Examples of common false beliefs in mathematics

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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

Answer 4

Probably my fault for not paying enough attention in analysis, but:

Any continuous function on the interval that has derivative equal to zero almost everywhere is constant.

Answer 5

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?

Answer 6

Here is a short list of some false beliefs I had when I was studying mathematics, I suppose they may be common but I have never checked:

• I was in the last year of high school and studying university-level math in advance. I remember trying for a week to prove that a continuous injective map from an open subset of $\mathbb{R}^2$ to $\mathbb{R}^2$ that preserves "being aligned" (I mean that maps aligned triples to aligned triples) must be the restriction of an affine map (over $\mathbb{R}$). That is disproved by restrictions of projective transformations... Which I knew of but I was not able to see they contradicted my belief. When my teacher told me "What about projective transformations?"... I felt dumb.
• I was in the 1st year of PhD studies. My advisor, Adrien Douady, had an idea to build polynomial Julia sets with positive Lebesgue measure. Julia sets are fractals, often with complicated topological structures at every scale. Surely that must be the source of measure? So as an exercise, I tried for a week to prove that Jordan curves are necessarily of Lebesgue measure 0. I told Adrien about my attempts. He gave me a counter-example. I felt dumb.
• Learning that there are closed subsets of the interval with positive Lebesgue measure but no interior did not surprise me as much, as the construction is very simple, but still that's a bit counterintuitive.
• When you zoom on the Mandelbrot set, you see all that round components with smooth boundaries. They look so round. Surely they must be circles, for otherwise the difference would be visible. Well... they are not (except one). Guess how I felt when I learned.
• Frankly, when learning the first time about complex numbers, did anybody here expect that, adding the square root of -1 to the reals would add the roots of all other polynomials?
• I was giving a lecture to math teachers about sensitivity to initial condition (call it chaos) and showing strange attractors on the computer, one told me that by the very presence of chaos, what we see may be quite far from the actual behaviour of the equation, save reality. It turns out hyperbolic systems are stable, so I believe this is still representative (it does not prove it but it is an encouraging hint).
• ... Chaos in deterministic systems. I won't develop on that.
• Surely before hearing of set theory and Cantor's argument, you will believe that all sets are countable. Then after learning that this is not the case, you will think that $\mathbb{R}^2$ must be bigger than $\mathbb{R}$, right?
• You have a $C^\infty$ function on the right half plane, all of whose derivatives have a continuous extension to the boundary line. Surely, it must be easy to extend it to a $C^\infty$ function of the whole plane, isn't it? Well... You can but I would not call it easy.
• Short statements have short proofs. Disproved by Fermat's last theorem (among others).
• I was quite disappointed to learn that there cannot be a finite non-commutative field (division algebra).

I have a few other examples, that I would not term "common false beliefs" but rather "fun and surprising math facts". Is there already a MO question about that?

Answer 7

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 $4$ ;-)

Answer 8

Maybe I am a bit late to the party, but here is something that I falsely believed in for a while:

False Belief: If $f$ is a continuously differentiable function with a horizontal asymptote, i.e. $\lim_{x\to +\infty}f(x)=L$, then $f'(x)\to 0$ as $x$ goes to infinity.

Of course, this is incorrect even for smooth functions; take $g$ to be a smooth $L^1$ function with oscillatory behavior so that $\lim_{x\to +\infty}g(x)$ does not exist and then set $f(x)=\int_{0}^xg(s)ds$.

I came to this misconception during an ODE course I took in my undergrad. In the study of the asymptotic behavior of solutions of an autonomous equation $$ \frac{dy}{dx}=F(y) $$ one argues that if a solution $y=y(x)$ tends to a limit $L$ as $x\to +\infty$, that limit must be a zero of the vector field. We then proceeds with determining which zeros of $F$ are asymptotically stable to find $L$. The point is that here not only $\lim_{x\to +\infty}y$, but also $\lim_{x\to +\infty}\frac{dy}{dx}$ exists due to the ODE. Hence:

Correct Statement: If $f$ is a continuously differentiable function with a horizontal asymptote, and if $f'$ admits a limit as $x$ goes to infinity, that limit must be zero.

This readily follows from the mean value theorem.

Answer 9

For some reason, I believed until recently that every compact operator on a Hilbert space must have eigenvalues. It's obviously true for finite-rank operators and Hermitian operators, so I must have subconsciously generalized.

This was despite being well aware of (say) the Volterra operator, say; I simply had two contradictory ideas in my head.

Answer 10

Taylor's Formula and displacement operator: I (too often) see in papers (mathematical physics but a recent paper (a) by mathematicians also) the statement

False belief 1 : a) Let $D=\frac{d}{dx}$ be the derivation operator. Then, for all $f\in C^\infty(\mathbb{R})$, $$ e^{tD}[f](x)=f(x+t) $$

which is false (take any $\phi\in C^\infty(\mathbb{R})$ with compact support, for instance).

False belief 2 : b) In the same vein, for formal power series (“if our object is formal then we do not have to ensure convergences”). Let $S(x)\in \mathbb{R}[[x]]$ ($x$ is a formal variable) then for $t\in \mathbb{R}$, one has $$ e^{tD}[S](x)=S(x+t) $$

which is false as we must have $t$ in the domain of convergence of $S$.

Remarks (i) The function $f\in C^\infty(\mathbb{R})$ is analytic over $\mathbb{R}$ iff $$ (\forall x\in \mathbb{R})(\exists R>0)(\forall t\in ]-R,R[) (\sum_{n\geq 0}\frac{t^n}{n!}D^n[f](x)=f(x+t))\tag{1}\label{275319_1} $$ (ii) Even if $f\in C^\omega(\mathbb{R})$, it can happen that the left hand side of eq. \eqref{275319_1} do not converge otherwise $f$ would be the restriction of an entire function (which e.g. $\frac{1}{1+x^2}$ is not, for example).

(iii) Even if the LHS of \eqref{275319_1} converges for all $x,t\in \mathbb{R}$, $f$ need not be analytic. Consider the following function (classic in theory of distributions)
$$ f(x)=0\text{ if } x\notin \mathopen]-1,1\mathclose[\text{ and } f(x)=e^{\frac{1}{1-x^2}} \text{ if } x\in \mathopen]-1,1[\mathclose. $$ (iv) In the (b) case $S=\sum_{n\geq 0}n!\, x^n$ for example cannot be displaced.

(v) As @vanxoo says false belief 2 could be cured by considering, in appropriate spaces, $t$ as a formal variable. This is the point of view of Bourbaki in Functions of a real variable (see Ch VI Generalized Taylor expansions).

Answer 11

"The set A = {a, b} has two elements..."

It's quite simple to notice that a can be the same as b, but after 5 years of university there were people still believing it...

Answer 12

For a bounded subset of a metric space the diameter is two times the radius!

Let $S\subset X$ be bounded. The definitions are:

$\mathrm{diameter}(S):=\sup\{d(x,y)\,|\,x,y\in S\}$

$\mathrm{radius}(S):=\inf\{r>0\,|\,\exists x\in X:\,S\subset B(x,r)\}$

where $B(x,r)$ denotes the open ball of radius $r$ around $x$.

Answer 13

I saw many students using the "fact" that for a subset $S$ of a group one has $SS^{-1}=\{e\}$

Answer 14

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?

Answer 15

A somewhat common belief among students starting out in cryptography:

Breaking RSA requires factoring the modulus.

Although it is not quite known to be a "false" belief, there is no known reduction showing that breaking RSA implies finding the prime factors of the modulus. This in contrast to e.g. Rabin's cryptosystem, and various cryptographic schemes built on other hard problems, whose security provably relies on the underlying hard problems.

Answer 16

Uncorrelatedness implies independence

This statement is indeed false. Suppose $X \sim U[-1, 1]$. Then $Cov(X, X^2) = EX^3 - EXEX^2 = 0$, but $X$ and $X^2$ are clearly not independent. However, that mistake is quite popular...

Answer 17

Heisenberg-Weyl and enveloping algebras

I have heard (and have read) the following belief even among distinguished mathematical physicists.

Let $HW_\mathbb{C}$ be the (associative with unit) algebra generated by two elements $\{a,a^\dagger\}$ subjected to the relation $[a,a^\dagger]=1$. It is easy to check that $$ \mathfrak{g}=span_\mathbb{C}\{a,a^\dagger,1_{HW_\mathbb{C}}\} $$ is a Lie subalgebra.

False belief The algebra $HW_\mathbb{C}$ is the universal enveloping algebra of $\mathfrak{g}$ i.e. $$ HW_\mathbb{C}=U(\mathfrak{g})\ . $$

One way to see that this is false at once is to observe that any enveloping algebra $U(\mathfrak{g})$ possesses (at least) a character $\varepsilon$, but there is none on $HW_\mathbb{C}$ (one would have indeed $1=\varepsilon([a,a^\dagger])=0$).

Late edit (After Darij's post) Indeed, the enveloping algebra of $$ \mathfrak{g}=span_\mathbb{C}\{a,a^\dagger,1_{HW_\mathbb{C}}\} $$ is the algebra (associative with unit) generated by $\{a,a^\dagger,e\}$ ($e$ is a clone of $1_{HW_\mathbb{C}}$) subjected to the relations $$ [a,a^\dagger]=e ; [a,e]=[a^\dagger,e]=0\qquad \mbox{[Rel 1]} $$ and the canonical arrow $U(\mathfrak{g})\to HW_\mathbb{C}$ transforms an expression in $a,a^\dagger,e$, already reduced by [Rel 1] into its image in $HW_\mathbb{C}$. For relations [Rel 1] one can take, for instance, the normal form: all $a^\dagger$ (creations) on the left, all $a$ (annihilations) on the right and $e$ anywhere (as they are central).

Combinatorially, in the reduced expressions of $U(\mathfrak{g})$, the power of e counts the number of times Wick commutations have been applied. I'd include this in my post if it were not out of place.

Answer 18

"Frac" and "Const" commute: A differential ring is simply a pair $(R,\partial)$ where $R$ is a ring and $\partial$ is a derivation in $R$ (i.e. $\partial\in End_{\mathbb{Z}}(R)$ follows identically Liebniz rule $\partial(ab)=\partial(a)b+a\partial(b)$). The constants of $R$, $\ker(\partial)$ form a subring $Const(R)\subset R$ called subring of constants.

If $R$ is a commutative domain (i.e. without zero divisors), it is standard to consider the field of fractions $Frac(R)$ and to extend $\partial$ by the formula of calculus $\partial(1/g)=-\partial(g)/g^2$.

False belief: $Const(Frac(R))=Frac(Const(R))$, in other words, every constant in $Frac(R)$ is of the form $\alpha/\beta$, where $\alpha,\beta\in Const(R)$.

Until yesterday morning, I postponed to prove this (false) lemma (thinking it was an easy exercise). Then, preparing a talk, I could not prove this and searched for a counterexample. Finally I found a simple one in MSE (see below).

See there for a discussion and counterexamples.

Answer 19

A common misconception in analysis is that if two $C^1$-functions are linearly independent then their Wronskian is non-zero at some point.

This is the result of carelessly reversing a true implication. The mistake was first made in 1882 by Thomas Muir (who is responsible for the name "Wronskian") in his "Treatise on the Theory of Determinants" and Giuseppe Peano pointed out a counterexample as early as 1889 (it suffices to take $y=x^2$ and $y=x|x|$).

For more information on the history of this mistake see "Peano on Wronskians: A Translation". by Susannah M. Engdahl and Adam E. Parker.

Answer 20

True: Given a graded algebra $A$, there is a notion of a "homogeneous" ideal of $A$. It is a property that connects an ideal of $I$ with the grading and is often necessary to require. For example, if $I$ is a homogeneous ideal of $A$, then the algebra $A / I$ is graded again. If $I$ is not homogeneous, then it is not graded in general (since the projections of different graded components of $A$ onto $A / I$ might have nonzero intersection).

False: Given a filtered algebra $A$, there is a notion of a "filtered" ideal of $A$.

There is no such notion. We can require $I$ to be generated by $I\cap A_n$ for some $n$, or actually to lie inside $A_n$ for some $n$, but in most cases none of these is actually needed. (Correct me if I am wrong.) Formulations like "Let $I$ be an ideal compatible with (or respecting) the filtration" are cargo cult.

But: Given a filtered algebra $A$ and a generating set $G$ of an ideal $I$ of $A$, it is an important question whether $I\cap A_n$ is generated by $G\cap A_n$ for every $n\in \mathbb N$. This is not always satisfied, often nontrivial (in many cases it can be proved by using the diamond lemma to show that every element of $A_n$ has a unique "remainder" modulo $I$ in a certain sense, and this remainder can be obtained by repeated subtraction multiples of elements of $G\cap A_n$) and used tacitly in various texts.

Answer 21

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.

Answer 22

Here are some various examples (I hope that some of them weren't already mentioned):
1. If a space $X$ have two different norms $\| \cdot \|_i, i=1,2$ such that $\| \cdot \|_1 \leq \| \cdot \|_2$ then the completion with respect to $\| \cdot \|_1$ is contained in the completion with respect to $\| \cdot \|_2$.
2. If $M_1,M_2$ are isomorphic modules and $N_1,N_2$ are isomorphic submodules then $M_1/N_1$ and $M_2/N_2$ are isomorphic.
3. If $A,B$ are subsets of topological spaces $X,Y$ (resp.) and $A,B$ are homeomorphic then the closures $\overline{A}$ and $\overline{B}$ are also homeomorphic.
4. The standard construction of adjoining unit to the Banach algebra $A$ yields nothing new if $A$ already was unital.
5. The phrase "a function is almost everywhere continuous" means the same as: "the function is almost everywhere equal to the continuous function".
6. Suppose you are trying to prove that some function space $F$ is complete (say that functions are defined on $X$ and real valued): you take a Cauchy sequence $\{f_n\}_n$ and prove that for each point $x \in X$ the sequence $\{f_n(x)\}_n$ is Cauchy. Then form the completeness of $\mathbb{R}$ you obtain a function $f$. The false belief is that it is now enough to show that $f$ belong to $F$.
7. If you have an ascending family $\{A_i\}_i$ then to obtain it's union $\bigcup_{i}A_i$ it is enough to take some countable subfamily
8. A convergent net $\{x_i\}_i$ in a metric space is bounded and the set $\{x_i\}_i \cup \{x\}$ is compact (where $x$ is the limit).
9. If $D$ is an open dense subset of a topological space $X$ then $card \; D= card \; X$

Answer 23

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).

Answer 24

Every matrix is the sum of a symmetric and an antisymmetric matrix. Hence:

If $V$ is a vector space and $k$ is a number, then the $k$-th tensor product of $V$ with itself decomposes as a direct sum into symmetric and antisymmetric tensors: $$ \underbrace{V \otimes ... \otimes V}_{k\text{ times}} = \Lambda^kV \oplus \mathrm{Sym}^kV $$

Recall (in the finite-dimensional case) the dimensions: $$ \dim \Lambda^k V = \binom{n}{k} \quad\text{ and }\quad \dim\mathrm{Sym}^kV = \binom{n+k-1}{k} $$

Looking at $k=1$ shows that we have non-trivial intersection.

Looking at $n=k=3$ shows that the sum is not exhausting.

Answer 25

This may not count as a false belief, but it is an amusing misconception I had. In college I had a numerical analysis professor who had both a strong accent and messy handwriting, so it was hard to know exactly what he was talking about sometimes.

I was not yet familiar with the Greek letter $\xi$, and that was the variable he always used to represent the error of a computation, but with his handwriting it just looked like a purposeful scribble. So he would say,"Here we have the calculated value and then of course with some error" (scribble).

I thought he was just being dismissive about the error and trying to represent it in a pejorative way.

To be fair, the letter $\xi$ is not one of the easier ones to draw by hand.

Answer 26

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.

Answer 27

It’s tempting to believe that a knot and its mirror-image, if tied in succession on a rope, can cancel. View https://youtu.be/lwWeRMmXIoU to see John Conway’s lucid explanation of why this is false.

I would be interested in knowing whether Lord Kelvin was under this misapprehension when he proposed his vortex theory of the atom. (Knots don’t cancel but vortices can.)

I once went to a conference on recreational mathematics in which a speaker claimed that a knot and its mirror-image can be cancelled; I was kind enough not to ask for a demonstration. (There is a way to pretend to make them cancel, but I don’t know the details of the trick.)

Answer 28

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.

Answer 29

Maybe that's a cute one:

If $f$ is a continuous real function, then $f^{-1}(x)$ is at most countable unless $f$ is constant on some interval.

Answer 30

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

---

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

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