# What are your favorite instructional counterexamples?

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.

• @Ben Linowitz. It is in the small and delightful book, J. W. S. Cassels, Lectures on elliptic curves. – Regenbogen Mar 2 '10 at 14:39
• @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/… – Regenbogen Mar 2 '10 at 17:38
• 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. – benblumsmith Jul 22 '11 at 12:55

I like the Sorgenfrey line. It's finer than the metric topology on R, and hereditarily Lindelöf, hereditarily separable, first countable, but not second countable. It's non-orderable, but generalised orderable, etc. It's a popular example for metrisation theorems, e.g. All its compact subsets are at most countable.

In question #14739, I asked whether the product of two ideals of a commutative ring $R$ could be defined lattice-theoretically the same way the sum and intersection can. Bjorn Poonen gave a great counterexample that shows the answer is no! This supports a point fpqc had been trying to make to me earlier that the relationship between $R$ and the Zariski topology on $\text{Spec } R$ was more subtle than I had thought: in particular, it has more structure than just the Galois connection.

The following are, I think, the "worst possible" counterexamples in measure theory. They would benefit from a nice list of properties -- I have a feeling that I'm forgetting a lot. Feel free to improve!

The Cantor set and its friend the Cantor function are standard counterexamples. Keeps increasing regardless of the zero derivative almost everywhere... Also, the corresponding measure $\mu$, defined so that the measure of the interval [a,b] is f(a)-f(b) where f is the Cantor function is supported on a Lebesgue-zero set.

Another good source of examples is the measurable set $A \subset [0,1]$ such that for any interval I, $\lambda(I\cap A) > 0$ and $\lambda(I\cap A^c) > 0$. ($\lambda$ is the Lebesgue measure, c denotes complement).

Here's a construction of A that I heard from Ulrik Buchholtz. Instead of just constructing A, we'll make two disjoint sets A and B which have intersection of positive measure with any interval. Consider the set of all subintervals of [0, 1] with rational endpoints. It is countable, so let In be the n-th interval in the list. Put two fat (positive-measure) disjoint Cantor sets (one for A and one for B) inside I1. (We can just put the second inside some gap of the first). By the main property of Cantor sets, every interval In minus the Cantor sets is a non-empty union of intervals. So, we can put two fat disjoint Cantor sets (also disjoint from the previous ones) inside I2, and keep going forever. Every time, we add one Cantor set to A and one to B.

Now, each subinterval of [0,1] will contain one of the In-s, and therefore its intersection with both A and B has positive measure. Both A and B are countable unions of measurable sets, and therefore measurable. We are done.

• What is a fat Cantor set? – Sune Jakobsen Apr 4 '10 at 15:08
• @Sune: a fat Cantor set is a variation on the Cantor set that has positive measure. Just like the usual Cantor set is closed, nowhere dense, and uncountable. The difference is that the usual Cantor set has measure zero (the total length of all the intervals you remove when constructing it is 1). You construct the fat Cantor set the same way as the usual Cantor set, but you carefully vary the sizes of the intervals you remove. Apparently, it's also called the Smith-Volterra-Cantor Set. More details: en.wikipedia.org/wiki/Smith%E2%80%93Volterra%E2%80%93Cantor_set – Ilya Grigoriev Apr 4 '10 at 20:26
• @llya The fat Cantor set is one of the great teaching examples of both analysis and topology.Most professors just go over the plain vanilla Cantor set. This is really doing the class a disservice because they don't really get the depth of the sheer diversity of pathology that can occur the real line simply by varying the details of the method of construction of the set, – The Mathemagician Jul 29 '10 at 18:36
• I also like the proof of the existence of the set $A$ you mentioned above using the Baire category. Consider the metric on measurable sets you get from the inclusion into L^{1}, to prove that such an $A$ exists, it suffices by Baire category to show that for any interval $I$ with rational endpoints the collection of sets $A$ which have $0<\lambda(A\cap I)<\lambda(I)$ are open and dense. But this is essentially a triviality. – Benjamin Hayes May 2 '11 at 19:53
• I guess I should note that the collection of measurable sets with this metric is complete, but it is fairly clearly closed in $L^{1},$ while not as constructive as your construction it's very pain free. – Benjamin Hayes May 2 '11 at 19:55

The Schoenflies conjecture was asserting that the two connected components of the complement of an embedded $2$-sphere $S^2$ in $S^3$ were simply connected. A kind of generalized Jordan theorem.

Antoine's necklaces gave a first counterexample, and that counterexample was reworked by Alexander to obtain the horned sphere :

In this counterexample, the set of singular points of the embedding is a Cantor set, so is quite big. Later, Artin and Fox developped the notion of wild arcs, and found the following simpler counterexample, where there are only two singular points :

• How does Antoine's necklace give a counterexample? It's not an embedded sphere. – Akiva Weinberger Mar 2 '16 at 14:35
• @AkivaWeinberger Antoine's Necklace (AN) gives an example of two homeomorphic subsets of $\mathbf R^3$ whose complements are not homeomorphic (AN is a Cantor set, the complement of AN is not simply connected). I think I read in Moïse that it implies, after a detour, that the Schoenflies conjecture is false ... I'm gonna check. – few_reps Mar 2 '16 at 17:04
• I found a paper that constructs "Antoine's Horned Disks", which, after identifying the boundary, creates a horned sphere. I think that that's probably what Antoine's Horned Sphere is. jstor.org/stable/2686463 – Akiva Weinberger Mar 6 '16 at 2:19

"Every finitely-branching tree with infinitely many nodes has an infinite branch" is constructively false, as witnessed by the following counterexample:

Andrej Bauer's exposition (above) is especially nice; most textbooks take a far less direct route to the result, which makes it harder to see what's really going on past the level of "yeah, the proof is correct step-by-step."

As a counter-example for Fatou's lemma in measure theory: strict inequality can occure! Just take the measure space $\mathbb{N}$ with the counting measure and consider the functions $$f_n(k) = \delta_{nk}$$ Then the sum of $f_n$ is always $1$ while the pointwise limit of the $f_n$ will be the zero function having zero integral. If you have this counter-example then you do not need fancy measures and integrals at al to produce examples that in Fatou's lemma strict inequality may happen...

I occasionally use the following "counterexample" to unique factorization in Z in an introduction to math course: (1003)(1007)=(901)(1121). Once the students figure out what's going on, I think they learn something from it.

• $(17\cdot59)(19\cdot53)=(17\cdot53)(19\cdot59)$ – Akiva Weinberger Sep 2 '15 at 2:06
• Is the point that, despite @AkivaWeinberger's explicit factorisation, all of the multiplicands 'look prime'? (That is, is it the same (counter)example as $6\cdot35 = 14\cdot15$, but for people with some number sense?) – LSpice May 13 '16 at 4:48

This is an easy one, but one I've found useful in the past to keep in mind, and which I've passed on to many younger students who are new to homological algebra. These students sometimes struggle with the idea of a non-free projective module because if you're new to modules and you still think of them via analogy to vector spaces then it's natural to think direct summands of free modules should be free.

A nice counter-example to keep in mind is the ring $\mathbb{Z}/6\mathbb{Z}$ and the projective but not free module $\mathbb{Z}/3\mathbb{Z}$ (projective because $\mathbb{Z}/6\mathbb{Z} \cong \mathbb{Z}/3\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z}$)

Coefficients of cyclotomic polynomials over $\mathbb{Q}$.

If you look at the factorization of $X^n-1$ over the integers, for $2 \leq n \leq 104$, you would "notice" that all nonzero coefficients of all factors are $\pm 1$. Indeed, $105$ is the first counterexample to this conjecture, with the 105th cyclotomic polynomial having coefficients of $2$ in its expansion. This can happen because $105$ has three distinct odd prime factor. The conjecture and the counterexample, however, are accessible even to high school students.

A quick Internet search suggests the following book as a reference:

McClellan, J. H. and Rader, C. Number Theory in Digital Signal Processing. Englewood Cliffs, NJ: Prentice-Hall, 1979.

I admit I have not read it - I first saw the counterexample while teaching high school, and it came up again in an advanced undergraduate course on Galois theory.

• – Gerry Myerson Jul 15 '13 at 5:44

Volterra's function has a derivative everywhere which is bounded, discontinuous, and cannot be Riemann-integrated. It depends on the Cantor sets, of course, already mentioned.

Possible reference: Bernard R. Gelbaum, John M. H. Olmsted: Counterexamples in Analysis.

The statement S "every injective endomap is also surjective" can be formalized in terms of second-order logic (and, of course, precisely states that the strcture in question is finite). This is a counterexample to any kind of compactness result for second-order logic, because if such a result existed, one would be able to get infinite sets satisfying S.

Any classical counter-example to inversion of a limit and an integral, $f_n:[0,1[\to\mathbb{R} ; x\mapsto n^2 x^n$ say. Basic, but important to motivate the dominated convergence theorem.

Here is a useful example of counter-examples in commutative ring theory;

Let $R=P(\mathbb{N})$ be the power set of $\mathbb{N}.$ It has a ring structure $(R, +, \times)$ where $+$ is the symmetric difference of sets and $\times$ is the intersection of sets.

Applications:

Obviously, $R$ is a commutative ring with $1$, ($\mathbb{N}$ is the $1$).

1) Let $R$ be a commutative ring with $1$ and a multiplicative closed set of $R$. If $R$ is Noetherian (Artinian) ring then $S^{-1}R$ is Noetherian (Artinian). Does the converse hold?

No, it doesn't.

Using the above example, for any prime ideal $p$ of $R$, $R_p$ (the localization at $p$) is Noetherian (Artinian) while, $R$ is not Noetherian (Artinian).

Outline:

Consider P({1}) $\subset$ P({1,2}) $\subset...$ and $P(\mathbb{N}) \supset$ P($\mathbb{N} \setminus${1}) $\supset$ P($\mathbb{N} \setminus${1,2}) $\supset ...$ showing that $R$ is neither Noetherian nor Artinian ring.

It is easy to verify that $R_p$ is isomorphic to $\mathbb{Z}/2$, hence it is both Noetherian & Artinian. (Every element of $R_p$ is either $0/1$ or a invertible.)

2) Let $R$ be an integral domain (also commutative with $1$), then for every multiplicative closed set of $R$, $S^{-1}R$ is an integral domain, hence for every $R_p.$ Does the converse hold?

By the above example, it doesn't, since $(P(\mathbb{N}),+,\times)$ is not an integral domain.

• It may be worth noticing that this ring $R$ is nothing but $(\mathbb Z/2)^{\mathbb N}$ in disguise. Also, I am surprised with your statement that localizations $R_p$ are all isomorphic to $\mathbb Z/2$. – ACL Mar 17 '11 at 9:10
• @ACL: Good point. Is it possible to understand $Spec(R)$ in this example? The point is that $P(A)$ where $A \subset \mathbb{N}$ is an ideal but not all ideals are being as such! – Ehsan M. Kermani Mar 17 '11 at 10:16
• The prime ideals in this ring are the complements of the ultrafilters on $\mathbb N$, so the spectrum is the Stone-Cech compactification of the discrete space $\mathbb N$. – Andreas Blass Mar 17 '11 at 13:53

Ackermann function (http://en.wikipedia.org/wiki/Ackermann_function) defined as

$A(m, n) = \begin{cases} n+1 & \mbox{if } m = 0 \\ A(m-1, 1) & \mbox{if } m > 0 \mbox{ and } n = 0 \\ A(m-1, A(m, n-1)) & \mbox{if } m > 0 \mbox{ and } n > 0. \end{cases}$

is total recursive but not primitive recursive. To see this we could prove by induction on the complexity of primitive recursive functions that each primitive recursive function is eventually dominated by this function (we need a bit coding to keep the number of arguments consistent). Essentially this function manages to capture the fast-growing'' property. Note that the index set for recursive functions is not recursive while that for primitive recursive function is.

• The Ackermann function also serves as a counterexample to the following false statement: "A total function is primitive recursive iff its graph (considered as a relation) is primitive recursive". (Here I call a relation pr.rec. iff its characteristic function is pr.rec.) – Goldstern Feb 17 '19 at 12:44

Here is some simple counterexample in commutative algebra, which I found really cute when I first meet it:

Let $k$ be a field, $A = k[X_{1},X_{2},X_{3}\ldots],$ $I = (X_{1}, X_{2}^{2}, X_{3}^{3},\ldots)$ and $R = A/I.$ Then $\text{Spec}(R)$ consists of one point (because $\text{rad}(I)$ is maximal ideal of $A$); in particular $\text{Spec}(R)$ is a noetherian space, and $\dim R = 0$; although $R$ is not noetherian ring (since $\text{nil}(R)^{n}\neq 0$ for every $n$).

• I don't know; I think it's pretty intuitive that there exist local rings that aren't Noetherian. Nothing in the definition of a local ring suggests that they need be Noetherian. – Qiaochu Yuan Apr 4 '10 at 21:21
• Nevertheless, I think that it's not obvious that there exist commutative rings with only one prime (not only with one maximal) ideal that are not noetherian. – ifk Apr 4 '10 at 21:35
• @ifk: There are simpler examples of that: Consider the direct sum $R=k\oplus V$ of a field $k$ and an infinite dimensional vector space $V$, made into a ring so that $V$ is an ideal which squares to zero, $k$ and $V$ multiply as you expect, and $k$ is a subring (this is called a trivial extension, in some contexts) Then $R$ is commutative, has only one prime, and it is not noetherian. – Mariano Suárez-Álvarez Apr 5 '10 at 6:05
• Nice, thank You Mariano. However I don't think it's really much simpler than above. – ifk Apr 5 '10 at 9:47

I'm shocked that noone has mentioned the Quaternion group! This thing is a counterexample to lots of basic questions you'd come up with while learning (finite) group theory.

For example (although not really a counterexample to a specific question), if you know the semidirect product construction and Sylow theorems and are trying to classify groups of low order, the quaternion group is the first group you can't construct as a semidirect product of cyclic groups. This can be an entry point for the extension problem for groups and cohomology of groups.

Rotations $\rho_\alpha$ of the unit circle by an angle $2\pi\alpha$ are nice examples in the theory of discrete dynamical systems.

If $\alpha=m/n$ is rational, then every point on the circle is periodic of prime period $n$ for $\rho_\alpha$, but has no fixed points. This shows that Sharkowskii's theorem does not hold in general for functions continuous $f\colon X\to X$ if $X$ is not the real line or an interval of the real line.

If $\alpha$ is irrational, then the orbit under $\rho_\alpha$ of every point of the circle is dense, but $\rho_\alpha$ has nor sensitive dependence on initial conditions, and in particular is not caotic.

The elliptic curve 960d1 in Cremona's tables is the smallest conductor example of an optimal elliptic curve with nontrivial Shafarevich-Tate group which is isogenous to an elliptic curve with trivial Shafarevich-Tate group.

• I had been concerned about wether this was proven to be the smallest example. It is now, thanks to the work of Robert Miller. – Jamie Weigandt Apr 30 '11 at 2:38

The 5-cycle $C_5$ is a great counterexample. It's the smallest imperfect graph, it's self-complementary, it has chromatic number $>\Delta$, it has no stable set meeting every maximum clique and yet satisfies $\omega = \frac{2}{3}(\Delta+1)$, it has chromatic number $> \frac 1 2 (\Delta+\omega+1)$, meaning that Reed's $\chi, \omega, \Delta$ conjecture is somehow tight.

And when you blow up each vertex into a clique or stable set of size $k$, the fun continues. For $k=3$ this gives you Catlin's counterexample to Hajos' Conjecture.

$$\textbf{Algebra.}$$

• The symmetric group $$S_{3}$$ is the first $$\text{non-abelian}$$ group and also this group has a fascinating property that $$S_{3} \cong \mathscr{I}(S_{3})$$ where $$\mathscr{I}$$ denotes the $$\text{Inner - Automorphism}$$ group.

• Example of a group which is $$\textbf{isomorphic}$$ to it's proper subgroup. $$\mathsf{Answer:}$$ Take $$G=(\mathbb{Z},+)$$ and take $$H= 2\mathbb{Z}$$. Then $$G \cong H$$.

• Example of a free module in which a linearly independent subset cannot be extended to a basis. $$\textbf{Answer.}$$ As a $$\mathbb{Z}$$ module $$\mathbb{Z}$$ is free with basis $$\{1\}$$ and $$\{-1\}$$. Now $$\{2\}$$ is linearly independent over $$\mathbb{Z}$$. Note that $$2$$ cannot generate $$\mathbb{Z}$$ over $$\mathbb{Z}$$. If at all there is a basis $$\mathscr{B}$$ containing $$2$$, $$\mathscr{B}$$ should have atleast one more element, say $$b$$. We then have $$b\cdot 2 - 2\cdot b =0$$, i.e $$\{2,b\}$$ is linearly dependent subset of $$\mathscr{B}$$ which is absurd.

$$\textbf{Analysis.}$$

• The function defined by $$f(x) = x^{2} \cdot \sin\frac{1}{x}$$ for $$x \neq 0$$ and $$f(x) =0$$ for $$x=0$$. This is example of a function whose derivatives are not continuous.

• Set that is not Lebesgue measurable. Example given by Vitali.

Assume given three projective systems $\{A_n,\alpha_{nm}\}_{n\in\mathbb{N}}$, $\{B_n,\beta_{nm}\}_{n\in\mathbb{N}}$ and $\{C_n,\kappa_{nm}\}_{n\in\mathbb{N}}$ of abelian groups (modules over some ring would equally do), endowed with arrrows $$0\rightarrow A_n\xrightarrow{f_n}B_n\xrightarrow{g_n}C_n\rightarrow 0$$ making the above sequences exact for every $n$ and satisfying the commutativity conditions $\beta_{nm}\circ f_n=f_m\circ\alpha_{nm}$ and $\kappa_{nm}\circ f_n=f_m\circ\beta_{nm}$. Then one can form the projective limits of the system to find a sequence $$0\rightarrow \varprojlim A_n\xrightarrow{f}\varprojlim B_n \xrightarrow{g}\varprojlim C_n$$ and a classical result says that, in order for this sequence to be right-exact, one needs the system $A_n$ to be stationary - meaning that $\alpha_{nm}(A_n)=\alpha_{n'm}(A_{n'})\subseteq A_m$ for all $n,n'\gg m$.

A classical counterexample showing the necessity of this condition is to take $A_n=p^n\mathbb{Z}$ with $\alpha_{nm}$ given by inclusions, $B_n=\mathbb{Z}$ for all $n$ with identity maps $\beta_{nm}=\mathrm{id}$, and $C_n=\mathbb{Z}/p^n\mathbb{Z}$ with the obvious maps. The system $A_n$ is non-stationary because the image of $A_n$ in $A_m$ is $p^n\mathbb{Z}\subseteq p^m\mathbb{Z}$ which becomes smaller and smaller as $n\rightarrow \infty$: the corresponding sequence of projective limits is $$0\rightarrow 0\rightarrow \mathbb{Z}\rightarrow\mathbb{Z}_p$$ which is clearly not right exact.

[Later remark]: After typing all down, I remarked that everything can be found in Wikipedia at http://en.wikipedia.org/wiki/Inverse_limit Moreover, the stationary condition quoted above, usually referred to as Mittag-Leffler condition, is enough to prove right-exactness of $\varprojlim$ in Ab, but there is a counterexample due to Deligne and Neeman showing that in other categories this is not enough, see http://www.springerlink.com/content/aeem2yx884nnufxn/

The matrix pencil $$\left\{ \begin{bmatrix} 0 & 1 & x\\ 1 & 0 & 0\\ x & 0 & 0 \end{bmatrix} : x \in \mathbb{R} \right\}.$$ The matrices composing it are all singular, but they have no common left or right kernel (which is a property that one expects when first diving into the theory of matrix pencils). Singular pencils are difficult (or impossible) to handle for algorithms to solve generalized eigenvalue problems. For instance, Matlab's eig([0 1 0; 1 0 0; 0 0 0],[0 0 1; 0 0 0; 1 0 0]) returns 0 NaN 0 instead of something like NaN NaN NaN which would make more sense (no zero eigenvalues here), since the algorithm is not designed to handle this kind of singular problems.

My favorite counter-example is given in the short paper, "Almost Commuting Unitaries," by R. Exel and T. Loring.

Here is a little background. Two $n \times n$ matrices $A$ and $B$ are said to be "almost-commuting" if there commutator, $[A, B]$, is small in some matrix norm. In the paper, the authors exhibit a family of unitary matrices, $U_n$ and $V_n$ that "almost-commute" in the sense that given $\epsilon > 0$ there exists an $N \in \mathbb{N}$ with $|| [U_n, V_n] || < \epsilon$ for all $n \geq N$, yet for any commuting $n \times n$ matrices, $X, Y$ $(XY = YX)$ there exists an absolute constant $C > 0$ such that $\max(||X - U_n||, ||Y - V_n||) > C > 0$. This was one of the first counter-examples in a research paper that I understood because the authors method of proof is very elementary. The most technical fact used is that the winding number of a closed curve around the origin is a homotopy invariant.

Writing $$H$$ for the Heaviside function, I was surprised when I first realised that the map $$t \mapsto H(\cdot - t)$$ is not Borel measurable as a map from $$\mathbf{R}$$ to $$L^\infty(\mathbf{R})$$. This illustrates that the intuition that "all unambiguously defined maps are measurable" really only works when the target space is separable.

• This is also a nice illustration of the difference between the norm and weak-* topologies, and the necessity of using the latter in certain situations. – Robert Furber Feb 17 '19 at 13:43

A nice counterexample to the statement "$L^p$ convergence to $0$ implies pointwise a.e. convergence to $0$" is obtained by taking characteristic functions of length $\frac{1}{n}$ wrapping around the interval $[0,1]$. These integrate to $\frac{1}{n}$, but converge nowhere to $0$ because the harmonic series diverges.

A counterexample to the converse is easier: just take $f_n = n(n+1)\chi_{[\frac{1}{n+1},\frac{1}{n}]}$. These integrate to $1$ and converge everywhere to $0$.

• I'm sorry, could I ask you to clarify what you mean by "characteristic functions of length $1/n$ wrapping around the interval $[0, 1]$". I ask as I thought $L^p$ convergence to zero did imply pwae convergence to $0$, so I would like to understand your example. – Arch Stanton May 19 '12 at 13:53
• $\chi_{[0,1/2]}, \chi_{[1/2,5/6]}, \chi_{[5/6,1] \cup [0,1/12]}, \chi_{[1/12,17/60]}...$. – Douglas Zare Jul 10 '12 at 4:58
• I should also mention here that $L^p$ convergence of $\{f_k\}$ to 0 implies that a subsequence converges pointwise a.e. to zero. To see this take a subsequence with $\int |f_k|^p < 2^{-k}$ (or any summable series) and use the monotone convergence theorem to conclude that $\int \sum |f_k|^p < \infty$. – Connor Mooney Aug 2 '12 at 12:10

I am surprised no one mentioned the unilateral shift on $l^2 ({\bf N})$, that is, $Te_i = e_{i+1}$ where $\{e_i\}$ is the standard orthonormal basis.

It provides examples for the following.

(a) a ring with elements $x,y$ such that $xy = 1 \neq yx$;

(b) an isometry from a Hilbert space to itself that is not a unitary;

(c) a one to one BLT that is not invertible;

(d) its adjoint is an onto BLT with nontrivial kernel;

and lots of other stuff (for example, its spectrum).

• Where is the ring? – Gerry Myerson Feb 9 '17 at 22:01
• @GerryMyerson For (a) one doesn't have to be so fancy, but it could be the ring of bounded linear maps taking $l^2$ to itself, with multiplication given by composition. Here $y$ would be given by $T$, and $x$ by $e_0 \mapsto e_0$ and $e_{i+1} \mapsto e_i$. – Todd Trimble Feb 9 '17 at 22:46
• Easier is $T^* : e_i \mapsto e_{i-1}$ if $i > 1$, and $T^*e_1 = 0$ (there is no $e_0$). – David Handelman Feb 10 '17 at 14:08
• Well, I say there is an $e_0$ because $0$ is a natural number (belongs to $\mathbf{N}$). But it's not important here. – Todd Trimble Feb 11 '17 at 14:13
• This boils down to the use of ${\bf N}$ to denote the positive integers (my use) or the nonnegative integers (your use); I prefer using $\bf Z^+$ for the latter, which avoids some ambiguity ... (and although it is nonstandard, $\bf Z^{++}$ for the positive integers avoids more ambiguity). – David Handelman Feb 11 '17 at 14:21

My favorite example in discrete mathematics is the sequence $1,2,4,8,16,31,..$. That is, number of regions in a circle after drawing all the chords between $n$ points on the boundary of the circle.

It shows that a simple pattern might be wrong, and that we do need formal proofs, no matter how many examples we've checked.

• I feel like this is a also good example of the power of intuition - we 'should' expect this sort of configuration to be polynomially sized, because of the geometry, and so one should innately be suspicious that the natural pattern can continue indefinitely. – Steven Stadnicki Feb 5 '18 at 20:17

Not every epi is surjective

Recall that if $$\mathcal{C}$$ is a category, then an arrow $$p:x\to y$$ in $$\mathcal{C}$$ is said to be an epi if for every pair of parallel arrows $$f,g:y\to z$$ such that $$f\circ p = g\circ p$$, we have $$f=g$$. Of course, every surjective map is epi, but the converse is not true.

Consider the category of rings and ring homomorphism and the canonical injection of $$\mathbb{Z}$$ into $$\mathbb{Q}$$, call it $$\iota$$. Assume that $$f,g:\mathbb{Q}\to R$$ are two ring homomophisms such that $$f\circ \iota=g\circ \iota$$. Then $$f\left(\frac{n}{m}\right) = f\left(\frac{n}{1}\right)f\left(\frac{m}{1}\right)^{-1} =(f\circ\iota)(n)(f\circ\iota)(m)^{-1} = g\left(\frac{n}{m}\right),$$ whence $$\iota$$ is an injective epi.

From an earlier post: "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."

Although the character tables for the dihedral group D of order 8 and the quaternion group Q of order 8 may seem the same, they are not. Using Adams operations on the representation rings for D and Q, it is possible to show that these representation rings are different as rings with operations (either lambda or Adams operations). These Adams operations are defined in a paper by Aityah and Tall, where it is shown how to calculate them directly from character tables.

• It can't be possible to define lambda operations directly from the character table; you need to know some of the multiplication table as well. – Qiaochu Yuan Jul 31 '11 at 18:07

In topology, The comb space is an example of a path connected space which is not locally path connected. see http://en.wikipedia.org/wiki/Comb_space.