# 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.

• I have to say this is proving to be one of the more useful CW big-list questions on the site... Commented May 6, 2010 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. Commented May 22, 2010 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. Commented Sep 20, 2010 at 12:39
• It's a thought -- I might consider it. Commented Oct 4, 2010 at 20:13
• Meta created tea.mathoverflow.net/discussion/1165/…
– user9072
Commented Oct 8, 2011 at 14:27

I'm not sure how common this is, but it confused me for years. Let $f : \mathbb{C} \to \mathbb{C}$ be an analytic function and $\gamma$ a path in $\mathbb{C}$. In your first class in complex analysis, you define the integral $\int_{\gamma} f(z) dz$.

Now let $a(x,y) dx + b(x,y) dy$ be a $1$-form on $\mathbb{R}^2$ and let $\gamma$ be a path in $\mathbb{R}^2$. In your first class on differential geometry, you define the integral $\int_{\gamma} a(x,y) dx + b(x,y) dy$.

It took me at least three years after I had taken both classes to realize that these notations are consistent. Until then, I thought there was a "path integral in the sense of complex analysis", and I wasn't sure if it obeyed the same rules as the path integral from differential geometry. (By way of analogy, although I wasn't thinking this clearly, the integral $\int \sqrt{dx^2 + dy^2}$, which computes arc length, is NOT the integral of a $1$-form, and I thought complex integrals were something like this.)

For the record, I'll spell out the relation between these notions. Let $f(x+iy) = u(x,y) + i v(x,y)$. Then $$\int_{\gamma} f(z) dz = \int_{\gamma} \left( u(x,y) dx - v(x,y) dy \right) + i \int_{\gamma} \left( u(x,y) dy + v(x,y) dx \right)$$ The right hand side should be thought of as multiplying out $\int_{\gamma} (u(x,y) + i v(x,y)) (dx + i dy)$, a notion which can be made rigorous.

• I think it is customary to say "contour integral" for the complex analysis gadget, "line integral" for the multivariable calculus gadget, and "path integral" for a (not necessarily rigorously defined) integral over a space of fields. Commented Jan 13, 2011 at 2:47
• Complex analysis does have $\int \lvert d z \rvert$, which also computes the arclength. Commented Apr 8, 2019 at 5:30

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.

• maybe not one of the best answers here, but why the down votes? Commented Feb 23, 2011 at 15:08
• @downvoters: Kindly provide a reason for the down votes.
– user11000
Commented Feb 23, 2011 at 15:54
• To my knowledge, there is only one sine function, and it is a map from $\mathbb{R}$ to $\mathbb{R}$ (or from $\mathbb{C}$ to $\mathbb{C}$, if you insist). What does it mean for a real number to be "in degrees"? And if you redefine the sine function as a map from $A$ to $\mathbb{R}$ where $A$ is the space of "angles" (whatever this means), then $\frac{\sin\,(x)}{x}$ is meaningless. Commented Mar 6, 2011 at 16:13
• @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
Commented Mar 11, 2011 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. Commented Apr 10, 2011 at 20:41

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

• True for abelian groups, though.
– Mark
Commented Mar 3, 2011 at 21:40
• Also true for finite index subgroups, of finitely generated groups. Commented Mar 4, 2011 at 21:18
• I've already mentioned this in my answer mathoverflow.net/questions/23478/…, but nevermind. ;) Commented Apr 12, 2011 at 8:44
• I had this false belief until the 2nd year of grad school. Commented Dec 3, 2013 at 1:21

The product of two symmetric matrices is symmetric!

• 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). Commented Mar 6, 2016 at 7:29

The standard projection map in a first course in topology is open. How could it not be closed? I always forget the standard homework exercise in which people first try to use this non-fact.

• The standard counter-example is $\mathbb{R} \times \mathbb{R} \to \mathbb{R}$, with the closed hyperbola $xy = 1$ mapping to the open set $x \neq 0$. I remember making this mistake, but not what problem prompted me to do so. Commented May 5, 2010 at 12:44
• this is another of the standard algebraic geometry examples from the "red book", whereby one shows the affine open set f(x) ≠ 0 is isomorphic to the closed set yf(x) = 1, in one dimension higher ambient space. Commented Apr 14, 2011 at 18:45
• @roysmith, I remember how puzzled I was when first learning about algebraic groups to be forced to think of $\mathrm{GL}_n$ as a closed subvariety of an affine space, when it so manifestly wants to be open …. Commented Jan 5, 2018 at 21:07

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

• is this not true? Commented Nov 29, 2014 at 22:27
• The uncountably many elements $1/(x-a)$ for all $a \in \mathbb C$ are linearly independent. Commented Nov 29, 2014 at 22:47
• By the way, this fact is the basis for a beautiful proof of Hilbert's Nullstellensatz.
– ACL
Commented Apr 21, 2016 at 6:40
• @ACL I learned this not very long ago myself, from here: mathoverflow.net/a/15232/2926 Commented Apr 21, 2016 at 7:45
• Wow, good one. For by me now it is very viscerally known that "passage to the field of fractions does not change the size." This is true in the sense of cardinality, but.... Commented Sep 6, 2017 at 22:16

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 $$x$$.

• I've just added the missing html to fix the link. Commented Jul 3, 2011 at 21:38

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 $AB \neq BA \text{ and} \exp(A)=\exp(B) = \exp(A+B) = 1$.

• 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. Commented Sep 27, 2010 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 $$\left(\exp\int_0^tA(s)ds\right)x(0).$$ Commented Oct 20, 2010 at 10:31

As a teaching assistant in an elementary number theory course, I've seen the following quite often :

If $a$ divides $bc$ and $a$ does not divide $b$, then $a$ divides $c$.

That's of course true if $a$ is prime, but people seem to forget that hypothesis.

• the correct assumption is $gcd(a,b)=1$. Commented Oct 5, 2010 at 8:23
• i was never able to eradicate this mistake, even by my best students, in spite of naming it the prime divisibility property. Commented May 9, 2011 at 2:15
• @MartinBrandenburg : What does it mean for an assumption to be incorrect or correct? Commented Feb 7, 2014 at 18:40
• Easy counterexample: $6 = 2 \times 3$ but $6$ does not divide $2$ or $3$... Commented Dec 28, 2017 at 9:41

I'm not sure that anyone holds this as a conscious belief but I have seen a number of students, asked to check that a linear map $\mathbb{R}^k \to \mathbb{R}^{\ell}$ is injective, just check that each of the $k$ basis elements has nonzero image.

• Higher-level version: $n$ vectors are linearly independent iff no two are proportional. I've seen applied mathematicians do that. Commented Apr 10, 2011 at 18:45

In measure-theoretic probability, I think there is sometimes an idea among beginners that independent random variables $X,Y$ should be thought of as having "disjoint support" as measurable functions on the underlying probability space $\Omega$. Of course this is the opposite of the truth.

I think this may come from thinking of measure theory as generalizing freshman calculus, so that one's favorite measure space is something like $[0,1]$ with Lebesgue measure. This is technically a probability space, but a really inconvenient one for actually doing probability (where you want to have lots of random variables with some amount of independence).

• A student this last semester made precisely this mistake, and it was a labor of three people to convince him otherwise. Commented May 17, 2010 at 0:28
• This disjoint support misconception reinforces the incorrect idea that pairwise independent implies independent. Commented Oct 20, 2010 at 18:47

"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.

• 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.) Commented Jun 7, 2010 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. Commented Jul 7, 2010 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. Commented Jul 20, 2010 at 14:51
• @roy: if there are no primes, there are finitely many a fortiori. So there's nothing to check there. Commented Mar 18, 2013 at 0:04
• @MichaelHardy This "myth", exactly as stated, is itself a myth. It conflates two different statements: In your article you repeatedly use the phrase "Euclid's proof of the infinitude of primes" for a proof of the statement "given any finite set of primes, there exists a prime not in that set" (let's call this statement $F$). But the phrase "infinitude of primes" means, of course, that there are infinitely many primes (call this statement $I$). Admittedly $F$ and $I$ appear almost trivially similar, but to actually prove $I$ from $F$ you need to use a proof by contradiction. Commented Feb 6, 2020 at 14:30

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.

• 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
Commented Jun 12, 2010 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.) Commented Jun 12, 2010 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. Commented Jun 12, 2010 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
Commented Jun 12, 2010 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
Commented Jun 12, 2010 at 22:02

Yet another one:

Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be differentiable. If $f'(x_0) > 0$, then there exists an interval $I$ containing $x_0$ such that $f$ is increasing in $I$.

• I sort of find it hard to believe that amongst the nearly 200 answers on this thread (and just over 20 deleted ones), no one has posted this. Commented Aug 10, 2015 at 6:07
• A counter-example is necessarily with $f'$ discontinuous in $x_0$, right? For example $f(x)=x^2 sin (1/x)+x/2$ and $x_0 = 0$. Commented Aug 10, 2015 at 8:05
• @SébastienPalcoux Yes, I think if $f'$ is continuous in $x_0$ then the statement is true. Commented Aug 10, 2015 at 15:16
• Bc continuous and positive means positive in a whole interval. Then MVT kicks in. I was asked this by some friends who were my seniors and taking analysis 3 while I was in analysis 1. I did find the counter-example on my own. One little math discovery I was proud of :) Commented Sep 28, 2022 at 8:38

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 $$g(z)=\gamma z$$.

• I think I remember reading it is the only one that is convex over $(0,+\infty)$ or some similar condition. Commented Oct 18, 2015 at 12:45
• It is the only one that is log convex (ie, $\log(\Gamma(x))$ is a convex function). Commented Mar 14, 2017 at 19:07
• The magic words here are "Bohr-Mollerup theorem"... Commented Feb 9, 2021 at 9:59
• One can multiply $\Gamma(x)$ e.g. by $\exp(\sin(2\pi x))$ and get a solution of the same functional equation, I think everybody knows that. Commented Aug 8, 2023 at 6:07
• This false belief is perhaps caused by the fact that continuity does imply sequential continuity, and sequential adherent points are adherent points. Commented Sep 27, 2010 at 5:53

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 $$\left(\matrix{i& 0 \cr 0 & -i\cr } \right), \left(\matrix{0& i \cr i & 0\cr } \right).$$

• 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'). Commented Dec 12, 2013 at 23:09

"Some real numbers are not definable, by Cantor's diagonal argument."

There are subtleties involved in formalizing the statement "some real numbers are not definable", as explained in Joel's answer to this question. The statement can be seen to hold in some models and fail in other models. However, the claim that the statement follows from Cantor's diagonal argument is clearly false, yet seems to be fairly common.

The false reasoning typically proceeds in three steps:

1. There are only countably many definitions of real numbers: $$\varphi_0(x),\varphi_1(x),\ldots$$ (this part is ok.)

2. Consider the countably many real numbers so defined: $$x_0,x_1,\ldots$$ (this part is problematic for subtle reasons.)

3. Use Cantor's diagonal argument to obtain a real number $$y$$ that is not in the sequence from step 2, and is therefore not definable.

For the moment, let us assume that step 2 succeeds in the way that one might naively think it would. Then we have defined a sequence $$x_0,x_1,\ldots$$ containing all definable real numbers. Therefore Cantor's diagonal argument in step 3 defines, from this sequence, a real number $$y$$ that is not in the sequence. So $$y$$ is both definable and not definable, and we obtain a contradiction outright! Clearly, something is wrong (and it turns out to be in step 2.)

• Is it because $\{x_0,x_1,\dots\}$ isn't necessarily a set? Commented Sep 1, 2015 at 0:17
• @columbus8myhw It's related to that, although it's possible for $\{x_0, x_1, \ldots\}$ to be a set for unrelated reasons. (For example, given a pointwise definable model of set theory, we can consider $\{x_0, x_1, \ldots\}$ from the outside, and see that it happens to equal a set in the model, namely $\mathbb{R}$ itself.) But in this case $(x_0,x_1,\ldots)$ will fail to be a sequence (of the model) and the argument will still fail (in the model). Commented Sep 1, 2015 at 18:30
• That being said, understanding why $\{x_0,x_1,\ldots\}$ isn't necessarily a set is similar to understanding why the argument isn't valid. Commented Sep 1, 2015 at 18:31

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...

• Commented May 6, 2010 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. Commented May 6, 2010 at 11:43

That Darboux functions are continuous is certainly a widely held belief among students, at least in France where it is induced by the way continuity is taught in high school.

I remember having gone through all the five "stages of grief" when shaken from this false belief with the $$\sin(1/x)$$ example : denial, anger ( "then the definition of continuity must be wrong ! Let's change it !), bargaining ("Ok, but a Darboux function must surely be continuous except at exceptional points. Let's prove that..."), depression (when shown a nowhere continuous Darboux function), acceptance ("Hey guys, you really think the intermediate value theorem has a converse? C'mon, you're smarter than that...")

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.

• Indeed; the key property is uniform continuity. Commented Oct 14, 2010 at 14:37
• How about this one: $T=[-1,1]$, $X=T-\{0\}$, $f(x)=$ sign of $x$. Commented Oct 19, 2010 at 7:23
• Nice! That's certainly a much simpler example. Commented Oct 19, 2010 at 10:45

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.

• Commented Oct 3, 2011 at 13:12
• A counterexample to the polynomial thing is $\frac12x^2+\frac12x$. Commented Aug 31, 2015 at 22:27
• Worth noting: The correct condition for a polynomial to take integer values at integers is for it to be an integer combination of the polynomials $\binom{x}{k}$. Commented Jan 31, 2022 at 8:45

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.

• 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
Commented Oct 6, 2011 at 0:50
• This example already appears on this very page. mathoverflow.net/questions/23478/… Commented Oct 6, 2011 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. Commented Oct 7, 2011 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. Commented Oct 19, 2011 at 23:03
• I suspect much confusion stems originally from George Gamow's book "One Two Three ... Infinity" — which got a number of things wrong about the continuum and the continuum hypothesis, which it implied was settled. None of the reprintings of this book fixed the error — to this day. Commented May 29, 2017 at 16:22

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 $$\frac {\partial y} {\partial x} \cdot \frac {\partial z} {\partial y} \cdot \frac {\partial w} {\partial z} \cdot \frac {\partial x} {\partial w} = 1$$

• 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). Commented Apr 7, 2011 at 12:56
• But this notation does not help one to understand that the above expression is actually $-1$. Commented Apr 10, 2011 at 11:05
• Can you help us understand it? Or is there no better way than computation? Commented Apr 10, 2011 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. Commented Dec 12, 2013 at 23:17

In descriptive set theory, we study properties of Polish spaces, typically not considered as topological spaces but rather we equip them with their "Borel structure", i.e., the collection of their Borel sets. Any two uncountable standard Borel Polish spaces are isomorphic, and the isomorphism map can be taken to be Borel. In practice, this means that for most properties we study it is irrelevant what specific Polish space we use as underlying "ambient space", it may be ${\mathbb R}$, or ${\mathbb N}^{\mathbb N}$, or ${\mathcal l}^2$, etc, and we tend to think of all of them as "the reals".

In Lebesgue Sur les fonctions representables analytiquement, J. de math. pures et appl. (1905), Lebesgue makes the mistake of thinking that projections of Borel subsets of the plane ${\mathbb R}^2$ are Borel. In a sense, this mistake created descriptive set theory.

Now we know, for example, that in ${\mathbb N}^{\mathbb N}$, projections of closed sets need not be Borel. Since we usually call reals the members of ${\mathbb N}^{\mathbb N}$,

it is not uncommon to think that projections of closed subsets of ${\mathbb R}^2$ are not necessarily Borel.

This is false. Note that closed sets are countable union of compact sets, so their projections are $F_\sigma$. The actual results in ${\mathbb R}$ are as follows: Recall that the analytic sets are (the empty set and) the sets that are images of Borel subsets of $\mathbb R$ by Borel measurable functions $f:\mathbb R\to\mathbb R$.

• A set is Borel iff it and its complement are analytic.

• A set is analytic iff it is the projection of the complement of the projection of a closed subset of ${\mathbb R}^3$.

• A set is analytic iff it is the projection of a $G_\delta$ subset of $\mathbb R^2$.

• There is a continuous $g:\mathbb R\to\mathbb R$ such that a set is analytic iff it is $g(A)$ for some $G_\delta$ set $A$.

• A set if analytic iff it is $f(\mathbb R\setminus\mathbb Q)$ for some continuous $f:\mathbb R\setminus\mathbb Q\to\mathbb R$. (Note that if $f$ is actually continuous on $\mathbb R$, then $f(\mathbb R\setminus\mathbb Q)$ is Borel.)

The quaternions $$\{x+yi+zj+wk\mid x,y,z,w\in \mathbb{R}$$} is a complex Banach algebra (with usual operations). Hence it is apparently a counterexample to the Gelfand-Mazur theorem

So, what is the error?

The error is the following:

However the quaternions, being a skew field extention of the field of complex numbers, is a vector space over the field of complex number and it is also a ring, but there is no compatibility between scalar multiplication and quaternion multiplication). So it is not a complex algebra. This shows that in the definition of a complex algebra $$A$$, the commutative condition $$\lambda (ab)=(a)(\lambda b),\;\;\lambda \in \mathbb{C},\;\;a,b\in A$$, is very essential.

• This is not a common false belief except among people who do not understand the definition of an algebra over a field Commented Nov 12, 2014 at 23:43
• Moreover, surely the quaternions are a real vector space, not a complex vector space Commented Nov 13, 2014 at 1:34
• @YemonChoi the field of complex number is a subring of the ring of quaternions. So quaternions is a complex vec. space.More generaly if a ring R contains a field F then R is a F-vector space,Ok? Commented Nov 13, 2014 at 5:31
• @YemonChoi I think this example is perhaps interesting unless a participant do not read it carefully. please think again to the main motivation and aim of the question of "common false..." Commented Nov 13, 2014 at 5:44
• @AliTaghavi You're right that $R$-multiplication induces an $F$-module ($F$-vector space) structure via the evident composite $F \times R \to R \times R \to R$. To be fair to both you and Yemon: a very common slip even among professionals is in knowing that for commutative algebras an $F$-algebra is tantamount to a homomorphism $F \to R$, but temporarily forgetting this doesn't apply in the noncommutative setting (except of course when $F$ is central in $R$) -- not rising to the level of false belief so much as a temporary slip-up. I've made that slip myself! Commented Nov 13, 2014 at 11:58

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.)

• 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. Commented Jun 6, 2010 at 21:13
• Sadly, people rely on technology so much nowadays that it gets increasingly unlikely that it will $\textit{ever}$ be proved. Commented Jun 10, 2010 at 6:40
• "Reduced" as in reduced row echelon form? Commented Feb 27, 2023 at 3:03
• "Reduced" as in "reduced scheme", i.e., every polynomial function that vanishes on the set of commuting pairs is a combination of the entries of AB-BA. Commented Feb 28, 2023 at 4:28

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.)

• Very interesting! Commented Apr 12, 2011 at 8:48

If $E$ is a contractible space on which the (Edit: topological) group $G$ acts freely, then $E/G$ is a classifying space for $G$.

A better, but still false, version:

If $E$ is a free, contractible $G$-space and the quotient map $E\to E/G$ admits local slices, then $E/G$ is a classifying space for $G$.

(Here "admits local slices" means that there's a covering of $E/G$ by open sets $U_i$ such that there exist continuous sections $U_i \to E$ of the quotient map.)

The simplest counterexample is: let $G^i$ denote $G$ with the indiscrete topology (Edit: and assume $G$ itself is not indiscrete). Then G acts on $G^i$ by translation and $G^i$ is contractible (for the same reason: any map into an indiscrete space is continuous). Since $G^i/G$ is a point, there's a (global) section, but it cannot be a classifying space for $G$ (unless $G=\{1\}$). The way to correct things is to require that the translation map $E\times_{E/G} E \to G$, sending a pair $(e_1, e_2)$ to the unique $g\in G$ satisfying $ge_1 = e_2$, is actually continuous.

Of course the heart of the matter here is the corresponding false belief(s) regarding when the quotient map by a group action is a principal bundle.

• I'm a little confused. How does requiring that $(e_1, e_2) \mapsto g$ be continuous fix things? In the indiscrete case, this map is continuous (since every map to the group is). And why isn't $G^i \to G^i/G$ a principal $G^i$--bundle? Commented Mar 6, 2011 at 17:52
• The group in this example starts out with some topology. (I guess I didn't specify that I was thinking of a topological group.) If G started with the indiscrete topology, then your commment makes sense, and we would have a principal bundle for this indiscrete group. But if G is not indiscrete, then the map $(e_1, e_2) \mapsto g$ is not continuous as a map into the topological group G. The proof that continuity of the translation map forces this to be a principal bundle can be found in Husemoller's book on fiber bundles (it's not hard). Let me know if this didn't answer your questions. Commented Mar 6, 2011 at 19:57
• Oh! You're saying that a point is not a classifying space for G with some other topology. I thought you were saying that $G^i/G$ wasn't $BG^i$. Thanks for the clarification! Commented Mar 6, 2011 at 20:01
• Yes, precisely. It's an odd little example, but helpful when people forget to include the proper conditions... Commented Mar 6, 2011 at 21:06
• Maybe even more amazing wrong belief in this field: $\dim(E/G)\le\dim E$ (there are counterexamples by A.N. Kolmogorov) Commented Jun 9, 2011 at 14:52

A common false belief is that all Gödel sentences are true because they say of themselves they are unprovable. See Peter Milne's "On Goedel Sentences and What They Say", Philosophia Mathematica (III) 15 (2007), 193–226. doi:10.1093/philmat/nkm015.