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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

247 Answers 247

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For awhile, I used to think:
If $depth\ M\ge depth\ N$ then $depth\ M_p\ge depth\ N_p$; for any prime ideal $p$ and finite R-modules $M$ and $N$ (Which is not true).

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Taylor's Formula and displacement operator: I (too often) see in papers (mathematical physics but a recent paper (a) by mathematicians also) the statement

False belief 1 : a) Let $D=\frac{d}{dx}$ be the derivation operator. Then, for all $f\in C^\infty(\mathbb{R})$, $$ e^{tD}[f](x)=f(x+t) $$

which is false (take any $\phi\in C^\infty(\mathbb{R})$ with compact support, for instance).

False belief 2 : b) In the same vein, for formal power series (``our object is formal then we do not have to ensure convergences''). Let $S(x)\in \mathbb{R}[[x]]$ ($x$ is a formal variable) then for $t\in \mathbb{R}$, one has $$ e^{tD}[S](x)=S(x+t) $$

which is false as we must have $t$ in the domain of convergence of $S$.

Remarks (i) The function $f\in C^\infty(\mathbb{R})$ is analytic over $\mathbb{R}$ iff $$ (\forall x\in \mathbb{R})(\exists R>0)(\forall t\in ]-R,R[) (\sum_{n\geq 0}\frac{t^n}{n!}D^n[f](x)=f(x+t))\qquad (1) $$ (ii) Even if $f\in C^\omega(\mathbb{R})$, it can happen that the left hand side of eq. (1) do not converge otherwise $f$ would be the restriction of an entire function (which e.g. $\frac{1}{1+x^2}$ is not, for example).

(iii) Even if the LHS of (1) converges for all $x,t\in \mathbb{R}$, $f$ need not be analytic. Consider the following function (classic in theory of distributions)
$$ f(x)=0\mbox{ if } x\notin ]-1,1[\mbox{ and } f(x)=e^{\frac{1}{1-x^2}} \mbox{ if } x\in ]-1,1[ $$ (iv) In the (b) case $S=\sum_{n\geq 0}n!\, x^n$ for example cannot be displaced.

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  • $\begingroup$ What is the correct statement? $\endgroup$ – ಠ_ಠ Jul 13 '17 at 7:33
  • $\begingroup$ @ಠ_ಠ: it is true for real analytic functions defined on the entire real number line. That is one possible choice of correct statement. $\endgroup$ – Ben McKay Jul 13 '17 at 8:07
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    $\begingroup$ Maybe you could point out the error in my reasoning: for a Lie group $G$, $G$ has a canonical left action on $C^\infty(G, \mathbb{R})$ by $(g.f)(x) =f(g^{-1}.x)$. Since $D=\frac{d}{dx}$ is a left-invariant vector field on $(\mathbb{R}, +)$, then $(e^{tD}.f) (x)=f(e^{-tD}.x)=f(x-t)$. $\endgroup$ – ಠ_ಠ Jul 13 '17 at 8:35
  • $\begingroup$ @ಠ_ಠ I added it. The fact that the LHS converges for some $t$ means that $f$ coincides with the sum of its Taylor series in the open disk of center $x$ and radius $|t|$. $\endgroup$ – Duchamp Gérard H. E. Jul 13 '17 at 19:52
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    $\begingroup$ @ಠ_ಠ I think I see what you meant. In the (Fréchet) space $C^\infty(\mathbb{R})$, the evolution equation $$ y(0)=f ; y′(t)=D[y(t)] $$ has a (unique) solution $y(t)[x]=f(x+t)$ which we could (legitimately ?) note $y(t)=e^{tD}[f]$, but one must keep in mind that, in this case $e^{tD}$ cannot be developed without caution. $\endgroup$ – Duchamp Gérard H. E. Jul 16 '17 at 14:17
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Here's another howler some people commit: If $m$, $n$ are integers such that $m$ divides $n^2$ then $m$ divides $n$.

It's true sometimes, for example if $m$ is prime (or more generally squarefree, i.e. a product of distinct primes). But in general all one can conclude is that there exists integers $p$, $q$, $r$ with $p$ squarefree such that $ m = p q^2 $ and $ n = p q r $

The usual counterexample is that $8$ divides $4^2$ but not $4$ ;-)

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    $\begingroup$ An even more trivial counterexample is that 4 divides 2^2 but not 2 :-P $\endgroup$ – Peter LeFanu Lumsdaine Feb 23 '11 at 9:40
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Here are some various examples (I hope that some of them weren't already mentioned):
1. If a space $X$ have two different norms $\| \cdot \|_i, i=1,2$ such that $\| \cdot \|_1 \leq \| \cdot \|_2$ then the completion with respect to $\| \cdot \|_1$ is contained in the completion with respect to $\| \cdot \|_2$.
2. If $M_1,M_2$ are isomorphic modules and $N_1,N_2$ are isomorphic submodules then $M_1/N_1$ and $M_2/N_2$ are isomorphic.
3. If $A,B$ are subsets of topological spaces $X,Y$ (resp.) and $A,B$ are homeomorphic then the closures $\overline{A}$ and $\overline{B}$ are also homeomorphic.
4. The standard construction of adjoining unit to the Banach algebra $A$ yields nothing new if $A$ already was unital.
5. The phrase "a function is almost everywhere continuous" means the same as: "the function is almost everywhere equal to the continuous function".
6. Suppose you are trying to prove that some function space $F$ is complete (say that functions are defined on $X$ and real valued): you take a Cauchy sequence $\{f_n\}_n$ and prove that for each point $x \in X$ the sequence $\{f_n(x)\}_n$ is Cauchy. Then form the completeness of $\mathbb{R}$ you obtain a function $f$. The false belief is that it is now enough to show that $f$ belong to $F$.
7. If you have an ascending family $\{A_i\}_i$ then to obtain it's union $\bigcup_{i}A_i$ it is enough to take some countable subfamily
8. A convergent net $\{x_i\}_i$ in a metric space is bounded and the set $\{x_i\}_i \cup \{x\}$ is compact (where $x$ is the limit).
9. If $D$ is an open dense subset of a topological space $X$ then $card \; D= card \; X$

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As a sequel of this famous answer on $\dim(U+V+W)$, the following inequality is not true $\forall n \ge 4$:
$$ \dim(\sum_{i = 1}^{n} U_i) \le \sum_{r=1}^{n} (-1)^{r+1} \sum_{i_1 < i_2 < \dots < i_r} \dim(\bigcap_{s=1}^{r}U_{i_s}) = \alpha$$
Darij Grinberg has found a counter-example (see this post).

Same flavor: for $n \le 5$, it is true that $\alpha \ge 0$ (see this proof), but it's false for $n>5$ (see this comment).

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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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I didn't notice this in the long list. A student beginning to learn group theory may believe that the converse of Lagrange's Theorem is true, because it is true for subgroups of prime power. They may also believe that a Sylow subgroup is normal because it has a special name. A counter example to both is $A_{4}$ of order $12$ which has no subgroup of order $6$ and whose four different Sylow $3$-subgroups are all conjugates of one another.

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True: Given a graded algebra $A$, there is a notion of a "homogeneous" ideal of $A$. It is a property that connects an ideal of $I$ with the grading and is often necessary to require. For example, if $I$ is a homogeneous ideal of $A$, then the algebra $A / I$ is graded again. If $I$ is not homogeneous, then it is not graded in general (since the projections of different graded components of $A$ onto $A / I$ might have nonzero intersection).

False: Given a filtered algebra $A$, there is a notion of a "filtered" ideal of $A$.

There is no such notion. We can require $I$ to be generated by $I\cap A_n$ for some $n$, or actually to lie inside $A_n$ for some $n$, but in most cases none of these is actually needed. (Correct me if I am wrong.) Formulations like "Let $I$ be an ideal compatible with (or respecting) the filtration" are cargo cult.

But: Given a filtered algebra $A$ and a generating set $G$ of an ideal $I$ of $A$, it is an important question whether $I\cap A_n$ is generated by $G\cap A_n$ for every $n\in \mathbb N$. This is not always satisfied, often nontrivial (in many cases it can be proved by using the diamond lemma to show that every element of $A_n$ has a unique "remainder" modulo $I$ in a certain sense, and this remainder can be obtained by repeated subtraction multiples of elements of $G\cap A_n$) and used tacitly in various texts.

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  • $\begingroup$ Good point, but "cargo cult"? $\endgroup$ – Tom Goodwillie Mar 15 '11 at 14:32
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    $\begingroup$ What I mean is: People use these formulations as a protective charm against a danger they don't see but intuitively feel is there, although closer inspection shows that it is pure superstition. $\endgroup$ – darij grinberg Mar 15 '11 at 17:26
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A set is compact iff it is closed and bounded.

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    $\begingroup$ This is perhaps a common false belief among undergraduates, but one that is dispelled by just a superficial acquaintance with functional analysis. $\endgroup$ – Todd Trimble Dec 9 '13 at 2:45
  • $\begingroup$ @ Todd Trimble: true, but then also the belief about $sin$ suggested by the OP is only common among people who have not completed a course in complex analysis. $\endgroup$ – Delio Mugnolo Dec 13 '13 at 8:34
  • $\begingroup$ I thought "bounded" is only defined on metric spaces, and this is true on metric spaces. Is that wrong? $\endgroup$ – Akiva Weinberger Sep 1 '15 at 2:48
  • $\begingroup$ I have seen analysis textbooks take this as a definition. I hope they realize that they are contributing to future confusion in their readers once they move on to topology or even metric spaces. @AkivaWeinberger, The Heine-Borel theorem stated in this way makes sense for arbitrary metric spaces, but it is only true for complete metric spaces for which balls are totally bounded. The correct statement of H-B for general metric spaces is "a metric space is compact iff it is complete and totally bounded". $\endgroup$ – Mario Carneiro Oct 20 '15 at 21:27
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    $\begingroup$ @AkivaWeinberger: Yes, it is wrong. The closed unit ball of an normed vector space is compact if and only if the space is finite dimensional. $\endgroup$ – ACL Apr 21 '16 at 13:43
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Some undergraduate common false beliefs that I found

(1) If $H$ is a subgroup of $\mathbb{Z}$ and $H$ and $\mathbb{Z}$ are isomorphic, then $H = \mathbb{Z}$;

(2) In a metric space every two open balls are homeomorphic;

(3) For $p \in [1, \infty]$, $L^p(X, \mathfrak{M}, \mu) = \left\{ f \in \mathbb{C}^X : \int_X |f|^p \, d \mu < \infty \right\}$ is a $\mathbb{C}$-normed vector space, with the norm $\lVert f \rVert_p = (\int_X |f|^p \, d \mu)^{1/p}$.

Belief (1) is very naive, for every nontrivial subgroup of $\mathbb{Z}$ is of the form $n \mathbb{Z}$, all of them isomorphic with $\mathbb{Z}$. For (2) people tend to think of normed vector spaces and forgets the discrete metric spaces. For (3) some people just forget that one have to consider the quotient space, where the classes $[f]=[g]$ iff $f=g$ $\mu$-almost everywhere.

Belief (1) is very naive, because every nontrivial subgroup of $\mathbb{Z}$ is of the form $n \mathbb{Z}$, all of them isomorphic to $\mathbb{Z}$. For (2) people tend to think of normed vector spaces and they forget the discrete metric spaces. For (3) some people just forget that one have to consider the quotient space, where the classes $[f]=[g]$ iff $f=g$ $\ \mu$-almost everywhere.

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  • $\begingroup$ Well, in (1) I think I can replace $\mathbb{Z}$ by an arbitrary group $G$, because $\mathbb{Z}$ do not come in mind so quickly. $\endgroup$ – Gustavo Jan 8 '16 at 4:03
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Anytime I wanted to write an answer to this question, I doubted maybe it is not as common as worthy of mentioning here. In fact, I am also not sure how common is the false belief that I observed today in a PDE class. I didn't observe that in many years of teaching calculus, but today four or five students in a small PDE class when calculating a definite integral by parts only applied the limits of the integral to the "second" integral, that is:

$$\int_{a}^b{f(x) g'(x) dx}=f(x) g(x) - \int_{a}^b{f'(x) g(x) dx}$$

Haven't I observed well enough in my calculus classes?

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    $\begingroup$ Definitely integrals are numbers and $f(x)g(x)$ is a function of variable $x$. Formula as written is something very strange. $\endgroup$ – Fedor Petrov Apr 20 '16 at 18:58
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    $\begingroup$ @FedorPetrov More strange is that most students don't see such a very strange something :) $\endgroup$ – Amir Asghari Apr 20 '16 at 19:50
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This example is similar to this earlier answer.

If $k$ is a field, then $k[x] \otimes_k k[y] \cong k[x,y]$. Therefore also $k[[x]] \otimes_k k[[y]] \cong k[[x,y]]$, right?

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Here are two beliefs. I think everybody will agree that one of them, at least, is false. I adhere to the second one.

Belief 1. There is no simple generalization of the Hodge Theorem to noncompact manifolds.

Belief 2. The most naive statement which would, if true, generalize the Hodge Theorem to noncompact manifolds is this.

The inclusion of the complex of coclosed harmonic forms into the de Rham complex of a riemannian manifold is a quasi-isomorphism.

This statement happens to be true.

Here is a reference:

http://www.iecl.univ-lorraine.fr/~Pierre-Yves.Gaillard/DIVERS/Hodgegaillard/

The simplest example is that of the real line with its standard metric. In degree zero the complex of coclosed harmonic forms is $\mathbb C\oplus\mathbb Cx$, and in degree one it is $\mathbb Cdx$, which gives the right cohomology.

Here is the (trivial) algebra background.

Let $A$ be a module over some unnamed ring, and let $d,\delta$ be two endomorphisms of $A$ satisfying $d^2=0=\delta^2$. Put $\Delta:=d\delta+\delta d$. Assume $A=\Delta A+A_{d,\delta}$ where $A_{d,\delta}$ stands for $\ker d\cap\ker\delta$. Write $A_{\delta,\Delta}$ for $\ker\Delta\cap\ker\delta$.

We claim that the natural map $$H(A_{\delta,\Delta},d)\to H(A,d)$$ between homology modules is bijective.

Injectivity. Assume $\delta da=0$ form some $a$ in $A$. We must find an $x$ in $A_{\delta,\Delta}$ such that $dx=da$. We have $a=\Delta b+c$ for some $b\in A$ and some $c\in A_{d,\delta}$. One easily checks that $x:=\delta db+c$ does the trick.

Surjectivity. Let $a$ be in $\ker d$. We must find $x\in A$, $y\in A_{d,\delta}$ such that $a=dx+y$. We have $a=\Delta b+c$ for some $b\in A$ and some $c\in A_{d,\delta}$. One easily checks that $x:=\delta b$, $y:=\delta db+c$ works.

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A somewhat common belief among students starting out in cryptography:

Breaking RSA requires factoring the modulus.

Although it is not quite known to be a "false" belief, there is no known reduction showing that breaking RSA implies finding the prime factors of the modulus. This in contrast to e.g. Rabin's cryptosystem, and various cryptographic schemes built on other hard problems, whose security provably relies on the underlying hard problems.

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    $\begingroup$ I would put this on the same level as the "false belief" that NP-complete problems really are hard. Nobody knows for sure if it actually is false or not, but given the current state of affairs it is a somewhat reasonable conjecture. And the very fact that nobody knows how to do it means that in practical applications NP-complete problems really are hard to solve and breaking RSA really does require factorisation of the modulus. $\endgroup$ – Johannes Hahn Mar 16 '18 at 22:56
  • $\begingroup$ Unless you say what the term "breaking RSA" is supposed to mean then it's impossible to evaluate whether that is a false belief or not. For example, say $m = pq$ is an RSA modulus with two different primes $p$ and $q$ and $e$ and $d$ are encryption and decryption exponents for that modulus (so $e$ an$d$ are positive integers such that $ed \equiv 1 \bmod \varphi(m)$). In practice $e > 1$ and $d > 1$, and also in practice $p$ and $q$ are both odd. If "Breaking RSA" means somehow determining $\varphi(m)$ from $m$ and $e$ then you can factor $m$ from knowing $\varphi(m)$ and $m$. If (cont.) ... $\endgroup$ – KConrad Feb 25 at 11:33
  • $\begingroup$ ... "Breaking RSA" means somehow determining $d$ from $m$ and $e$, then you know $ed - 1$, which is a positive multiple of $\varphi(m)$, and there is a probabilistic algorithm that with over 50% probability of success each time will lead to a nontrivial factor of $m$ starting with a random choice of an integer from 1 to $m-1$. See Theorem 5.6 of kconrad.math.uconn.edu/blurbs/ugradnumthy/RSAnotes.pdf. If "Breaking RSA" means "a wizard tells you how to decode each message without explaining how it is done" then that's not math and thus can't be judged as being a false belief in math. $\endgroup$ – KConrad Feb 25 at 11:36
  • $\begingroup$ @KConrad This answer is referring to the fact that, unlike for the mentioned Rabin variant, the RSA problem is not known to be as hard as factoring: given an oracle that decrypts ciphertexts, there is no efficient way known to use this oracle to recover the factorization of the modulus. This in contrast to many other schemes and protocols, for which such an oracle would immediately yield the solution for the hard problem it is presumed to rely on. (And many would disagree with you that such oracle reductions are "not math"...) $\endgroup$ – TMM Apr 2 at 23:46
  • $\begingroup$ @JohannesHahn I'd say the difference is that in cryptography, there do exist plenty reductions for other schemes proving that breaking these schemes implies solving the underlying well-studied hard problems. The whole hardness-hierarchy at least gives us some faith that some problems are probably really hard, but the RSA decoding problem is not connected to factoring in this hierarchy, let alone to any NP-hard problems. $\endgroup$ – TMM Apr 4 at 1:24
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These are 2 instances which i have seen to happen with my friends. If $A$ and $B$ are 2 matrices, then they believe that $(A+B)^{2}=A^{2}+ 2 \cdot A \cdot B +B^{2}$.

Another mistake is if one i asked to solve this equation, $ \displaystyle\frac{\sqrt{x}}{2}=-1$, people generally square both the sides and do get $x$ as $4$.

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  • $\begingroup$ What "people"? Non-mathematicians? $\endgroup$ – Todd Trimble May 4 '11 at 0:03
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    $\begingroup$ @Todd: No i was talking of high school students. $\endgroup$ – crskhr May 4 '11 at 4:08
  • $\begingroup$ @S.C.:if squarring both sides will not give the solution then how can second problem be solved? $\endgroup$ – P.Styles Oct 25 '17 at 16:08
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"It cannot be shown without some form of AC that the union (or disjoint union) of countably many countable sets is countable. I have a countably infinite set X of countably infinite sets. Therefore, the union of X cannot be shown to be countable without Choice."

The fallacy is that in many cases of interest, it is possible to exhibit an explicit counting of every element of X. In such a case a counting of X by antidiagonals is easily constructed. The usual counting of the rationals is an example of this.

I think this may even be an example of a more general phenomenon of "people think AC is necessary for a certain construction, but in fact it turns out not to be necessary for the example they have in mind". For example, AC is necessary to find a maximal ideal in an arbitrary ring ... but it isn't if you're prepared to assume the ring is Noetherian.

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    $\begingroup$ If "Noetherian" is defined by the ascending chain condition or by requiring all ideals to be finitely generated, then in order to deduce the existence of maximal ideals, you still need a weak form of the axiom of choice. The usual argument uses the axiom of dependent choice. (Of course, if you define "Noetherian" to mean that every set of ideals has a maximal element, then deducing the existence of maximal ideals is a choiceless triviality.) A good reference is "Six impossible rings" by Wilfrid Hodges (J. Algebra 31 (1974) 218-244). $\endgroup$ – Andreas Blass Oct 22 '10 at 15:29
  • $\begingroup$ Thanks Andreas! I had a feeling there was a technicality somewhere there, but couldn't remember what it was. As a philosophical point I personally think that of course in the absence of AC you want to define Noetherian so that my original statement is true, but admittedly that's a harder sell than my countable-sets example. $\endgroup$ – Karol Nov 16 '10 at 21:06
  • $\begingroup$ @AndreasBlass's reference, clickably: Hodges - 6 impossible rings. $\endgroup$ – LSpice Feb 5 at 1:09
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Hopefully this isn't a repeat answer. False belief: a matrix is positive definite if its determinant is positive.

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    $\begingroup$ Is this really a common(!) false belief? $\endgroup$ – Martin Brandenburg Oct 3 '11 at 7:23
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In ${\mathbb F}_p^\times$, the non-squares are the opposite of the squares. In other words, $a$ is square iff $-a$ is not a square.

This is a confusion with the facts that the kernel of $x\mapsto x^2$ is $\{1,-1\}$ and the subgroup of squares has index $2$.

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A few mistakes I remember:

  • The quotient groups $\frac{G}{N}$ and $\frac{H}{K}$ are isomorphic if $G \thicksim H$ and $N\thicksim K$.
  • A closed interval of a complete lattice is a complete sublattice.
  • Two homeomorphic topologies on a set are the same.
  • The set of all compatible uniformities of a topological group forms a complete lattice.
  • The trace of the identity matrix is 1.
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    $\begingroup$ A closed interval of a complete lattice does form a lattice that is complete, right? So that the mistake is that sups and infs in the interval (particularly the sup and inf over the empty set) are not necessarily computed as they would be in the ambient complete lattice; is that what you have in mind? $\endgroup$ – Todd Trimble Sep 6 '15 at 1:47
  • $\begingroup$ Yes‌‌‌‌‌‌‌‌‌‌‌‌. $\endgroup$ – user47958 Sep 6 '15 at 1:55
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Most people that study Riemannian geometry for their first time make the following assumption at some point: "Let $(e_1,\dots,e_n)$ be a local orthonormal frame of $TM$ such that all Lie brackets $[e_i,e_j]$ vanish..."

This one is not so common (maybe special to me), but here we go: "$\mathbb{RP}^\infty$ and $\mathbb{CP}^\infty$ are Eilenberg-Mac Lane spaces, so $\mathbb{HP}^\infty$ is one, too."

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(1) All Lebesgue-null sets are countable, or are strongly measure zero. (2) The following,verbatim, was a Q in American Mathematical Monthly : " A student asserted that any uncountable real set has a closed uncountable subset. Is this true ?" .

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I'm seven years late to the game, but here is mine:

False belief: The irrational numbers, in their usual topology as a subset of $\mathbb{R}$, are not a complete metric space.

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    $\begingroup$ Could you write that more carefully? You mean the false belief to be that the irrationals, as a topological space, can't be complete for some metric. What you wrote can be easily confused with saying the irrational numbers are not complete for the usual metric coming from $\mathbf R$, which is true rather than false. Consider the "false belief" that $(-1,1)$ with its topology from $\mathbf R$ can't be made into a complete metric space for some metric. Certainly it's not complete for the usual metric, but it is if we use $\tan(\pi x/2)$ to identify $(-1,1)$ with $\mathbf R$ topologically. $\endgroup$ – KConrad Jul 15 '17 at 3:26
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    $\begingroup$ "are not completely metrizable" is the wording you want. $\endgroup$ – Andrés E. Caicedo Jul 15 '17 at 3:28
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    $\begingroup$ @KConrad and Andres: The wording is part of what made that false belief so believable! At the time I didn't think about the fact that there could be multiple metrics, much less that the completeness of those metrics wasn't a topological property. I only realized my mistake when I was introduced to the ideas contained in your two comments. (That, and picturing the irrationals as complete is HARD!) $\endgroup$ – Pace Nielsen Jul 15 '17 at 4:15
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    $\begingroup$ @DuchampGérardH.E. A topological subspace of a completely metrizable topological space is completely metrizable if and only if it is a $G_\delta$, that is a countable intersection of open sets. One can use Baire's category theorem to show that $\mathbb{Q}$ is not a $G_\delta$. All this can be found at: en.wikipedia.org/wiki/G%CE%B4_set $\endgroup$ – Michael Greinecker Aug 11 '17 at 14:41
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    $\begingroup$ @DuchampGérardH.E. Exactly, it is the countable intersection $\bigcap_{q\in\mathbb{Q}}\mathbb{R}\setminus\{q\}$ of open sets. $\endgroup$ – Michael Greinecker Aug 11 '17 at 17:05
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People seem to believe that conventional computation (for example, running a chaotic irreversible cellular automaton) can be as efficient as one wants simply with good engineering, but this is not the case. Landauer's principle states that erasing a bit of information always takes $\ln(2)\cdot k\cdot T$ energy where $k$ is Boltzmann's constant ($k=1.38065\cdot 10^{-23}$ Joules/Kelvin) and $T$ is the temperature. Landauer's principle is a consequence of the second law of thermodynamics since if Landauer's principle were violated, then entropy would decrease. Landauer's principle means that conventional irreversible computation always must take $\ln(2)\cdot k\cdot T$ energy per bit erased (and one can erase data just by running it through AND and OR gates, so every irreversible gate must take a minimum amount of energy by Landauer's principle). However, Landauer's principle does not apply to reversible computation since reversible computers are not allowed to erase data.

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    $\begingroup$ Okay, I see what you are aiming at with this last edit. The idea that ordinary computation can be made arbitrarily efficient is a reasonably common false belief about our physical world (and may even have a somewhat solid mathematical interpretation). I withdraw my initial objection. $\endgroup$ – S. Carnahan Aug 11 '17 at 14:04
  • $\begingroup$ All of the downotes to this post are illegitimate. $\endgroup$ – Joseph Van Name Apr 3 at 11:59
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Here are two beliefs. I think everybody will agree that one of them, at least, is false. I adhere to the second one.

Belief 1. The simplest way to compute the exponential $e^A$ of a complex square matrix $A$ is to use the Jordan decomposition.

Belief 2. It's simpler and more efficient to use the following fact.

Let $f(z)$ be the minimal polynomial of $A$, let $g(z)$ be $f(z)$ times the singular part of $e^z/f(z)$, and observe $e^A=g(A)$.

(By abuse of notation $z$ is at the same time an indeterminate and a complex variable.) (The problems of computing the exponential of $A$ and that of computing the Jordan decomposition of $A$ have the same difficulty level. But, to solve one of them, there is no need to refer to the other.) Here are two references

http://en.wikipedia.org/wiki/Matrix_exponential#Alternative

http://www.iecl.univ-lorraine.fr/~Pierre-Yves.Gaillard/DIVERS/Constant_coefficients/

Jordan decomposition is often mentioned in relation with matrix exponentials. I'm convinced (rightly or wrongly) that the association of these notions in this context is purely irrational. I think somebody once made this association by accident, and then many people repeated it mechanically.

Here is another attempt to describe the situation.

Put $B:=\mathbb C[A]$. This is a Banach algebra, and also a $\mathbb C[X]$-algebra ($X$ being an indeterminate). Let $$\mu=\prod_{s\in S}\ (X-s)^{m(s)}$$ be the minimal polynomial of $A$, and identify $B$ to $\mathbb C[X]/(\mu)$. The Chinese Remainder Theorem says that the canonical $\mathbb C[X]$-algebra morphism $$\Phi:B\to C:=\prod_{s\in S}\ \mathbb C[X]/(X-s)^{m(s)}$$ is bijective. Computing exponentials in $C$ is trivial, so the only missing piece in our puzzle is the explicit inversion of $\Phi$. Fix $s$ in $S$ and let $e_s$ be the element of $C$ which has a one at the $s$ place and zeros elsewhere. It suffices to compute $\Phi^{-1}(e_s)$. This element will be of the form $$f=g\ \frac{\mu}{(X-s)^{m(s)}}\mbox{ mod }\mu$$ with $f,g\in\mathbb C[X]$, the only requirement being $$g\equiv\frac{(X-s)^{m(s)}}{\mu}\mbox{ mod }(X-s)^{m(s)}$$ (the congruence taking place in the ring of rational fractions defined at $s$). So $g$ is given by Taylor's Formula.

This can be summarized as follows:


There is a unique polynomial $E$ such that $\deg E<\deg\mu$ and $e^A=E(A)$. Moreover $E$ can be uniquely written as $$E=\sum_{s\in S}\\ E_s\\ \frac{\mu}{(X-s)^{m(s)}}$$ with (for all $s$) $\deg E_s < m(s)$ and $$E_s\equiv e^s\ e^{X-s}\\ \frac{(X-s)^{m(s)}}{\mu}\mbox{ mod }(X-s)^{m(s)},$$ the congruence taking place in $\mathbb C[[X-s]]$.


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    $\begingroup$ "$\emptyset$ is a basis for {0}" is an immediate consequence of the definitions. There is no false belief in this point. $\endgroup$ – Johannes Hahn May 12 '10 at 15:14
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    $\begingroup$ Dear Johannes, please reread my post. $\endgroup$ – Pierre-Yves Gaillard May 12 '10 at 15:18
  • $\begingroup$ Even a cursory examination of Nick Higham's book amazon.co.uk/Functions-Matrices-Computation-Nicholas-Higham/dp/… will show that both these opinions on the evaluation of matrix exponentials are hopelessly naive. $\endgroup$ – Robin Chapman May 15 '10 at 9:17
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    $\begingroup$ Your opinions are normative statements: "one should" and "it is better". It is naive to suppose that there is one best method that one should use to compute the matrix exponential. $\endgroup$ – Robin Chapman May 15 '10 at 14:07
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    $\begingroup$ I don't think the OP wants examples of normative statements. As I read it, the question is about conceptual errors regarding non-normative mathematical statements. $\endgroup$ – Qiaochu Yuan May 17 '10 at 6:19
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(Cauchy and/or ordinary) product of two summable families. Until recently, I thought that, in a topological ring (i.e. a ring $R$ with topology $\tau$ such that, the maps $x\mapsto -x;\ (x,y)\mapsto x+y;\ (x,y)\mapsto x.y$ are continuous), products of two summable families were summable. In the following contexts, were my (false) beliefs

  • $(a_i)_{i\in I},\ (b_j)_{j\in J}$ supposed summable and then $(a_ib_j)_{(i,j)\in I\times J}$ is summable
  • Same situation with $I=J=\mathbb{N}$ and $c_n=\sum_{p+q=n}a_pb_q$ (Cauchy product).

But, I found this question and discussion (which proved me that this belief was false in general), returned to Bourbaki General Topology Chapter III, § 6, and there were Exercises 4-5 which proved me that this question was very delicate. Then I could debunk it.

Late addition: See also discussions and the beautiful answer by Robert Furber here.

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If $a$ is a real zero of a cubic polynomial with rational coefficients then $a$ can be written as a combination of cube roots of rational numbers.

More generally if $a$ is a real zero of an irreducible polynomial with rational coefficients that is solvable by radicals then students expect the following:

  1. Any expression inside a radical evaluates to a real number
  2. Any sub-expression of the expression for $a$ evaluates to an algebraic number of order less than or equal to the order of $a$

Of course the problem is that from Cardan's solution to the cubic we can have negative rational numbers inside a square root. Let $c$ = $4*(-1 + \sqrt{-3})$.

$a$ = $\frac{\sqrt[3]{c}}{4} + \frac{1}{\sqrt[3]{c}}$

$f(x) = 4x^3 - 3x + \frac{1}{2}$.

So while $a$ is an algebraic number of degree three, it can not be written as combination of cube roots of rational numbers. Indeed, it is counter-intuitive that $\sqrt[3]{c}$ has degree 6 over the rational numbers yet we can use this number and simple arithmetic to produce an algebraic number of degree 3.

Also $a$ = $\sin(50^{\circ})$. For many values of $\theta$, $\sin \theta$ is a radical number. See also radical values for sine and cosine

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Something I was sure about until earlier today:

Suppose $\kappa$ is an $\aleph$ number, then $AC_\kappa$ is equivalent to $W_\kappa$, namely the universe holds that the product of $\kappa$ many sets is non-empty if and only if every cardinality is either of size less than $\kappa$ or has a subset of cardinality $\kappa$.

In fact this is only true if you assume full $AC$, and $(\forall \kappa) AC_\kappa$ doesn't even imply $W_{\aleph_1}$, I was truly shocked.

Furthermore, $W_\kappa$ doesn't even imply $AC_\kappa$ in most cases.

The strongest psychological implication is that most people actually think of the well-ordering principle as a the "correct form" of choice, when it is actually Dependent Choice (limited to $\kappa$, or unbounded) which is the "proper" form, that is $DC_\kappa$ implies both $AC_\kappa$ and $W_\kappa$.

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    $\begingroup$ How common is this misconception? $\endgroup$ – Thierry Zell Apr 17 '11 at 3:08
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    $\begingroup$ @Thierry: For the past couple of weeks I spent a lot time considering models without choice, not only I held that misconception but not once anyone corrected me about it - grad students and professors alike. $\endgroup$ – Asaf Karagila Apr 17 '11 at 6:09
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Coordinates on a manifold do not have an immediate metric meaning. Until becoming familiar with differential geometry one tends to think they do. (Einstein wrote that he took seven years to free himself from this idea.)

For example, linear control theory is for the most part metric with variables in $R^n$. When moving away from linear control theory, variables are represented as coordinates on a manifold. Nevertheless, much of the literature tends to either abandon metric notions altogether, or to keep using an Euclidean metric though it is no longer very useful.

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The "conditional Vitali convergence theorem": Let $X_n$ be a uniformly integrable sequence of random variables with $X_n \to X$ almost surely, and $\mathcal{G}$ a sub-$\sigma$-field. Then $\mathbb{E}[X_n \mid \mathcal{G}] \to \mathbb{E}[X \mid \mathcal{G}]$ almost surely (FALSE).

I believed this one until I read Uniformly integrable sequence such that a.s. limit and conditional expectation do not commute. It is particularly startling because the conditional versions of the monotone convergence theorem, the dominated convergence theorem, and Fatou's lemma are all true!

What is true is that $\mathbb{E}[X_n \mid \mathcal{G}] \to \mathbb{E}[X \mid \mathcal{G}]$ in $L^1$, so you do have a subsequence converging almost surely.

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Here is a false belief I had. Let $f:X \to Y$ be a map of topological spaces having the property that for every finite CW complex $K$, the induced map $f_{\ast}:[K,X] \to [K,Y]$, on unpointed homotopy classes of maps, is a bijection. Then $f$ is a weak homotopy equivalence (that is, it induces isomorphisms on all homotopy groups relative to all basepoints). A counterexample is given by the stabilization map $B \Sigma_{\infty}\xrightarrow{+1} B \Sigma_{\infty}$, which is not an isomorphism on $\pi_1$.

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    $\begingroup$ Although the original intent of this question seems to have long since evaporated, I can't help asking: is this really a "common false belief"? $\endgroup$ – Yemon Choi Feb 17 '15 at 1:24
  • $\begingroup$ how about: if two CW complexes have all homotopy groups isomorphic, then they are homotopy equivalent? as i recall, you need those isomorphisms to be induced by a single continuous map. $\endgroup$ – roy smith Apr 22 '17 at 0:01
  • $\begingroup$ @roysmith Yes. You can even have two non weakly equivalent spaces having all Postnikov stages weakly equivalent $\endgroup$ – Ilan Barnea May 8 '17 at 10:47

protected by François G. Dorais Oct 15 '13 at 2:34

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