# Examples of common false beliefs in mathematics

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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I have to say this is proving to be one of the more useful CW big-list questions on the site... – Qiaochu Yuan May 6 '10 at 0:55
The answers below are truly informative. Big thanks for your question. I have always loved your post here in MO and wordpress. – Unknown May 22 '10 at 9:04
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. – Suvrit Sep 20 '10 at 12:39
It's a thought -- I might consider it. – gowers Oct 4 '10 at 20:13
Meta created tea.mathoverflow.net/discussion/1165/… – user9072 Oct 8 '11 at 14:27

False belief: Every commuting pair of diagonalizable elements of $PSL(2,\mathbb{C})$ are simultaneously diagonalizable. The truth: I suppose not many people have thought about it, but it surprised me. Look at $$\left(\matrix{i& 0 \cr 0 & -i\cr } \right), \left(\matrix{0& i \cr i & 0\cr } \right).$$

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To me, it is marvellous that the failure of this fact (as opposed to the truth of the corresponding fact for $\operatorname{SL}(2, \mathbb C)$) is a matter of topology; that is, from the point of view of algebraic groups, it comes from the fact that $\operatorname{SL}(2, \mathbb C)$ is simply connected, whereas $\operatorname{PSL}(2, \mathbb C)$ (which I had rather call $\operatorname{PGL}(2, \mathbb C)$) is not (it is at the opposite extreme---`adjoint'). – L Spice Dec 12 '13 at 23:09

After learning that the Witt vectors of a finite field of size $p^n$ is the ring of integers of the unramified extension of ${\mathbf Q}_p$ of degree $n$, I think lots of people then think that the Witt vectors of $\overline{\mathbf F}_p$ (the algebraic closure of ${\mathbf F}_p$) is the ring of integers of the maximal unramified extension of ${\mathbf Q}_p$. It isn't: the integers of the maximal unramified extension is the union of the Witt vectors of the finite fields of $p$-power size whereas the Witt vectors of $\overline{\mathbf F}_p$ is the $p$-adic completion of the integers of the maximal unramified extension; the distinction turns on being able to write Witt vectors over $\overline{\mathbf F}_p$ as series with coefficients that are prime-to-$p$ roots of unity of increasingly large degree instead of having bounded degree.

I was at a conference last fall where a famous mathematician was confused by this point, although to be fair he really never worked seriously with Witt vectors before.

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If you're going to hold a false belief, this is a good one to have. While the maximal unramified extension of $\mathbb{Q}_p$ is not the fraction field of $W(\overline{\mathbb{F}_p})$ -- the latter is complete, the former isn't -- one is a subfield of the other and the two fields have very similar properties. One way to express/justify this is by saying that the inclusion map is an elementary embedding in the sense of model theory. For arithmetic geometers, this is related to work of Greenberg on rational points over Henselian fields. – Pete L. Clark May 5 '10 at 2:25
Pete: Witt vectors are confusing enough when you're trying to learn about them that I think this is a false belief that is important to clear up if, say, your thesis depends on understanding Witt vectors. – KConrad May 5 '10 at 4:13
@K: Well, sure, believing false things is never desirable. All I'm saying is that there are some things working underneath the surface to prevent this particular false belief from really screwing you up. I seem to remember once writing something like "where $\mathbb{Q}_{p^{\infty}}$ is, according to the reader's preference, either the maximal unramified extension of $\mathbb{Q}_p$ or its completion." Not that this is especially good mathematical writing, but the point was that it manifestly didn't matter which. – Pete L. Clark May 5 '10 at 4:57
P.S.: $W(\overline{\mathbb{F}_p})$ does appear in my thesis...I hope I got it right. :) – Pete L. Clark May 5 '10 at 6:39
Pete: It can screw you up very badly. For example, F-isocrystals are very different over the maximal unramified extension of Q_p and its completion. – JS Milne May 8 '10 at 16:44

I and several of my friends made our living off exploiting the fact that a space doesn't really have just one universal cover.

It has one at each basepoint. If the space is decent and connected, then these are all isomorphic, but the isomorphism requires a choice of a path connecting the points. You can clearly see the point if you make a bundle of universal covers over a moebius strip.

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You make a living off it?? Awesome. What do you do? – David Roberts Aug 31 '10 at 21:43
Related to this, here is a cool trick question which I heard from Allen Knutson: Consider the Hopf fibration, which is an $S^1$ bundle over $S^2$. Take the universal cover of each fiber, to get an $\mathbb{R}$ bundle over $S^2$. Is this line bundle nontrivial? – David Speyer Sep 27 '10 at 13:38
@DavidRoberts : What he does for a living is that he makes a bundle off universal covers. $\qquad$ – Michael Hardy Mar 10 at 5:13
@MichaelHardy how droll :-) – David Roberts Mar 10 at 6:11

The gamma function is not the only meromorphic function satisfying $$f(z+1)=z f(z),\qquad f(1)=1,$$ with no zeroes and no poles other than the points $z=0,-1,-2\dots$.

In fact, there is a whole bunch of such functions, which, in general, have the form $$f(z)=\exp{(-g(z))}\frac{1}{z\prod\limits_{n=1}^{\infty} \left(1+\frac{z}{m}\right)e^{-z/m}},$$ where $g(z)$ is an entire function such that $$g(z+1)-g(z)=\gamma+2k\pi i,\qquad g(1)=\gamma+2l\pi i,\qquad k,l\in\mathbb Z,$$ ($\gamma$ is Euler's constant). The gamma function corresponds to the simplest choice $g(z)=\gamma z$.

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I think I remember reading it is the only one that is convex over $(0,+\infty)$ or some similar condition. – Arnaud Chéritat Oct 18 '15 at 12:45

A subgroup of a finitely generated group is again finitely generated.

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True for abelian groups, though. – Mark Mar 3 '11 at 21:40
Also true for finite index subgroups, of finitely generated groups. – Michalis Mar 4 '11 at 21:18
I've already mentioned this in my answer mathoverflow.net/questions/23478/…, but nevermind. ;) – Martin Brandenburg Apr 12 '11 at 8:44
I had this false belief until the 2nd year of grad school. – Michael Dec 3 '13 at 1:21

Here's one that bugged me from point set topology: "A subnet of a sequence is a subsequence".

See here for the definitions. Using this one gives a great proof that compactness implies sequential compactness in any topological space:

Let $X$ be a compact space. Let $(x_n)$ be a sequence. Since a sequence is a net and it's a basic theorem of point set topology that in a compact topological space, every net has a convergent subnet (proof in the above link), there is a convergent subnet of the sequence $(x_n)$. Using the above belief, the sequence $(x_n)$ has a convergent subsequence and hence $X$ is sequentially compact.

For a counterexample to this "theorem", consider the compact space $X= \lbrace 0,1 \rbrace ^{[0,1]}$ with $f_n(x)$ the $n$th binary digit of $x$.

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I've just added the missing html to fix the link. – gowers Jul 3 '11 at 21:38

My example is $G_{1}$ isomorphic to $G_{2}$'s subgroup and $G_{2}$ isomorphic to $G_{1}$'s subgroup implies $G_{1}$ and $G_{2}$ are isomorphic...

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– Qiaochu Yuan May 6 '10 at 0:53
I'd think of that not as a belief that people are likely to have so much as a statement that looks moderately plausible and doesn't have an obvious counterexample. In other words, it's not something that people unthinkingly assume, because they don't tend to think about it at all. But perhaps I'm wrong about that and it is routinely used as a lemma by inexperienced algebraists. – gowers May 6 '10 at 11:43

A common misbelief for the exponential of matrices is $AB=BA \Leftrightarrow \exp(A)\exp(B) = \exp(A+B)$. While the one direction is of course correct: $AB=BA \Rightarrow \exp(A)\exp(B) = \exp(A+B)$, the other direction is not correct, as the following example shows: $A=\begin{pmatrix} 0 & 1 \\ 0 & 2\pi i\end{pmatrix}, B=\begin{pmatrix} 2 \pi i & 0 \\ 0 & -2\pi i\end{pmatrix}$ with $AB \neq BA \text{ and} \exp(A)=\exp(B) = \exp(A+B) = 1$.

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A more elementary, and I would bet more common, mistake is to believe that exp(A+B)=exp(A) exp(B) with no hypotheses on A and B. – David Speyer Sep 27 '10 at 13:40
Related to the mistake mentionned by David, the fact that the solution of a vector ODE $x'(t)=A(t)x(t)$ should be $$\left(\exp\int_0^tA(s)ds\right)x(0).$$ – Denis Serre Oct 20 '10 at 10:31

"Euclid's proof of the infinitude of primes was by contradiction."

That is a very widespread false belief.

"Prime Simplicity", Mathematical Intelligencer, volume 31, number 4, pages 44--52, by me and Catherine Woodgold, debunks it. The proof that Euclid actually wrote is simpler and better than the proof by contradiction often attributed to him.

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And you'd be surprised how many quite knowledgable PHD's spend decades repeating this mistake to thier students,Micheal. – The Mathemagician Jun 7 '10 at 0:07
Actually, if you read our paper on this, you'll find that I won't be surprised at all. (BTW, my first name is spelled in the usual way, not the way you spelled it.) – Michael Hardy Jun 7 '10 at 3:28
@ BlueRaja: I'm assuming "Euler" is a typo and you meant Euclid. Euclid said if you take any arbitrary finite set of prime numbers, then multiply them and add 1, and factor the result into primes, you get only new primes not already in the finite set you started with. The proof that they're not in that set is indeed by contradiction. But the proof as a whole is not, since it doesn't assume only finitely many primes exist. – Michael Hardy Jul 7 '10 at 21:55
Note indeed the original Euclid's statement: Prime numbers are more than any previously assigned finite collection of them (my translation). This reflects a remarkable maturity and consciousness, if we think that mathematicians started speaking of infinite sets a long time before a well founded theory was settled and paradoxes were solved. Euclid's original proof in my opinion is a model of precision and clearness. It starts: Take e.g. three of them, A, B and Γ . He takes three prime numbers as the first reasonably representative case to get the general construction. – Pietro Majer Jul 20 '10 at 14:51
@roy: if there are no primes, there are finitely many a fortiori. So there's nothing to check there. – Toink Mar 18 '13 at 0:04

Perhaps the most prevalent false belief in math, starting with calculus class, is that the general antiderivative of f(x) = 1/x is F(x) = ln|x| + C. This can be found in innumerable calculus textbooks and is ubiquitous on the Web.

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Well, the false belief is correct under the (frequently unspoken) condition that we only speak of antiderivatives over intervals on which the function we're antidifferentiating is "well-behaved" (and I'm not 100% sure what the right technical condition there is; "continuous"?). – JBL Jun 12 '10 at 0:57
Really? What about the function F(x) given by ln(x) + C_1, x > 0 F(x) = ln(-x) + C_2, x < 0 for arbitrary reals C_1, C_2 ? (The appropriate technical condition is that an antiderivative be differentiable on the same domain as the function it's the antiderivative of is defined on.) – Daniel Asimov Jun 12 '10 at 4:25
In case that wasn't clear: F(x) = ln(x) + C_1 for x > 0, and F(x) = ln(-x) + C_2 for x < 0, where C_1 and C_2 are arbitrary real constants. – Daniel Asimov Jun 12 '10 at 4:29
That function is not "nice" on any interval containing 0; on any interval not containing 0, it is of the form you are complaining about. This is exactly my point -- the word "interval" is important to what I wrote! – JBL Jun 12 '10 at 19:33
Sorry, I'm surely communicating less clearly than would be ideal. Starting over from scratch: suppose we have some differential equation, with unknown function $y(x)$. We say that for some function $f(x)$, $y = f(x)$ is a solution of the differential equation if there exists an interval $I$ on which $f(x)$ has all the requisite derivatives and the equation is satisfied on this interval. Antidifferentiation is the particular case $y' = g(x)$; implicit in the statement "$F(x)$ is an antiderivative of $f(x)$" is the condition "on some interval for which $F(x)$ is differentiable." – JBL Jun 12 '10 at 22:02
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This false belief is perhaps caused by the fact that continuity does imply sequential continuity, and sequential adherent points are adherent points. – Terry Tao Sep 27 '10 at 5:53

A projection of a measurable set is measurable. Not only students believe this. I was asked once (the quote is not precise): "Why do you need this assumption of a measurability of projection? It follows from ..."

A polynomial which takes integer values in all integer points has integer coefficients.

Another one seems to be more specific, I just recalled it reading this example. A sub-$\sigma$-algebra of a countably generated $\sigma$-algebra is countably generated.

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– Nate Eldredge Oct 3 '11 at 13:12
A counterexample to the polynomial thing is $\frac12x^2+\frac12x$. – Akiva Weinberger Aug 31 '15 at 22:27

"Suppose that two features $[x,y]$ from a population $P$ are positively correlated, and we divide $P$ into two subclasses $P_1$, $P_2$. Then, it cannot happen that the respective features ( $[x_1,y1]$ and $[x_2,y_2]$) are negatively correlated in both subclasses

Or more succintly:

"Mixing preserves the correlation sign."

This seems very plausible - almost obvious. But it's false - see Simpon's paradox

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This seems like much more than a mathematical mistake - really it's a cognitive bias which incidentally can be turned into a mathematically false statement. – Qiaochu Yuan May 9 '11 at 22:53
There were many cases of this error made by reporters and even courts. – Michael Dec 3 '13 at 1:18

To my knowledge, noone has proven that the scheme of pairs of matrices (A,B) satisfying the equations AB=BA is reduced. But whenever I mention this to people someone says "Surely that's known to be reduced!"

(Similar-sounding problem: consider matrices M with $M^2=0$. They must be nilpotent, hence have all eigenvalues zero, hence $Tr(M)=0$. But that linear equation can't be derived from the original homogeneous quadratic equations. Hence this scheme is not reduced.)

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I remember Etingof thinking it was open as recently as 2006 in his lectures on Calogero-Moser systems. Also, it's fairly easy to check that in small cases the commutator scheme is reduced, though getting it in general seems to be beyond current technology. – Charles Siegel Jun 6 '10 at 21:13
Sadly, people rely on technology so much nowadays that it gets increasingly unlikely that it will $\textit{ever}$ be proved. – Victor Protsak Jun 10 '10 at 6:40

This is (I think) a fairly common misconception about maths that arises in connection with quantum mechanics. Given a Hermitian operator A acting on a finite dimensional Hilbert space H, the eigenvectors of A span H. It's easy to think that the infinite dimensional case is "basically the same", or that any "nice" operator that physicists might want to consider has a spanning eigenspace. However, neither the position nor the momentum operator acting on $L^2(\mathbb{R})$ have any eigenvectors at all, and these are certainly important physical operators! Based on an admittedly fairly small sample size, it seems that it's not uncommon to simultaneously believe that Heisenberg's uncertainty relation holds and that the position and momentum operators possess eigenvectors.

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Yeah, for some reason many physicists are taught exactly no functional analysis... In fact, I know of no "quantum mechanics for physicists" books which use much more than a beginning undergrad level of analysis. Though admittedly these details are not so important for doing simple calculations, though they can be important in doing more sophisticated calculations, or understanding, e.g., why field theory works the way it does... – jeremy Jun 1 '10 at 23:33
Reciprocally, many mathematicians are taught no quantum mecha... make it, no physics at all! This is shocking, since the biggest impetus to the development of PDEs and functional analysis was given by what? You guessed it, physics. – Victor Protsak Jun 10 '10 at 6:56

Another false belief which I have been asked thrice so far in person is $$\lim_{x \rightarrow 0} \frac{\sin(x)}{x} = 1$$ even if $x$ is in degrees. I was asked by a student a year and half back when I was a TA and by couple of friends in the past 6 months.

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maybe not one of the best answers here, but why the down votes? – Yaakov Baruch Feb 23 '11 at 15:08
@downvoters: Kindly provide a reason for the down votes. – user11000 Feb 23 '11 at 15:54
+1. The limit when $x$ is in degrees is an exercise in many calculus textbooks (or equivalently, the derivative of $\sin (x degrees)$. Yet, it seems people are slow to pick up on it. Your point was made by Deane Yang in this answer: mathoverflow.net/questions/40082/… (and no one found anything wrong with it then...) – Thierry Zell Feb 27 '11 at 14:45
@JBL: Well, there are also some universities outside the US ;) This is not standard, yet not unusual though becoming rarer, in certain parts of Europe: In HS one learns about trig. func. in a geom. way; about diff./int. without a formal notion of limit, mainly rat. funct; in any case that limit wouldn't show up explictly. (Maybe 'invisibly' if derivative of trig. functions are mentioned.) Then, at univ. at the very start you take (real) analysis: constr. of the reals, basic top. notions(!), continuity,...,series of functions as application powerseries, and as appl exp and trig. func. – user9072 Mar 11 '11 at 17:47
@Laurent Moret-Bailly: The definition of $\frac{\sin x}x$ in degrees is the number you get when you type it into your calculator while forgetting to push deg/rad/grad first. – user11235 Apr 10 '11 at 20:41

By definition, an asymptote is a line that a curve keeps getting closer to but never touches. The teaching of this false belief at an elementary level is standard and nearly universal. Everybody "knows" that it is true. A tee-shirt has a clever joke about it. In the course of describing the function $f(x) = \dfrac{5x}{36 + x^2}$, I mentioned about an hour ago before a class of about 10 students that its value at 0 is 0 and that it has a horizontal asymptote at 0. One of them accused me of contradicting myself. What of $y = \dfrac{\sin x}{x}$? And even with simple rational functions there are exceptions, although there the curve can touch or cross the asymptote only finitely many times. And $3 - \dfrac{1}{x}$ gets closer to 5 as $x$ grows, and never reaches 5, so by the widespread false belief there would be a horizontal asymptote at 5.

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+1: I actually had this exact misconception several years ago!! I cannot remember the particular example, but I know I went to my teacher and told him the function could not have an asymptote since it crossed the line. – Eric Naslund Mar 6 '11 at 6:58
For this to be a false definition, it would have to be a definition in the first place. And this means you have to define a "curve" first, and then define "get closer" and "touch". – Laurent Moret-Bailly Mar 6 '11 at 16:01
@Laurent: It's hard to imagine a comment more irrelevant to what happens in classrooms than yours. – Michael Hardy Mar 7 '11 at 4:38
It happens to be the literal meaning of the word asymptote "not together falling". You could say that it is a bad choice of name, but for hyperbolas it worked just fine and then it was mercilessly generalized. – user11235 Apr 8 '11 at 14:35

Regard a reasonably nice surface in $\mathbb R^3$ that can locally be expressed by each of the functions $x(y,z)$, $y(x,z)$ and $z(x,y)$, then obviously

$\frac {dy} {dx} \cdot \frac {dz} {dy} \cdot \frac {dx} {dz} = 1$

(provided everything exists and is evaluated at the same point).

After all, this kind of reasoning works in $\mathbb R^2$ when calculating the derivative of the inverse function, it works for the chain rule and it works for separation of variables.

Note that this product is in fact $-1$ which can either be seen by just thinking about what happens to the equation $ax+by+cz=d$ of a plane / tangent plane or by looking at the expression coming out of the implicit function theorem.

I recall someone claiming that this example proves that $dx$ should be regarded as linear function rather than infinitesimal, but I cannot reconstruct the argument at the moment as this discussion was 15 years ago.

In particular, it is true under appropriate conditions in $\mathbb R^4$ that $\frac {\partial y} {\partial x} \cdot \frac {\partial z} {\partial y} \cdot \frac {\partial w} {\partial z} \cdot \frac {\partial x} {\partial w} = 1$

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This is an example of the principle that naïve reasoning with Leibniz notation works fine for total derivatives but not for partial derivatives. This is one reason why I would always write the left-hand side as $\frac{\partial{y}}{\partial{x}} \cdot \frac{\partial{z}}{\partial{y}} \cdot \frac{\partial{x}}{\partial{z}}$ if not $\left(\frac{\partial{y}}{\partial{x}}\right)_z \cdot \left(\frac{\partial{z}}{\partial{y}}\right)_x \cdot \left(\frac{\partial{x}}{\partial{z}}\right)_y$ (notation that I learnt from statistical physics, where the independent variables are otherwise not clear). – Toby Bartels Apr 7 '11 at 12:56
But this notation does not help one to understand that the above expression is actually $-1$. – user11235 Apr 10 '11 at 11:05
Can you help us understand it? Or is there no better way than computation? – darij grinberg Apr 10 '11 at 18:27
@TobyBartels, I remember an analyst colleague talking about a problem in a paper of his that he resolved by noticing (if I remember the particular example correctly) that $\partial/\partial r$ means something different in cylindrical and spherical co\"ordinates. – L Spice Dec 12 '13 at 23:17

Just today I came across a mathematician who was under the impression that $\aleph_1$ is defined to be $2^{\aleph_0}$, and therefore that the continuum hypothesis says there is no cardinal between $\aleph_0$ and $\aleph_1$.

In fact, Cantor proved there are no cardinals between $\aleph_0$ and $\aleph_1$. The continuum hypothesis says there are no cardinals between $\aleph_0$ and $2^{\aleph_0}$.

$2^{\aleph_0}$ is the cardinality of the set of all functions from a set of size $\aleph_0$ into a set of size $2$. Equivalently, it is the cardinality of the set of all subsets of a set of size $\aleph_0$, and that is also the cardinality of the set of all real numbers.

$\aleph_1$, on the other hand, is the cardinality of the set of all countable ordinals. (And $\aleph_2$ is the cardinality of the set of all ordinals of cardinality $\le \aleph_1$, and so on, and $\aleph_\omega$ is the next cardinal of well-ordered sets after all $\aleph_n$ for $n$ a finite ordinal, and $\aleph_{\omega+1}$ is the cardinality of the set of all ordinals of cardinality $\le \aleph_\omega$, etc. These definitions go back to Cantor.

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Do you think this is common? – user9072 Oct 5 '11 at 16:40
I retract my above question to my suprise it indeed seems to be common. Yet, this answer is a dublicate see an answer of April 16. – user9072 Oct 6 '11 at 0:50
This example already appears on this very page. mathoverflow.net/questions/23478/… – Asaf Karagila Oct 6 '11 at 12:41
One of the deficiencies of mathoverflow's software is that there is no easy way to search through the answers already posted. Even knowing that the date was April 16th doesn't help. – Michael Hardy Oct 7 '11 at 20:26
@Michael Hardy: You can sort the answers by date by clicking on the "Newest" or "Oldest" tabs instead of the "Votes" tab. – Douglas Zare Oct 19 '11 at 23:03

Here's a little factoid: (The Mean-value theorem for functions taking values in $\mathbb{R} ^n$.) If $\alpha : [a,b]\rightarrow \mathbb{R}^n$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then there exists a $c\in (a,b)$ such that $\frac{\alpha (b)-\alpha (a)}{b-a}=\alpha '(c)$

A counterexample is the helix $(\cos (t),\sin (t), t)$ with $a=0$, $b=2\pi$.

Another common misunderstanding (although not mathematical) is about the meaning of the word factoid. In fact, the common mistaken definition of the word factoid is factoidal.

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On the other hand, perhaps the most useful corollary of the mean value theorem is the "mean value inequality": that $|\alpha(b) - \alpha(a)| \le (b-a) \sup_{t \in [a,b]} |\alpha'(t)|$. If you look carefully, most applications of the MVT in calculus are really using this "MVI". The MVI remains true for absolutely continuous functions taking values in any Banach space, and so is probably the right generalization to keep in mind. – Nate Eldredge May 6 '10 at 14:37
According to at least one dictionary, there are two different definitions of factoid: (1) an insignificant or trivial fact, and (2) something fictitious or unsubstantiated that is presented as fact, devised especially to gain publicity and accepted because of constant repetition. I am not convinced that the multi-d mean value “theorem” fits either definition. – Harald Hanche-Olsen May 8 '10 at 19:09
Related to the M-V Thm is the following fact. If $f:I=(a,b)\rightarrow{\mathbb R}$ is differentiable (not necessarily ${\mathcal C}^1$), then $f'(I)$ is connected (i.e. is an interval). This is false when $f:I=(a,b)\rightarrow{\mathbb R}^n$ is differentiable, and $n\ge2$. – Denis Serre Oct 20 '10 at 10:48

In the past I have found myself making this mistake (probably fueled by the fact that you can indeed extend bounded linear operators), and I think it is common in students with a not-deep-enough topology background:

"Let $T$ be a compact topological space, and $X\subset T$ a dense subset. Take $f:X\to\mathbb{C}$ continuous and bounded. Then $f$ can be extended by continuity to all of $T$ ".

The classical counterexample is $T=[0,1]$, $X=(0,1]$, $f(t)=\sin\frac1t$ . It helps to understand how unimaginable the Stone-Cech compactification is.

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Indeed; the key property is uniform continuity. – Nate Eldredge Oct 14 '10 at 14:37
How about this one: $T=[-1,1]$, $X=T-\{0\}$, $f(x)=$ sign of $x$. – Laurent Moret-Bailly Oct 19 '10 at 7:23
Nice! That's certainly a much simpler example. – Martin Argerami Oct 19 '10 at 10:45

The product of two symmetric matrices is symmetric!

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Surprising! I explored a little and discovered that if the matrices are instead radially symmetric then the product is as well. That is, a_ij = a_(n - i)(n - j). – Whosyourjay Mar 6 at 7:29

The ring $\mathbb{C}[x]$ has countable dimension over $\mathbb{C}$; therefore its field of fractions $\mathbb{C}(x)$ also has countable dimension over $\mathbb{C}$.

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is this not true? – Turbo Nov 29 '14 at 22:27
The uncountably many elements $1/(x-a)$ for all $a \in \mathbb C$ are linearly independent. – darij grinberg Nov 29 '14 at 22:47
By the way, this fact is the basis for a beautiful proof of Hilbert's Nullstellensatz. – ACL Apr 21 at 6:40
@ACL I learned this not very long ago myself, from here: mathoverflow.net/a/15232/2926 – Todd Trimble Apr 21 at 7:45
• The category of commutative C*-algebras is equivalent to the opposite category of locally compact Hausdorff spaces.

It's actually not quite that simple! There is some discussion on math.SE.

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Very interesting point. – Ali Taghavi Nov 12 '14 at 12:42
Whoops! Thanks for posting this and particularly the link, which shows that the nLab got this one wrong as well. – Todd Trimble Nov 12 '14 at 13:03

In his answer above, Martin Brandenburg cited the false belief that every short exact sequence of the form

$$0\rightarrow A\rightarrow A\oplus B\rightarrow B\rightarrow 0$$

must split.

I expect that a far more widespread false belief is that such a sequence can fail to split, when A, B and C are finitely generated modules over a commutative noetherian ring.

(Sketch of relevant proof: We need to show that the identity map in $Hom(A,A)$ lifts to $Hom(A\oplus B,A)$. Thus we need to show exactness on the right of the sequence

$$0\rightarrow Hom(B,A)\rightarrow Hom(A\oplus B,A)\rightarrow Hom(A,A)\rightarrow 0$$

For this, it suffices to localize and then complete at an arbitrary prime $P$. But completion at $P$ is a limit of tensorings with $R/P^n$, so to check exactness we can replace the right-hand $A$ in each Hom-group with $A/P^nA$. Now we are reduced to looking at modules of finite length, and the sequence is forced to be exact because the lengths of the left and right terms add up to the length in the middle. This is due, I think, to Miyata.)

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Very interesting! – Martin Brandenburg Apr 12 '11 at 8:48
• Many students have the false belief that if a topological space is totally disconnected, then it must be discrete (related to examples already given). The rationals are a simple counter-example of course.

• It is common to imagine rotation in an n-dimensional space, as a rotation through an "axis". this is of course true only in 3D, In higher dimensions there is no "axis".

• In calculus, I had some troubles with the following wrong idea. A curve in a plane parametrized by a smooth function is "smooth" in the intuitive sense (having no corners). the curve that is defined by $(t^2,t^2)$ for $t\ge0$ and $(-t^2,t^2)$ for $t<0$ is the graph of the absolute value function with a "corner" at the origin, though the coordinate functions are smooth. the "non-regularity" of the parametrization resolves the conflict.

• When first encountering the concept of a spectrum of a ring, the belief that a continuous function between the spectra of two rings must come from a ring homomorphism between the rings.

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What do you mean by "In higher dimensions there is no "axis" but a n-2 dimensional subspace instead" ? Whenever n is even, there are rotations without real eigenvectors. – Johannes Hahn Apr 14 '11 at 13:12
Unfortunately, "smooth" is a word which means whatever its utterer does not want to specify. Differentiable, C^infty, continuous, everything is mixed. – darij grinberg Apr 14 '11 at 15:12
@Zsban: thanks for noticing. I edited the post. now I believe it is correct (though less impressive perhaps...). – KotelKanim May 3 '11 at 15:25
+1 for the discrete $\neq$ totally disconnected example. – Jim Conant May 4 '11 at 15:12
Discrete $\ne$ totally disconnected is a good one that I thought of today and just had to check to see if it was posted already. It adds to the confusion that every finite subset of a totally disconnected space must have the discrete topology, and that in most topological spaces encountered "in nature," the connected components are open sets. – Timothy Chow Oct 20 '11 at 14:30

It always confused me as an undergraduate that $\mathbb{Q}\subset\mathbb{R}$ has an open neighborhood $N_\epsilon \supset\mathbb{Q}$ of arbitrarily small measure $\epsilon$, because $\mathbb{Q}$ is dense in $\mathbb{R}$.

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In geometric combinatorics, there is a widespread belief that polytopes of equal volume are not scissor congruent (as in Hilbert's third problem) only because their dihedral angles are incomparable. The standard example is a cube and a regular tetrahedron, where dihedral angles are in $\Bbb Q\cdot \pi$ for the cube, and $\notin \Bbb Q\cdot \pi\$ for the regular tetrahedron. In fact, things are rather more complicated, and having similar dihedral angles doesn't always help. For example, the regular tetrahedron is never scissor congruent to a union of several smaller regular tetrahedra (even though the dihedral angles are obviously identical). This is a very special case of a general result due to Sydler (1944).

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Oh, I previously thought that eight regular tetrahedra with side length 1 fit in one with side length 2. That might be a more common false belief. – Junyan Xu May 5 '12 at 7:45

A stunning, ignorance-based false belief I have witnessed while observing a class of a math education colleague is that there is no general formula for the n-th Fibonacci number. I wonder if this false belief comes from conflating the (difficult) lack of formulas for prime numbers with something that is just over the horizon of someone whose interests never stretch beyond high-school math.

Behind a number of the elementary false beliefs listed here there is a widespread tendency among people to give up too easily (maybe when having to read at least to page 2 in a book), or to nourish an ego that allows to conclude that something is impossible if they cannot do it themselves.

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I hope at least your colleague had it right! There is another one along these lines: there is no formula for the sequence $1,0,1,0,1,0,... .$ Your second paragraph is right on target, but I also think that the specific beliefs you and I mentioned have a lot to do with a very limited understanding of what is a "formula". – Victor Protsak Jun 10 '10 at 7:02
Perhaps the people who believe this are using the meta-reasoning that the sequence would not be interesting as an example of recursion if it could be solved exactly. Since it is a popular example of recursion, then... – Ryan Reich Oct 5 '11 at 17:01
When, as an undergrad, I couldn't solve a problem given to me by the advisor, and asserted that it's "unsolvable", the advisor replied that "solvability of a problem is a function of two arguments: the problem and the solver." – Michael Dec 3 '13 at 1:14
@Michael (The constant function is still a function. :P ) – Akiva Weinberger Sep 1 '15 at 0:03

The distinction between convergence and uniform convergence. It even got Cauchy in its time.

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

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