Related: question #879, Most interesting mathematics mistake. But the intent of this question is more pedagogical.

In many branches of mathematics, it seems to me that a good counterexample can be worth just as much as a good theorem or lemma. The only branch where I think this is explicitly recognized in the literature is topology, where for example Munkres is careful to point out and discuss his favorite counterexamples in his book, and Counterexamples in Topology is quite famous. The art of coming up with counterexamples, especially minimal counterexamples, is in my mind an important one to cultivate, and perhaps it is not emphasized enough these days.

So: what are your favorite examples of counterexamples that really illuminate something about some aspect of a subject?

Bonus points if the counterexample is minimal in some sense, bonus points if you can make this sense rigorous, and extra bonus points if the counterexample was important enough to impact yours or someone else's research, especially if it was simple enough to present in an undergraduate textbook.

As usual, please limit yourself to one counterexample per answer.

  • $\begingroup$ @Ben Linowitz. It is in the small and delightful book, J. W. S. Cassels, Lectures on elliptic curves. $\endgroup$ – Regenbogen Mar 2 '10 at 14:39
  • 1
    $\begingroup$ @Ben Linowitz. Oh I am sorry for saying irrelevant things. I must confess I do not know anything at all. Maybe the following MSRI video might interest you(if you were not already aware of it)... msri.org/communications/vmath/VMathVideos/VideoInfo/3821/… $\endgroup$ – Regenbogen Mar 2 '10 at 17:38
  • 3
    $\begingroup$ From a pedagogical standpoint, sometimes the minimal counterexample isn't the best one; in particular if it is "too small" to exhibit important general features of what's going on. $\endgroup$ – benblumsmith Jul 22 '11 at 12:55

69 Answers 69


The Fabius function, everywhere $C^\infty$, nowhere analytic.


see... sci.math post by G.A. Edgar, in thread “integral3”, Sep 9 2003

J. Fabius, "A probabilistic example of a nowhere analytic $C^\infty$-function". Z. Wahrsch. Verw. Geb. 5 (1966) 173--174.

K. Stromberg, PROBABILITY FOR ANALYSTS (Chapman & Hall, 1994), pp. 117--120.


The matrix $\left(\begin{smallmatrix}0 & 1\\ 0 & 0\end{smallmatrix}\right)$ has the following wonderful properties. (Feel free to add or edit; I can't remember all the reason I loathed it when I was learning linear algebra. It's funny how unexciting they all now seem, but it's a counterexample for almost every wrong linear algebra proof I tried to give.)

  • Only zeroes as eigenvalues, but non-zero minimal polynomial (in particular, the minimal polynomial has bigger degree than the number of eigenvalues). Probably my favorite way to state this fact: the minimal polynomial is not irreducible or square-free. The same thing in a fancier language: the Jordan canonical form is not diagonal.

  • Not diagonalizable, even over an algebraically closed field.

  • Not divisible over $\mathbb C$. There are no matrices $M$ and integers $n\ge2$ so that $M^n = \left(\begin{smallmatrix}0 & 1\\\ 0 & 0\end{smallmatrix}\right).$ All diagonalizable and most non-diagonalizable complex matrices have $n$th roots.

    (This is because, if there was a square root, it'd have minimal polynomial x4, but since it's a two-by-two matrix, Cayley-Hamilton implies that the characteristic polynomial has degree 2).

  • The matrix is nilpotent but not zero.

  • It's one of the best examples when you need to remember why matrix multiplication is not commutative.

  • Thinking of k2 as a k[x]-module where x acts as this matrix should give wonderful (counter)-examples of modules for all the same reasons.

Also, $\left(\begin{smallmatrix}1 & 1\\ 0 & 1\end{smallmatrix}\right)$ is an example of an invertible matrix with the first three properties above. Its action on k2 is in some sense the simplest example of a representation of a group ($\mathbb{Z}$) which is indecomposable but not irreducible.

  • 2
    $\begingroup$ "the minimal polynomial is not irreducible" I am not sure I get your point. Minimal polynomials are usually not irreducible. Did you mean "does not have to be multiplicity-free"? $\endgroup$ – Vladimir Dotsenko Mar 3 '10 at 14:38
  • $\begingroup$ @Vladimir: Yes and no. In other context (Galois theory) minimal polynomials are always irreducible. But you're right, I was a bit careless; I've edited it now. $\endgroup$ – Ilya Grigoriev Mar 4 '10 at 0:07
  • $\begingroup$ I don't follow your comment that the invertible matrix has only zero eigenvalues (one of the first three properties you list). Isn't that impossible? $\endgroup$ – Nick Salter Jul 29 '10 at 16:31
  • 2
    $\begingroup$ You mean that most non-diagonalizable square matrices have $n$th roots. $\endgroup$ – Kevin O'Bryant Mar 27 '11 at 23:50
  • 16
    $\begingroup$ I think it's better to remember that functions are not commutative: if f puts pants on a person and g puts underpants on them, then f(g(naked person)=half-naked person, and g(f(naked person)=half-naked superman. $\endgroup$ – Michał Masny Mar 8 '15 at 15:23

A polynomial $p(x) \in \mathbb{Z}[x]$ is irreducible if it is irreducible $\bmod l$ for some prime $l$. This is an important and useful enough sufficient criterion for irreducibility that one might wonder whether it is necessary: in other words, if $p(x)$ is irreducible, is it necessarily irreducible $\bmod l$ for some prime $l$?

The answer is no. For example, the polynomial $p(x) = x^4 + 16$ is irreducible in $\mathbb{Z}[x]$, but reducible $\bmod l$ for every prime $l$. This is because for every odd prime $l$, one of $2, -2, -1$ is a quadratic residue. In the first case, $p(x) = (x^2 + 2 \sqrt{2} x + 4)(x^2 - 2 \sqrt{2} x + 4)$. In the second case, $p(x) = (x^2 + 2 \sqrt{-2} x - 4)(x^2 - 2 \sqrt{-2} x - 4)$. In the third case, $p(x) = (x^2 + 4i)(x^2 - 4i)$. This result can be thought of as a failure of a local-global principle, and the counterexample is minimal in the sense that the answer is yes for quadratic and cubic polynomials.

  • 13
    $\begingroup$ Similarly, if the units modulo n are not cyclic, then the nth cyclotomic polynomial \Phi_n(x) will be reducible mod p for all p. $\endgroup$ – Ben Linowitz Mar 2 '10 at 6:15
  • 54
    $\begingroup$ Even better, the polynomial $x^4-72x^2+4$ is irreducible in $\mathbb{Z}[x]$, but reducible modulo every <I>integer</I>. (Dummit and Foote, 3rd edition, page 309) $\endgroup$ – Alfonso Gracia-Saz Mar 2 '10 at 16:41
  • 23
    $\begingroup$ The polynomial $(x^2+31)(x^3+x+1)$ has a root modulo every prime but no roots in Q. No polynomial of degree < 5 has this property. $\endgroup$ – AVS Apr 4 '10 at 20:21
  • $\begingroup$ Good example! I think it's also important to note that no irreducible polynomials have this property by the Frobenius density theorem. $\endgroup$ – Qiaochu Yuan Apr 4 '10 at 21:32
  • 1
    $\begingroup$ @Qiaochu Yuan: Yes. Here one also uses the fact that every transitive permutation group contains an element that fixes no points. $\endgroup$ – AVS Apr 5 '10 at 11:40

I like the double sequence $a_{nm} = \frac{n}{n+m}$ to show that $\lim_{n\to\infty}\lim_{m\to\infty} a_{nm}\neq \lim_{m\to\infty}\lim_{n\to\infty} a_{nm}$ .

  • 14
    $\begingroup$ The sequence which is 0 or 1 depending on which of m and n is greater also works. See the fifth example from tricki.org/article/Just-do-it_proofs and its accompanying discussion. $\endgroup$ – aorq Mar 19 '11 at 18:18
  • 30
    $\begingroup$ Yeah, that works, too! However, I experienced that undergraduates sometimes feel a bit uncomfortable with such "piecewise" definitions and are more happy with a more "natural" example (whatever that means). $\endgroup$ – Dirk Mar 19 '11 at 19:58
  • 2
    $\begingroup$ @aorq Iverson bracket: $[m>n]$ $\endgroup$ – Akiva Weinberger Sep 2 '15 at 0:39
  • $\begingroup$ Would $a_{nm}=\frac{n}{m}$ also be such a sequence? $\endgroup$ – fierydemon Jan 23 '17 at 20:29
  • 3
    $\begingroup$ In principle, yes, but then you also have "improper limits" (i.e. convergence to plus infinity), and I in Germany we teach these separately. $\endgroup$ – Dirk Jan 23 '17 at 21:31

The Cantor set is a nice source of counterexamples:

The first measure zero sets you meet are usually countable. However, the Cantor set is uncountable and measure zero.

It is totally disconnected, yet it is not a discrete space. In particular, this shows that connected components of a topological space need not be open sets.

  • 8
    $\begingroup$ This is a very good answer. I would go so far as to say that if you're studying general topology but haven't encountered the Cantor set, your ideas of what a topological space can be are fundamentally incomplete. $\endgroup$ – Pete L. Clark Dec 29 '10 at 6:49
  • 7
    $\begingroup$ “It is totally disconnected, yet it is not a discrete space.” As a professional matter, I like $\mathbb Q_p$ even better as an example of this behaviour. (Of course, topologically it's nearly the same!) If I may piggy-back on Pete's comment, if you haven't encountered $\mathbb Q_p$, then your ideas of what a complete metric space can be are fundamentally incomplete. :-) $\endgroup$ – LSpice Mar 28 '11 at 16:03
  • 8
    $\begingroup$ also positive measure cantor set is a very nice example to difference betwean meagre and null sets $\endgroup$ – Ostap Chervak Apr 10 '11 at 9:57
  • 1
    $\begingroup$ Take its union with the set of midpoints of the Cantor set's "gaps" (gap = connected component of Cantor set's complement in $[0,1]$), to get an interesting space: An uncountable set with a countable, dense and isolated subset! (The subset being the set of gap-midpoints.) // I found this after a class on the topology on the ultrafilters on $\Bbb N$. A homework problem was to show that the principle ultrafilters were dense in the space, but each was isolated. It continued, "Try to draw this, and become sad." So I was surprised to find out that I actually could draw this situation. $\endgroup$ – Akiva Weinberger Sep 2 '15 at 0:38
  • $\begingroup$ The Cantor set is also a complete metric space, in a few interesting ways. $\endgroup$ – Toby Bartels Feb 9 '17 at 21:24

A counter-example in graph theory - the Petersen graph.

alt text
(source: sdu.dk)

In many ways it is the most simple graph with many strange properties. See the article on Wiki.

Quote from our professor who teaches graph theory:

If you think you've proved any lemma about graphs, try Petersen first!

  • $\begingroup$ That last drawing, it looks like some vertices have degree 5. $\endgroup$ – Gerry Myerson Feb 17 at 5:28

The 8-element quaternion group. It can't be reconstructed from its character table (D_4 has the same one), and every subgroup is normal but it's not abelian.

  • 25
    $\begingroup$ And it has quaternionic representations - something that serves as counterexample to many beginner's conjectures in representation theory (a la "any representation can be constructed in the smallest field where its character lives"). $\endgroup$ – darij grinberg Apr 4 '10 at 15:59
  • $\begingroup$ +1, darij: this is perhaps worthy of its own answer. $\endgroup$ – Pete L. Clark Dec 29 '10 at 6:52
  • 2
    $\begingroup$ Also it has nontrivial inner and outer automorphisms, and (for minimality) is the smallest group with this property. This makes it a good example for distinguishing conjugate subgroups from the broader class of subgroups that are carried to each other by an automorphism. The klein 4-group can also be used for this, but the quaternion group is more pedagogically satisfying because it has a nontrivial characteristic subgroup as well as the 3 that are normal but not characteristic. (It is minimal for this property as well.) $\endgroup$ – benblumsmith Jul 22 '11 at 12:52

I'm surprised no one mentioned the Hawaiian Earring:

alt text

It's path-connected but not semi-locally simply connected (because any small neighborhood of the origin must contain a non-contractible loop). This implies many interesting properties, which make it a great counter-example. For instance...

  • The Hawaiian Earring cannot have a universal cover.
  • The Hawaiian Earring is not a CW-complex, although it is a compact, complete metric space
  • An example of a space which is semi-locally simply connected and simply connected but is not locally simply connected is the cone on the Hawaiian Earring.
  • For many years people thought the fundamental group was always a topological group. This turns out to be false, thanks to the Hawaiian Earring. There's a nice post about this here on MO
  • This question is Community Wiki for a reason. I'm sure there are other examples of conjectures the Hawaiian Earring has disproven, so please add them!
  • 4
    $\begingroup$ I like the fact, that it's fundamental group is uncountable. This is a vivid example for showing students, which are new to algebraic topology, that the fundamental group are not just "some" loops in the space. $\endgroup$ – archipelago Mar 23 '13 at 13:39
  • 9
    $\begingroup$ The analogs in higher dimensions have nonzero homology in arbitrarily high dimensions! $\endgroup$ – Jeff Strom Mar 23 '13 at 14:48

The Weierstrass function - which I guess is a counterexample to the conjecture that a function which is continuous everywhere must be differentiable somewhere. I remember being pretty amazed when I first encountered it. It made me realize that continuity and differentiability are really different notions.

  • $\begingroup$ I was really glad when my analysis professor first showed this example; I had the same realization that you did. $\endgroup$ – Cory Knapp Mar 4 '10 at 3:15
  • $\begingroup$ I think most of us do.I actually was shown several variamts,including van der Wearden's. A very good thing indeed to show newbies in real analysis. $\endgroup$ – The Mathemagician Apr 4 '10 at 19:21

The basic fact that there are smooth non-analytic functions on $\mathbb R$, and that there are compactly supported smooth functions, is important in real analysis and functional analysis.

$f(x) =\begin{cases} \exp(-1/(1-x^2)),& x \in (-1,1) \\\ 0& \text {otherwise} \end{cases}$

alt text

The usual examples of these functions often seem contrived. Here are examples of smooth nowhere analytic functions.

  • $\begingroup$ Hey, more cool information on smooth, nowhere-analytic functions! $\endgroup$ – Vectornaut Apr 4 '10 at 22:10
  • $\begingroup$ How is this compactly supported? I thought the support is where it's nonzero, and this is nonzero on $(-1,1)$ which isn't compact. $\endgroup$ – Akiva Weinberger Sep 2 '15 at 1:16
  • $\begingroup$ @columbus8myhw: The support is the closure of where it is nonzero. en.wikipedia.org/wiki/Support_(mathematics) $\endgroup$ – Douglas Zare Sep 3 '15 at 7:09
  • $\begingroup$ @columbus8myhw The support is the closure of the set where the function is not zero. $\endgroup$ – Dirk Sep 3 '15 at 18:15
  • $\begingroup$ Ah, that makes sense. $\endgroup$ – Akiva Weinberger Sep 3 '15 at 18:16

The matrices $A=\begin{pmatrix} 17\times 11 + 1 & 25\times 11\\ 11^2 & 16\times 11 + 1 \end{pmatrix}$ and $B = \begin{pmatrix} 17\times 11 + 1 & 11 \\ 25\times 11^2 & 16\times 11 + 1 \end{pmatrix}$ are similar modulo $m$ for every positive integer $m$ but are not similar over the integers.

In other words, there exist matrices $X_m\in GL_2(\mathbf Z/m\mathbf Z)$ such that $X_mA \equiv BX_m \mod m$ for every $m$, but there does not exist a matrix $X\in GL_2(\mathbf Z)$ such that $XA = BX$.

This is due to Stebe, Conjugacy separability of groups of integer matrices. Proc. Amer. Math. Soc., 32:1–7, 1972.


Counterexamples are very important when a student learns how to think in intuitionistic logic (and he has already been "spoiled" by classical logic). The counterexamples destroy the classical intuition, and when properly explained they help the student understand how to think intuitionistically. Some that seem to work praticularly well in my experience involve finite sets. Intuitionistically the following are not provable:

  1. A subset of a finite set is finite.
  2. The powerset of a finite set is finite.
  3. If a subset of $\mathbb{N}$ is not finite then it is infinite.
  4. The elements of a finite set may be listed without repetition.

All of these can be rescued with the additional assumption that the sets involved have decidable equality and that the subsets involved have decidable membership.

However, it does not really help the student to just know that certain "obvious" facts are not provable. He really needs to see how the "facts" can be false. The ones listed above are all false in the effective topos, but that's a complicated gadget for a beginner. It turns out informal explanations work well enough because most students know a little bit of programming. They just needs to know that the Halting Oracle does not exist.

My favorite counterexample in intuitionistic logic is that it is consistent to assume the so-called Axiom of Enumerability, which says that there are countably many countable subsets of $\mathbb{N}$. (Explanation: in the effective topos this just means that there is an effective enumeration of computably enumerable subsets of $\mathbb{N}$.) Many basic theorems of computability theory can be proved, phrased in a suitable form, from the axiom of enumerability using just constructive logic and no mention of machines of any kind.

  • 5
    $\begingroup$ Which definitions of "finite set" and "infinite set" are you using? $\endgroup$ – aorq Mar 2 '10 at 14:33
  • 3
    $\begingroup$ A set $A$ is finite if there exists a natural number $n$ and an onto map $e : \lbrace{1, ..., n\rbrace} \to A$. A set $B$ is infinite if there exists 1-1 map $m : \mathbb{N} \to B$, i.e., $B$ contains an infinite sequence without repetitions. $\endgroup$ – Andrej Bauer Mar 2 '10 at 21:34
  • $\begingroup$ Thanks, Andrej; These are all really fun... now to pick them apart and internalize them. $\endgroup$ – Cory Knapp Mar 4 '10 at 3:11

I've always been fond of the popcorn function (aka Thomae's Function), which is given by $f\colon \mathbb{R} \to \mathbb{R}$ via

$f(x) = \begin{cases} \frac{1}{n} & \mbox{if } x = \frac{m}{n} \in \mathbb{Q} \\ 0 & \mbox{if } x \notin \mathbb{Q}. \end{cases}$

This function has a couple of amusing properties.

(1) It is upper semicontinuous on $\mathbb{R}$, yet has a dense set of discontinuities (every one of which is removable) (namely $\mathbb{Q})$.

(2) Since it is bounded and has a set of measure zero as its set of discontinuities, it is Riemann integrable. So if we consider $g(x) = \int_0^x f(t)\ dt$, we see that $g \equiv 0$, so that $g'(x) \not \hskip 2pt = f(x)$ on a dense set.

References: http://en.wikipedia.org/wiki/Thomae%27s_function and of course "Counterexamples in Analysis" (Sec 2.15-2.17)

  • 4
    $\begingroup$ John H. Conway called that the "Stars over Babylon" function in a class I took. Finally I know another name. $\endgroup$ – Elizabeth S. Q. Goodman Mar 17 '11 at 7:35
  • $\begingroup$ And its "reciprocal" ($g(x)=\begin{cases}n,&x=m/n\\0,&x\notin\Bbb Q\end{cases}$) is also interesting, in that it's unbounded on every interval. Not quite as weird as Conway's base 13 function, which has an image of $\Bbb R$ on every interval, but still a nice little pathology. $\endgroup$ – Akiva Weinberger Sep 2 '15 at 1:27

The Baumslag--Solitar groups have presentations

$BS(p,q)=\langle a,b\mid a^p=b^{-1}a^q b\rangle$.

They have the following nice properties:

  1. they're two generator, one relator groups;
  2. they can be written as an HNN extension of $\mathbb{Z}$ over $\mathbb{Z}$. (This means that they're constructed by 'gluing' $\mathbb{Z}$ to itself in some way.)

So from the point of view of combinatorial group theory, they could hardly be simpler. And yet, for suitable values of $p$ and $q$ (typically $p,q$ relatively prime integers greater than 1 will do), we find that:

  1. they're non-Hopfian, meaning that they admit a self-epimorphism with non-trivial kernel;
  2. hence they're not even residually finite;
  3. they have exponential Dehn function (meaning that the word problem can be solved, but only very slowly);
  4. their virtual first Betti number is one (meaning that every finite-index subgroup has abelianisation of rank one)...

I could go on.

  • 4
    $\begingroup$ And for p != q they can't be the fundamental group of a 3-manifold -- see mathoverflow.net/questions/6132/… for some references. $\endgroup$ – Steven Sivek Mar 2 '10 at 20:32
  • $\begingroup$ That the Dehn function is big does not mean that the word problem can be solved only slowly. $\endgroup$ – YCor Aug 22 '15 at 12:20
  • $\begingroup$ @YCor - of course that is strictly correct. I was trying to give some rough idea of what it meant. $\endgroup$ – HJRW Aug 28 '15 at 21:38

Let $P dx + Q dy$ be a one-form, or if you're using the terminology of an introductory multivariable calculus course, a "vector field" that you can take line integrals of. Then students learn Green's Theorem, which says that if some countour $C$ bounds a region $D$, then $$\int_C P dx + Q dy = \int_D \left(\frac{dQ}{dx}-\frac{dP}{dy}\right) dx dy.$$

From this, one deduces that if the expression on the right hand side vanishes, then the integral around any contour is $0$. In particular, this allows one to define a primitive for $P dx + Q dy$.

Many students (myself included, a long way back) don't pay enough attention to the hypotheses in Green's Theorem and then assume that this is true of the following one-form (or "vector field"), which is my fundamental counterexample:

$$\frac{-ydx}{x^2+y^2} + \frac{xdy}{x^2+y^2}$$

Eventually a student discovers that the integral of this around the origin is $2 \pi$ and then wonders what went wrong. The problem is that the hypothesis of Green's Theorem requires that the form be defined everywhere in $D$.

In other words, this is a fundamental counterexample to the claim that a one-form in the plane with zero curl (where by "curl" I just mean the right hand side of the above) has a primitive.

Furthermore, this is a fundamental example of a nontrivial element in a de Rham cohomology group. In this case, the one-form above generates $H^1_{\text{dR}}(\mathbb{R}^2\setminus \{(0,0)\},\mathbb{R})$.


The Poincaré homology sphere, a spherical 3-manifold with fundamental group the binary isosahedral group, was Poincaré's counterexample to the original formulation (in terms of homology) of his conjecture. Due to its countless descriptions -- as a spherical 3-manifold, via Dehn surgery, as the configuration space of an isosahedron, etc -- it's still a motivational example in geometry and topology.


Homotopy groups do not, in general, commute with sequential colimits, even for nice maps between nice spaces.

I just learned this beautiful example from Bill Dwyer.
Take the sequence


Here $n$ denotes the $n$th power map on $S^1$. Thinking of $S^1$ as $\mathbb{R}/\mathbb{Z}$, one finds that the colimit of this sequence (in the category of topological spaces) is the quotient group $\mathbb{R}/\mathbb{Q}$. Note that this quotient group, topologized as a quotient space of $\mathbb{R}$ by the relation $x\sim y$ if $x-y\in \mathbb{Q}$, has the indiscrete topology. In particular, the colimit of this sequence is a contractible topological space and has trivial homotopy groups.

On the other hand, the colimit of the corresponding sequence of fundamental groups is the group $\mathbb{Q}$ (checking this is a fun exercise).

(There's something sort of odd here, because one might have guessed that $\mathbb{R}/\mathbb{Q}$ would be a model for $K(\mathbb{Q}, 1)$, since after all $\mathbb{R}$ is a free $\mathbb{Q}$-space. But there are no interesting open sets in the quotient and hence no chance of local triviality.)

  • $\begingroup$ Just a remark that this does work for sufficiently nice maps, namely, cofibrations between, say, CW complexes. In this case the colimit becomes homotopy equivalent to the mapping telescope, which is a CW complex, so that maps from spheres and homotopies between them all factor through some finite stage, as doesn't happen with this example. $\endgroup$ – Kevin Carlson Apr 16 '18 at 0:21
  • 2
    $\begingroup$ Indeed. In another direction, homotopy groups commute with filtered colimits of simplicial sets (the earliest reference I know of is Quillen's article "Algebraic K-theory I" in LNM 341). The above is not a counterexample to this statement, because there is no simplicial structure on S^1 for which all of the maps out of the first term in the above sequence can be simplicial (e.g. to make the squaring map simplicial, one needs at least two 1-cells in the domain circle; to make the 6th power map simplicial, one needs at least six 1-cells in the domain; etc.). $\endgroup$ – Dan Ramras Apr 16 '18 at 3:05
  • $\begingroup$ Well, why not allow infinitely many 1-simplices, as is necessary in any case to get a Kan fibrant circle? $\endgroup$ – Kevin Carlson Apr 16 '18 at 4:11
  • $\begingroup$ @KevinCarlson There's no simplicial structure on the topological space $S^1$ with infinitely many simplices (by compactness). What you're suggesting is replacing the spaces in the sequence with homotopy equivalent spaces. That will change the colimit of the sequence. $\endgroup$ – Dan Ramras Apr 16 '18 at 14:54
  • 4
    $\begingroup$ Of course, the mapping telescope of the system of maps that Dan describes is a model for $K(\mathbb{Q},1)$. One way to look at this example is as a case where hocolim is not homotopy equivalent to colim. $\endgroup$ – IJL May 21 '18 at 16:11

The Moulton plane is a projective plane that is a counterexample to the Desargues theorem, the little Desargues theorem, and just about every "nice" property of projective planes.

Its discoverer, F.R. Moulton, is best known as an astronomer. He apparently came up with the Moulton plane after sitting in on a projective geometry course as a graduate student.

  • 1
    $\begingroup$ Part of why Desargues "Theorem" is intriguing is that it holds in some projective planes and not in others. There are finite planes where it holds and finite planes where it does not hold. If there is a way to introduce coordinates for the plane with numbers from a division ring then then Desargues Theorem will hold. It also holds for projective planes sitting in higher dimensional projective spaces. In the real projective plane the theorem holds. The Moulton plane is a fascinating example. $\endgroup$ – Joseph Malkevitch Mar 3 '10 at 14:05

A basic result in commutative algebra asserts that direct limits commute with tensor products. My favourite counterexample to the statement obtained by replacing "direct" with "inverse" is the following. Let $p$ be a prime number; then

$\bigl(\varprojlim_n\mathbb Z/p^n\mathbb Z\bigr)\otimes_{\mathbb Z}\mathbb Q\cong\mathbb Q_p$,

the field of $p$-adic numbers (completion of $\mathbb Q$ with respect to the metric induced by the $p$-adic valuation), while

$\varprojlim_n\bigl((\mathbb Z/p^n\mathbb Z)\otimes_{\mathbb Z}\mathbb Q\bigr)=0$,

since every $\mathbb Z/p^n\mathbb Z$ is torsion and $\mathbb Q$ is divisible.


Added in edit. To understand this answer, one needs a definition that is current in French but for which I have not found a clearly equivalent word in English (causing misunderstandings, as can be seen in the comments). One says that a function $f$ defined near $0$ (say) has a "développement limité" (DL, which could be translated by "restricted expansion") of order $n$ at $0$ if there are numbers $a_0,a_1,\dots, a_n$ such that

$$ f(x) = a_0+a_1 x+a_2 x^2 + \dots + a_n x^n + o_{x\to 0}(x^n). $$

My instructional counter-example is: (end of addition)

The function $x\mapsto x^3\sin(1/x)$ has a second-order DL (before edit: "Taylor series") but is not twice differentiable at $0$. The circumstances where I came across this example are too embarrassing to tell here...

  • 2
    $\begingroup$ I flunked an exam in ordinary differential equations getting that one wrong-trust me,the circumstances can't be worse then that. $\endgroup$ – The Mathemagician Jul 29 '10 at 18:28
  • 1
    $\begingroup$ What is the second-order Taylor series for this function? $\endgroup$ – Mehrdad Mar 30 '15 at 8:15
  • 6
    $\begingroup$ @Mehrdad That's precisely my point: the Taylor-Peano formula deduces a Taylor series from the derivatives, but a Taylor series may exist even when the higher-order derivative don't. $\endgroup$ – Benoît Kloeckner Mar 30 '15 at 18:50
  • 6
    $\begingroup$ Somehow it seems less snappy when you phrase this as saying that $x \mapsto x\sin(1/x)$ has a $0$th-order Taylor series at $0$ …. $\endgroup$ – LSpice May 13 '16 at 4:41
  • 1
    $\begingroup$ Would you also say that the function $f(x) = x^2 \cdot I_{\mathbb{Q}}(x)$ (where $I_\mathbb{Q}$ is the indicator function of the rationals) has a second order Taylor expansion at $x=0$? $\endgroup$ – Steven Gubkin Jan 11 '18 at 19:17

A standard result in introductory calculus classes is that, if a function has positive derivative on an open interval, then it's increasing there.

Based on this, students tend to think that, if $f'(a)>0$, then $f$ must be increasing "near $a$."

However, the example $f(x) = 2x^2\sin(1/x)+x$ (set $f(0)=0$) shows that this is quite false!


I'd say the Tutte Graph, which is a counterexample to

Tait's conjecture: Every 3-connected cubic planar graph has a Hamiltonian cycle.

Initially, I thought this counterexample was extremely non-instructive, since I assumed that Tutte discovered it via some ingenious trial and error. But, after seeing a talk by Bill Cunningham, I discovered how Tutte came up with his counterexample and why it is a counterexample (it's unclear from looking at Tutte's graph that it is not Hamiltonian). The idea is quite simple but useful. Tutte assumed that Tait's conjecture was true and proceeded to prove a sequence of stronger (yet equivalent) conjectures. He then found a very small counterexample to the strongest conjecture, and then deconstructed the sequence of proofs to obtain a counterexample to Tait's conjecture.

I really like this method because it shows that there is hope for a mathematical caveman like me as long as I use my brain. That is, the counterexample was actually not pulled out of thin air like I initially thought.

  • 24
    $\begingroup$ Very few results are pulled out of thin air. The problem is, mathematical culture has not favored showing where they do come from. $\endgroup$ – Mariano Suárez-Álvarez Mar 3 '10 at 16:56
  • 8
    $\begingroup$ I'd add that one of the reason for this is that journals have a strong preference for short articles, although today, with internet distribution, I do not see any reason for this. As a result, mathematicians cut out the why from the proofs, leaving only the necessary steps. But this is of course discussed in other threads... $\endgroup$ – Andrea Ferretti Mar 3 '10 at 17:09

The blowup of $\mathbb{P}^2$ in the 9 points of intersection of two generic cubics admits infinitely many $(-1)$ curves. This example is very important in getting rid of the naif picture of algebraic surfaces.

  • 4
    $\begingroup$ Even though the idea of "blowing down" is to get rid "a" (-1) curve. You can always blow down finitely many times to get rid of all of them, even if you start infinitely many. This is a nice example of that case. $\endgroup$ – Matt Apr 4 '10 at 17:38

The alternating group on 4 letters is nice because it provides a counterexample to the converse of Lagrange's theorem: It has order 12, but it does not have a subgroup of order 6.

  • 3
    $\begingroup$ Similarly, the symmetric group on 5 letters has no subgroup of order 15, since (up to isomorphism) the only group of order 15 is the cyclic group, and $S_5$ has no element of order 15. $\endgroup$ – Gerry Myerson Mar 27 '11 at 23:35

Finite topological spaces often provide nice and simple counterexamples in topology, including algebraic topology (check J. Barmak's thesis). After getting familiar with those spaces one easily comes up with examples of phenomena such as weakly homotopy equivalent spaces which are not homotopy equivalent (spaces consisting of 4 points and 6 points suffice) or homomorphisms between homology/homotopy groups that are not induced by continuous maps.

Of course, other counterexamples are available, but finite ones are certainly minimal in a sense.


Small's Example from noncommutative algebra...

The triangular ring $T = \pmatrix{\mathbb{Z} & \mathbb{Q} \\\ 0 & \mathbb{Q}}$ has the following properties:

  • It's right noetherian but not left noetherian
  • It's right hereditary but not left hereditary
  • The right global dimension is 1 but the left global dimension is 2
  • This generalizes to give an example of a ring with right global dimension $n$ and left global dimension $n+1$ by replacing $\mathbb{Z}$ by $R$, a commutative noetherian domain of global dimension $n$, then replacing $\mathbb{Q}$ by $K = Frac(R)$
  • A similar example gives a ring which is noetherian but neither left nor right Ore. Just take $R = \pmatrix{S & 0 \\\ S & I}$ where $S = \pmatrix{\mathbb{Z} & 0 \\\ \mathbb{Z}_p & \mathbb{Z}_p}$ and $I = \pmatrix{\mathbb{Z} & 0 \\\ 0 & \mathbb{Z}_p}$ is an $S$-ideal.

Having been trained to think in a commutative world, I found the existence of an example for any one of these to be surprising. The fact that they were all (basically) the same example is even more amazing.


There exists a $3$-dimensional smooth projective variety $X$ which cannot be birational to a smooth variety with nef canonical bundle. This is because $K_X$ is big; if it was also nef it would have no cohomology and we could compute its self-intersection with Riemann-Roch by looking at the number of sections of its powers. It turns out that the self-intersection would be $3/2$.

This example (by Reid, I think) shows that if you want to have minimal models you have to allow singular varieties, so that $K_X$ can still be defined, but is not a Cartier divisor. This has led to the whole branch of birational geometry studying the type of singularities which are allowed in the minimal model program, like terminal, canonical, log-terminal, KLT and so on.


My favourite counterexample is purely academic: it does not have any applications, but I think it is pretty.

Let $X = \mathbb{N} \times \mathbb{N}$. Define a non-empty set $U \subseteq X$ to be open if for cofinitely many $x \in \mathbb{N}$ the set $\{ y \in \mathbb{N} \vert (x,y) \in U\}$ is cofinite.

Construct a sequence in $X$ that hits every point in $X$ exactly once. In other words, take a bijection $\mathbb{N} \rightarrow X$. Then:

  • $X$ is countable;
  • every point in $X$ is an accumulation point of this sequence, but
  • the sequence has no convergent subsequences.

In particular, this is an example in a countable set that accumulation point of a sequence does not have to be a limit of a subsequence. I call this the Herreshoff topology for the (high-school) student of mine who came up with it. (I could not find it anywhere else, although I do not discard that I did not look hard enough.)

  • 3
    $\begingroup$ This is sort of Arens-Fort space: en.wikipedia.org/wiki/Arens-Fort_space. Here you have the neighbourhoods of (0,0) in that space (considering (0,0) to be not in N x N), where there N x N is discrete. Your sequence then has the same properties. $\endgroup$ – Henno Brandsma Mar 6 '10 at 8:40

The Warsaw circle $W$ http://en.wikipedia.org/wiki/Continuum_%28topology%29 is a counterexample for quite a number of too naive statements.

The Warsaw circle can be defined as the subspace of the plane $R^2$ consisting of the graph of $y = \sin(1/x)$, for $x\in(0,1]$, the segment $[−1,1]$ in the $y$ axis, and an arc connecting $(1,\sin(1))$ and $(0,0)$ (which is otherwise disjoint from the graph and the segment).

Some observations: $W$ is weakly contractible (because a map from a locally path connected space cannot ''go over the bad point'').

Let $I$ denote the segment $[−1,1]$ in the $y$ axis. Then $W/I\cong S^1$ is just the usual circle, and thus we have a natural projection map $g:W \to S^1$. The point-preimages of $g$ are either points or, for a single point on $S^1$, a closed interval.

Thus the assumptions of the Vietoris-Begle mapping theorem hold for $g$, proving that $g$ induces an isomorphism in Cech cohomology. Thus the Cech cohomology of $W$ is that of $S^1$, but it has the singular homology of a point, by Hurewicz.

Since $I\to W$ is an embedding of compact Hausdorff spaces, we have an induced long exact sequence in (reduced) topological $K$-theory (see, for example, Atiyah's $K$-theory Proposition 2.4.4). Since $I$ is contractible, we get that $W$ and $S^1$ also have the same topological $K$-theory.

Note that the Warsaw circle is a compact metrizable space, being a bounded closed subspace of $R^2$. By looking on points on $I$ one sees that $W$ is not locally path-connected (and, in particular, not locally contractible).

The above observations imply:

  1. A map with contractible point-inverses does not need to be a weak homotopy equivalence, even if both, source and target, are compact metric spaces. Assuming that the base and the preimages are finite CW complexes does not help.

  2. The Vietoris-Begle Theorem is false for singular cohomology (in particular, the wikipedia version of that Theorem is not quite correct).

  3. The embedding $I\to W$ cannot be a cofibration in any model structure on $Top$, where the weak equivalences are the weak homotopy equivalences and the interval $I$ is cofibrant. Because then we would have a cofiber sequence $I\to W\to S^1$ and thus also a long exact sequence in singular cohomology.

  4. $W$ does not have the homotopy type of a CW complex (since it is not contractible).

  5. Even though the map $g$ is trivial on fundamental groups, it does not lift to the universal cover $p: \mathbb{R} \to S^1$, because $g$ cannot be nullhomotopic. Thus the assumption of local path connectivity in the lifting theorem is necessary.


Two common misconceptions that students have about the concept of independence in probability are

1) Thinking "$X$ and $Y$ are independent" means "$X$ and $Y$ don't affect each other".
2) Thinking "A set of variables is independent" means the same thing as "each pair of variables are independent.

A useful counterexample that addresses both of these misconceptions at once is as follows:

Let $X$ and $Z$ be independent random variables, each equally likely to be $1$ or $-1$. Let $Y=XZ$.

In this example, changing the value of $X$ clearly changes the value of $Y$, but it does not change the distribution of $Y$. So $X$ and $Y$ are independent even though they "affect each other" in some sense. Furthermore, the variables $X, Y, Z$ are pairwise independent, but not independent.


protected by Qiaochu Yuan Sep 21 '17 at 7:05

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.