What are your favorite instructional counterexamples?

Question

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.

Answer 1

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 $0$.

Answer 2

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.

Answer 3

In statistics not all distributions worth studying have mean, variance, skewness, or higher moments. Thus applications of the Central Limit Theorem may break down in such cases.

For example we have the Cauchy distribution:

$$f(x; x_0,\gamma) = \frac{1}{\pi\gamma \left[1 + \left(\frac{x - x_0}{\gamma}\right)^2\right]} = { 1 \over \pi \gamma } \left[ { \gamma^2 \over (x - x_0)^2 + \gamma^2 } \right],$$

with location parameter $x_0$ (=median/mode) and scale parameter $\gamma$ (=half width at half maximum).

The Cauchy distribution is also the distribution of the ratio of two variables $U$ and $V$, both of which are $\sim N(0,1)$. The moments do not converge.

Image from Wikipedia:

Answer 4

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.

Answer 5

I am quite keen of a counterexample in Strichartz estimates by Thomas Wolff that I think satisfies the requirements in your question. The problem was whether or not there are $L^p$ Strichartz estimates that lose no derivatives. Wolff gave a counterexample that, assuming the best control possible, then this is not possible for $p>2$. It uses the result by Charles Fefferman that the disk multiplier $S_1$ is not bounded on $L^p,~p\neq 2$. This is brilliantly explained by Professor Tao in the Notice of the AMS article "From Rotating Needles to Stability of Waves: Emerging Connections Between Combinatorics, Analysis, and PDE". It goes (more-or-less) like this:

Let $B(0,1)$ denote the unit ball of centre $0$ and radius $1$. Wolff showed that the inequality \begin{equation*} ||u||_{L^p([1,2]\times \mathbb{R}^n)}\leq C||f||_{L^{\infty}(B(0,1))} \end{equation*} fails, for $f$ bounded on $B(0,1)$.

Wolff used a mixture of Kakeya tube constructions, and the manipulation of specially constructed waves consisting of bump and exponential functions adapted to the tubes.

Answer 6

The tensor product over a ring is not left exact

Consider the short exact sequence $0\to2\mathbb{Z}\to\mathbb{Z}\to\mathbb{Z}_2\to0$ and tensor it by $\mathbb{Z}_2$ over $\mathbb{Z}$. It is well-known that this produces the exact sequence $$2\mathbb{Z}\otimes_{\mathbb{Z}}\mathbb{Z}_2 \to \mathbb{Z}_2 \to \mathbb{Z}_2\otimes_{\mathbb{Z}}\mathbb{Z}_2 \to 0$$ where the first arrow is the $0$ map even if $2\mathbb{Z}\otimes_{\mathbb{Z}}\mathbb{Z}_2$ is non-zero (in particular, it cannot be injective).

Answer 7

The group algebras of non-isomorphic groups could be isomorphic.

The simplest such example is as follows. If $G_1$ and $G_2$ are the two non-isomorphic non-abelian groups of order $p^3$ for an odd prime $p$, then $\mathbb{Q}[G_i]$ for $i=1,2$ are isomorphic. This is indirectly mentioned in an earlier answer to a question regarding counterexamples in algebra.

If one wants examples with $\mathbb{Q}$ replaced by finite fields or even integers, then such groups have been found but are significantly more complicated.

Answer 8

For a function $f:\mathbb R^n\to\mathbb R^m$, it is possible for the directional derivatives in all directions to exist at a point without $f$ being continuous there, let alone differentiable. Let $f:\mathbb R^2\to\mathbb R$ be the characteristic function of the set $E=\{(x,y):y=x^2,x\neq0\}$, and consider the behaviour of $f$ near the origin.

I learnt this example from Andrew D. Hwang in this post.

Answer 9

I like "the deleted Tychonov plank" which is described in "Counterexamples in Topology".

This space provide us a pure algebraic counterexample:

A commutative ring is called a $Z$-ring if all its elements are zero divisors. At first glance, it seems that an extension of a $Z$-ring by a $Z$-ring is necessarily a $Z$ ring. But this space provides a counterexample. The reason is explained here.

https://arxiv.org/abs/1307.5836

https://link.springer.com/article/10.1007/s41980-023-00771-x

To be honest, I spent a few weeks to give a (positive) proof for this ring-extension statment, but finaly I found this counterexample in the book "Counterexamples in Topology" which excited me a lot.

As a consequence of this post we ask:

Are there two $Z$-rings $R_{1}$ and $R_{2}$ such that for every short ring exact sequence $$0\to R_{1} \to S \to R_{2} \to 0$$ $S$ must be a $Z$-ring?

Answer 10

Another one of my favorite counter examples is $2\mathbb{Z}$ which is a RNG, or a ring without identity.

Answer 11

One might expect that for any function $f$ in the Hilbert space $L^2$ with orthonormal basis $\{ f_i \}$, the generalized Taylor series $$\sum_{j=1}^\infty \langle f, f_j \rangle f_j$$ converges pointwise to $f$ almost everywhere. Carleson's theorem gives that this is true for the standard Fourier basis of $L^2([0,1])$, but there exist uncountably many pairs of functions $g$ and orthonormal (ordered) bases $\{ f_i \}$ such that the generalized Fourier series above diverges pointwise almost everywhere, as discussed here.

Answer 12

Square root and square power are not inverse operations

In fact, $\sqrt{x^2}=|{x}|$, which is defined for every $x\in\mathbb{R}$, while $({\sqrt{x}})^2=x$ is defined only for $x\geq 0$.