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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… –  quid Oct 8 '11 at 14:27

201 Answers 201

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.

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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.)

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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. –  Andrés Caicedo Dec 2 '13 at 23:22

That Darboux functions are continuous is certainly a widely held belief among students, at least in France where it is induced by the way continuity is taught in high school.

I remember having gone through all the five "stages of grief" when shaken from this false belief with the $sin(1/x)$ example : denial, anger ( "then the definition of continuity must be wrong ! Let's change it !), bargaining ("Ok, but a Darboux function must surely be continuous except at exceptional points. Let's prove that..."), depression (when shown a nowhere continuous Darboux function), acceptance ("Hey guys, you really think the intermediate value theoem has a converse ? C'mon, you're smarter than that...")

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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

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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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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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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

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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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 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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1- A very common mistake that 1st year students (but not even a single mathematician) think that it is true is "a transitive and symmetric relation on a set is reflexive". But as the empty set is a transitive and symmetric relation but not reflexive on any non-empty set. Of course there lots of non-trivial examples also.

2- Another common mistake is that the expression "countable union of countable sets is again countable" is independent of axiom of choice (AC). Many people make the proof of this statement without mentioning axiom of choice. Indeed, in his holly book Algebra, Lang proves this statement just by taking an ordering from each countable set and continues without the mentioning AC.

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For big-list questions, it's usually best to post independent answers as separate answers. –  Nate Eldredge Dec 2 '10 at 15:13
+1 for #2. Baby Rudin is another offender. And many authors use so-called "diagonalization tricks" for proving compactness theorems like Arzela-Ascoli and Prohorov, which typically reduce to the compactness of $[0,1]^\mathbb{N}$. –  Nate Eldredge Dec 2 '10 at 15:21

I have heard the following a few times :

"If $f$ is holomorphic on a region $\Omega$ and not one-to-one, then $f'$ must vanish somewhere in $\Omega$."

$f(z)=e^z$ of course is a counterexample.

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No true. Take $f(z)=z^3-3z$ and restrict it to the complement of $\lbrace 1,-1\rbrace$ so that $f'(z)$ is never $0$. It maps this domain onto $\mathbb C$. –  Tom Goodwillie May 4 '11 at 0:16

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

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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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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 $\dim_h(A) = \dim_H(B) = 0$. –  JavaMan Apr 5 '11 at 6:22

This is a common error made by mature mathematicians in many books and papers in analysis, especially in differential equations: If $X$ is a closed subspace of a Banach space $Y$, then the $Y^*$ (the dual of $Y$) is isomorphic to a subspace of $X^*$ (the dual of $X$). It is false (of course) since Euclidian space $\mathbb R$ is a subspace of $\mathbb R^2$, yet the dual of $\mathbb R^2=\mathbb R^2$ is not isomorphic to a subspace of the dual of $\mathbb R=\mathbb R$. I guess, sometimes they really, really want it to be true. Cheers Boris

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I would also be shocked if this really gets believed often! It seems to be a sort of “mis-dualisation”: they dualise “$X$ is a subobject of $Y$” to “$Y^*$ is a subobject of $X^*$”, where the correct dual is “$X^*$ is a quotient of $Y^*$”. –  Peter LeFanu Lumsdaine Dec 1 '10 at 15:19

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.

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Why not $x \sin(1/x)$ as example? –  quid 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 Caicedo Jan 2 '14 at 23:44
  • The category of commutative C*-algebras is equivalent to the opposite category of locally compact Hausdorff spaces.

Let me know in case that this has been mentioned before; I haven't been able to find it. There is some discussion on math.SE.

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Whoops! Thanks for posting this and particularly the link, which shows that the nLab got this one wrong as well. –  Todd Trimble Nov 12 '14 at 13:03

I confess that I didn't carefully comb through all the answers, although I've read through this thread a few times in the past. So maybe these are repeats.

  • "The category of compact Hausdorff spaces is complete but not cocomplete; for example, it doesn't have all coproducts."

  • "The category of torsion abelian groups is cocomplete but not complete; for example, it doesn't have all products."

One of my professors in graduate school (quite a well-known and strong mathematician actually) insisted on the first, and quite a few people here at MO have mistakenly believed the second before the error was pointed out.

The moral of the story: sometimes categorical limits/colimits aren't computed the way you might first think of, e.g., colimits of compact Hausdorff spaces aren't always computed as colimits in $\mathrm{Top}$, and limits of torsion abelian groups aren't always computed as limits in $\mathrm{Ab}$.

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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.

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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.

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Duality reverses inclusions of vector spaces.

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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

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.

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A Banach space $X$ is reflexive if it is isomorphic to its double dual ${X^*}^*$.

(Couldn't find this is the list…)

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Even isometric fails. (Lindenstrauss & Tzafriri, in the '60s I believe.) –  Hachino May 12 at 8:19

An elementary false belief in elementary number theory: for $a, b, c\hspace{.1cm}\varepsilon\hspace{.1cm} \mathbb{N}$

$LCM\left(a,b\right)\times GCF\left(a,b\right) = ab$ .

Thus, $LCM\left(a,b,c\right)\times GCF\left(a,b,c\right) = abc$.

In general, $\left(a_1,a_2,\ldots,a_n\right)[a_1,a_2,\ldots,a_n] = a_1a_2\ldots a_n$.

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This kind of stuff is easy to rule out, though; it's dimensionally inconsistent. Replacing a, b, c by ka, kb, kc leads to a quick contradiction. –  Qiaochu Yuan Feb 24 '11 at 21:08
@Qiaochu, that is a nice quick check. Let the RCF(remnant common factor) be the leftover factor that would make the above second equality true. There seems to be no interesting way of determining $RCF\left(a,b,c\right)$ so that $LCM\left(a,b,c\right)\times RCF\left(a,b,c\right) \times GCF\left(a,b,c\right) = abc$ –  Unknown Feb 24 '11 at 22:05
@Elohemahab: actually, the correct generalization is $\gcd(a, b, c) \text{lcm}(ab, bc, ca) = \text{lcm}(a, b, c) \gcd(ab, bc, ca) = abc$. –  Qiaochu Yuan May 8 '11 at 23:47

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

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This is an interesting example, because it addresses the mistakes that come from the all-too frequent confusion with notations. But we need our shortcuts, our $f^{-1}(x)$ versus $x^{-1}$, etc. Obtaining concise notations while avoiding confusion: a tricky proposition! –  Thierry Zell Apr 14 '11 at 15:50

Let $B(r_1) \subset B(r_2)$ be two open balls of radius $r_1$ and $r_2$ respectively. Then $r_1 \leq r_2$.

Bounded metric spaces give trivial counterexamples. Also, $B \left( \frac{1}{6}, \frac{2}{3} \right) \subsetneq B \left( \frac{1}{2}, \frac{1}{2} \right)$ in $(0,+ \infty)$.

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Many students believe that every abelian subgroup is a normal subgroup.

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"A real symmetric matrix is positive-definite iff all the leading principal minors are positive, and positive-semidefinite iff all the leading principal minors are nonnegative."

This paper collects some evidence that this belief is "common", and presents a counterexample (of size $3\times 3$. Exercise: find an example of size $2\times 2$).

(Related to, but not the same as this answer.)

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$$\pmatrix{0&0\cr0&-1\cr}$$ –  Gerry Myerson Jul 29 at 3:22

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

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