# 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... – 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/… – user9072 Oct 8 '11 at 14:27

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.

• I'm a little confused. How does requiring that $(e_1, e_2) \mapsto g$ be continuous fix things? In the indiscrete case, this map is continuous (since every map to the group is). And why isn't $G^i \to G^i/G$ a principal $G^i$--bundle? – Autumn Kent Mar 6 '11 at 17:52
• The group in this example starts out with some topology. (I guess I didn't specify that I was thinking of a topological group.) If G started with the indiscrete topology, then your commment makes sense, and we would have a principal bundle for this indiscrete group. But if G is not indiscrete, then the map $(e_1, e_2) \mapsto g$ is not continuous as a map into the topological group G. The proof that continuity of the translation map forces this to be a principal bundle can be found in Husemoller's book on fiber bundles (it's not hard). Let me know if this didn't answer your questions. – Dan Ramras Mar 6 '11 at 19:57
• Oh! You're saying that a point is not a classifying space for G with some other topology. I thought you were saying that $G^i/G$ wasn't $BG^i$. Thanks for the clarification! – Autumn Kent Mar 6 '11 at 20:01
• Yes, precisely. It's an odd little example, but helpful when people forget to include the proper conditions... – Dan Ramras Mar 6 '11 at 21:06
• Maybe even more amazing wrong belief in this field: $\dim(E/G)\le\dim E$ (there are counterexamples by A.N. Kolmogorov) – mikhail skopenkov Jun 9 '11 at 14:52

"The universal cover of $SL_2(R)$ is a universal central extension" (which I believed until recently...)

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. – Toby Bartels 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 $\dim_h(A) = \dim_H(B) = 0$. – David Apr 5 '11 at 6:22
• By the way, is there a simple counter-example with $A=B$? – Benoît Kloeckner 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. – Arnaud Chéritat Oct 18 '15 at 13:25

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.

• 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
• should be $(x_1+x_2)(y_1+y_2)$. – Junyan Xu May 6 '12 at 4:09
• @Junyan Xu : Thanks, I corrected it. – user11235 May 9 '12 at 18:18

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. – Izhar Oppenheim 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. – Andrés E. Caicedo 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

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/

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. – Todd Trimble 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

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.

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. – nullUser Jul 8 '13 at 15:46

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)? – 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

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.) – Michael Hardy 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)? – LSpice 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. – Michael Hardy Dec 13 '13 at 0:58

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 ;). – Martin Brandenburg 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. – Marc Palm 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. – Benjamin Hayes 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 $C_{0}(X)? – Benjamin Hayes Oct 12 '11 at 3:47 • Okay, excuse my false claim, I was overlooking that this holds for the subset$M^+(T)$of positive Radon measure, and does not generalize to the complex linear span. – Marc Palm Oct 16 '11 at 10:24 A random$k$-coloring of the vertices of a graph$G$is more likely to be proper than a random$(k-1)$-coloring of the same graph. (A vertex coloring is proper if no two adjacent vertices are colored identically. In this case, random means uniform among all colorings, or equivalently, that each vertex is i.i.d. colored uniformly from the space of colors.) • ...wait, what's the truth then? – Harry Altman May 10 '11 at 0:06 • It sounds plausible. – Michael Hardy May 10 '11 at 0:34 • For some graphs$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. – Michael Hardy May 10 '11 at 15:36 False belief: A function being continuous in some open interval implies that it is also differentiable on some point in that interval: Counterexample: The Weierstrass function is an example of a function that is continuous everywhere but differentiable nowhere:$f(x) = \sum_{n=0}^\infty a^n \cos(b^n \pi x)$Where$a \in (0, 1)$,$b$is a positive odd integer, and$ab > 1 + \frac{3\pi}{2}$. The function has fractal-like behavior, which leads to it not being differentiable. This notion is rather disheartening to most calculus students, though. • Related: if f is continuous on the interval I, there must be an interval J in I on which f is monotone. Easily believed by the beginner. – Thierry Zell Aug 31 '10 at 2:34 • Did you mean "differentiable on some point in that interval:" ? – Rasmus Sep 18 '13 at 19:13 • Haha, figures that I edit a 3 year old post to introduce an even worse typo. Yes, that is what I meant. – Jon Paprocki Sep 18 '13 at 22:53 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.) – Hachino May 12 '15 at 8:19$\mathbb{R}^2$has a unique complex manifold structure; it's just$\mathbb{C}$right? 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. – Noam D. Elkies Mar 4 '16 at 4:08 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$colim_n lim_k R[x_1,\ldots,x_n]/\mathfrak{m}_n^k \not\cong \lim_k colim_n R[x_1,\ldots,x_n]/\mathfrak{m}_n^k$. – Johannes Hahn Mar 16 '18 at 23:07 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... – 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 • 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 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. 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. • In the statistics text at the college where I teach, (ii) is universal among the examples given, so I formulated the conjecture; but when I tried to prove it and thought about what the mode really is, I realised how badly behaved that can be and found immediate counterexamples. (Then this gets me wondering why anybody would bother using the mode as a statistic for anything, since it's pretty much meaningless, but never mind.) – Toby Bartels Apr 4 '11 at 9:24 • Toby: sure, you use the mode for cases when the domain of the measurement is not an ordered set but just a set without structure and so the median wouldn't make sense. – Zsbán Ambrus Apr 7 '11 at 12:01 The fundamental group of the Klein bottle is$D_\infty$, the infinite dihedral group (which is$\mathbb Z \rtimes \mathbb Z_2$). I believed this for some time, and I seem to recall some others having the same confusion. The group that has been mistaken for$D_\infty$is in fact$\mathbb Z \rtimes\mathbb Z$, which can also be written with the presentation$x^2y^2=1$. The former abelianizes to$\mathbb Z_2\oplus \mathbb Z_2$, the latter to$\mathbb Z\oplus \mathbb Z_2$. A 2-dimensional Lie group is a product of circles and lines, in particular it is abelian. I don't know if anyone else suffered this one. The mistake is (a) in forgetting that the classification of surfaces doesn't apply since homeomorphic Lie groups are not necessarily isomorphic (e.g., the (bijective, orientation preserving) affine transformations$x\mapsto ax+b$, where$a>0, b\in \mathbb R$are homeomorphic to$\mathbb R^2$, though not isomorphic) and (b) that Lie groups aren't necessarily connected, in particular$\mathbb R^2$cross any finite non-abelian group is non-abelian. • Count me in for the 2nd fallacy. – Michael Dec 3 '13 at 0:41 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. • 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 • You are right, I should have specified open ball, thanks. I think it is just barely false for the open ball. Honestly, I held this false belief until a couple of days ago, and I haven't thought much about correcting my belief. Probably the real open epsilon ball is the union of all functions that fit between dashed curves a distance strictly less than epsilon away from f? At any rate, I think the above picture is the right way to think about it most of the time. But it gives results such as$tan^{-1}$being in the open ball of radious pi/2 centered at 0 if you interpret it literally. – user4977 Oct 10 '10 at 19:24 • Hmm, very nice (once clarified to the open ball)! Easily dispelled as soon as you question it, but I could easily imagine using it without thinking and missing the alternation of quantifiers that’s going on under the surface. – Peter LeFanu Lumsdaine Dec 1 '10 at 15:30 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. – Michael Hardy Dec 1 '10 at 22:35 • ....sorry; I meant "$\langle f,g \rangle = 0$for every[....]" – Michael Hardy Dec 1 '10 at 22:36 • @Michael: all answers are CW; so if we think some wording needs clarifying, we can do it ourselves! – Peter LeFanu Lumsdaine Dec 2 '10 at 0:43 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. – Thierry Zell 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. – David E Speyer 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. – Manuel Bärenz Sep 4 '15 at 15:42 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.) • If$n$is even and$x \in \pi_{2n-1}(S^n)$and$f$a degree$k$map and$H$the Hopf invariant, then$H(f_* (x)) = k^2 H(x)$. A related misbelief: if$M$is a framed manifold and$N\to $M a finite cover, of degree$d$. Then the framed bordism classes satisfy$[N]=d [M]\$. Completely wrong. – Johannes Ebert Apr 14 '11 at 9:04