# Examples of common false beliefs in mathematics

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.

• I have to say this is proving to be one of the more useful CW big-list questions on the site... 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. 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. Sep 20 '10 at 12:39
• It's a thought -- I might consider it. Oct 4 '10 at 20:13
• Meta created tea.mathoverflow.net/discussion/1165/…
– user9072
Oct 8 '11 at 14:27

In a finite abelian $p$-group, every cyclic subgroup is contained in a cyclic direct summand.

Added for Gowers: Maybe one reason why people fall into this error goes something like this: First you learn linear algebra, so you know about vector spaces, bases for same, splittings of same. Then you run into elementary abelian $p$-groups and recognize this as a special case of vector spaces. Then you learn the pleasant fact that all finite abelian $p$-groups are direct sums of cyclic $p$-groups, and a corresponding uniqueness statement. You notice that all of the cyclic subgroups of order $p^2$ in $\mathbb Z/p^2\times \mathbb Z/p$ are summands, and if you have a certain sort of inquiring mind then you also notice that not every subgroup of order $p$ is a summand: one of them is contained in a copy of $\mathbb Z/p^2$, in fact in all of those copies of it. Having learned so much, both positive and negative, from the example of $\mathbb Z/p^2\times \mathbb Z/p$, you may think that it shows all the interesting basic features of the general case and overlook the fact that in $\mathbb Z/p^3\times \mathbb Z/p$ there is a $\mathbb Z/p^2$ not contained in any $\mathbb Z/p^3$.

In any case, reputable people sometimes make this blunder; it happened to somebody here at MO just the other day.

• Can you sketch the "proof" that makes this plausible? Jul 7 '10 at 16:40
• Finite abelian $p$-groups are direct sum of cyclic subgroups so they look a bit like vector spaces. Therefore, you expect them to behave the same way, i.e. every subspace should have a complement. In other words, take a minimal generating set for your subgroup and complete it to a minimal generating set for the whole group. This fails since your generating set for the subgroup might be depended modulo the Fratinni subgroup of the whole group. (A set is a minimal generating set for a finite $p$-group iff it is abasis for the group modulo the Fratinni subgroup). Jul 7 '10 at 17:58
• Is there an easily stated classification of the ways one can place a subgroup inside a finite abelian p-group (up to automorphisms of the larger group)?
– T..
Jul 7 '10 at 22:31
• I once worked out a classification of the ways one can place an element inside a finitely generated abelian group (up to automorphisms of the larger group), but I don't recall how it went exactly. Jul 8 '10 at 0:24
• This is related to a somewhat subtle issue of characterizing inclusions between the closures of the conjugacy classes of matrices. Suppose $A$ is a nilpotent $n\times n$ matrix of type $\lambda$ (i.e. with Jordan blocks of sizes $\lambda_1\geq \lambda_2\geq\ldots$ adding up to $n$) and $B$ is ... $\mu.$ Can $B$ be obtained as a limit of the conjugates of $A$? This is clearly possible if $\lambda$ is componentwise greater or equal than $\mu$, but the necessary and sufficient condition is given by the dominance order, en.wikipedia.org/wiki/Dominance_order. Jul 9 '10 at 4:05

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.

• 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. Jul 7 '10 at 23:56
• 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... 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
• @kow: I disagree. That is the "wrong" definition of transitivity for empty G-sets. See the discussion at qchu.wordpress.com/2010/12/03/empty-sets . Dec 16 '10 at 23:08
• 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. Mar 14 '12 at 22:01

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.)

• ...wait, what's the truth then? May 10 '11 at 0:06
• It sounds plausible. May 10 '11 at 0:34
• 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. May 10 '11 at 15:36

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:

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

• Well, it's disappointing that this fails, although it hadn't occurred to me to conjecture it. Apr 4 '11 at 9:53
• 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 $\dim_h(A) = \dim_H(B) = 0$. Apr 5 '11 at 6:22
• By the way, is there a simple counter-example with $A=B$? May 9 '11 at 7:51
• Nice, I did not know that, though Hausdorff dimension is part of my mathematical life! But the sets I study (Julia sets in complex dimension one) usually are uniform enough that this does not occurr, I guess. Here's what happens, morally, in the example given here: the scales epsilon at which you have good covers of A and the scales at which you have good covers of B are disjoint. The products of these good covers are extremely distorted : they are thin rectangles, instead of squares. Oct 18 '15 at 13:25

The following seems not to be here yet.

Misconception.

$R[[x_1,x_2,x_3,\dotsc]]/(x_2,x_3,\dotsc)$ isomorphic to $R[[x_1]]$ ${}\hspace{118pt}$ (f)

Source of the misconception. A fallacy of type false generalization: for any $n\in\mathbb{N}$ it is true that

$R[[x_1,x_2,x_3,\dotsc,x_n]]/(x_2,x_3,\dotsc,x_n)\cong R[[x_1]]$ ${}\hspace{125pt}$ (t)

but to conclude from this that (f) was true by passing to the limit $n\to\infty$ is fallacious.

Reason for why the misconception is false. E.g. the formal power series $f:=x_2+x_3+\dotsm$ is an element of $R[[x_1,x_2,x_3,...]]$, but by the standard definition of $I:=(x_2,x_3,\dotsc)$, which after all means nothing more than the $R[[x_1,x_2,x_3,\dotsc]]$-module generated by the infinite set $\{x_i\colon i\in \omega,\ i\geq 2\}$, the ideal $I$ does not contain $f$. (Having coefficients from the huge power series ring $R[[x_1,x_2,x_3,\dotsc]]$ does not help.)

Reason for including the example. I saw this misconception in a dissertation. For obvious reasons, I won't give the source.

Further remarks. In the above, $R$ can be any commutative unital ring, and $R[[x_1,x_2,x_3,\dotsc]]$ as usual means the projective limit in the category of commutative unital rings of the diagram $\dotsm\twoheadrightarrow R[[x_1,x_2,x_3]]\twoheadrightarrow R[[x_1,x_2]]\twoheadrightarrow R[[x_1]]$ consisting of the canonical projections.

• Maybe this is just a misunderstanding of / disagreement over the "correct" definition of the symbol $R[[x_1,x_2,\ldots]]$. If one believes that it denotes the completion of the localisation of $R[x_1,x_2,\ldots]$ at its maximal ideal $(x_1,x_2,\ldots)$, then this misconception becomes a true statement. Or phrased differently: Maybe this misconception is a failure of recognising that $colim_n lim_k R[x_1,\ldots,x_n]/\mathfrak{m}_n^k \not\cong \lim_k colim_n R[x_1,\ldots,x_n]/\mathfrak{m}_n^k$. Mar 16 '18 at 23:07

"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.

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.

• 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)? Jul 28 '10 at 13:39
• I don't know, but this could be a good questions for MO! 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 $\mathcal{B}[0,1] \times \mathcal{B}[0,1]$. Jul 8 '13 at 15:39

Here's a mistake I've seen from students taking a first course in linear analysis. For a vector $g$ in a Hilbert space $H$, it is true that $\langle f,g\rangle=0$ for every $f\in H$ implies $g=0$. This leads us to the mistaken:

“Let $(g_n)$ be a sequence in $H$. If, for every $f\in H$, $\langle f,g_n\rangle\to0$, then $g_n\to 0$.”

• You wrote: "Here's a mistake I've seen from students taking a first course in linear analysis." Then you wrote: "For a vector $g$ in a Hilbert space, $\langle f,g\rangle$ for every $f \in H$ implies $g = 0$." At this point the reader could be wondering what that is a mistake. Dec 1 '10 at 22:35
• ....sorry; I meant "$\langle f,g \rangle = 0$ for every[....]" Dec 1 '10 at 22:36
• @Michael: all answers are CW; so if we think some wording needs clarifying, we can do it ourselves! Dec 2 '10 at 0:43

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 $J=X_1R$.

• Omg, I will have to be careful about that. Thanks Darij ;). Apr 12 '11 at 8:45

(*) "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.

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.

• Minor nitpick: Consider a locally compact Hausdorff space $T$. The $*$ topology on the dual of the $C^*$ algebra $C_0(T)$ is metrizable, if and only if $X$ is second countable. That is a theorem in Choquet's book on functional analysis. So your claim, that the first statement is never true in infinite dimensional situations, is false. Take e.g. $T$ being a circle. Oct 6 '11 at 13:38
• I think $M(T)$ is not metrizable in the weak$^\ast$ 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 $X^\ast$ is weak$^\ast$ metrizable, then a first countabliity at the origin argument implies that $X^\ast$ has a translation invariant metric given the weak$^\ast$ topology. 2. One can characterize completeness topologically for translation-invariant metrics, and see directly that if $X^\ast$ had a translation-invariant metric given the weak$^\ast$ topology it would be complete. Oct 12 '11 at 3:42
• $X^{∗}$ in the weak∗ topology is a countable union of $\{\phi\in X^{*}:\|\phi\|\leq N\}$, which have empty weak∗ interior. Hence, if the weak∗ topology were metrizable, we get a contradiction to the Baire Category Theorem. Are you sure you don't mean the weak∗ topology on the state space of $C_{0}(X)? Oct 12 '11 at 3:47 • Okay, excuse my false claim, I was overlooking that this holds for the subset$M^+(T)$of positive Radon measure, and does not generalize to the complex linear span. Oct 16 '11 at 10:24 A common trap which sometimes I see people fall is that a Hermitian matrix$M$is negative definite if and only if its leading principal minors are negative. What is true is the Sylvester's criterion, which says that$M$is positive definite if and only if its principal minors are positive. Thus, the true statement is that$M$is negative definite if and only if the principal minors of$-M$are positive. I guess you don't want commonly held beliefs of students that for every real number there is a next real number, or that convergent sequences are eventually constant. A version I saw in a book asked whether points on a line "touch." Understanding the topology of a line is a challenge for many students, although presumably not for most mathematicians. Here is a more esoteric belief that I have even seen in some books: "The Banach-Tarski Paradox says that a ball the size of a pea can be cut into 5 pieces and reassembled to make a ball the size of the sun." As a consequence of the Banach-Tarski paradox, a ball the size of a pea can be partitioned (not really "cut") into a finite number of pieces which can be reassembled into a ball the size of the sun, but a simple outer measure argument implies that the number of pieces must be very large (I roughly estimate at least$10^{30}$). The number 5 probably comes from the fact that the basic Banach-Tarski paradox is that a ball of radius 1 can be partitioned into 5 pieces which can be reassembled into two disjoint balls of radius 1. (It can almost, but not quite, be done with four pieces; one of the five pieces can be taken to be a single point.) • Since points do not touch, this was an objection to the set theoretic view of the geometric continuum as a set of points, for example by Veronese. A decent account of this can be found in Debates about infinity in mathematics around 1890: The Cantor-Veronese controversy, its origins and its outcome, by Detlef Laugwitz. Dec 2 '13 at 23:22 • Jan 2 '14 at 4:03 • Convergent sequences are eventually constant! With the discrete topology/metric/norm, that is. Sep 1 '15 at 0:19 • Of course I meant sequences in$\mathbb{R}$with the usual topology. Hopefully by the time students study general metric spaces or topological spaces they understand the topology of$\mathbb{R}$. Jan 7 '16 at 19:35 • The link in my first comment does not seem to work anymore. Try this one. Aug 30 '20 at 13:19 Let$(X,\tau)$be a topological space. The false belief is: "Every sequence$(x_n)$in$X$with an accumulation point$a\in X$has a subsequence that converges to$a$". I subscribed to this intuitively until I stumbled over a counterexample, see http://dominiczypen.wordpress.com/2014/10/13/accumulation-without-converging-subsequence/ A Banach space$X$is reflexive if it is isomorphic to its double dual${X^*}^*$. (Couldn't find this is the list…) • Even isometric fails. (Lindenstrauss & Tzafriri, in the '60s I believe.) May 12 '15 at 8:19 • Here's a counterexample: en.wikipedia.org/wiki/James%27_space Nov 15 '20 at 5:12 I would like to turn the attention of mathematical community to a false beliefs related to the direct limit topologies. Many years ago in the theory of topological groups there was a false belief that for every space$X$the free topological group carries the topology of direct limit of the sequence$F_n(X)$of words of length$\le n$. This illusion was broken up by Fay, Ordman and Thomas who showed that even for the space of rational numbers the free topological group$F(\mathbb Q)$is not a$k$-space. The problems with direct limit topologies is that for the direct limit$X=lim X_n$of an increasing sequence$(X_n)$of topological spaces the topology on$X\times X$does not coincide with the direct limit topology of the sequence$ (X_n\times X_n)$. Now specialists in General Topology and Topological Algebra are conscious of pathological behaviour of direct limit topologies and are careful with this delicate topic. On the other hand, I was quite surprised lerning that in Algebraic Geometry this misbelief still is alive. For example, in this paper posted to arxiv (maybe it is already published) in the very introduction (on page 3) it is written that for any topological space$X$the Ran space (of all non-empty finite subsets of$X$, endowed with the topology of direct limit of the sequence$R_n(X)$of sets of cardinality$\le n$in$X$) is a topological semilattice. But this is not true in general, see Proposition 4 here. So, some false beliefs that have died in some areas of mathematics can be still alive in others. By the way, this situation also explains why mathematicians should not neglect general topology. Let $$X$$ be a "nice" path-connected topological space, say a connected manifold or CW-complex. False belief: "A universal covering $$Y\to X$$ of $$X$$ is unique up to unique isomorphism" and therefore can be called "the" universal covering. The isomorphism far from unique in general (there are as many as elements in "the" fundamental group). However uniqueness (and the universal property) holds in the category of coverings of pointed topological spaces. (In particular, for topological groups there's a canonical choice.) Browsing I found several textbooks teaching the above "false belief" (I saw several too that are careful with this issue). • Likewise 'the' algebraic closure of a field. Apr 23 '20 at 7:10 • @OscarCunningham Yes and no: yes it's similar, but it seems to me many more people are aware and careful (e.g. not saying "the algebraic closure")– btw I wrote this answer after reading this comment. – YCor Apr 25 '20 at 14:53 • ncatlab.org/nlab/show/torsor Sep 28 '20 at 0:08 Some things from pseudo-Riemannian geometry are a bit hard to swallow for students who have had previous exposure to Riemannian geometry. Aside from the usual ones arising from sign issues (like, in a two dimensional Lorentzian manifold with positive scalar curvature, time-like geodesics will not have conjugate points), an example is that in Riemannian manifolds, connectedness + geodesic completeness implies geodesic connectedness (every two points is connected by a geodesic). This is not true for Lorentzian manifolds, and the usual example is the pseudo-sphere. 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!). • You meant: I believe this behaviour can happen even with (varieties birationally isomorphic to) abelian varieties. Nice example although perhaps too technical for MO. – VA. May 5 '10 at 3:27 • Incredibly common? The number of people who can even understand the statement, let alone believe it, isn't all that large... May 5 '10 at 6:57 • Yes, but among those, almost all believe that the wrong version is true. 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... May 5 '10 at 11:41 • 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... – user5117 May 6 '10 at 7:24 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$A \to C, B \to C$. "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. • The key here is that quadratic variation is defined as a limit in probability, not a limit almost surely. Jul 8 '13 at 15:46 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.) • Let's see if you TeX code can be improved: $$f(z) := \int_0^z \exp(\zeta^2)\,d\zeta$$ (The backslash in \exp not only should prevent italicization but should also result in proper spacing in things like "a \exp b", and the space before d\zeta seems appropriate.) Jul 8 '10 at 15:19 • @MichaelHardy, if we're going to {\TeX}pick, then surely it should be something like${\mathrm d}\zeta$(rather than$d\zeta$), since the$\mathrm d$is an operator (rather than a variable)? Dec 12 '13 at 23:20 • @LSpice : I understand the case for that usage; in particular, it allow the use of$d$as a variable, so that one can write$\dfrac{\mathrm{d}f}{\mathrm{d}d}$, etc. However, the usage with the$d$italicized as if it were a variable is standard although not universal. Dec 13 '13 at 0:58 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...) • But only finite fields... Oct 19 '10 at 8:42 Duality reverses inclusions of vector spaces. • That's funny, because I don't imagine this kind of idea would occur to someone who has just learned the definition of a dual space. That would be a strangely sophisticated mistake to make. Apr 7 '11 at 0:21 • And, once you learn that this is wrong, you can make the opposite mistake. See my comments here sbseminar.wordpress.com/2011/02/22/sobolev-spaces-on-manifolds on how surprised I was that duality DOES reverse the inclusions between Soboloev spaces. Apr 11 '11 at 12:07 • The mistake is somehow in the wording. The dual of the inclusion morphism is reversed, it's just not an inclusion anymore. Sep 4 '15 at 15:42 Common false belief: a space that is locally homeomorphic to$\mathbb{R}^n$must be Hausdorff. More generally, many people forget that the usual definition of a manifold contains the Hausdorff and paracompact conditions. There are of course examples that show that forgetting this assumption leads to unexpected result, and they are in fact much wilder than I knew a few weeks ago. Notably, among examples of (Hausdorff) non-paracompact "manifolds" are the well-known long line, but also the Prüfer manifold constructed from a closed half-plane by attaching to it a half plane at each boundary point. Added: Let me give a particular case of this false belief to illustrate what kind of weird things can happen that most people would not realize when they are sloppy with the paracompact hypothesis: there exists a path-connected, locally contractible, simply-connected space that admits non-trivial locally trivial bundles with fiber$[0,1]$. Indeed, the first octant in the product of two long line is not homeomorphic to a product a long ray with an interval, but has a natural bundle structure over a long ray. Teaching introduction to analysis, I had students using the "fact" that if$f: [a,b] \rightarrow \mathbb{R}$is continuous, then$[a,b]$can be divided to subintervals$[a,c_1],[c_1,c_2],...,[c_n,b]$such that$f$is monotone on every subinterval. For instance you can use this "fact" to "prove" the (true) fact that$f$must be bounded on$[a,b]$. Also, some students used the same "fact", but with countably many subintervals. I found this mistake hard to explain to students, because constructing a counterexample (such as the Weierstrass function) is impossible at the knowledge level of an introduction course. • Why not$x \sin(1/x)$as example? – user9072 Jan 2 '14 at 17:33 • It is in the case of finitely many subintervals, but not in the case of countably many subintervals. Jan 2 '14 at 19:17 • You can surely discuss fractal shapes without needing to go into the details of a technical counterexample. The point seems to be that it is hard to imagine that "increasing at a point" and "increasing in a neighborhood of a point" are not the same for continuous functions. You can give easy examples showing that indeed they disagree, locally, and fractals suggest that you can make the disagreement happen everywhere. You can revisit this later, once more technology has been set in place. Jan 2 '14 at 23:44 • While technically it is true one can do it with countably many for the function I gave (if one includes degenerate intervals) I would be surprised if not at least some (or rather most) of the confusion of the students could be addressed by the example (possibly continuing with discussion along the lines suggested by @AndresCaicedo). – user9072 Jan 5 '14 at 16:50 False belief: Any orthonormal basis of a subvectorspace$W\subset V$of an inner product space$V$can always be extended to an ONB of$V$. Counterexample: Let$V$be$\bigoplus_{i\ge 1} \mathbb{R}$with the inner product given by$\langle a_*,b_*\rangle =\sum_{i\ge 1} a_ib_i$and let$W$be the subvectorspace of$V$spanned by$e_1+e_i$for$i\ge 2$. The given set is basis and we can apply Gram-Schmidt to obtain an ONB. However$W^\perp = 0$so there is no way to complete it. Related false belief:$(W^\perp)^\perp=W$. These beliefs are all true in finite dimensions, but false in general. • That's why we like Hilbert spaces (inner product spaces that are complete w.r.t. the inner-product norm) much better than arbitrary inner-product spaces. Mar 4 '16 at 4:08 If$H$and$K$are subgroups of$G$, then$HK$is a subgroup of$G$. • Hm, wonder how common that false belief actually is. It seems obviously implausible in the nonabelian case. Sep 6 '15 at 15:39 • Its common, specially when undergraduates use product formula :$|HK|=\frac{|H||K|}{| H \cap K | }$Because all of them are subgroup, except$HK$probably. – user68208 Apr 10 '16 at 17:30 True: The solution operator of the linear one-dimensional time-dependent ordinary differential equation (ODE)$x' = a(t) x$is $$\exp\Bigl(\int\limits_{t_0}^{t} a(s) \, ds\Bigr).$$ True: The solution operator of the linear multi-dimensional time-independent ODE$x' = A x$is $$\exp\,(A(t - t_0)).$$ A quite popular misconception, even among research mathematicians: The solution operator$\Phi(t;t_0)$of the linear multi-dimensional time-dependent ODE$x' = A(t) x\$ is $$\exp\Bigl(\int\limits_{t_0}^{t} A(s) \, ds \Bigr),$$

perhaps strengthened by Liouville's formula:

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

• I am holding this belief right now. Apr 27 '18 at 23:26
• Is there a simple expression that is true?
– Hans
May 3 '18 at 18:11
• @Hans The Peano-Baker series is, in my opinion, simple, and converges where it should. There is a nice paper by Baake and Schlägel The Peano-Baker series, Proceedings of the Steklov Institute of Mathematics 275 (1) (2011), 155-159 (the paper is behind a paywall on the Publisher's page, but the authors put a copy on ResearchGate (researchgate.net/publication/47702535_The_Peano-Baker_series)). May 4 '18 at 8:15
• @Hans In contrast, the Magnus expansion is very complicated and may not converge except close to the initial time. May 4 '18 at 8:20
• @Hans Physicists refer to the time ordered exponential: en.wikipedia.org/wiki/Ordered_exponential Aug 27 '18 at 14:58

"Every 1-dimensional knot in $$R^n, n\ge 4,$$ is trivial." This is true for tame knots and false for wild knots. See here.