathbb{Z}$. I realized this mistake when a colleague asked me about factorization over a non-commutative ring, and I realized that I did not even know what would happen in the presence of zero-divisors.

This does motivate a question I have not found an answer to: is the number of solutions of $p(x)$ bounded by a function of the degree and the characteristic of the ring?

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    $\begingroup$ To answer your last question, $x^2-1$ has an infinite number of roots in $k^{\mathbb{N}}$ for any nonzero ring $k$. So, no. $\endgroup$
    – Gro-Tsen
    Apr 20 '16 at 17:00
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    $\begingroup$ Doesn't it only have one solution in $k^\mathbb{N}$ if $k$ has characteristic 2, @Gro-Tsen? $\endgroup$ Apr 20 '16 at 17:36
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    $\begingroup$ @OmarAntolín-Camarena Oh right, what I wanted to write was $x^2-x$, and I got confused between "idempotent" and "one-potent"(?). But of course $x^2-1$ also works provided, as you point out, that $1\neq -1$ in $k$. $\endgroup$
    – Gro-Tsen
    Apr 20 '16 at 20:31
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    $\begingroup$ A similar eample is $\mathbb Z/4 \mathbb Z[t]$. For each $n$ the element $2t^n$ is a root of $x^2=0$. $\endgroup$
    – Nick S
    Mar 3 '20 at 6:17
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Here's another howler some people commit: If $m$, $n$ are integers such that $m$ divides $n^2$ then $m$ divides $n$.

It's true sometimes, for example if $m$ is prime (or more generally squarefree, i.e. a product of distinct primes). But in general all one can conclude is that there exists integers $p$, $q$, $r$ with $p$ squarefree such that $ m = p q^2 $ and $ n = p q r $

The usual counterexample is that $8$ divides $4^2$ but not $4$ ;-)

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    $\begingroup$ An even more trivial counterexample is that 4 divides 2^2 but not 2 :-P $\endgroup$ Feb 23 '11 at 9:40
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