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 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 R must be the whole of 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

The following false belief enjoyed a certain success in the '70. (See R.S.Palais, Critical point theory and the minimax principle for an account.)

A second countable, Hausdorff, Banach manifold is paracompact.

Regular is necessary, otherwise there are counterexamples!

Answer 2

By googling one sees that each of the following statements has a significant number of believers:

(1) the vector space {0} has no basis,

(2) the empty set is a basis of {0} by convention,

(3) the statements "{0} has no basis" and "the empty set is a basis of {0}" are equivalent,

(4) the statements "{0} has no basis" and "the empty set is a basis of {0}" are NOT equivalent,

(5) the statement "the empty set is a basis of {0}" is an immediate consequence of the definitions of the terms involved.

I think that we'll all agree that the 5 beliefs are not ALL true. My personal religion is to believe in (4) and (5). I don't think I'll ever understand the arguments in favor of (1), (2) or (3).

Answer 3

A common misbelief for the exponential of matrices is $AB=BA \Leftrightarrow \exp(A)\exp(B) = \exp(A+B)$. While the one direction is of course correct: $AB=BA \Rightarrow \exp(A)\exp(B) = \exp(A+B)$, the other direction is not correct, as the following example shows: $A=\begin{pmatrix} 0 & 1 \\ 0 & 2\pi i\end{pmatrix}, B=\begin{pmatrix} 2 \pi i & 0 \\ 0 & -2\pi i\end{pmatrix} $ with $AB \neq BA \text{ and} \exp(A)=\exp(B) = \exp(A+B) = 1$.

Answer 4

A common belief of students in real analysis is that if $$ \lim_{x\to x_0}f(x,y_0),\qquad\lim_{y\to y_0}f(x_0,y) $$ exist and are both equal to $l$, then the function has limit $l$ in $(x_0,y_0)$. It is easly to show counter-examples. More difficult is to show that also the belief $$ \lim_{t\to 0}f(x_0+ht,y_0+kt)=l,\quad\forall\;(h,k)\neq(0,0)\quad\Rightarrow\quad\lim_{(x,y)\to(x_0,y_0)}f(x,y)=l $$ is false. For completeness's sake (presumably anybody who ever taught calculus has seen it, but it's easily forgotten) the standard counterexample is $$ f(x,y)=\frac{xy^2}{x^2+y^4} $$ at $(0,0$).

Answer 5

As is well known, if $V$ is a vector space and $S, T \subset V$ are subspaces, then $S \cup T$ is a subspace iff $S \subset T$ or viceversa. However, $S \cup T \cup U$ can be a subspace even if no two spaces are contained in each other (think finite fields...)

Answer 6

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.

Answer 7

By definition, an asymptote is a line that a curve keeps getting closer to but never touches. The teaching of this false belief at an elementary level is standard and nearly universal. Everybody "knows" that it is true. A tee-shirt has a clever joke about it. In the course of describing the function $f(x) = \dfrac{5x}{36 + x^2}$, I mentioned about an hour ago before a class of about 10 students that its value at 0 is 0 and that it has a horizontal asymptote at 0. One of them accused me of contradicting myself. What of $y = \dfrac{\sin x}{x}$? And even with simple rational functions there are exceptions, although there the curve can touch or cross the asymptote only finitely many times. And $3 - \dfrac{1}{x}$ gets closer to 5 as $x$ grows, and never reaches 5, so by the widespread false belief there would be a horizontal asymptote at 5.

Answer 8

Sequence ${a_n}$ has a limit $A$ in $\mathbb{R}$ and a limit $B$ in $\mathbb{Q}_p$. Then $A$ is rational iff $B$ is rational.

Answer 9

There are cases that people know that a certain naive mathematical thought is incorrect but largely overestimate the amount by which it is incorrect. I remember hearing on the radio somebody explaining: "We make five experiments where the probability for success in every experiment is 10%. Now, a naive person will think that the probability that at least one of the experiment succeed is five times ten, 50%. But this is incorrect! the probability for success is not much larger than the 10% we started with."

Of course, the truth is much closer to 50% than to 10%.

(Let me also mention that there are various common false beliefs about mathematical terms: NP stands for "not polynomial" [in fact it stands for "Nondeterministic Polynomial" time]; the word "Killing" in Killing form is an adjective [in fact it is based on the name of the mathematician "Wilhelm Killing"] etc.)

Answer 10

I just realized yesterday that, given $A \to C, B \to C$ in an abelian category, the kernel of $A \oplus B \to C$ is not the direct sum of the kernels of $A \to C, B \to C$.

Answer 11

"A 'random' number field has large class number"

I've heard this belief quite a few times. Usually random means taking a not-too-small degree (7?) and then somehow taking integer coefficients (around 10,000?).

But in fact class number tend to be much smaller than one expects. Usually they are logarithmic in the size of the discriminant.

The main reasons for the belief are the common examples of fields given in undergraduate and early graduate courses - imaginary quadratic fields and cyclotomic fields. In more advanced courses students see abelian extensions and CM-fields, which also have special arithmetic properties that make their class groups somewhat larger. In the courses I have taken the actual size of 'random' number fields was not addressed, and, say, the Cohen-Lenstra heuristics were not mentioned.

Answer 12

A common false belief is that all Gödel sentences are true because they say of themselves they are unprovable. See Peter Milne's "On Goedel Sentences and What They Say", Philosophia Mathematica (III) 15 (2007), 193–226. doi:10.1093/philmat/nkm015

Answer 13

Here's one from basic set theory. Let k be a cardinal and consider the operation "adding k", meaning

l |--> k+l

on cardinals. We know that this operation "stabilizes" to the identity after k, that is, for any l>k, we have l+k = l. Similarly, the "multiplying by k" operation,

l |--> l * k

stabilizes to the identity after k.

Everyone also knows that if l is an infinite cardinal then l^2 is equipotent to l, and more generally l^n is equipotent to l for every natural number n. I.e. all the finite power functions stabilize to the identity at omega.

Well, obviously "exponentiation by omega" also stabilizes at some point, right? Like, l^omega is equal to l for sufficiently large l? Look, we probably already have the stabilization point at 2^omega.

Right?

Answer 14

Before reading about it, I really thought that if $f \colon [0,1] \times [0,1] \to [0,1]$ is a function with the following properties:

1. for any $x \in [0,1]$ the function $f_x\colon [0,1] \to [0,1]$ defined by $f_x(y)=f(x,y)$ is Lebesgue measurable, and also the function $f^y \colon [0,1]\to[0,1]$ defined by $f^y(x)=f(x,y)$ is Lebesgue measurable, for all $y \in [0,1]$;
2. both $\varphi(x)=\int_0^1 f_x d\mu$ and $\psi(y)=\int_0^1 f_y d\mu$ are Lebesgue measurable.

Then the two iterated integrals $$ \int_0^1\varphi(x)dx \mbox{ and } \int_0^1\psi(y)dy $$ should be equal. This is false (see Rudin's "Real and Complex Analysis", pag. 167), at least if you assume the continuum hypothesis.

Answer 15

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.

Answer 16

Complex variables: "An entire function that is onto and locally one-to-one is globally one-to-one."

Counterexample: $f(z) := \int_0^z \exp(\zeta^2)\,d\zeta$

I'll leave the proof that this is indeed a counterexample as a pleasant exercise.

(I believe this example is due to Lawrence Zalcman.)

Answer 17

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.

Answer 18

If $E$ is a contractible space on which the (Edit: topological) group $G$ acts freely, then $E/G$ is a classifying space for $G$.

A better, but still false, version:

If $E$ is a free, contractible $G$-space and the quotient map $E\to E/G$ admits local slices, then $E/G$ is a classifying space for $G$.

(Here "admits local slices" means that there's a covering of $E/G$ by open sets $U_i$ such that there exist continuous sections $U_i \to E$ of the quotient map.)

The simplest counterexample is: let $G^i$ denote $G$ with the indiscrete topology (Edit: and assume $G$ itself is not indiscrete). Then G acts on $G^i$ by translation and $G^i$ is contractible (for the same reason: any map into an indiscrete space is continuous). Since $G^i/G$ is a point, there's a (global) section, but it cannot be a classifying space for $G$ (unless $G={1}$). The way to correct things is to require that the translation map $E\times_{E/G} E \to G$, sending a pair $(e_1, e_2)$ to the unique $g\in G$ satisfying $ge_1 = e_2$, is actually continuous.

Of course the heart of the matter here is the corresponding false belief(s) regarding when the quotient map by a group action is a principal bundle.

Answer 19

(*) "Let $(I,\leq)$ be a directed ordered set, and $E=(f_{ij}:E_i\to E_j)_{i\geq j}$ be an inverse system of nonempty sets with surjective transition maps. Then the inverse limit $\varprojlim_I\,E$ is nonempty."

This is true if $I=\mathbb{N}$ ("dependent choices"), and hence more generally if $I$ has a countable cofinal subset. But surprisingly (to me), those are the only sets $I$ for which (*) holds for every system $E$. (This is proved somewhere in Bourbaki's exercises, for instance).

Of course, other useful cases where (*) holds are when the $E_i$'s are finite, or more generally compact spaces with continuous transition maps.

Answer 20

It took me a bit too long to realize that these two beliefs are contradictory:

• Period 3 $\Rightarrow$ chaos: if a continuous self-map on the interval has a period-3 orbit, then it has orbits of all periods.
• The black dots on each horizontal slice of this picture above $x=a$ show the location of the periodic points of the logistic map $f_a(y) = ay(1-y)$:

You can clearly see a 3-cycle in the light area towards the right; yet we know that if there is a 3-cycle in that slice then there must be a cycle of any period in that slice... so where are they?

(The other cycles are there of course, but they are repelling and hence are not visible. You can see artifacts from these repelling cycles near the period-doubling bifurcations in this picture)

Answer 21

Here are mistakes I find surprisingly sharp people make about the weak$^{*}$ topology on the dual of $X,$ where $X$ is a Banach space.

-It is metrizable if $X$ is separable.

-It is locally compact by Banach-Alaoglu.

-The statement $X$ is weak$^{*}$ dense in the double dual of $X$ proves that the unit ball of $X$ is weak$^{*}$ dense in the unit ball of the double dual of $X.$

The first two are in fact never true if $X$ is infinite dimensional. While both statements in the third claim are true, the second one is significantly stronger, but a lot of people believe you can get it from the first by just "rescaling the elements" to have norm $\leq 1.$ (Although the proof of the statements in the third claim is not hard). The difficulty is that if $X$ is infinite dimensional then for any $\phi$ in the dual of $X,$ there exists a net $\phi_{i}$ in the dual of $X$ with $\|\phi_{i}\|\to \infty$ and $\phi_{i}\to \phi$ weak$^{*},$ so this rescaling trick cannot be uniformly applied. Really these all boil down to the following false belief:

-The dual of $X$ has a non-empty norm bounded weak$^{*}$ open set.

Again when $X$ is infinite dimensional this always fails.

Answer 22

The cost of multiplying two $n$-digit numbers is of order $n^2$ (because each digit of the first number has to be multiplied with each digit of the second number).

---

A lot of information is found on http://en.wikipedia.org/wiki/Multiplication_algorithm .

The first faster (and easily understandable) algorithm was http://en.wikipedia.org/wiki/Karatsuba_algorithm with complexity $n^{log_2 3} \sim n^{1.585}$.

Basic idea: To multiply $x_1x_2$ and $y_1y_2$ where all letters refer to $n/2$-digit parts of $n$-digit numbers, calculate $x_1 \cdot y_1$, $x_2\cdot y_2$ and $(x_1+x_2)\cdot(y_1+y_2)$ and note that this is sufficient to calculate the result with three such products instead of four.

Answer 23

"the quadratic variation of a Brownian motion between $0$ and $T$ is equal to $T$"

this is only true that if $\mathcal{D}^N$ is a nested sequence of partitions of $[0,T]$ (with mesh size going to $0$) then the quadratic variation of a Brownian motion along these partitions converges towards $T$, almost surely. If we define the quadratic variation of a continuous function $f$ as we would like to, $$Q(f,[0,T]) = \sup_{0=t_0<\ldots, t_n=T } \sum |f(t_k)-f(t_{k+1})|^2,$$ then the Brownian paths have almost surely infinite quadratic variation.

This was something I had never noticed until I read the wonderful book "Brownian motion" by Peter Morters and Yuval Peres.

Answer 24

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.

Answer 25

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.

Answer 26

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.)

Answer 27

"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.

Answer 28

Inversion is an automorphism of a group. ('Cause it, like, preserves the conjugacy classes and all that...)

Answer 29

I don't know how common this is, but I've noticed it half an hour ago in some notes I had written: If $J$ is a finitely generated right ideal of a not necessarily commutative ring $R$, and $n$ is natural, then $J^n$ is finitely generated, isn't it?

No, it isn't. For an example, try $R=\mathbb Z\left\langle X_1,X_2,X_3,...\right\rangle $ (ring of noncommutative polynomials) and $J=X_1R$.

Answer 30

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.)