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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.

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  • $\begingroup$ @Ben Linowitz. It is in the small and delightful book, J. W. S. Cassels, Lectures on elliptic curves. $\endgroup$
    – Regenbogen
    Mar 2, 2010 at 14:39
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    $\begingroup$ @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, 2010 at 17:38
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    $\begingroup$ 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. $\endgroup$ Jul 22, 2011 at 12:55

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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:

Cauchy Distribution

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The four-color theorem is often expressed in different ways which are loosely claimed to be equivalent. However, not all of these formulations are actually equivalent, and not all of them are even true. This means that there are "counterexamples" to certain (incorrect) formulations of the four-color theorem.

The mathematically true statement of the theorem is:

(1) Any loopless planar graph can be colored with (at most) four colors such that no edge connects two vertices of the same color.

However, the theorem is usually introduced in the context of coloring maps, and is loosely framed as saying that "Any map can colored such that no two adjacent countries have the same color." However, making this formulation mathematically precise is surprisingly challenging. For example, here is a seemingly reasonable attempt to formalize the "map" version of the four-color theorem:

(2) Let $D$ be an open subset of the plane, and consider an arbitrary partition of $D$ into path-connected open subsets $S_i$ and their shared boundaries. It is possible to color each subset $S_i$ with one of four colors, such that the shared boundary of any two subsets $S_i$ that are assigned the same color consists only of isolated points.

However, it turns out that not only is proposition (2) not equivalent to proposition (1), but in fact (1) is true and (2) is false. Indeed, even even we further require that the boundaries of the subsets $S_i$ described in (2) consist only of straight line segments and right angles, the claim is still false. A counterexample - a partition of a rectangle into six subsets satisfying the requirements of (2) that cannot be four-colored - is given in https://www.jstor.org/stable/3647828.

This "counterexample" to the four-color theorem - really a counterexample to the incorrect version (2) - demonstrates the utility of formulating the theorem in terms of graph theory, where its statement is quite simple, rather than in terms of the motivating "map" version (which can be done, but requires a large number of fairly complex conditions).

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

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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.

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    $\begingroup$ 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. $\endgroup$ Jul 31, 2011 at 18:07
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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.

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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.

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

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    $\begingroup$ I also use this as a standard example where unique factorization fails. $\endgroup$
    – Nick S
    Feb 11, 2021 at 3:35
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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.

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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

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?

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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.

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    $\begingroup$ +1 I first saw this in Michael Spivak's Calculus on Manifolds, Probem 1-26. $\endgroup$ Apr 23, 2023 at 10:58
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    $\begingroup$ @TheAmplitwist: Thanks for the reference. It seems that in Spivak, he instead considers the set $A=\{(x,y):x>0\text{ and }0<y<x^2\}$, but the idea is the same: for every straight line $\ell$ that passes through the origin, there is an open disc $N$ centred at $\mathbf 0$ such that $N\cap \ell$ and $A$ are disjoint, meaning that the directional derivatives vanish. $\endgroup$
    – Joe
    Apr 23, 2023 at 16:30
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    $\begingroup$ Yes, that's right, Spivak's exercise is slightly different, but the idea remains the same. An instructional example! $\endgroup$ Apr 23, 2023 at 16:51
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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).

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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 $x\geq 0$.

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