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The first thing to say is that this is not the same as the question about interesting mathematical mistakes. I am interested about the type of false beliefs that many intelligent people have while they are learning mathematics, but quickly abandon when their mistake is pointed out -- and also in why they have these beliefs. So in a sense I am interested in commonplace mathematical mistakes.

Let me give a couple of examples to show the kind of thing I mean. When teaching complex analysis, I often come across people who do not realize that they have four incompatible beliefs in their heads simultaneously. These are

(i) a bounded entire function is constant;
(ii) $\sin z$ is a bounded function;
(iii) $\sin z$ is defined and analytic everywhere on $\mathbb{C}$;
(iv) $\sin z$ is not a constant function.

Obviously, it is (ii) that is false. I think probably many people visualize the extension of $\sin z$ to the complex plane as a doubly periodic function, until someone points out that that is complete nonsense.

A second example is the statement that an open dense subset $U$ of $\mathbb{R}$ must be the whole of $\mathbb{R}$. The "proof" of this statement is that every point $x$ is arbitrarily close to a point $u$ in $U$, so when you put a small neighbourhood about $u$ it must contain $x$.

Since I'm asking for a good list of examples, and since it's more like a psychological question than a mathematical one, I think I'd better make it community wiki. The properties I'd most like from examples are that they are from reasonably advanced mathematics (so I'm less interested in very elementary false statements like $(x+y)^2=x^2+y^2$, even if they are widely believed) and that the reasons they are found plausible are quite varied.

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    $\begingroup$ I have to say this is proving to be one of the more useful CW big-list questions on the site... $\endgroup$ – Qiaochu Yuan May 6 '10 at 0:55
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    $\begingroup$ The answers below are truly informative. Big thanks for your question. I have always loved your post here in MO and wordpress. $\endgroup$ – Unknown May 22 '10 at 9:04
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    $\begingroup$ wouldn't it be great to compile all the nice examples (and some of the most relevant discussion / comments) presented below into a little writeup? that would make for a highly educative and entertaining read. $\endgroup$ – Suvrit Sep 20 '10 at 12:39
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    $\begingroup$ It's a thought -- I might consider it. $\endgroup$ – gowers Oct 4 '10 at 20:13
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    $\begingroup$ Meta created tea.mathoverflow.net/discussion/1165/… $\endgroup$ – user9072 Oct 8 '11 at 14:27

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Let $B(r_1) \subset B(r_2)$ be two open balls of radius $r_1$ and $r_2$ respectively. Then $r_1 \leq r_2$.

Bounded metric spaces give trivial counterexamples. Also, $B \left( \frac{1}{6}, \frac{2}{3} \right) \subsetneq B \left( \frac{1}{2}, \frac{1}{2} \right)$ in $(0,+ \infty)$.

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  • $\begingroup$ What is $B(a,b)$ here? $\endgroup$ – 1.. Dec 3 '13 at 10:09
  • $\begingroup$ Here, $B(a,b)$ is the open ball of radius $b$ centered at $a$. $\endgroup$ – Seirios Dec 3 '13 at 21:41
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"A real symmetric matrix is positive-definite iff all the leading principal minors are positive, and positive-semidefinite iff all the leading principal minors are nonnegative."

This paper collects some evidence that this belief is "common", and presents a counterexample (of size $3\times 3$. Exercise: find an example of size $2\times 2$).

(Related to, but not the same as this answer.)

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    $\begingroup$ $$\pmatrix{0&0\cr0&-1\cr}$$ $\endgroup$ – Gerry Myerson Jul 29 '15 at 3:22
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While this example was already mentioned by Skopenkov in a comment, here it is, explicitly:

If $G\times X\to X$ is a free continuous action of a compact metrizable group on a compact metrizable space $X$, then $X\to X/G$ is a principal $G$-bundle.

This is true if $G$ is a Lie group (Gleason) and, more generally, for proper Lie group actions on locally compact spaces (Palais), but false in general (the example is essentially due to Kolmogorov), see:

R. F. Williams, A useful functor and three famous examples in topology. Trans. Amer. Math. Soc. 106 (1963) 319–329.

In this example, $\dim(X/G)=2, \dim(X)=1$.

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A false belief that I held until very recently, was that for a surface embedded in a three-manifold, orientability was the same as two-sidedness. However, it turns out that if the ambient manifold is not orientable, you can embed a torus so that it is one-sided, and a Möbius strip or a Klein bottle so that it is two-sided.

This picture, from Weeks' wonderful The shape of space, shows a two-sided Klein bottle embedded in a product of a Klein bottle and a circle, by identifying sides as indicated by the arrows.

Two-sided Klein Bottle

This one shows a torus embedded single-sidedly in the same 3-manifold.

One-sided Torus

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  • $\begingroup$ What's the definition of an embedding being one-sided or two-sided? $\endgroup$ – Oscar Cunningham Apr 21 at 7:21
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    $\begingroup$ @OscarCunningham Informally: if you imagine the surface to be made of iron and you and someone esle are wearing magnetic shoes, if you start out at the same location at opposite sides of the surface, and however you move you will never meet, it is two-sided, otherwise one-sided. More formally, it is two-sided if the normal bundle of the surface has a nowhere vanishing section. $\endgroup$ – doetoe Apr 21 at 9:08
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    $\begingroup$ Thanks! I only convinced myself you were right when I thought of a simpler example: the circle is orientable, but if you embed it in a Möbius strip in the obvious way then it's one-sided. $\endgroup$ – Oscar Cunningham Apr 21 at 9:47
  • $\begingroup$ The simplest example of a two-sided nonorientable surface is probably the embedding of the Klein bottle $K$ into $K\times\mathbb R$ by $k\mapsto (k,0)$. Then its impossible for the point $(k,r)$ to go from $r>0$ to $r<0$ without passing through $r=0$. $\endgroup$ – Oscar Cunningham Apr 23 at 7:46
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Let $X$ be a "nice" path-connected topological space, say a connected manifold or CW-complex.

False belief: "A universal covering $Y\to X$ of $X$ is unique up to unique isomorphism" and therefore can be called "the" universal covering.

The isomorphism far from unique in general (there are as many as elements in "the" fundamental group). However uniqueness (and the universal property) holds in the category of coverings of pointed topological spaces. (In particular, for topological groups there's a canonical choice.)

Browsing I found several textbooks teaching the above "false belief" (I saw several too that are careful with this issue).

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    $\begingroup$ Likewise 'the' algebraic closure of a field. $\endgroup$ – Oscar Cunningham Apr 23 at 7:10
  • $\begingroup$ @OscarCunningham Yes and no: yes it's similar, but it seems to me many more people are aware and careful (e.g. not saying "the algebraic closure")– btw I wrote this answer after reading this comment. $\endgroup$ – YCor Apr 25 at 14:53
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Consider the following well-known result: Let $(E,\leq)$ be an ordered set. Then the following are equivalent: (i) Every nonempty subset of $E$ has a maximal element. (ii) Every increasing sequence in $E$ is stationary.

It is immediate that (i) implies (ii). To prove the converse, one assumes that (i) is false and then "constructs step by step" a strictly increasing sequence.

The common mistake (which I have seen in textbooks) is to describe the latter construction as a proof by induction. In fact, the construction uses the axiom of choice (or at least the dependent choice axiom).

(As a special case, I don't think ZF can prove that every PID is a UFD.)

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    $\begingroup$ It’s not exactly wrong to call it a proof by induction. In ZFC, the proof of dependent choice — or of just about any instance of it, eg the one here — works by combining induction and choice. So I’d agree it’s wrong to sweep the choice under the carpet; but if you’re not explicitly invoking DC, then you will be using induction as well. $\endgroup$ – Peter LeFanu Lumsdaine Dec 1 '10 at 15:34
  • $\begingroup$ Peter, let's state DC as follows: "If $(p_n:X_{n+1}\to X_n)$ is an $\mathbb{N}$-projective system of nonempty sets with all $p_n$ surjective , then projlim($X_n$) is nonempty." Proof from AC: put $X:=\coprod_{n\geq0}X_n$ and $X^+=\coprod_{n>0}X_n$ with obvious map $p:X^+\to X$. Then $p$ is onto, so has a section $s$ (family of sections of all $p_n$'s). Given $x_0\in X_0$, sequence $(s^n(x_0))$ is an element of projlim($X_n$). I agree that we do need induction to define $s^n$. But iteration of a map is such a basic tool that I don't agree to call any proof using it a "proof by induction". $\endgroup$ – Laurent Moret-Bailly Dec 7 '10 at 11:49
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1- A very common mistake that 1st year students (but not even a single mathematician) think that it is true is "a transitive and symmetric relation on a set is reflexive". But as the empty set is a transitive and symmetric relation but not reflexive on any non-empty set. Of course there lots of non-trivial examples also.

2- Another common mistake is that the expression "countable union of countable sets is again countable" is independent of axiom of choice (AC). Many people make the proof of this statement without mentioning axiom of choice. Indeed, in his holly book Algebra, Lang proves this statement just by taking an ordering from each countable set and continues without the mentioning AC.

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    $\begingroup$ For big-list questions, it's usually best to post independent answers as separate answers. $\endgroup$ – Nate Eldredge Dec 2 '10 at 15:13
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    $\begingroup$ +1 for #2. Baby Rudin is another offender. And many authors use so-called "diagonalization tricks" for proving compactness theorems like Arzela-Ascoli and Prohorov, which typically reduce to the compactness of $[0,1]^\mathbb{N}$. $\endgroup$ – Nate Eldredge Dec 2 '10 at 15:21
  • $\begingroup$ Isn't the more right statement that a transitive and irreflexive relation is assymetric? $\endgroup$ – Zsbán Ambrus Dec 4 '10 at 22:46
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This is more of a false philosophy than a clear mistake, but nevertheless it is very common:

A compact topological space must be "small" in some sense: it should be second countable or separable or have cardinality $ \le 2^{\aleph_0}$, etc.

This is all true for compact metric spaces, but in the general case, Tychonoff's theorem gives plenty of examples of compact spaces which are "huge" in the above sense.

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  • $\begingroup$ More or less along the same lines, one is usually lead to think that "every topological space is Hausdorff". $\endgroup$ – Marco Golla May 4 '11 at 14:43
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A possible false belief is that "a maximal Abelian subgroup of a compact connected Lie group is a maximal torus". Think of the $\mathbf Z_2\times\mathbf Z_2$-subgroup of $SO(3)$ given by diagonal matrices with $\pm1$ entries.

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    $\begingroup$ Fu... I just "proved" that again as an exercise a few days ago. $\endgroup$ – Johannes Hahn Mar 6 '13 at 0:02
  • $\begingroup$ Or, which is essentially the same example, $\left\langle\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}, \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\right\rangle$ inside $\operatorname U(2)/\operatorname U(1)$. $\endgroup$ – LSpice Aug 5 at 21:09
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This is a common error made by mature mathematicians in many books and papers in analysis, especially in differential equations: If $X$ is a closed subspace of a Banach space $Y$, then the $Y^*$ (the dual of $Y$) is isomorphic to a subspace of $X^*$ (the dual of $X$). It is false (of course) since Euclidian space $\mathbb R$ is a subspace of $\mathbb R^2$, yet the dual of $\mathbb R^2=\mathbb R^2$ is not isomorphic to a subspace of the dual of $\mathbb R=\mathbb R$. I guess, sometimes they really, really want it to be true. Cheers Boris

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  • $\begingroup$ I'll take your word for it, but since that statement is false even without introducing norms and topology, it staggers me that people could even believe that. They might say it without thinking, I guess $\endgroup$ – Yemon Choi Oct 20 '10 at 2:58
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    $\begingroup$ I would also be shocked if this really gets believed often! It seems to be a sort of “mis-dualisation”: they dualise “$X$ is a subobject of $Y$” to “$Y^*$ is a subobject of $X^*$”, where the correct dual is “$X^*$ is a quotient of $Y^*$”. $\endgroup$ – Peter LeFanu Lumsdaine Dec 1 '10 at 15:19
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I don't know how common this mistake is, but I think it's worth mentioning. I used to think that existence of non-measurable sets is guaranteed by the axiom of choice only.

In the presence of AC, there cannot be a $\sigma$-additive measure on $\mathcal{P}(\mathbb{R})$ that extends the usual Lebesgue measure.

It is true that we cannot extend the Lebesgue measure in a translation-invariant way by various Vitali set constructions. On the other hand, if you do not insist that the extension is translation-invariant, it might be possible to do this relative to a real-valued measurable cardinal assumption.

Theorem (Ulam): If there exists a cardinal $\kappa$ such that there exists an atomless $\kappa$-additive probability measure on $\mathcal{P}(\kappa)$, then there exists a $\sigma$-additive measure on $\mathcal{P}(\mathbb{R})$ extending the Lebesgue measure.

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  • $\begingroup$ I think you need $\kappa\leq\frak c$, no? $\endgroup$ – Asaf Karagila Jan 22 '15 at 14:44
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    $\begingroup$ @AsafKaragila: I believe the assumption that our measure is atomless already implies that $\kappa \leq 2^{\omega}$. $\endgroup$ – Burak Jan 22 '15 at 14:45
  • $\begingroup$ Take any measurable cardinal, then there is an atomless probability measure on its power set. It's just that an event is either improbable or its negation is improbable. Unless by probability measure you mean it obtains many values, not just two. $\endgroup$ – Asaf Karagila Jan 22 '15 at 14:48
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    $\begingroup$ Isn't that measure atomic if you are deriving it from the ultrafilter? (By an atom, I mean any $A$ of positive measure such that for any $B \subseteq A$ either $\mu(B)=0$ or $\mu(B)=\mu(A)$). I will have to catch a course now but the theorem I referred to should be in Kanomori (indeed, I checked the pdf and it is Theorem 2.5) $\endgroup$ – Burak Jan 22 '15 at 14:53
  • $\begingroup$ Ohhhh, right. I was thinking about atoms in the sense of Boolean algebra, as minimal positive elements. Thanks for the clarification! $\endgroup$ – Asaf Karagila Jan 22 '15 at 14:55
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Many students believe that every abelian subgroup is a normal subgroup.

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    $\begingroup$ And many other students believe every normal subgroup is an abelian subgroup. $\endgroup$ – Gerry Myerson Jul 3 '18 at 22:49
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    $\begingroup$ and many believe that a subgroup of an abelian group is normal (which I also believe). $\endgroup$ – user137767 Apr 16 '19 at 21:17
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"Let $E$ be a complete locally convex topological vector space (or a complete topological vector space or a complete topological group) and let $F$ be closed vector subspace (or a closed subgroup). Then the quotient $E/F$ is complete."

This just has to be true. One can almost see the proof. And in fact it is true for Banach spaces. So it has to be true for locally convex spaces as well.

Another one with completions:

"Every topological group is a dense subgroup of a complete topological group." True for abelian groups but false in general (take the homeomorphism group of $[0,1]$ with the compact open topology)

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I would like to turn the attention of mathematical community to a false beliefs related to the direct limit topologies.

Many years ago in the theory of topological groups there was a false belief that for every space $X$ the free topological group carries the topology of direct limit of the sequence $F_n(X)$ of words of length $\le n$. This illusion was broken up by Fay, Ordman and Thomas who showed that even for the space of rational numbers the free topological group $F(\mathbb Q)$ is not a $k$-space.

The problems with direct limit topologies is that for the direct limit $X=lim X_n$ of an increasing sequence $(X_n)$ of topological spaces the topology on $X\times X$ does not coincide with the direct limit topology of the sequence $ (X_n\times X_n)$.

Now specialists in General Topology and Topological Algebra are conscious of pathological behaviour of direct limit topologies and are careful with this delicate topic.

On the other hand, I was quite surprised lerning that in Algebraic Geometry this misbelief still is alive. For example, in this paper posted to arxiv (maybe it is already published) in the very introduction (on page 3) it is written that for any topological space $X$ the Ran space (of all non-empty finite subsets of $X$, endowed with the topology of direct limit of the sequence $R_n(X)$ of sets of cardinality $\le n$ in $X$) is a topological semilattice. But this is not true in general, see Proposition 4 here.

So, some false beliefs that have died in some areas of mathematics can be still alive in others. By the way, this situation also explains why mathematicians should not neglect general topology.

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True: The solution operator of the linear one-dimensional time-dependent ordinary differential equation (ODE) $x' = a(t) x$ is $$ \exp\Bigl(\int\limits_{t_0}^{t} a(s) \, ds\Bigr). $$ True: The solution operator of the linear multi-dimensional time-independent ODE $x' = A x$ is $$ \exp\,(A(t - t_0)). $$ A quite popular misconception, even among research mathematicians:

The solution operator $\Phi(t;t_0)$ of the linear multi-dimensional time-dependent ODE $x' = A(t) x$ is $$ \exp\Bigl(\int\limits_{t_0}^{t} A(s) \, ds \Bigr), $$

perhaps strengthened by Liouville's formula:

$$ \det{\Phi(t;t_0)} = \exp\Bigl(\int\limits_{t_0}^{t} \operatorname{tr}{A(s)} \, ds\Bigr). $$

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  • $\begingroup$ I am holding this belief right now. $\endgroup$ – Michael Apr 27 '18 at 23:26
  • $\begingroup$ Is there a simple expression that is true? $\endgroup$ – Hans May 3 '18 at 18:11
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    $\begingroup$ @Hans The Peano-Baker series is, in my opinion, simple, and converges where it should. There is a nice paper by Baake and Schlägel The Peano-Baker series, Proceedings of the Steklov Institute of Mathematics 275 (1) (2011), 155-159 (the paper is behind a paywall on the Publisher's page, but the authors put a copy on ResearchGate (researchgate.net/publication/47702535_The_Peano-Baker_series)). $\endgroup$ – user539887 May 4 '18 at 8:15
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    $\begingroup$ @Hans In contrast, the Magnus expansion is very complicated and may not converge except close to the initial time. $\endgroup$ – user539887 May 4 '18 at 8:20
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    $\begingroup$ @Hans Physicists refer to the time ordered exponential: en.wikipedia.org/wiki/Ordered_exponential $\endgroup$ – Phil Tosteson Aug 27 '18 at 14:58
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When I was studying Banach spaces, I was confused with the following: We know that, in any Banach Space $V$, the closed unit ball is compact in the topology generated by the norm if, and only if, the dimension of $V$ is finite. But thinking about $\mathbb R$ as a vector space over $\mathbb Q$, we have an infinite-dimensional vector space which is complete in the norm (given by the modulus) but the closed unit ball is, of course, compact in topology generated by the norm.

I took some time to discover that my mistake was that I thought about $\mathbb R$ over $\mathbb Q$ as a Banach space. In fact, this vector space is a complete metric space (in the sense of Cauchy sequences), but I realized later that the word Banach space is reserved only for vector spaces defined over the fields $\mathbb R$ or $\mathbb C$.

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    $\begingroup$ You can define Banach spaces over any complete field. For example, one can define p-adic Banach spaces. But Q isn't complete with respect to any of its norms. $\endgroup$ – Qiaochu Yuan May 4 '10 at 22:14
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    $\begingroup$ In fact the Theorem I mentioned in my answer, is based on the Riesz lemma and this lemma is not valid if the scalar fields is not complete. $\endgroup$ – Leandro May 4 '10 at 22:34
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    $\begingroup$ @QiaochuYuan ...unless you use the trivial norm. $\endgroup$ – Mario Carneiro Oct 20 '15 at 21:16
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Here's one I was reminded recently during lunch in the common room.

A maximal abelian subalgebra of a semisimple Lie algebra is a Cartan subalgebra.

This is true for compact real forms of semisimple Lie algebras, but fails in general. The missing condition is that the subalgebra should equal its normaliser.

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  • $\begingroup$ For example, the usual proof of Schur's theorem on commutative algebras of matrices produces abelian Lie subalgebras of $\mathfrak{sl}_n$ much larger than the rank of $\mathfrak{sl}_n$. $\endgroup$ – Mariano Suárez-Álvarez Aug 4 '10 at 23:41
  • $\begingroup$ Yes, my favourite example (being a physicist) is the stabiliser in $\mathfrak{so}(1,n)$ of a nonzero zero-norm vector in $\mathbb{R}^{1,n}$ for $n>4$. The stabiliser contains an abelian ideal (infinitesimal null rotations) of dimension $n-1$. $\endgroup$ – José Figueroa-O'Farrill Aug 5 '10 at 0:02
  • $\begingroup$ I meant to write $n>3$ above. $\endgroup$ – José Figueroa-O'Farrill Aug 5 '10 at 0:03
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    $\begingroup$ ``The missing condition is that the subalgebra should equal its normaliser'', or that the subalgebra consists of semisimple elements, no? (That provides another perspective on why it's true for compact real forms.) $\endgroup$ – LSpice Dec 12 '13 at 23:26
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"If a field $K$ has characteristic 0 and $G$ is a group, then all $KG$-modules are completely reducible."

True for finite groups but very false in general.

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I have heard the following a few times :

"If $f$ is holomorphic on a region $\Omega$ and not one-to-one, then $f'$ must vanish somewhere in $\Omega$."

$f(z)=e^z$ of course is a counterexample.

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    $\begingroup$ is it true though if the image is simply-connected? $\endgroup$ – KotelKanim Apr 17 '11 at 11:33
  • $\begingroup$ thanks. that's great. I never thought about it before, but it just sounded right... $\endgroup$ – KotelKanim Apr 19 '11 at 7:46
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    $\begingroup$ No true. Take $f(z)=z^3-3z$ and restrict it to the complement of $\lbrace 1,-1\rbrace$ so that $f'(z)$ is never $0$. It maps this domain onto $\mathbb C$. $\endgroup$ – Tom Goodwillie May 4 '11 at 0:16
  • $\begingroup$ @TomGoodwillie: what if both domain and image are simply-connected? $\endgroup$ – Michael Dec 3 '13 at 0:45
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If $\alpha>0$ is not an integer, the set of functions $f:[a,b]\rightarrow\mathbb R$ such that $$\sup_{y\ne x}\frac{|f(y)-f(x)|}{|y-x|^\alpha}<+\infty$$ is ${\mathcal C}^\alpha([a,b])$.

False for $\alpha>1$, because this set contains only constant functions.

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Multiplication of differential forms is inherently anti-commutative. Thus, if $x$ and $y$ are coordinates on a surface, then $dx \wedge dy$ makes sense but $(dx)^2+(dy)^2$ is either nonsense or, if it means anything, is $0$.

I'm not sure why I believed this, but I did for several years. I tried my best to avoid creating this impression in my students, but I think it still happened in some of them, simply because the curriculum spends a lot of time on integration and Stokes theorem and very little time on metrics, curvature, etc.

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Probably my fault for not paying enough attention in analysis, but:

Any continuous function on the interval that has derivative equal to zero almost everywhere is constant.

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False belief. If a family $(x_n)_{n\geq 1}$ is commutatively convergent (i.e. summable) in a normed space $(V,\|\ \|)$ then $$ \sum_{n\geq 1} \|x_n\|<+\infty\ . $$

This is true in finite dimensions and has counterexamples in infinite dimensions. Details and counterexamples can be found there.

Recall that a family $(x_i)_{i\in I}$ is called summable with sum $S$ iff $$ (\forall \epsilon>0)(\exists F\subset_{finite} I)(\forall F_1\subset_{finite} I)(F\subset F_1\Longrightarrow \|S-\sum_{i\in F_1}x_i\|<\epsilon) $$ This is equivalent with commutative convergence in case $I\subset \mathbb{N}$ is infinite.

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    $\begingroup$ I particularly like Grothendieck's version of this. He proves that unconditional convergence is the same as absolute convergence for a locally convex space if and only if it is nuclear, i.e. every continuous linear map to a normed space is a nuclear map. The falsity of this belief then follows from the fact that a nuclear normed space is finite dimensional. $\endgroup$ – Robert Furber Feb 17 '18 at 7:44
  • $\begingroup$ @RobertFurber Thank you for this learned description (+1) $\endgroup$ – Duchamp Gérard H. E. Feb 17 '18 at 8:03
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    $\begingroup$ In fact, a commonly neglected fact is that the representation of an element $x$ of a Hilbert space as sum of its Fourier components in a Hilbert basis $\{u_\lambda\}_{\lambda\in\Lambda}$, namely $x=\sum_{\lambda\in\Lambda} (x\cdot u_\lambda)u_\lambda$, is always a summable family in $H$. Recalling this, the above false belief would just reduce to the very false "$\ell_1(\Lambda)=\ell_2(\Lambda)$" for any set $\Lambda$, that hopefully not many believe! $\endgroup$ – Pietro Majer Aug 1 '19 at 10:33
  • $\begingroup$ @ good hint (+1) $\endgroup$ – Duchamp Gérard H. E. Aug 1 '19 at 17:06
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I do not think anybody mentioned this example:

If $M$ is a $C^k$-smooth manifold then the tangent space $T_pM$ is isomorphic to the space of derivations of germs (at $p$) of $C^k$-smooth functions on $M$.

This holds for $k=\infty$ but not otherwise.

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  • $\begingroup$ +1. Does it isomorphic to the space of derivations of germs of $C^{k-1}$-smooth functions? $\endgroup$ – C.F.G Aug 27 at 11:30
  • $\begingroup$ @C.F.G: No, this does not help. $\endgroup$ – Moishe Kohan Aug 28 at 21:03
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An elementary false belief in elementary number theory: for $a, b, c\hspace{.1cm}\varepsilon\hspace{.1cm} \mathbb{N}$

$LCM\left(a,b\right)\times GCF\left(a,b\right) = ab$ .

Thus, $LCM\left(a,b,c\right)\times GCF\left(a,b,c\right) = abc$.

In general, $\left(a_1,a_2,\ldots,a_n\right)[a_1,a_2,\ldots,a_n] = a_1a_2\ldots a_n$.

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  • $\begingroup$ Does GCF mean gcd (greatest common divisor) here? $\endgroup$ – Zsbán Ambrus Nov 27 '10 at 19:33
  • $\begingroup$ Yes. GCF means the same: Greatest Common Factor. $\endgroup$ – Unknown Nov 29 '10 at 6:46
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    $\begingroup$ This kind of stuff is easy to rule out, though; it's dimensionally inconsistent. Replacing a, b, c by ka, kb, kc leads to a quick contradiction. $\endgroup$ – Qiaochu Yuan Feb 24 '11 at 21:08
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    $\begingroup$ @Qiaochu, that is a nice quick check. Let the RCF(remnant common factor) be the leftover factor that would make the above second equality true. There seems to be no interesting way of determining $RCF\left(a,b,c\right)$ so that $LCM\left(a,b,c\right)\times RCF\left(a,b,c\right) \times GCF\left(a,b,c\right) = abc$ $\endgroup$ – Unknown Feb 24 '11 at 22:05
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    $\begingroup$ @Elohemahab: actually, the correct generalization is $\gcd(a, b, c) \text{lcm}(ab, bc, ca) = \text{lcm}(a, b, c) \gcd(ab, bc, ca) = abc$. $\endgroup$ – Qiaochu Yuan May 8 '11 at 23:47
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I saw many students using the "fact" that for a subset $S$ of a group one has $SS^{-1}=\{e\}$

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    $\begingroup$ This is an interesting example, because it addresses the mistakes that come from the all-too frequent confusion with notations. But we need our shortcuts, our $f^{-1}(x)$ versus $x^{-1}$, etc. Obtaining concise notations while avoiding confusion: a tricky proposition! $\endgroup$ – Thierry Zell Apr 14 '11 at 15:50
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Every matrix is the sum of a symmetric and an antisymmetric matrix. Hence:

If $V$ is a vector space and $k$ is a number, then the $k$-th tensor product of $V$ with itself decomposes as a direct sum into symmetric and antisymmetric tensors: $$ \underbrace{V \otimes ... \otimes V}_{k\text{ times}} = \Lambda^kV \oplus \mathrm{Sym}^kV $$

Recall (in the finite-dimensional case) the dimensions: $$ \dim \Lambda^k V = \binom{n}{k} \quad\text{ and }\quad \dim\mathrm{Sym}^kV = \binom{n+k-1}{k} $$

Looking at $k=1$ shows that we have non-trivial intersection.

Looking at $n=k=3$ shows that the sum is not exhausting.

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    $\begingroup$ Is this a common false belief? $\endgroup$ – Jim Conant Oct 18 '15 at 2:52
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    $\begingroup$ @JimConant: I believed this once. Of course, if you count dimensions it's obviously false. But if $V$ is an infinite-dimensional Hilbert space it sure seems natural to decompose the full tensor product into its Bosonic and Fermionic parts, and you might not think right away to ask whether it works in finite dimensions. That's my excuse, anyway! $\endgroup$ – Nik Weaver Mar 4 '16 at 4:22
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A polynomial $p(x)$ of degree $n$, with coefficients in a commutative ring $R$, has at most $n$ roots, counting multiplicity. This is true if $R$ is an integral domain, but it can fail in the presence of zero divisors.

For instance, $p(x) = x^2+5x$ has four solutions when $R = \mathbb{Z}/6\mathbb{Z}$. I realized this mistake when a colleague asked me about factorization over a non-commutative ring, and I realized that I did not even know what would happen in the presence of zero-divisors.

This does motivate a question I have not found an answer to: is the number of solutions of $p(x)$ bounded by a function of the degree and the characteristic of the ring?

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    $\begingroup$ To answer your last question, $x^2-1$ has an infinite number of roots in $k^{\mathbb{N}}$ for any nonzero ring $k$. So, no. $\endgroup$ – Gro-Tsen Apr 20 '16 at 17:00
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    $\begingroup$ Doesn't it only have one solution in $k^\mathbb{N}$ if $k$ has characteristic 2, @Gro-Tsen? $\endgroup$ – Omar Antolín-Camarena Apr 20 '16 at 17:36
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    $\begingroup$ @OmarAntolín-Camarena Oh right, what I wanted to write was $x^2-x$, and I got confused between "idempotent" and "one-potent"(?). But of course $x^2-1$ also works provided, as you point out, that $1\neq -1$ in $k$. $\endgroup$ – Gro-Tsen Apr 20 '16 at 20:31
  • $\begingroup$ A similar eample is $\mathbb Z/4 \mathbb Z[t]$. For each $n$ the element $2t^n$ is a root of $x^2=0$. $\endgroup$ – Nick S Mar 3 at 6:17
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"This algebraic variety is a $C^\infty$-smooth manifold, therefore it must be non-singular". This sounds obvious (and in fact it is true over $\mathbb{C}$) however it is false in general (for instance over $\mathbb{R}$). See the discussion here for many details.

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If $A$ and $D$ are $n \times n$ matrices and $D$ is diagonal, then $A \cdot D=D \cdot A$.

Many of my Linear Algebra students believe this, also because it's written in the textbook in some form, but it's only true if $D$ is a multiple of the identity matrix or if both $A$ and $D$ are diagonal.

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    $\begingroup$ Grading their work on similar matrices must be interesting. $\endgroup$ – nombre May 1 '19 at 21:01
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