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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 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.

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I have to say this is proving to be one of the more useful CW big-list questions on the site... –  Qiaochu Yuan May 6 '10 at 0:55
The answers below are truly informative. Big thanks for your question. I have always loved your post here in MO and wordpress. –  Unknown May 22 '10 at 9:04
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. –  Suvrit Sep 20 '10 at 12:39
It's a thought -- I might consider it. –  gowers Oct 4 '10 at 20:13
Meta created tea.mathoverflow.net/discussion/1165/… –  quid Oct 8 '11 at 14:27

179 Answers 179

Piggybacking on one of Pierre's answers, I once had to teach beginning linear algebra from a textbook wherein the authors at one point stated words to the effect that the the trivial vector space {0} has no basis, or that the notion of basis for the trivial vector space makes no sense. It is bad enough as a student to generate one's own false beliefs without having textbooks presenting falsehoods as facts.

My personal belief is that the authors of this text actually know better, but they don't believe that their students can handle the truth, or perhaps that it is too much work or too time-consuming on the part of the instructor to explain such points. Whatever their motivation was, I cannot countenance such rationalizations. I told the students that the textbook was just plain wrong.

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Bjorn Poonen once gave a lecture at MIT about the empty set; it really opened my eyes. If someone wrote a textbook or something on the matter I think everyone would be a lot less confused. –  Qiaochu Yuan Jul 7 '10 at 23:56
For most of the history of civilization, zero was very controversial... –  Victor Protsak Jul 9 '10 at 4:12
I can combine Qiaochu's and Victor's remarks in this memory I have of a coffee break conversation between two colleagues, who were arguing on whether it made sense to say that the 1-element group acts on the empty set. I wisely decided to stay out of the controversy... –  Thierry Zell Aug 31 '10 at 2:24
Thierry: of course it makes sense. But the action is not transitive. –  ACL Dec 1 '10 at 22:53
I once taught abstract algebra from a book that adopted the artificial convention that the domain of a map of sets must be nonempty. I eventually figured out that the reason was in order to be able to say that every one-to-one map has a left inverse. And I have many times taught topology from a book that adopts the artificial convention that when speaking of the product of two spaces we require both spaces to be nonempty. I eventually figured out that the reason was in order to be able to say that $X\times Y$ is compact if and only if both $X$ and $Y$ are compact. –  Tom Goodwillie Mar 14 '12 at 22:01

The distinction between convergence and uniform convergence. It even got Cauchy in its time.

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I'm pretty sure I've heard both of the following multiple times:

  1. Transfinite induction requires the axiom of choice. False, though many applications of transfinite induction require axiom of choice (either in the form of the well-ordering theorem, or directly (though using transfinite induction together with choice directly is essentially the same as just using Zorn's Lemma)).

  2. Transfinite induction requires the axiom of foundation. I guess some people get transfinite induction mixed up with epsilon-induction?

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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!

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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).

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I feel like there are a lot of areas in mathematics in which the empty set is interpreted in a certain way (for example, the empty product is one, the empty sum is zero, the empty set has one map into any non-empty set, etc). Given each of these particular situations locally, I might agree that it is a convention in each case. However, given the ubiquity of such "conventions," one might think that there is a uniform description of what the empty set really "means" in these contexts. If this becomes the case, then I might argue for (5), which would follow from this conception of the empty set. –  David Corwin Jul 7 '10 at 23:48

"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.

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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.)

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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 $f_a(y) = ay(1-y)$: Bifurcation diagram for the logistic map

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)

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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.

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I think $M(T)$ is not metrizable in the weak$^{}$ topology, and in fact my claim that this fails for every infinite dimensional Banach space i also think is true. The rough outline of the proof I saw was this: 1. If $X^{*}$ is weak$^{}$ metrizable, then a first countabliity at the origin argument implies that $X^{*}$ has a translation invariant metric given the weak$^{}$ topology. 2. One can characterize completeness topologically for translation-invariant metrics, and see directly that if $X^{*}$ had a translation-invariant metric given the weak$^{}$ topology it would be complete. –  Benjamin Hayes Oct 12 '11 at 3:42

A random $k$-coloring of the vertices of a graph $G$ is more likely to be proper than a random $(k-1)$-coloring of the same graph.

(A vertex coloring is proper if no two adjacent vertices are colored identically. In this case, random means uniform among all colorings, or equivalently, that each vertex is i.i.d. colored uniformly from the space of colors.)

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...wait, what's the truth then? –  Harry Altman May 10 '11 at 0:06
For some graphs $G$ and integers $k$, the opposite. The easiest example is the complete bipartite graph $K_{n,n}$ with $k=3$. The probability a $2$-coloring is proper is about $(1/4)^n$ while the same for a $3$-coloring is about $(2/9)^n$, where I've ignored minor terms like constants. The actual probabilities cross at $n=10$, so as an explicit example, a random $2$-coloring of $K_{10,10}$ is more likely to be proper than a random $3$-coloring. –  aorq May 10 '11 at 0:37
This seems like a good example of a counterintuitive statement, but to call it a common false belief would mean that there are lots of people who think it's true. The question would probably never have occurred to me it I hadn't seen it here. The false belief that Euclid's proof of the infinitude of primes, on the other hand, actually gets asserted in print by mathematicians---in some cases good ones. –  Michael Hardy May 10 '11 at 15:36

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.

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It would be better if these misconceptions would come with explanations how things really are... –  darij grinberg Apr 10 '11 at 18:28
Along these lines: there is a widespread misapprehension that multiplication is the same thing as a multiplication algorithm (whichever one the speaker learned in elementary school). –  Thierry Zell Apr 10 '11 at 19:25
At least it's better than people thinking multiplication is constant-time. :P –  Harry Altman Apr 10 '11 at 19:35

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.

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Or: if a sequence has a rational limit in Q_p and in Q_r, then they're the same. –  Qiaochu Yuan May 5 '10 at 4:16
But if a rational sequence has a limit in all Q_p, including Q_\infty ... –  Gerald Edgar May 5 '10 at 12:17

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 $A \to C, B \to C$.

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"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.

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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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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.

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I really like this example from Rudin's book. Do you know if there exist such an example that does not use the continuum hypothesis (or if it's even possible to find one)? –  Malik Younsi Jul 28 '10 at 13:39
I don't know, but this could be a good questions for MO! –  Ricky Jul 28 '10 at 14:28
For others reading, the hypothesis left off here is that one must assume $f$ is measurable with respect to the product $\mathcal{B}[0,1] \times \mathcal{B}[0,1]$. –  nullUser Jul 8 '13 at 15:39

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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``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.) –  L Spice Dec 12 '13 at 23:26

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...)

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But only finite fields... –  darij grinberg Oct 19 '10 at 8:42

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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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... –  Nate Eldredge Oct 10 '10 at 18:26

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.

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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 $J=X_1R$.

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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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False statement: If $A$ and $B$ are subsets of $\mathbb{R}^d$, then their Hausdorff dimension $\dim_H$ satisfies

$$\dim_H(A \times B) = \dim_H(A) + \dim_H(B). $$

EDIT: To answer Benoit's question, I do not know about a simple counterexample for $d = 1$, but here is the usual one (taken from Falconer's "The Geometry of Fractal Sets"):

Let $(m_i)$ be a sequence of rapidly increasing integers (say $m_{i+1} > m_i^i$). Let $A \subset [0,1]$ denote the numbers with a zero in the $r^{th}$ decimal place if $m_j + 1 \leq r \leq m_{j+1}$ and $j$ is odd. Let $B \subset [0,1]$ denote the numbers with a zero in the $r^{th}$ decimal place if $m_{j} + 1 \leq r \leq m_{j+1}$ and $j$ is even. Then $\dim_H(A) = \dim_B(A) = 0$. To see this, you can cover $A$, for example, by $10^k$ covers of length $10^{- m_{2j}}$, where $k = (m_1 - m_0) + (m_3 - m_2) + \dots + (m_{2j - 1} - m_{2j - 2})$.

Furthermore, if $\mathcal{H}^1$ denotes the Hausdorff $1$-dimensional (metric) outer measure of $E$, then the result follows by showing $\mathcal{H}^1(A \times B) > 0$. This is accomplished by considering $u \in [0,1]$ and writing $u = x + y$, where $x \in A$ and $y \in B$. Let $proj$ denote orthogonal projection from the plane to $L$, the line $y = x$. Then $proj(x,y)$ is the point of $L$ with distance $2^{-1/2}(x+y)$ from the origin. Thus, $proj( A \times B)$ is a subinterval of $L$ of length $2^{-1/2}$. Finally, it follows:

$$ \mathcal{H}^1(A \times B) \geq \mathcal{H}^1(proj(A \times B)) = 2^{-1/2} > 0. $$

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Actually, the situation is worse than I say: there exist sets $A, B \subset \mathbb{R}$ with $dim_H(A \times B )= 1$, and yet $\dim_h(A) = \dim_H(B) = 0$. –  JavaMan Apr 5 '11 at 6:22

It always confused me as an undergraduate that $\mathbb{Q}\subset\mathbb{R}$ has an open neighborhood $N_\epsilon \supset\mathbb{Q}$ of arbitrarily small measure $\epsilon$, because $\mathbb{Q}$ is dense in $\mathbb{R}$.

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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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Incredibly common? The number of people who can even understand the statement, let alone believe it, isn't all that large... –  Victor Protsak May 5 '10 at 6:57
Yes, but among those, almost all believe that the wrong version is true. –  Andrea Ferretti May 5 '10 at 10:13
And about 50% of the large community who cannot understand the point will believe that the right version is true! Rather high percentage... –  Wadim Zudilin May 5 '10 at 11:41

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

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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.


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Why not? As an algebraist, my reaction already after "addition of k stabilizes" would be "if THAT holds, than WHATEVER". –  Victor Protsak Jun 10 '10 at 6:45
Victor, I held this belief for a good while when first learning set theory. I tried proving it a couple of times and failed, but I was in that stage just after I'd gotten the hang of basic cardinality arguments and they all seemed simple, so I figured it was just a matter of small details. –  Pietro KC Jun 10 '10 at 9:01
But it turns out that k^l is intimately linked with the cofinality of k, which is the length of the shortest unbounded sequence in k. For example, cof(omega) = omega, since sequences of length less than omega are finite, and thus bounded in omega. Similarly, cof(aleph_1) is aleph_1, since any countable sequence in aleph_1 is bounded. It's not immediately obvious that some cardinal k has cof(k) < k, but aleph_omega does! Anyway, the relevant theorem is that k^cof(k) > k, so there are arbitrarily large k s.t. k^omega > k. –  Pietro KC Jun 10 '10 at 9:06

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

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protected by François G. Dorais Oct 15 '13 at 2:34

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