What are your favorite instructional counterexamples? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T11:03:42Z http://mathoverflow.net/feeds/question/16829 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/16829/what-are-your-favorite-instructional-counterexamples What are your favorite instructional counterexamples? Qiaochu Yuan 2010-03-02T05:57:45Z 2013-04-12T13:54:59Z <p>Related: question #879, <a href="http://mathoverflow.net/questions/879/most-interesting-mathematics-mistake" rel="nofollow">Most interesting mathematics mistake</a>. But the intent of this question is more pedagogical.</p> <p>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 <em>Counterexamples in Topology</em> 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.</p> <p>So: what are your favorite examples of counterexamples that really illuminate something about some aspect of a subject? </p> <p>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. </p> <p>As usual, please limit yourself to <strong>one counterexample per answer.</strong> </p> http://mathoverflow.net/questions/16829/what-are-your-favorite-instructional-counterexamples/16830#16830 Answer by Qiaochu Yuan for What are your favorite instructional counterexamples? Qiaochu Yuan 2010-03-02T06:00:37Z 2010-03-02T06:00:37Z <p>In <a href="http://mathoverflow.net/questions/14739/how-can-i-define-the-product-of-two-ideals-categorically" rel="nofollow">question #14739</a>, 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.</p> http://mathoverflow.net/questions/16829/what-are-your-favorite-instructional-counterexamples/16834#16834 Answer by Qiaochu Yuan for What are your favorite instructional counterexamples? Qiaochu Yuan 2010-03-02T06:12:13Z 2011-07-30T10:47:37Z <p>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 $l$?</p> <p>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.</p> http://mathoverflow.net/questions/16829/what-are-your-favorite-instructional-counterexamples/16835#16835 Answer by Andrej Bauer for What are your favorite instructional counterexamples? Andrej Bauer 2010-03-02T06:43:52Z 2010-03-02T06:53:34Z <p>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 <strong>not</strong> provable:</p> <ol> <li>A subset of a finite set is finite.</li> <li>The powerset of a finite set is finite.</li> <li>If a subset of $\mathbb{N}$ is not finite then it is infinite.</li> <li>The elements of a finite set may be listed without repetition.</li> </ol> <p>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.</p> <p>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.</p> <p>My favorite counterexample in intuitionistic logic is that it <strong>is</strong> consistent to assume the so-called <em>Axiom of Enumerability</em>, 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.</p> http://mathoverflow.net/questions/16829/what-are-your-favorite-instructional-counterexamples/16836#16836 Answer by Hsien-Chih Chang for What are your favorite instructional counterexamples? Hsien-Chih Chang 2010-03-02T06:58:03Z 2010-03-02T06:58:03Z <p>A counter-example in graph theory - the <strong>Petersen graph</strong>.</p> <p><img src="http://www.imada.sdu.dk/~btoft/GT2009/PetersenGraphEmbeddings_800.gif" alt="alt text"></p> <p>In many ways it is the most simple graph with many strange properties. See the article on <a href="http://en.wikipedia.org/wiki/Petersen_graph" rel="nofollow">Wiki</a>.</p> <p>Quote from our professor who teaches graph theory:</p> <blockquote> <p>If you think you've proved any lemma about graphs, try Petersen first!</p> </blockquote> http://mathoverflow.net/questions/16829/what-are-your-favorite-instructional-counterexamples/16841#16841 Answer by Ilya Grigoriev for What are your favorite instructional counterexamples? Ilya Grigoriev 2010-03-02T09:31:49Z 2010-03-04T00:06:32Z <p>The matrix <code>$\left(\begin{smallmatrix}0 &amp; 1\\ 0 &amp; 0\end{smallmatrix}\right)$</code> 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.)</p> <ul> <li><p>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: <em>the minimal polynomial is not irreducible or square-free</em>. The same thing in a fancier language: the Jordan canonical form is not diagonal. </p></li> <li><p>Not diagonalizable, even over an algebraically closed field. </p></li> <li><p>Not divisible over $\mathbb C$. There are no matrices $M$ and integers $n\ge2$ so that $M^n = \left(\begin{smallmatrix}0 &amp; 1\\ 0 &amp; 0\end{smallmatrix}\right).$ All diagonalizable and most non-diagonalizable complex matrices have $n$th roots. </p> <p>(This is because, if there was a square root, it'd have minimal polynomial <em>x<sup>4</sup></em>, but since it's a two-by-two matrix, Cayley-Hamilton implies that the characteristic polynomial has degree 2).</p></li> <li><p>The matrix is nilpotent but not zero.</p></li> <li><p>It's one of the best examples when you need to remember why matrix multiplication is not commutative.</p></li> <li><p>Thinking of <em>k<sup>2</sup></em> as a <em>k[x]</em>-module where <em>x</em> acts as this matrix should give wonderful (counter)-examples of modules for all the same reasons.</p></li> </ul> <p>Also, <code>$\left(\begin{smallmatrix}1 &amp; 1\\ 0 &amp; 1\end{smallmatrix}\right)$</code> is an example of an <em>invertible</em> matrix with the first three properties above. Its action on <em>k<sup>2</sup></em> is in some sense the simplest example of a representation of a group ($\mathbb{Z}$) which is indecomposable but not irreducible. </p> http://mathoverflow.net/questions/16829/what-are-your-favorite-instructional-counterexamples/16847#16847 Answer by John Stillwell for What are your favorite instructional counterexamples? John Stillwell 2010-03-02T10:33:10Z 2010-03-02T10:33:10Z <p>The <a href="http://en.wikipedia.org/wiki/Moulton_plane" rel="nofollow">Moulton plane</a> 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.</p> <p>Its discoverer, <a href="http://www-history.mcs.st-andrews.ac.uk/Biographies/Moulton.html" rel="nofollow">F.R. Moulton</a>, 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.</p> http://mathoverflow.net/questions/16829/what-are-your-favorite-instructional-counterexamples/16854#16854 Answer by Andrea Ferretti for What are your favorite instructional counterexamples? Andrea Ferretti 2010-03-02T13:16:15Z 2010-03-02T13:16:15Z <p>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.</p> http://mathoverflow.net/questions/16829/what-are-your-favorite-instructional-counterexamples/16855#16855 Answer by Andrea Ferretti for What are your favorite instructional counterexamples? Andrea Ferretti 2010-03-02T13:21:12Z 2010-03-02T13:21:12Z <p>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 $3/2$.</p> <p>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.</p> http://mathoverflow.net/questions/16829/what-are-your-favorite-instructional-counterexamples/16878#16878 Answer by Alfonso Gracia-Saz for What are your favorite instructional counterexamples? Alfonso Gracia-Saz 2010-03-02T17:11:45Z 2010-03-04T01:08:29Z <p>My favourite counterexample is purely academic: it does not have any applications, but I think it is pretty.</p> <p>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 <code>$\{ y \in \mathbb{N} \vert (x,y) \in U\}$</code> is cofinite.</p> <p>Construct a sequence in $X$ that hits every point in $X$ exactly once. In other words, take a bijection $\mathbb{N} \rightarrow X$. Then: <UL> <LI> $X$ is countable; <LI> every point in $X$ is an accumulation point of this sequence, but <LI> the sequence has no convergent subsequences. </UL></p> <p>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.)</p> http://mathoverflow.net/questions/16829/what-are-your-favorite-instructional-counterexamples/16897#16897 Answer by HW for What are your favorite instructional counterexamples? HW 2010-03-02T19:00:14Z 2010-03-02T19:00:14Z <p>The Baumslag--Solitar groups have presentations</p> <p><code>$BS(p,q)=\langle a,b\mid a^p=b^{-1}a^q b\rangle$</code>.</p> <p>They have the following nice properties:</p> <ol> <li>they're two generator, one relator groups;</li> <li>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.)</li> </ol> <p>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:</p> <ol> <li>they're non-Hopfian, meaning that they admit a self-epimorphism with non-trivial kernel;</li> <li>hence they're not even residually finite;</li> <li>they have exponential Dehn function (meaning that the word problem can be solved, but only very slowly);</li> <li>their virtual first Betti number is one (meaning that every finite-index subgroup has abelianisation of rank one)...</li> </ol> <p>I could go on.</p> http://mathoverflow.net/questions/16829/what-are-your-favorite-instructional-counterexamples/16909#16909 Answer by Michał Kukieła for What are your favorite instructional counterexamples? Michał Kukieła 2010-03-02T20:17:05Z 2010-03-02T20:17:05Z <p>Finite topological spaces often provide nice and simple counterexamples in topology, including algebraic topology (check J. Barmak's <a href="http://www.math.kth.se/~jbarmak/tesisfinal2.pdf" rel="nofollow">thesis</a>). 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.</p> <p>Of course, other counterexamples are available, but finite ones are certainly minimal in a sense.</p> http://mathoverflow.net/questions/16829/what-are-your-favorite-instructional-counterexamples/16919#16919 Answer by Matthew Stover for What are your favorite instructional counterexamples? Matthew Stover 2010-03-02T21:07:52Z 2010-03-02T21:07:52Z <p>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.</p> http://mathoverflow.net/questions/16829/what-are-your-favorite-instructional-counterexamples/16947#16947 Answer by Dinakar Muthiah for What are your favorite instructional counterexamples? Dinakar Muthiah 2010-03-03T02:54:55Z 2010-09-02T18:51:55Z <p>The <a href="http://en.wikipedia.org/wiki/Cantor_set" rel="nofollow">Cantor set</a> is a nice source of counterexamples:</p> <p>The first measure zero sets you meet are usually countable. However, the Cantor set is uncountable and measure zero.</p> <p>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.</p> http://mathoverflow.net/questions/16829/what-are-your-favorite-instructional-counterexamples/16948#16948 Answer by Ilya Grigoriev for What are your favorite instructional counterexamples? Ilya Grigoriev 2010-03-03T02:57:49Z 2010-03-03T04:38:50Z <p>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!</p> <p>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 [<em>a,b</em>] is <em>f(a)-f(b)</em> where <em>f</em> is the Cantor function is supported on a Lebesgue-zero set.</p> <p>Another good source of examples is the measurable set $A \subset [0,1]$ such that for any interval <em>I</em>, $\lambda(I\cap A) > 0$ and $\lambda(I\cap A^c) > 0$. ($\lambda$ is the Lebesgue measure, <em><sup>c</sup></em> denotes complement). </p> <hr> <p>Here's a construction of <em>A</em> that I heard from Ulrik Buchholtz. Instead of just constructing <em>A</em>, we'll make two disjoint sets <em>A</em> and <em>B</em> 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 <em>I<sub>n</sub></em> be the <em>n</em>-th interval in the list. Put two <em>fat</em> (positive-measure) disjoint Cantor sets (one for <em>A</em> and one for <em>B</em>) inside <em>I<sub>1</sub></em>. (We can just put the second inside some gap of the first). By the main property of Cantor sets, every interval <em>I<sub>n</sub></em> 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 <em>I<sub>2</sub></em>, and keep going forever. Every time, we add one Cantor set to <em>A</em> and one to <em>B</em>.</p> <p>Now, each subinterval of [0,1] will contain one of the <em>I<sub>n</sub></em>-s, and therefore its intersection with both <em>A</em> and <em>B</em> has positive measure. Both <em>A</em> and <em>B</em> are countable unions of measurable sets, and therefore measurable. We are done.</p> http://mathoverflow.net/questions/16829/what-are-your-favorite-instructional-counterexamples/16965#16965 Answer by Akhil Mathew for What are your favorite instructional counterexamples? Akhil Mathew 2010-03-03T10:52:53Z 2010-03-03T10:52:53Z <p>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.</p> http://mathoverflow.net/questions/16829/what-are-your-favorite-instructional-counterexamples/16980#16980 Answer by Tony Huynh for What are your favorite instructional counterexamples? Tony Huynh 2010-03-03T16:50:57Z 2010-03-03T16:50:57Z <p>I'd say the <a href="http://mathworld.wolfram.com/TuttesGraph.html" rel="nofollow">Tutte Graph</a>, which is a counterexample to </p> <p><strong>Tait's conjecture:</strong> Every 3-connected cubic planar graph has a Hamiltonian cycle.</p> <p>Initially, I thought this counterexample was extremely <em>non-instructive</em>, 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 <em>is</em> 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. </p> <p>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. </p> http://mathoverflow.net/questions/16829/what-are-your-favorite-instructional-counterexamples/16986#16986 Answer by Tony Huynh for What are your favorite instructional counterexamples? Tony Huynh 2010-03-03T18:12:05Z 2010-03-03T18:37:00Z <p>The <a href="http://en.wikipedia.org/wiki/Weierstrass_function" rel="nofollow">Weierstrass function</a> - 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 <em>really</em> different notions. </p> http://mathoverflow.net/questions/16829/what-are-your-favorite-instructional-counterexamples/17038#17038 Answer by Douglas Zare for What are your favorite instructional counterexamples? Douglas Zare 2010-03-04T01:43:58Z 2010-03-04T01:43:58Z <p>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. </p> <p>$f(x) =\begin{cases} \exp(-1/(1-x^2)),&amp; x \in (-1,1) \\ 0&amp; \text {otherwise} \end{cases}$</p> <p><img src="http://upload.wikimedia.org/wikipedia/commons/5/53/Mollifier_illustration.png" alt="alt text"></p> <p>The usual examples of these functions often seem contrived. Here are examples of <a href="http://www.math.niu.edu/~rusin/known-math/99/nowhere_analy" rel="nofollow">smooth nowhere analytic functions</a>.</p> http://mathoverflow.net/questions/16829/what-are-your-favorite-instructional-counterexamples/17043#17043 Answer by Will Jagy for What are your favorite instructional counterexamples? Will Jagy 2010-03-04T02:17:48Z 2010-03-05T12:52:38Z <p><a href="http://en.wikipedia.org/wiki/Volterra%2527s_function%20" rel="nofollow">Volterra's function</a> has a derivative everywhere which is bounded, discontinuous, and cannot be Riemann-integrated. It depends on the Cantor sets, of course, already mentioned. </p> <p>Possible reference: Bernard R. Gelbaum, John M. H. Olmsted: Counterexamples in Analysis.</p> <p>See also <a href="http://mathoverflow.net/questions/6711/integrability-of-derivatives" rel="nofollow">MO:Integrability of derivatives</a>.</p> http://mathoverflow.net/questions/16829/what-are-your-favorite-instructional-counterexamples/17206#17206 Answer by Allen Knutson for What are your favorite instructional counterexamples? Allen Knutson 2010-03-05T19:51:40Z 2010-03-05T19:51:40Z <p>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.</p> http://mathoverflow.net/questions/16829/what-are-your-favorite-instructional-counterexamples/17270#17270 Answer by Henno Brandsma for What are your favorite instructional counterexamples? Henno Brandsma 2010-03-06T09:00:24Z 2010-04-04T10:27:53Z <p>I like the <a href="http://en.wikipedia.org/wiki/Sorgenfrey_line" rel="nofollow">Sorgenfrey line</a>. 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. </p> http://mathoverflow.net/questions/16829/what-are-your-favorite-instructional-counterexamples/17284#17284 Answer by Adam for What are your favorite instructional counterexamples? Adam 2010-03-06T14:10:05Z 2010-03-06T14:10:05Z <p>"Every finitely-branching tree with infinitely many nodes has an infinite branch" is constructively <strong>false</strong>, as witnessed by the following counterexample:</p> <p><a href="http://math.andrej.com/wp-content/uploads/2006/05/kleene-tree.pdf" rel="nofollow">http://math.andrej.com/wp-content/uploads/2006/05/kleene-tree.pdf</a></p> <p>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."</p> http://mathoverflow.net/questions/16829/what-are-your-favorite-instructional-counterexamples/17285#17285 Answer by Gerald Edgar for What are your favorite instructional counterexamples? Gerald Edgar 2010-03-06T14:38:41Z 2010-03-06T14:38:41Z <p>The Fabius function, everywhere $C^\infty$, nowhere analytic.</p> <p><img src="http://www.math.ohio-state.edu/~edgar/selfdiff/fb.gif" alt="alt text"></p> <p>see... <a href="http://groups.google.com/group/sci.math/msg/d54d31af074022f1" rel="nofollow">sci.math post</a></p> <p>references:<br> J. Fabius, "A probabilistic example of a nowhere analytic $C^\infty$-function". Z. Wahrsch. Verw. Geb. 5 (1966) 173--174. </p> <p>K. Stromberg, PROBABILITY FOR ANALYSTS (Chapman &amp; Hall, 1994), pp. 117--120.</p> http://mathoverflow.net/questions/16829/what-are-your-favorite-instructional-counterexamples/20307#20307 Answer by Jamie Weigandt for What are your favorite instructional counterexamples? Jamie Weigandt 2010-04-04T14:37:53Z 2010-04-04T14:37:53Z <p>The elliptic curve 960d1 in <a href="http://www.sagemath.org/doc/reference/sage/databases/cremona.html" rel="nofollow">Cremona's tables</a> 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.</p> http://mathoverflow.net/questions/16829/what-are-your-favorite-instructional-counterexamples/20312#20312 Answer by Benoît Kloeckner for What are your favorite instructional counterexamples? Benoît Kloeckner 2010-04-04T16:24:55Z 2010-04-04T16:24:55Z <p>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.</p> http://mathoverflow.net/questions/16829/what-are-your-favorite-instructional-counterexamples/20315#20315 Answer by Benoît Kloeckner for What are your favorite instructional counterexamples? Benoît Kloeckner 2010-04-04T16:31:11Z 2010-04-04T16:31:11Z <p>The function $x\mapsto x^3\sin(1/x)$ has a second-order Taylor series but is not twice differentiable at $0$. The circumstances where I came across this example are too embarrassing to tell here...</p> http://mathoverflow.net/questions/16829/what-are-your-favorite-instructional-counterexamples/20329#20329 Answer by ifk for What are your favorite instructional counterexamples? ifk 2010-04-04T20:07:09Z 2010-04-04T20:07:09Z <p>Here is some simple counterexample in commutative algebra, which I found really cute when I first meet it:</p> <p>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 $n$).</p> http://mathoverflow.net/questions/16829/what-are-your-favorite-instructional-counterexamples/20366#20366 Answer by Jesse Madnick for What are your favorite instructional counterexamples? Jesse Madnick 2010-04-05T05:59:01Z 2010-04-05T05:59:01Z <p>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</p> <p><code>$f(x) = \begin{cases} \frac{1}{n} &amp; \mbox{if } x = \frac{m}{n} \in \mathbb{Q} \\ 0 &amp; \mbox{if } x \notin \mathbb{Q}. \end{cases}$</code></p> <p>This function has a couple of amusing properties.</p> <p>(1) It is upper semicontinuous on $\mathbb{R}$, yet has a dense set of discontinuities (every one of which is removable) (namely $\mathbb{Q})$.</p> <p>(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.</p> <p>References: <a href="http://en.wikipedia.org/wiki/Thomae%27s_function" rel="nofollow">http://en.wikipedia.org/wiki/Thomae%27s_function</a> and of course "Counterexamples in Analysis" (Sec 2.15-2.17)</p> http://mathoverflow.net/questions/16829/what-are-your-favorite-instructional-counterexamples/33786#33786 Answer by David for What are your favorite instructional counterexamples? David 2010-07-29T13:04:52Z 2010-07-29T13:04:52Z <p>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.</p> http://mathoverflow.net/questions/16829/what-are-your-favorite-instructional-counterexamples/33802#33802 Answer by Julián Aguirre for What are your favorite instructional counterexamples? Julián Aguirre 2010-07-29T15:49:56Z 2013-03-23T18:58:02Z <p>Rotations $\rho_\alpha$ of the unit circle by an angle $2\pi\alpha$ are nice examples in the theory of discrete dynamical systems.</p> <p>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 <a href="http://en.wikipedia.org/wiki/Sharkovskii%2527s_theorem" rel="nofollow">Sharkowskii's theorem</a> 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.</p> <p>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 <a href="http://en.wikipedia.org/wiki/Chaos_theory" rel="nofollow">caotic</a>.</p> http://mathoverflow.net/questions/16829/what-are-your-favorite-instructional-counterexamples/37541#37541 Answer by Andrew D. King for What are your favorite instructional counterexamples? Andrew D. King 2010-09-02T21:10:35Z 2010-09-02T21:10:35Z <p>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.</p> <p>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.</p> http://mathoverflow.net/questions/16829/what-are-your-favorite-instructional-counterexamples/40557#40557 Answer by Dirk for What are your favorite instructional counterexamples? Dirk 2010-09-29T23:50:35Z 2010-09-29T23:50:35Z <p>I like the double sequence $a_{nm} = \frac{n}{n+m}$ to show that $\lim_{n\to\infty}\lim_{m\to\infty} a_{nm}\neq \lim_{m\to\infty}\lim_{n\to\infty} a_{nm}$ .</p> http://mathoverflow.net/questions/16829/what-are-your-favorite-instructional-counterexamples/58664#58664 Answer by Ehsan M. Kermani for What are your favorite instructional counterexamples? Ehsan M. Kermani 2011-03-16T17:23:17Z 2011-03-17T06:31:19Z <p>Here is a useful example of counter-examples in commutative ring theory;</p> <p>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. </p> <p><em>Applications</em>: </p> <p>Obviously, $R$ is a commutative ring with $1$, ($\mathbb{N}$ is the $1$).</p> <p>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?</p> <p>No, it doesn't.</p> <p>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). </p> <p>Outline: </p> <p>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. </p> <p>It is easy to verify that $R_p$ is isomorphic to $\mathbb{Z}/2$, hence it is both Noetherian &amp; Artinian. (Every element of $R_p$ is either $0/1$ or a invertible.)</p> <p>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?</p> <p>By the above example, it doesn't, since $(P(\mathbb{N}),+,\times)$ is not an integral domain.</p> http://mathoverflow.net/questions/16829/what-are-your-favorite-instructional-counterexamples/59776#59776 Answer by Stefano V. for What are your favorite instructional counterexamples? Stefano V. 2011-03-27T23:25:39Z 2011-03-28T09:12:38Z <p>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</p> <p>$\bigl(\varprojlim_n\mathbb Z/p^n\mathbb Z\bigr)\otimes_{\mathbb Z}\mathbb Q\cong\mathbb Q_p$,</p> <p>the field of $p$-adic numbers (completion of $\mathbb Q$ with respect to the metric induced by the $p$-adic valuation), while</p> <p>$\varprojlim_n\bigl((\mathbb Z/p^n\mathbb Z)\otimes_{\mathbb Z}\mathbb Q\bigr)=0$,</p> <p>since every $\mathbb Z/p^n\mathbb Z$ is torsion and $\mathbb Q$ is divisible.</p> http://mathoverflow.net/questions/16829/what-are-your-favorite-instructional-counterexamples/59783#59783 Answer by Dan Ramras for What are your favorite instructional counterexamples? Dan Ramras 2011-03-28T00:59:30Z 2011-03-28T00:59:30Z <p>Homotopy groups do not, in general, commute with sequential colimits, even for nice maps between nice spaces. </p> <p>I just learned this beautiful example from Bill Dwyer.<br> Take the sequence</p> <blockquote> <p><code>$S^1\stackrel{2}{\longrightarrow}S^1\stackrel{3}{\longrightarrow}S^1\stackrel{4}{\longrightarrow}\cdots.$</code></p> </blockquote> <p>Here <code>$n$</code> denotes the <code>$n$</code>th power map on <code>$S^1$</code>. Thinking of <code>$S^1$</code> as <code>$\mathbb{R}/\mathbb{Z}$</code>, one finds that the colimit of this sequence (in the category of topological spaces) is the quotient group <code>$\mathbb{R}/\mathbb{Q}$</code>. Note that this quotient group, topologized as a quotient space of <code>$\mathbb{R}$</code> by the relation <code>$x\sim y$</code> if <code>$x-y\in \mathbb{Q}$</code>, has the indiscrete topology. In particular, the colimit of this sequence is a contractible topological space and has trivial homotopy groups.</p> <p>On the other hand, the colimit of the corresponding sequence of fundamental groups is the group <code>$\mathbb{Q}$</code> (checking this is a fun exercise).</p> <p>(There's something sort of odd here, because one might have guessed that <code>$\mathbb{R}/\mathbb{Q}$</code> would be a model for <code>$K(\mathbb{Q}, 1)$</code>, since after all <code>$\mathbb{R}$</code> is a free <code>$\mathbb{Q}$</code>-space. But there are no interesting open sets in the quotient and hence no chance of local triviality.)</p> http://mathoverflow.net/questions/16829/what-are-your-favorite-instructional-counterexamples/63704#63704 Answer by Jon Bannon for What are your favorite instructional counterexamples? Jon Bannon 2011-05-02T15:26:26Z 2011-05-02T15:45:54Z <p>I'm shocked that noone has mentioned the <a href="http://en.wikipedia.org/wiki/Quaternion_group" rel="nofollow">Quaternion group</a>! This thing is a counterexample to lots of basic questions you'd come up with while learning (finite) group theory. </p> <p>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. </p> http://mathoverflow.net/questions/16829/what-are-your-favorite-instructional-counterexamples/63705#63705 Answer by Stefan Waldmann for What are your favorite instructional counterexamples? Stefan Waldmann 2011-05-02T15:52:49Z 2011-05-02T15:52:49Z <p>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...</p> http://mathoverflow.net/questions/16829/what-are-your-favorite-instructional-counterexamples/68861#68861 Answer by Anil P for What are your favorite instructional counterexamples? Anil P 2011-06-26T16:25:30Z 2011-06-26T16:25:30Z <p>the example which shows that exp(zw) is not equal to exp (exp(z),w)</p> <p>another one a continuous function of a complex variable need not have primitive in a region.the example is f(z) = square ( | z| ). </p> http://mathoverflow.net/questions/16829/what-are-your-favorite-instructional-counterexamples/71645#71645 Answer by Chandrasekhar for What are your favorite instructional counterexamples? Chandrasekhar 2011-07-30T08:49:49Z 2011-07-30T11:00:34Z <p>$\textbf{Algebra.}$</p> <blockquote> <ul> <li><p>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.</p></li> <li><p>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 $G \cong H$.</p></li> <li><p>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.</p></li> </ul> </blockquote> <p>$\textbf{Analysis.}$</p> <blockquote> <ul> <li><p>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. </p></li> <li><p>Set that is not <a href="http://en.wikipedia.org/wiki/Vitali_set" rel="nofollow">Lebesgue measurable.</a> Example given by Vitali.</p></li> </ul> </blockquote> http://mathoverflow.net/questions/16829/what-are-your-favorite-instructional-counterexamples/71662#71662 Answer by Connor Mooney for What are your favorite instructional counterexamples? Connor Mooney 2011-07-30T16:38:19Z 2011-07-30T16:38:19Z <p>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.</p> <p>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 $0$.</p> http://mathoverflow.net/questions/16829/what-are-your-favorite-instructional-counterexamples/71697#71697 Answer by paul Monsky for What are your favorite instructional counterexamples? paul Monsky 2011-07-31T00:42:47Z 2011-07-31T00:42:47Z <p>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.</p> http://mathoverflow.net/questions/16829/what-are-your-favorite-instructional-counterexamples/71726#71726 Answer by Ramsey for What are your favorite instructional counterexamples? Ramsey 2011-07-31T15:12:44Z 2011-07-31T15:12:44Z <p>A standard result in introductory calculus classes is that, if a function has positive derivative on an open interval, then it's increasing there.</p> <p>Based on this, students tend to think that, if $f'(a)>0$, then $f$ must be increasing "near $a$."</p> <p>However, the example $f(x) = 2x^2\sin(1/x)+x$ (set $f(0)=0$) shows that this is quite false!</p> http://mathoverflow.net/questions/16829/what-are-your-favorite-instructional-counterexamples/71744#71744 Answer by Paul Pearson for What are your favorite instructional counterexamples? Paul Pearson 2011-07-31T17:39:37Z 2011-07-31T17:39:37Z <p>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."</p> <p>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. </p> http://mathoverflow.net/questions/16829/what-are-your-favorite-instructional-counterexamples/74542#74542 Answer by David White for What are your favorite instructional counterexamples? David White 2011-09-04T23:51:33Z 2011-09-04T23:51:33Z <p>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. </p> <p>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 $\mathbb{Z}/6\mathbb{Z} \cong \mathbb{Z}/3\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z}$)</p> http://mathoverflow.net/questions/16829/what-are-your-favorite-instructional-counterexamples/74543#74543 Answer by David White for What are your favorite instructional counterexamples? David White 2011-09-05T00:08:18Z 2011-09-05T00:08:18Z <p>I'm surprised no one mentioned the Hawaiian Earring:</p> <p><img src="http://upload.wikimedia.org/wikipedia/en/9/9e/Hawaiian_Earring.PNG" alt="alt text"></p> <p>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...</p> <ul> <li>The Hawaiian Earring cannot have a universal cover. </li> <li>The Hawaiian Earring is not a CW-complex, although it is a compact, complete metric space</li> <li>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.</li> <li>For many years people thought the fundamental group was always a topological group. This turns out to be <a href="http://arxiv.org/abs/0909.3086" rel="nofollow">false</a>, thanks to the Hawaiian Earring. There's a nice post about this <a href="http://mathoverflow.net/questions/26680/fundamental-group-as-topological-group/26712#26712" rel="nofollow">here</a> on MO</li> <li>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!</li> </ul> http://mathoverflow.net/questions/16829/what-are-your-favorite-instructional-counterexamples/74546#74546 Answer by David White for What are your favorite instructional counterexamples? David White 2011-09-05T00:38:48Z 2011-09-05T00:38:48Z <p><a href="http://books.google.com/books?id=r9VoYbk-8c4C&amp;q=small%2527s+example#v=snippet&amp;q=small%2527s%2520example&amp;f=false" rel="nofollow">Small's Example</a> from noncommutative algebra...</p> <p>The triangular ring $T = \pmatrix{\mathbb{Z} &amp; \mathbb{Q} \\ 0 &amp; \mathbb{Q}}$ has the following properties:</p> <ul> <li>It's right noetherian but not left noetherian</li> <li>It's right hereditary but not left hereditary</li> <li>The right global dimension is 1 but the left global dimension is 2</li> <li>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 $K = Frac(R)$</li> <li>A similar example gives a ring which is <a href="http://books.google.com/books?id=r9VoYbk-8c4C&amp;q=%2522noetherian+ring+that+is+neither%2522#v=snippet&amp;q=%2522noetherian%2520ring%2520that%2520is%2520neither%2522&amp;f=false" rel="nofollow">noetherian but neither left nor right Ore</a>. Just take $R = \pmatrix{S &amp; 0 \\ S &amp; I}$ where $S = \pmatrix{\mathbb{Z} &amp; 0 \\ \mathbb{Z}_p &amp; \mathbb{Z}_p}$ and $I = \pmatrix{\mathbb{Z} &amp; 0 \\ 0 &amp; \mathbb{Z}_p}$ is an $S$-ideal.</li> </ul> <p>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.</p> http://mathoverflow.net/questions/16829/what-are-your-favorite-instructional-counterexamples/101812#101812 Answer by David Corwin for What are your favorite instructional counterexamples? David Corwin 2012-07-09T22:13:12Z 2012-07-09T22:13:12Z <p>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 $$\int_C P dx + Q dy = \int_D \left(\frac{dQ}{dx}-\frac{dP}{dy}\right) dx dy.$$</p> <p>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 $P dx + Q dy$.</p> <p>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:</p> <p>$$\frac{-ydx}{x^2+y^2} + \frac{xdy}{x^2+y^2}$$</p> <p>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 <em>everywhere</em> in $D$.</p> <p>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.</p> <p>Furthermore, this is a fundamental example of a nontrivial element in a de Rham cohomology group. In this case, the one-form above generates $H^1_{\text{dR}}(\mathbb{R}^2\setminus {(0,0)},\mathbb{R})$.</p> http://mathoverflow.net/questions/16829/what-are-your-favorite-instructional-counterexamples/101819#101819 Answer by Filippo Alberto Edoardo for What are your favorite instructional counterexamples? Filippo Alberto Edoardo 2012-07-10T01:55:32Z 2012-07-10T01:55:32Z <p>Assume given three projective systems <code>$\{A_n,\alpha_{nm}\}_{n\in\mathbb{N}}$</code>, <code>$\{B_n,\beta_{nm}\}_{n\in\mathbb{N}}$</code> and <code>$\{C_n,\kappa_{nm}\}_{n\in\mathbb{N}}$</code> 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 $n,n'\gg m$.</p> <p>A classical counterexample showing the necessity of this condition is to take <code>$A_n=p^n\mathbb{Z}$</code> 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 <code>$p^n\mathbb{Z}\subseteq p^m\mathbb{Z}$</code> 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 <em>not</em> right exact.</p> <p>[Later remark]: After typing all down, I remarked that everything can be found in Wikipedia at <a href="http://en.wikipedia.org/wiki/Inverse_limit" rel="nofollow">http://en.wikipedia.org/wiki/Inverse_limit</a> Moreover, the stationary condition quoted above, usually referred to as Mittag-Leffler condition, is enough to prove right-exactness of $\varprojlim$ in <strong>Ab</strong>, but there is a counterexample due to Deligne and Neeman showing that in other categories this is not enough, see <a href="http://www.springerlink.com/content/aeem2yx884nnufxn/" rel="nofollow">http://www.springerlink.com/content/aeem2yx884nnufxn/</a></p> http://mathoverflow.net/questions/16829/what-are-your-favorite-instructional-counterexamples/125363#125363 Answer by Amritanshu Prasad for What are your favorite instructional counterexamples? Amritanshu Prasad 2013-03-23T09:44:09Z 2013-03-23T13:14:46Z <p>The matrices <code>$A=\begin{pmatrix} 17\times 11 + 1 &amp; 25\times 11\\ 11^2 &amp; 16\times 11 + 1 \end{pmatrix}$</code> and <code>$B = \begin{pmatrix} 17\times 11 + 1 &amp; 11 \\ 25\times 11^2 &amp; 16\times 11 + 1 \end{pmatrix}$</code> are similar modulo $m$ for every positive integer $m$ but are not similar over the integers.</p> <p>In other words, there exist matrices $X_m\in GL_2(\mathbf Z/m\mathbf Z)$ such that $XA \equiv BX \mod m$ for every $m$, but there does not exist a matrix $X\in GL_n(\mathbf Z)$ such that $XA = BX$.</p> <p>This is due to Stebe, Conjugacy separability of groups of integer matrices. <em>Proc. Amer. Math. Soc.</em>, 32:1–7, 1972.</p> http://mathoverflow.net/questions/16829/what-are-your-favorite-instructional-counterexamples/125437#125437 Answer by Johannes Ebert for What are your favorite instructional counterexamples? Johannes Ebert 2013-03-24T05:44:05Z 2013-03-24T05:44:05Z <p>The Warsaw circle $W$ <a href="http://en.wikipedia.org/wiki/Continuum_%28topology%29" rel="nofollow">http://en.wikipedia.org/wiki/Continuum_%28topology%29</a> is a counterexample for quite a number of too naive statements.</p> <p>Some observations: $W$ is weakly contractible (because a map from a locally path connected space cannot ''go over the bad point''). There is a projection map $g:W \to S^1$ onto the usual circle. The point-preimages of $g$ are either points or, for a single point on $S^1$, a closed interval. </p> <p>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. These observations imply:</p> <ol> <li><p>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.</p></li> <li><p>The Vietoris-Begle Theorem is false for singular cohomology (in particular, the wikipedia version of that Theorem is not quite correct).</p></li> <li><p>$W$ does not have the homotopy type of a CW complex (since it is not contractible).</p></li> <li><p>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.</p></li> </ol> http://mathoverflow.net/questions/16829/what-are-your-favorite-instructional-counterexamples/125441#125441 Answer by Mustafa Said for What are your favorite instructional counterexamples? Mustafa Said 2013-03-24T07:22:29Z 2013-03-24T07:22:29Z <p>My favorite counter-example is given in the shore paper, "Almost Commuting Unitaries," by R. Exel and T. Loring.</p> <p>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 $|| [U_n, V_n] || &lt; \epsilon$ 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. </p> http://mathoverflow.net/questions/16829/what-are-your-favorite-instructional-counterexamples/127366#127366 Answer by GArunkumar for What are your favorite instructional counterexamples? GArunkumar 2013-04-12T13:54:59Z 2013-04-12T13:54:59Z <p>In topology, The comb space is an example of a path connected space which is not locally path connected. see <a href="http://en.wikipedia.org/wiki/Comb_space" rel="nofollow">http://en.wikipedia.org/wiki/Comb_space</a>.</p>