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... $\endgroup$ May 6, 2010 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, 2010 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, 2010 at 12:39
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    $\begingroup$ It's a thought -- I might consider it. $\endgroup$
    – gowers
    Oct 4, 2010 at 20:13
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    $\begingroup$ Meta created tea.mathoverflow.net/discussion/1165/… $\endgroup$
    – user9072
    Oct 8, 2011 at 14:27

292 Answers 292

2 3 4 5

For vector spaces, $\dim (U + V) = \dim U + \dim V - \dim (U \cap V)$, so $$ \dim(U +V + W) = \dim U + \dim V + \dim W - \dim (U \cap V) - \dim (U \cap W) - \dim (V \cap W) + \dim(U \cap V \cap W), $$ right?

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    $\begingroup$ Wait, that isn't true? $\endgroup$
    – Simon Rose
    May 4, 2010 at 23:19
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    $\begingroup$ Take three distinct lines in R^2 as U, V, W. All intersections have 0 dimensions. The LHS is 2, the RHS is 3. The problem is that $(U+V)\cap W \neq U\cap W + V\cap W$. $\endgroup$ May 4, 2010 at 23:38
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    $\begingroup$ Take 3 lines in $\mathbb{R}^2$... $\endgroup$
    – Tom Smith
    May 4, 2010 at 23:38
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    $\begingroup$ This is perhaps a shameful comment for math overflow, but: ROFL (in the best possible sense) :-) excellent answer! $\endgroup$ May 5, 2010 at 0:26
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    $\begingroup$ @Tilman: Only a remark not related to the topic: The identity $$\dim (U + V) = \dim U + \dim V - \dim (U \cap V)$$ is valid only for finite dimensional spaces, but if one writes it as follows $$\dim (U + V) + \dim (U \cap V)= \dim U + \dim V$$ it is valid for all vector spaces. $\endgroup$ Mar 25, 2016 at 7:23

Everyone knows that for any two square matrices $A$ and $B$ (with coefficients in a commutative ring) that $$\operatorname{tr}(AB) = \operatorname{tr}(BA).$$

I once thought that this implied (via induction) that the trace of a product of any finite number of matrices was independent of the order they are multiplied.

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    $\begingroup$ Indeed. I never thought much about this before, but clearly this only implies the trace of a product is invariant under cyclic permutations. I bet there is some fact from the representation theory of the symmetric group lurking here, but am too lazy to think about it... $\endgroup$ May 4, 2010 at 21:16
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    $\begingroup$ In fact Tr$(AB)=$Tr$(BA)$ holds also for non-square matrices $A,B$ for which both $AB$ and $BA$ are defined. Now for determinants, det$(AB)$=det$(BA)$ holds for square matrices, but of course not for non-square matrices (consider the case where $A$ is a column vector and $B$ a row vector). $\endgroup$
    – user2734
    May 4, 2010 at 21:46
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    $\begingroup$ @Nate: If you want high-powered generalities, the most general situation I know where one can prove this statement is in a ribbon category. These have a graphical calculus where tr(ABC...) corresponds to a closed loop on which A, B, C... sit as labels in order, which clearly shows that the only invariance one should expect is under cyclic permutation. See, for example, the beginning of Turaev's "Quantum Invariants of Knots and 3-Manifolds." $\endgroup$ May 4, 2010 at 22:32
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    $\begingroup$ @unknown: nonetheless, the characteristic polynomials of AB and BA are the same up to a power of $\lambda$ (A is m by n and B is n by m), which generalizes both properties $\endgroup$ May 5, 2010 at 6:54
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    $\begingroup$ @Victor Protsak: Nice! BTW, one way to get what you say is from det$(I_m+AB)=$det$(I_n+BA)$, which funnily doesn't hold for the trace in case of non-square matrices (there is a difference of $m-n$). $\endgroup$
    – user2734
    May 5, 2010 at 7:44

The closure of the open ball of radius $r$ in a metric space, is the closed ball of radius $r$ in that metric space.

In a somewhat related spirit: the boundary of a subset of (say) Euclidean space has empty interior, and furthermore has Lebesgue measure zero. (This false belief is closely related to Gowers' example of the belief that there are no non-trivial open dense sets.)

More generally, point set topology and measure theory abound with all sorts of false beliefs that only tend to be expunged once one plays with the canonical counterexamples (Cantor sets, bullet-riddled squares, space-filling curves, the long line, $\sin\left(\dfrac{1}{x}\right)$ and its variants, etc.).

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    $\begingroup$ I remember being assigned as an exercise to find a counterexample to the first statement, but I can't remember where. Rudin? $\endgroup$ Jun 6, 2010 at 23:39
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    $\begingroup$ What about a space with 2 points a distance 1 apart, and the open/closed ball having radius 1? I don't remember seeing this before, though. $\endgroup$ Jun 6, 2010 at 23:53
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    $\begingroup$ @Terry Really good examples. We can count on you to do anything but waste our time with a post, Terry. I hope you keep finding the time to post here and lend your support! $\endgroup$ Jun 7, 2010 at 0:11
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    $\begingroup$ Peter: actually the simplest counterexample is the open/closed ball of radius $0$, empty set vs a singleton. $\endgroup$ Aug 1, 2011 at 15:43
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    $\begingroup$ In response to Qiaochu's comment, I'm surprised nobody ever mentioned that a canonical counterexample to the first claim is given by the $p$-adics: there every ball is clopen, and if the "closed radius" of the ball is $p^{-n}$, the "open radius" is $p^{-n+1}$. This is because the image of the distance function is discrete (except at distance 0). $\endgroup$
    – Tim Campion
    Sep 6, 2015 at 3:02

Many students believe that 1 plus the product of the first $n$ primes is always a prime number. They have misunderstood the contradiction in Euclid's proof that there are infinitely many primes. (By the way, $2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13 + 1$ is not prime and there are many other such examples.)

Much later edit: As pointed out elsewhere in this thread, Euclid's proof is not by contradiction; that is another widespread false belief.

Much much later edit: Euclid's proof is not not by contradiction. This is another very widespread false belief. It depends on personal opinion and interpretation what a proof by contradiction is and whether Euclid's proof belongs to this category. In fact, if the derivation of an absurdity or the contradiction of an assumption is a proof by contradiction, then Euclid's proof is a proof by contradiction. Euclid says (Elements Book 9 Proposition 20): The very thing (is) absurd. Thus, G is not the same as one of A, B, C. And it was assumed (to be) prime.

Nb. The above edits were not added by the OP of this answer.

Edit on 24 July 2017: Euclid's proof was not by contradiction, but contains a small lemma in the middle of it that is proved by contradiction. The proof shows that if $S$ is any finite set of primes (not assumed to be the set of all primes) then the prime factors of $1+\prod S$ are not in $S$, so there is at least one more prime than those in $S.$ The proof that $\prod$ and $1+\prod$ have no common factors is the part that is by contradiction. All of this is shown in the following paper: M. Hardy and C. Woodgold, "Prime simplicity", Mathematical Intelligencer 31 (2009), 44–52.

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    $\begingroup$ When I was 11 y.o. I was screamed at by a teacher and thrown out of class for pointing this out when he claimed the false belief stated (it wasn't class material, but the teacher wanted to show he was smart). I found the counterexample later at home. I didn't let the matter drop either... I knew I was right and he was wrong, and really had a major fallout with that math teacher and the school; and flunked math that year. $\endgroup$ May 5, 2010 at 1:19
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    $\begingroup$ @Daniel: Sorry to hear that. When my daughter Meena was the same age (11), her teacher asserted that 0.999... was not equal to 1. Meena supplied one or two proofs that they were equal, but her teacher would not budge. Maybe this is another example of a common false belief. $\endgroup$ May 5, 2010 at 2:59
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    $\begingroup$ @Daniel: I've heard a worse story. A college instructor claimed in Number Theory class that there are only finitely many primes. When confronted by a student, her reply was: "If you think there are infinitely many, write them all down". She was on tenure track, but need I add, didn't get tenure. $\endgroup$ May 5, 2010 at 5:38
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    $\begingroup$ This false belief leads to a proof of the Twin Prime conjecture: For every $n$, $(p_1 p_2 \cdots p_n -1, p_1 p_2 \cdots p_n +1)$ are twin primes, right? $\endgroup$ May 6, 2010 at 15:50
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    $\begingroup$ Daniel, about the same age, I was asked to leave class for claiming that pi is not 22/7. The math teacher said that 3.14 is an approximation and while some people falsly believe that pi=3.14 but the true answer is 22/7. Years later an Israeli newspaper published a story about a person who can memorize the first 2000 digits of pi and the article contained the first 200 digits. A week later the newspaper published a correction: "Some of our readers pointed out that pi=22/7". Then the "corrected" (periodic) 200 digits were included. Memorizing digits of pi is a whole different matter if pi=22/7. $\endgroup$
    – Gil Kalai
    May 11, 2010 at 5:45

Here's my list of false beliefs 😉

  • If $U$ is a subspace of a Banach space $V$, then $U$ is a direct summand of $V$.
  • If $M/L$ and $L/K$ are normal field extensions, then the same is true for $M/K$.
  • Submodules / subgroups / subalgebras of finitely generated modules / groups / algebras are finitely generated.
  • For a subring $S \subseteq R$ of a commutative ring the Krull dimension satisfies $\dim(S) \leq \dim(R)$.
  • The Krull dimension of a noetherian integral domain is finite.
  • If $A \otimes B = 0$ for abelian groups $A,B$, then either $A=0$ or $B=0$.
  • If $f$ is a smooth function with $df=0$, then $f$ is constant.
  • If $X,Y$ are sets such that $P(X), P(Y)$ are equipotent, then $X,Y$ are equipotent.
  • Every short exact sequence of the form $0 \to A \xrightarrow{f} A \oplus B \xrightarrow{g} B \to 0$ splits.
  • $R[x]^{\times} = R^{\times}$ for any commutative ring $R$.
  • Every presheaf on a site has an associated sheaf.
  • (Co)limits may be computed in full subcategories. For example, $\mathrm{Spec}(\prod_i R_i) = \coprod_i \mathrm{Spec}(R_i)$ as schemes because $\mathrm{Spec}$ is an anti-equivalence between commutative rings and affine schemes.
  • Every finite CW-complex is compact, thus every CW-complex is locally compact.
  • The smash product of pointed spaces is associative, products of topological spaces commute with quotients, and so on.
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    $\begingroup$ +1: you had me at "Here's my list of false beliefs". $\endgroup$ May 5, 2010 at 23:41
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    $\begingroup$ $A \to A \oplus B$ does not have to be the inclusion; likewise $A \oplus B$ does not have to be the projection. Thus the error here is: Two chain complexes, which are isomorphic "pointwise", don't have to be isomorphic. This occurs sometimes. $\endgroup$ May 6, 2010 at 16:12
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    $\begingroup$ Your fifth example reminds me of an even more plausible false belief I once held: if $A \otimes A = 0$, then $A = 0$. $\endgroup$ May 11, 2010 at 2:12
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    $\begingroup$ @Regenbogen: Take the abelian group $\mathbb{Q}/\mathbb{Z}$. $\endgroup$
    – Steve D
    May 15, 2010 at 13:44
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    $\begingroup$ $f$ is just locally constant ;-) $\endgroup$ May 19, 2010 at 8:39

I don't know if this is common or not, but I spent a very long time believing that a group $G$ with a normal subgroup $N$ is always a semidirect product of $N$ and $G/N$. I don't think I was ever shown an example in a class where this isn't true.

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    $\begingroup$ Argh! Me too! What is a good example? $\endgroup$ May 4, 2010 at 21:28
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    $\begingroup$ umm Z/4Z contains Z/2Z? $\endgroup$ May 4, 2010 at 21:30
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    $\begingroup$ This suggests that we do a terrible job of talking about semi-direct products no? $\endgroup$ May 5, 2010 at 0:27
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    $\begingroup$ Schur--Zassenhaus says that this is true if $N$ and $G/N$ have coprime orders, so there is some intrinsic pressure in the subject towards this. Coupled with the fact that it is true for the first non-trivial non-abelian example ($A_3$ inside $S_3$), it's easy to see how this misconception arises. $\endgroup$
    – Emerton
    May 5, 2010 at 1:49
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    $\begingroup$ Remember being confused by this too. It became much clearer when I formally was taught about short exact sequences. Then you can see exactly the obstruction to such a decomposition. $\endgroup$ May 10, 2010 at 11:14

These are actually metamathematical (false) beliefs that many intelligent people have while they are learning mathematics, but usually abandon when their mistake is pointed out, and I am almost certain to draw fire for saying it from those who haven't, together with the reasons for them:

The results must be stated in complete and utter generality.

Easy examples are left as an exercise to the reader.

It is more important to be correct than to be understood.

(Applicable to talks as well as papers.)

Reasons: 1. Von Neumann is in the audience. 2. This is just a generalization of Lemma 1.2.3 in volume X of Bourbaki. 3. The results are impressive and speak for themselves.

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    $\begingroup$ IMO "It is more important to be correct than to be understood" is not a false belief. $\endgroup$
    – Michael
    Oct 16, 2014 at 18:32
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    $\begingroup$ I definitely agree with the OP that "It is more important to be correct than to be understood" is false - in the context of giving mathematical talks. Or perhaps, it's fairer to say that being understood is more important than being 100% correct. Talks are about the listener, not about the speaker. $\endgroup$ Mar 5, 2015 at 20:15
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    $\begingroup$ @GregMartin: When you are giving a talk, sure. When you are giving a lecture, maybe, but you should give an indication of where you are imprecise. When you are writing a paper, most definitely not. $\endgroup$
    – tomasz
    Sep 2, 2015 at 3:00
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    $\begingroup$ @Michael: Victor and you are both right: you should be correct $\mathbf{and}$ understandable. So that the audience can understand, that you are correct. $\endgroup$
    – co.sine
    May 29, 2017 at 5:17
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    $\begingroup$ One of the most important answers here. Every mathematician should think about this answer. $\endgroup$
    – shuhalo
    Nov 7, 2020 at 17:00

a student, this afternoon: "this set is open, hence it is not closed: this is why [...]"

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    $\begingroup$ The terminology is rather unfortunate. $\endgroup$ May 5, 2010 at 14:39
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    $\begingroup$ Either that or topologists need a sit-down about the facts of life in life, where they are told how unfortunate their notation is... $\endgroup$ May 5, 2010 at 20:25
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    $\begingroup$ Munkres is fond of saying "sets are not doors." $\endgroup$ May 5, 2010 at 20:53
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    $\begingroup$ On my office door I once put "clopen the door" $\endgroup$
    – hypercube
    May 26, 2010 at 22:52
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    $\begingroup$ Actually, topologists have studied spaces where every set is open or closed (or both, of course), and they're called "Door spaces".... $\endgroup$ Jun 6, 2010 at 11:46

Some false beliefs in linear algebra:

  • If two operators or matrices $A$, $B$ commute, then they are simultaneously diagonalisable. (Of course, this overlooks the obvious necessary condition that each of $A$, $B$ must first be individually diagonalisable. Part of the problem is that this is not an issue in the Hermitian case, which is usually the case one is most frequently exposed to.)

  • The operator norm of a matrix is the same as the magnitude of the most extreme eigenvalue. (Again, true in the Hermitian or normal case, but in the general case one has to either replace "operator norm" with "spectral radius", or else replace "eigenvalue" with "singular value".)

  • The singular values of a matrix are the absolute values of the eigenvalues of the matrix. (Closely related to the previous false belief.)

  • If a matrix has distinct eigenvalues, then one can find an orthonormal eigenbasis. (The orthonormality is only possible when the matrix is, well, normal.)

  • A matrix is diagonalisable if and only if it has distinct eigenvalues. (Only the "if" part is true. The identity matrix and zero matrix are blatant counterexamples, but this false belief is remarkably persistent nonetheless.)

  • If $\mathcal L: X \to Y$ is a bounded linear transformation that is surjective (i.e. $\mathcal Lu=f$ is always solvable for any data $f$ in $Y$), and $X$ and $Y$ are Banach spaces then it has a bounded linear right inverse. (This is subtle. Zorn's lemma gives a linear right inverse; the open mapping theorem gives a bounded right inverse. But getting a right inverse that is simultaneously bounded and linear is not always possible!)

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    $\begingroup$ Wow. I believed that second one until now. Which is ridiculous, of course, since the operator norm of a nilpotent matrix can't be zero or else it wouldn't be a norm! $\endgroup$ Jun 6, 2010 at 23:45
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    $\begingroup$ The parethentical comment in 2nd bulleted point is worded as if, $\textit{in general},$ the operator norm were equal both to the spectral radius and the largest singular value (or, perhaps, that $\|A\|=\rho(A)$ and $\lambda_1(A)=s_1(A).$) But for a nilpotent matrix the spectral radius is 0, whereas the operator norm and the largest singular values aren't. $\endgroup$ Jun 10, 2010 at 7:59
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    $\begingroup$ Fair enough; I've reworded the parenthetical. $\endgroup$
    – Terry Tao
    Jun 10, 2010 at 16:35
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    $\begingroup$ Yes, I meant right inverse, thanks. Getting a continuous right-inverse is actually a subtle question - the OMT only gets boundedness, which is not equivalent to continuity when one is not linear. I believe that the existence of a continuous right inverse may follow from a classical theorem of Bartle and Graves, but this is nontrivial. $\endgroup$
    – Terry Tao
    Jun 10, 2010 at 19:06
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    $\begingroup$ In the last clause, there is always a continuous (usually non-linear, of course) left inverse. This is the Bartle-Graves theorem. $\endgroup$ May 23, 2017 at 9:57

I once thought that if $A$, $B$, $C$, and $D$ were $n$-by-$n$ matrices, then the determinant of the block matrix $\pmatrix{A & B \\\ C & D}$ would be $\det(A) \det(D) - \det(B) \det(C)$.

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    $\begingroup$ This mistake's like assuming heavier objects fall faster then light ones. It's perfectly reasonable-just happens to be DEAD WRONG. $\endgroup$ Jun 7, 2010 at 0:08
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    $\begingroup$ Make each block a $2 \times 2$ matrix with exactly one 1, in a different position in each block. You can arrange the blocks so that the total matrix is a permutation, but each block has zero determinant. $\endgroup$ Jun 7, 2010 at 4:11
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    $\begingroup$ ...Also! The false formula has the wrong symmetry properties. You can swap the left and right blocks by passing the $n$ right columns over the $n$ left columns one at a time for a total of $n^2$ swaps, which affects the determinant by a factor of $(-1)^{n^2}$ but affects the proposed formula by a factor of $-1$. So at least if $n$ is even, the formula can't be right. $\endgroup$ Jun 7, 2010 at 4:53
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    $\begingroup$ It seems that the answer $det(AD-BC)$ is the correct one provided $A,B,C,D$ commute pairwise. A more general result (for block matrices consisting of $k^2$ square blocks) appears in jstor.org/stable/2589750 $\endgroup$ Jun 10, 2010 at 18:49
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    $\begingroup$ @senti_today: It is fascinating how the Amer. Math. Monthly spreads and reproves lemmata from Bourbaki as theorems since they are "not widely known". The whole point of this paper is lemma 1 in Algebra ch. III §9 no. 4. $\endgroup$ Nov 19, 2013 at 13:15

Here are two things that I have mistakenly believed at various points in my "adult mathematical life":

For a field $k$, we have an equality of formal Laurent series fields $k((x,y)) = k((x))((y))$.

Note that the first one is the fraction field of the formal power series ring $k[[x,y]]$. For instance, for a sequence $\{a_n\}$ of elements of $k$, $\sum_{n=1}^{\infty} a_n x^{-n} y^n$ lies in the second field but not necessarily in the first. [Originally I had $a_n = 1$ for all $n$; quite a while after my original post, AS pointed out that that this actually does lie in the smaller field!]

I think this is a plausible mistaken belief, since e.g. the analogous statements for polynomial rings, fields of rational functions and rings of formal power series are true and very frequently used. No one ever warned me that formal Laurent series behave differently!

[Added later: I just found the following passage on p. 149 of Lam's Introduction to Quadratic Forms over Fields: "...bigger field $\mathbb{R}((x))((y))$. (This is an iterated Laurent series field, not to be confused with $\mathbb{R}((x,y))$, which is usually taken to mean the quotient field of the power series ring $\mathbb{R}[[x,y]]$.)" If only all math books were written by T.-Y. Lam...]

Note that, even more than KConrad's example of $\mathbb{Q}_p^{\operatorname{unr}}$ versus the fraction field of the Witt vector ring $W(\overline{\mathbb{F}_p})$, conflating these two fields is very likely to screw you up, since they are in fact very different (and, in particular, not elementarily equivalent). For instance, the field $\mathbb{C}((x))((y))$ has absolute Galois group isomorphic to $\hat{\mathbb{Z}}^2$ -- hence every finite extension is abelian -- whereas the field $\mathbb{C}((x,y))$ is Hilbertian so has e.g. finite Galois extensions with Galois group $S_n$ for all $n$ (and conjecturally provably every finite group arises as a Galois group!). In my early work on the period-index problem I actually reached a contradiction via this mistake and remained there for several days until Cathy O'Neil set me straight.

Every finite index subgroup of a profinite group is open.

This I believed as a postdoc, even while explicitly contemplating what is probably the easiest counterexample, the "Bernoulli group" $\mathbb{B} = \prod_{i=1}^{\infty} \mathbb{Z}/2\mathbb{Z}$. Indeed, note that there are uncountably many index $2$ subgroups -- because they correspond to elements of the dual space of $\mathbb{B}$ viewed as a $\mathbb{F}_2$-vector space, whereas an open subgroup has to project surjectively onto all but finitely many factors, so there are certainly only countably many such (of any and all indices). Thanks to Hugo Chapdelaine for setting me straight, patiently and persistently. It took me a while to get it.

Again, I blame the standard expositions for not being more explicit about this. If you are a serious student of profinite groups, you will know that the property that every finite index subgroup is open is a very important one, called strongly complete and that recently it was proven that each topologically finitely generated profinite group is strongly complete. (This also comes up as a distinction between the two different kinds of "profinite completion": in the category of groups, or in the category of topological groups.)

Moreover, this point is usually sloughed over in discussions of local class field theory, in which they make a point of the theorem that every finite index open subgroup of $K^{\times}$ is the image of the norm of a finite abelian extension, but the obvious question of whether this includes every finite index subgroup is typically not addressed. In fact the answer is "yes" in characteristic zero (indeed $p$-adic fields have topologically finitely generated absolute Galois groups) and "no" in positive characteristic (indeed Laurent series fields do not, not that they usually tell you that either). I want to single out J. Milne's class field theory notes for being very clear and informative on this point. It is certainly the exception here.

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    $\begingroup$ Milne also, in his notes on field and Galois theory, takes the time to point out (and prove using Zorn's lemma and the group $\mathbb{B}$ above) that the absolute Galois group of $\mathbb{Q}$ has non-open subgroups of index $2^n$ for all $n>1$. He adds as a footnote a quote of Swinnerton-Dyer where he mentions the "unsolved [problem]" of determining whether every finite index subgroup of $G_\mathbb{Q}$ is open or not, observing that this problem seems "very difficult." $\endgroup$ May 5, 2010 at 12:45
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    $\begingroup$ Nice examples! Actually it is known that any finite groups arises as a Galois group over K=C((x,y)). Since K is Hilbertian, it is enough to prove it for K(t). Now, we know that if L is a large field (ie any smooth L-curve has infinitely many L-points as soon as it has one), then any finite groups arises as a Galois group over L(t) (see F.Pop, Embedding problems over large fields, Ann. of Math., 1996). And F. Pop recently proved that if R is a domain which is complete wrt a non-zero ideal (Henselian's enough), then its fraction field is large (see Henselian implies Large on his webpage). $\endgroup$ May 5, 2010 at 13:52
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    $\begingroup$ @Pete: I remember once reading a paper of Katz and being bewildered by what he was saying until I realised that Q_p[[x]] was much bigger than Z_p[[x]] tensor_{Z_p} Q_p. $\endgroup$ May 5, 2010 at 20:19
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    $\begingroup$ Kevin: I'm quite fond of this distinction myself. It's why $\mathbb{Q}_p[[x]]$, which is the $\mathbb{Q}_p$-pro-unipotent completion of $\mathbb{Z}$, almost never comes up in Iwasawa theory. There, you're far more likely to see the small algebra. $\endgroup$ May 13, 2010 at 9:26
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    $\begingroup$ It's funny that you are illustrating yourself how tricky the distinction between $k((x))((y))$ and $k((x,y))$ can be, by giving a wrong example: in fact $\sum_{i \geq 0}x^{−i}y^i \in k((x,y))$. (Isn't it just $x/(x - y)$? Think a bit about convergence issues.) But I believe that $\sum_{i \geq 0} x^{-i^2} y^i \not\in k((x,y))$ - and I think I can prove this using the Weierstrass preparation theorem for Laurent series over complete DVRs, or something like that. $\endgroup$
    – Wanderer
    Jul 8, 2010 at 18:07

"Any subspace of a separable topological space is separable, too." Sounds natural.

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    $\begingroup$ (and it is true of metric spaces, and natural generalizations...) $\endgroup$ May 4, 2010 at 23:31
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    $\begingroup$ This seems to be the fault of the "divorce" of second countability and separability in general topological spaces: they coincide for metrizable spaces, but for general spaces the favorable properties were split up in a custody hearing: see en.wikipedia.org/wiki/…. (I think I wrote this part of the article.) $\endgroup$ May 5, 2010 at 2:14
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    $\begingroup$ I think I kind of believed that until today. $\endgroup$
    – Olivier
    May 5, 2010 at 8:17
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    $\begingroup$ By an amusing coincidence, I came across this for the first time a couple of days ago. There was a Cambridge exam question in 2008 where you had to show that products and subspaces of separable metric spaces were separable, and then you were given a topological space and asked to show that its square was separable, and that a certain subspace of it was not separable. I had to stare at it for about a minute before I understood why I had not just proved a contradiction. $\endgroup$
    – gowers
    May 5, 2010 at 9:18
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    $\begingroup$ Perhaps some discontinuities between metric space theory and topology arise because when studying metric spaces, distances are mysteriously required to be symmetric, and the requirement is dropped when switching to topological spaces. So you can have a point x that has y as a limit (i.e. it is in all neighbourhoods of y) but y doesn't have x as a limit. With this idea in mind, you can make any topological space separable by adding a single point, and making it belong to all open sets. $\endgroup$ May 5, 2010 at 14:53

Here are a few more: (Everything between quotation marks is a false belief.)

Basic logic: Among students: "If A implies B then B implies A" (or "if A implies B then not A implies not B").

Even among mature mathematicians a frequent false belief is to forget that the conclusion of a theorem need not hold once the conditions of the theorem fail. Another common frequent belief is to assume that once the conditions fail then the conclusion must fail too.

Calculus: "The derivative of a differentiable function is continuous."

"An infinite series whose general term tend to 0 is convergent."

Geometry: "The circle is the only figure which has the same width in all directions." (Feynman regarded this mistake as one reason for the space shuttle Challenger disaster).

Polytopes: Often people believe that "given a convex polytope P you can slightly move the vertices to rational positions keeping the structure of the polytope unchanged."

(From Udi de Shalit): Some people believe that "if you hold a cube along a main diagonal, the remaining vertices all lie on a plane." Some even say that their number is 4.

Algebra (Also from Udi) "I have encountered many misconceptions about solvability by radicals. Some people think that 'the solution of an irreducible equation of degree 5 and higher, say over $\mathbb Q$, is never expressible by radicals'. Some amateur mathematicians even say that 'equations of degree 5 and higher have no solutions'."

Probability: "If you play the casino patiently and carefully you will win in the long run" (and "you do not believe that?, this is my own experience on the matter!" and "Indeed when I am calm and patient I win, but when I lose my temper I lose big time".)

"an event which may occur has positive probability": (not true for infinite probability spaces)

Various places: "If you want to prove that a certain infinite structure exists it is enough to show that there is no upper bound to the sizes of such structures."

Combinatorics: "This is a finite problem, surely you can solve it with a computer."

"Hall marriage theorem is very nice and I am surprised no combinatorialist bothered to extend it to a matching built from triples instead of pairs." (It is unlikely that a general characterization when a hypergraph built from triples has a perfect matching (of triangles) will be found.)

Computer science: "It is known that quantum computers can solve NP complete problems in polynomial time."

  • 17
    $\begingroup$ "If you want to prove that a certain infinite structure exists it is enough to show that there is no upper bound to the sizes of such structures." This is not necessarily false. In some important cases this does work! One uses the compactness theorem for such proofs. $\endgroup$ May 7, 2010 at 12:46
  • 13
    $\begingroup$ Probability: There are two opposite errors. Both are common. Say we are flipping a fair coin repeatedly. (1) if there have been more heads than tails, then tails is "overdue" and thus more likely on the next flip. (2) if there have been more heads than tails, then heads is "hot" and thus more likely on the next flip. $\endgroup$ May 7, 2010 at 15:15
  • 21
    $\begingroup$ I don't know anything about polytopes, but I'm having a hard time disbelieving this false result. Are we talking about finite polytopes here? $\endgroup$
    – Tom Ellis
    May 9, 2010 at 19:26
  • 38
    $\begingroup$ Entirely finite, Tom. There are 4-dimensional polytopes with 33 vertices that cannot be presented with rational coordinates. Here is a reference arxiv.org/PS_cache/arxiv/pdf/0710/0710.4453v2.pdf $\endgroup$
    – Gil Kalai
    May 9, 2010 at 20:46
  • 51
    $\begingroup$ You list the statement "Quantum computers can solve NP complete problems in polynomial time" as a false belief, but I don't believe you actually know this belief to be false. For example, the assertion that this belief is false implies $P\neq NP$. Perhaps the false belief that you intend to mention is: "It has been proved that Quantum computers can solve NP complete problems in polynomial time." $\endgroup$ May 17, 2010 at 12:34

From the Markov property of the random walk $(X_n)$ we have

$$P(X_4>0 \ |\ X_3>0, X_2>0) = P(X_4>0\ |\ X_3>0).$$

To paraphrase Kai Lai Chung in his book "Green, Brown, and Probability",

"The Markov property means that the past has no after-effect on the future when the present is known; but beware, big mistakes have been made through misunderstanding the exact meaning of the words "when the present is known"."

  • 24
    $\begingroup$ This is a nice one. I almost fell for it. The best way to see it's not true is perhaps to condition on $X_3> 0, X_2 \le 0$. Then that forces X_4 < 1, if the random walk has increment $\pm 1$. $\endgroup$
    – John Jiang
    Oct 10, 2010 at 19:40
  • 4
    $\begingroup$ It may easily happen that $X_2 X_3 <0$ always, so LHS probability may be even not defined. $\endgroup$ Apr 20, 2016 at 19:18
  • $\begingroup$ @JohnJiang If $\{w_k\}_{k}$ is a (discrete) Markov process on a filtered probability space $(\Omega, \mathcal{F},\{\mathcal{F}_k\}_{k}, \mathrm{P})$ and we define $x_{k+1} = x_k + w_k$, then we may say that $\{x_k\}_{k}$ is driven by $\{w_k\}_k$, but it is not a Markov process itself. Indeed, $$\mathrm{P}[x_{k}=\alpha\mid x_{k-1}=\beta, x_{k-2}=\gamma] = \mathrm{P}[w_{k-1}=\alpha-\beta\mid w_{k-2}=\beta-\gamma]$$ which is different from $\mathrm{P}[x_{k}=\alpha\mid x_{k-1}=\beta]$. But, $(x_k, w_k)$ is a Markov process. $\endgroup$ Jun 25, 2016 at 15:41
  • $\begingroup$ Related: math.stackexchange.com/q/1956127/631742 $\endgroup$ Mar 29, 2022 at 9:03
  • $\begingroup$ This can also be explained from the fact that $$\mathsf E(X\mid\mathcal F)(\omega)=\mathsf E(X\mid A) \text{ for almost all }\omega\in A$$ is, in general, false for an $L^1$-random variable $X$, a sigma-algebra $\mathcal F$ and $A\in \mathcal F$. It becomes more obvious when writing for instance $$P(X_4>0 \ |\ X_3\in\mathbb R, X_2>0) = P(X_4>0\ |\ X_3\in\mathbb R),$$ which is, of course, false in general. $\endgroup$ Mar 29, 2022 at 9:08

$$2^{\aleph_0} = \aleph_1$$

This is a pet peeve of mine, I'm always surprised at the number of people who think that $\aleph_1$ is defined as $2^{\aleph_0}$ or $|\mathbb{R}|$.

  • 78
    $\begingroup$ Show me a proof that it's not! $\endgroup$ Jun 15, 2011 at 21:39
  • 18
    $\begingroup$ Or: Show me a proof that it is correct! ;-) $\endgroup$ Oct 3, 2011 at 7:25
  • 19
    $\begingroup$ This error can be found in Gamow's famous book "One, Two, Three...Infinity", and even the Oxford English Dictionary contains a quote in their definition of "aleph null": "There is no infinite number between aleph-null (the number of positive integers) and aleph-one (the number of real numbers)." (apparently pulled from a Scientific American article) $\endgroup$ Apr 16, 2013 at 18:38
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    $\begingroup$ The key word is "defined". The false belief that annoys me is that people believe $\aleph_1$ is defined as $|\mathbb{R}|$. $\endgroup$ Jan 9, 2015 at 16:14
  • 20
    $\begingroup$ (This also appears in "The gods themselves" by Asimov but, it being science-fiction, I always "fix it" by imagining that in the world of the story, $\mathsf{CH}$ has been accepted as an axiom.) $\endgroup$ Jan 13, 2015 at 22:01

I think, there are different types of false beliefs. The first kind are statements which are quite natural to believe, but a moment of thought shows the contradiction. Of this type is the sin-example in the opening post or a favorite of mine (also occurred to me):

  • The underlying additive group of the field with $p^n$ elements is $\mathbb{Z}/p^n\mathbb{Z}$.

The other type is also quite natural to believe, but one has really to think to construct a counter example:

  • Every contractible manifold is homeomorphic to $\mathbb{R}^n$.
  • Every manifold is homotopy equivalent to a compact one.
  • Quotients commute with products in topological spaces.
  • Every connected component of a topological space is open and closed. Or related to this:
  • To give a continuous action of a topological group $G$ on a discrete space $X$ is the same as to give an action of the group of connected components of $G$ on $X$.
  • 48
    $\begingroup$ Along the same vein as the first example: the field with $p^2$ elements is a subfield of the field with $p^3$ elements (etc...) $\endgroup$
    – Sean Kelly
    May 5, 2010 at 17:06
  • 3
    $\begingroup$ Profinite groups are not the topological space, most topologists are most familiar with. $\endgroup$ Dec 3, 2010 at 14:42
  • 3
    $\begingroup$ en.wikipedia.org/wiki/Whitehead_manifold $\endgroup$ Mar 25, 2011 at 23:24
  • 3
    $\begingroup$ @Lennart: True, but topologists know the rationals. They are (usually) not the ones with this mistaken belief. $\endgroup$ May 4, 2011 at 13:14
  • 3
    $\begingroup$ The finite fields bit hits very close to home, especially when you start denoting them $\Bbb F_q$, because now the $p$ is gone from your notation, so it's very easy to forget that additive group isn't just cyclic of order $q$. $\endgroup$
    – Dylan Yott
    Feb 11, 2014 at 15:13

I used to believe that a continuous algebra homomorphism from $k[[x_1,\dots, x_m]]$ to $k[[y_1,\dots,y_n]]$, with $m > n$, could not be injective. Konstantin Ardakov set me straight on this.

  • 36
    $\begingroup$ Whoa, that's not true?! What's the counter-example? $\endgroup$ May 21, 2010 at 19:44
  • 30
    $\begingroup$ Not true and deeply frustrating. Take a map from $k[[x,y,z]]$ to $k[[u,v]]$ that sends $x$ to $u$, $y$ to $uv$ and $z$ to $uf(v)$ for some $f\in k[[v]]$ and think about what the kernel is. It isn't hard to see that only for countably many choices of $f$ can it possibly be zero. $\endgroup$ Jun 7, 2010 at 19:45
  • 4
    $\begingroup$ @Simon: Were you thinking of $k$ a countable field, and did you mean "can it possibly be non-zero"? (It seems the condition on $f$ is that it should not be a solution of any monic poly. whose coefficients are in $k(v)$.) $\endgroup$
    – fherzig
    Jun 7, 2010 at 22:00
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    $\begingroup$ Thanks, Simon, I had meant to get around to coming back and answering David! It's disturbing, isn't it? I had a paper ready to go apart from one lemma I was confused about that used needed this (false) statement, and indeed, not only is the statement wrong, the thing I thought I had almost proved turned out to be false! $\endgroup$
    – JSE
    Jun 8, 2010 at 22:15
  • 4
    $\begingroup$ Reading all of these has made me really glad that I am... "slower" than most people. To make all of these errors you would at least need to have a little intuition or understanding as to why they should be plausible. When I started learning algebraic geometry I was so lost and understood so little that I rigorously proved every little detail of every claim I made because I couldn't even figure out which ones were obvious enough not to prove completely and didn't want to look like a ding-dong in front of my advisor. $\endgroup$
    – Prince M
    Nov 19, 2017 at 5:17

Here are two group theory errors I've seen professionals make in public.

1) Believing that if $G_1 \subset G_2 \subset \cdots$ is an ascending union of groups such that $G_i$ is free, then $\bigcup_{i=1}^{\infty} G_i$ is free. Probably the vague idea they have is that any relation has to live in some $G_i$, so there are no nontrivial relations.

2) Consider a group $G$ acting on a vector space $V$ (over $\mathbb{C}$, say). Assume that $G$ acts as the identity on a subspace $W$ and that the induced action of $G$ on $V/W$ is trivial. Then I've seen people conclude that the action of $G$ on $V$ is trivial. Of course, this is true if $G$ is finite since then all short exact sequences of $G$-modules split, but it is trivial to construct counterexamples for infinite $G$.

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    $\begingroup$ So people think $\mathbb{Q}$ is a free group? Curious. $\endgroup$ May 5, 2010 at 23:39
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    $\begingroup$ Sadly enough, I suspect that many people who care about geometric/combinatorial group theory do not think of Q as a group... $\endgroup$ May 6, 2010 at 0:28
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    $\begingroup$ Pete, they don't think... they BELIEVE! $\endgroup$ May 14, 2010 at 6:56
  • 20
    $\begingroup$ Come on Steve -- aren't all abelian groups finitely generated? =) $\endgroup$ May 15, 2010 at 15:04
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    $\begingroup$ @yatima You probably mean $G_i=\{n/i!\}$. $\endgroup$
    – The User
    May 11, 2013 at 23:58

The field of $p$-adic numbers has characteristic $p$.

  • 6
    $\begingroup$ This seems to be an artifact of notation, since things with subscript $p$ have characteristic $p$ in early algebra. I've seen people make this mistake while simultaneously holding the (correct) beliefs that a) $\mathbb{Q}\subset \mathbb{Q}_p$ b) $\mathbb{Z}$ is dense in $\mathbb{Z}_p$ and c) $\mathbb{Z}_p/p\mathbb{Z}_p \cong \mathbb{Z}/p\mathbb{Z}$ Any one of which would be evidence against the false belief. $\endgroup$ May 3, 2011 at 19:44
  • 3
    $\begingroup$ Related: The unit sphere in the $p$-adics has measure zero. $\endgroup$
    – Phil Isett
    Jul 3, 2011 at 19:17
  • $\begingroup$ This one I heard from a reputed colleague in another field, but I presume he didn't really think of it and only got influenced by the notation. $\endgroup$
    – ACL
    Apr 21, 2016 at 14:41
  • $\begingroup$ There is a great problem that topologists sometimes denotes $\mathbb F_p$ by $\mathbb Z_p$. An example: Hatcher's textbook. $\endgroup$
    – Z. M
    May 10, 2021 at 21:02

"Either you can prove the statement, or you can find a counterexample."

This statement is usually applied to universal statements, those having the form $\forall x\ \varphi(x)$, where the concept of counterexample makes sense, but the general sentiment is the belief that every statement in mathematics is either provable or refutable.

The belief is false, because of the independence phenomenon.

  • 14
    $\begingroup$ Goedel's completeness theorem says that under certain hypotheses, there's either a proof or a counterexample. But "under certain hypotheses" needs to be understood. $\endgroup$ Jun 6, 2010 at 19:50
  • 3
    $\begingroup$ For people whoc categorize people in two classes: there are two kinds of mathematicians. Those, optimistic, who prove statements. Those, pessimistic, who construct counter-example. Which one are you ? $\endgroup$ Sep 30, 2010 at 15:42
  • 14
    $\begingroup$ To counter this false belief, exam questions should always start with: Prove, disprove or prove that neither is possible. :) $\endgroup$
    – user11235
    Apr 10, 2011 at 11:37
  • 56
    $\begingroup$ Thei, this would merely put off the dilemma one more step, since perhaps the statement is neither provable, nor refutable, not is its resulting independence provable! So one would need to add a fourth possibility, "or prove that none of the preceding is possible", and then repeat this process infinitely many times, with smaller and smaller fonts on the exam sheet. $\endgroup$ Oct 6, 2011 at 20:50
  • 3
    $\begingroup$ @TobyBartels : If you continue it through the uncountable set of all countable ordinals, will it fit on the exam sheet? $\endgroup$ Jun 12, 2019 at 22:04

I remember from my first analysis class thinking that if $\mathbb{Q}\subset E\subset\mathbb{R}$ with $E$ open, then $E$ would have to be all of $\mathbb{R}$ (at least more or less, maybe up to countably many points). And once we started measure theory I remember arguing with a friend over it for a good two hours.

  • 73
    $\begingroup$ The construction of a counterexample is really not too difficult: enumerate Q and take the union of intervals of size 1/3^n about the nth rational. Morals: Q is small and open sets are weird. $\endgroup$ May 6, 2010 at 0:49
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    $\begingroup$ @Qiaochu Yuan: That example as an indicator function is also what I use to really illustrate the power of Lebech integration over Riemann integration. The usual example of 0 if x is irrational and 1 if it is rational leaves me unsatisfied since later one proves that lebech integration only depends on the equivalence class of the function and under this equivalence this function is 0. $\endgroup$ May 6, 2010 at 8:58
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    $\begingroup$ Where do you get the spelling "Lebech"? I've only ever seen "Lebesgue," and the only "Lebech" in Wikipedia is a contemporary Danish politician. $\endgroup$ May 6, 2010 at 14:43
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    $\begingroup$ Measure theory? Enumerating rationals? Isn't $\mathbb{R}-\{\sqrt{2}\}$ open and contains the rationals?? $\endgroup$ May 20, 2010 at 15:30
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    $\begingroup$ Dror, note the parenthetical remark in Owen's post. $\endgroup$ May 25, 2010 at 21:21

If $f(x,y)$ is a polynomial with real coefficients, then the image of $f$ is a closed subset of $\mathbb{R}$. Note. Problem A1 on the 1969 Putnam exam asked to describe all possible images of $f$. I was told that the writers of this problem did not realize its subtlety.

  • 99
    $\begingroup$ One concrete counterexample: $(1-xy)^2+x^2$ $\endgroup$ Jul 8, 2010 at 5:35
  • 4
    $\begingroup$ So the examiners were expecting any unbounded closed interval (the same answer as for a polynomial in one variable), but the correct answer is any unbounded interval? $\endgroup$ Apr 8, 2019 at 14:20
  • 2
    $\begingroup$ Edit to my previous comment: and in both cases, we also need to allow singletons (for the constant polynomials). $\endgroup$ Aug 23, 2020 at 7:57

In order to show that a polynomial $P \in F[x_1,\ldots,x_n]$ vanishes, it suffices to show that $P(x_1,\ldots,x_n) = 0$ for all $x_1,\ldots,x_n \in F$. True in infinite fields, but very false for small finite fields.

Closely related: if two polynomials $P$, $Q$ agree at all points, then their coefficients agree. Again, true in infinite fields, but false for finite fields.

(This is ultimately caused by a conflation of the concept of a polynomial as a formal algebraic expression, and the concept of a polynomial as a function. Once one learns enough algebraic geometry to be comfortable with concepts such as "the $F$-points $V(F)$ of a variety $V$" then this confusion goes away, though.)

  • $\begingroup$ This seems related to the widely held belief that 1+1=2, (and how could 2 be equal to 0 ?!). On a psychological level this may connect with the assumption we all tend to make that a is different from b when asked to count the elements of set {a,b} (a related entry is somewhere else on this page), or that in general two different symbols are instinctively taken to denote two distinct objects. $\endgroup$ Jun 6, 2010 at 20:53
  • $\begingroup$ This is another example of intension/extension (the distinction I was trying to get at in question mathoverflow.net/questions/18848/…). $\endgroup$ Jun 8, 2010 at 0:04
  • $\begingroup$ Also closely related: Fermat's Little Theorem can be used to prove equality of polynomials. E.g., $X^2=X$ in $\mathbb{Z}_2[X]$ because $x^2=x$ in $\mathbb{Z}_2$. $\endgroup$
    – Jose Brox
    Apr 22, 2016 at 8:20
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    $\begingroup$ I once heard this used to "prove" that $\mathbb{F}_2$ is algebraically closed. The "idea" is that any non-constant polynomial over this field "must" take a value different from 1. However, the only other value available is 0! $\endgroup$
    – Kapil
    Oct 27, 2018 at 6:37
  • $\begingroup$ As a simple example: Consider the (formal) polynomial $P(X)=X^2+X \simeq (0,1,1,0,0,\dots)\in\mathbb Z_2^{\mathbb N_0}$, which indeed satisfies $P(0)=P(1)=0$ in $\mathbb Z_2$. $\endgroup$ Sep 30, 2021 at 16:46

"If any two of the $3$ random variables $X,Y,Z$ are independent, all three are mutually independent." In fact, they may be dependent; the simplest example is probably $(X, Y, Z)$ chosen uniformly from $\{(0, 0, 0), (1, 1, 0), (1, 0, 1), (0, 1, 1)\}$.

  • $\begingroup$ This is a false belief? $\endgroup$ May 5, 2010 at 0:26
  • $\begingroup$ Harald, I've edited to clarify. $\endgroup$
    – JBL
    May 5, 2010 at 0:42
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    $\begingroup$ or in words: take two fair $0-1$ coins and the random variables ""value first coin", "value second coin" and "sum modulo 2". $\endgroup$
    – Alekk
    May 5, 2010 at 8:25
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    $\begingroup$ A physics professor assigned proving this false statement as homework in a statistical mechanics course. $\endgroup$ May 5, 2010 at 17:43
  • 1
    $\begingroup$ A related thread: mathoverflow.net/questions/7998/… $\endgroup$
    – Yoo
    Jun 15, 2010 at 18:21

"Automorphisms of the symmetric group $S_n$ are inner (that is, each one is of the form $x \to axa^{-1}$ for some $a \in S_n$)" is a popular misconception, false for nontrivial reasons when $n=6$. That is both an easy mistake to make and important conceptually as an early hint of the complexities and special combinatorics that arise in finite group theory. Many people make it through a first class in group theory without understanding that something different happens for $S_6$ and in doing so have missed an important piece of the the big picture, as far as finite groups are concerned.

It is easy to implicitly or explicitly acquire this belief, because:

  1. those really are all the automorphisms for $n$ other than 6, and

  2. the inner automorphisms are used so often, for all values of $n$ (or $n>2$) without distinguishing any specific case as unusual.

  3. $S_n$ behaves in many ways as a family of similar groups rather than a list of individual groups with their own diverse features. A typical proof might show some property of $S_n$ by induction on $n$, starting from a small value such as $n=1$ for basic properties, or $n=3$ to assure noncommutativity. Apart from the classification of symmetric group automorphisms itself (exposure to which would be an explicit articulation and correction of the false belief), these arguments never start as high as $n=7$ and I don't know of any that distinguish $n=6$ or some equivalent case as a lone nontrivial exception. So it is easy to get the idea of more uniformity in the $S_n$ than really exists.

In essence, there are no obvious clues in the environment that $n=6$ might be special, and a number of indicators that no special case should exist at all.

  • 4
    $\begingroup$ Thanks for that example -- and the detailed diagnosis that accompanied it. $\endgroup$
    – gowers
    Jun 10, 2010 at 20:23
  • 6
    $\begingroup$ The main reason why one can fall for this is that any two permutations in $S_n$ with equal cycle type (and thus, any two permutations in $S_n$ which can be automorphed into each other) are conjugate, so one would expect that the same holds "globally". $\endgroup$ Jun 10, 2010 at 21:53
  • 7
    $\begingroup$ Cycles presume a given representation (group action) of $S_n$. The defining representation is permutations of an $n$-point set; "$S_n$-automorphisms arise from permuting the points" is a restatement of the false belief, and so cannot fully explain it. $S_n$ as abstract group carries only its regular (Cayley) representation, permuting a set of size $n!$. The possibility of non-conjugate elements in $S_n$ having the same cycle type in the regular representation opens the door for a nontrivial outer automorphism to exist. Missing these ideas may be the origin of the error, in "cycle" terms. $\endgroup$
    – T..
    Jun 10, 2010 at 23:31
  • 4
    $\begingroup$ @Nate: The outer automorphismen is induced from swapping the conjugation class of all transpositions (ab) with all tripeltranspositions (ab)(cd)(ef). $\endgroup$ Jun 11, 2010 at 14:30
  • 5
    $\begingroup$ Nate, automorphisms for $n=3,4,5$ and $n /geq 7$ fix the conjugacy class of transpositions (because no other class has the same size), thus are inner. The outer automorphism of $S_6$ is unique up to conjugation and several constructions are described at: en.wikipedia.org/wiki/… $\endgroup$
    – T..
    Jun 11, 2010 at 18:06

There are a couple of false beliefs regarding the $I$-adic completion functor, where $I$ is an ideal in a commutative ring $A$.

The first is that the completion of an $A$-module $M$ is complete, or in other words, that the completion functor is idempotent. This is true if $I$ is finitely generated (in particular when $A$ is Noetherian), but false in general. I find this quite unexpected - you take a module, "complete" it, and the result is not complete...

Another issue is the exactness of the completion functor. The completion functor is exact on the category of finitely generated modules over a Noetherian ring, but when you consider arbitrary modules, it is neither left-exact (this is easy to see) nor it is right-exact (this probably less known), even when $I$ is finitely generated and the modules in question are finitely presented.

  • 38
    $\begingroup$ Oh my god .... $\endgroup$ May 19, 2010 at 14:18
  • 1
    $\begingroup$ Can you give an example of an $A$-module $M$ with non complete completion? Thank you! $\endgroup$
    – Ricky
    Jul 23, 2010 at 11:53
  • 6
    $\begingroup$ @Ricky, see for example the paper arxiv.org/abs/0902.4378 - example 1.8. $\endgroup$
    – the L
    Jul 25, 2010 at 10:23
  • 4
    $\begingroup$ Let me just leave here that many (all?) of these issues can be solved if you work with the derived completion instead (whether you think it a good thing or not, I suppose, depends on how much the word derived makes you uncomfortable). For a principal ideal $I=(x)$ generated by a nonzerodivisor the derived completion of $M$ is just $\mathrm{Ext}(A/x^\infty,M)$, so it's not that bad (for a zerodivisor in general it will have a piece in degree 1) $\endgroup$ Apr 9, 2019 at 20:29
  • $\begingroup$ @DenisNardin By $A/x^\infty$ do you mean the inverse limit of the "obvious" tower? $\endgroup$
    – Pedro
    Sep 11, 2020 at 22:45

Some people have trouble understanding that (and why is) 0.999... = 1

  • 182
    $\begingroup$ And as Ehud de Shalit mentioned to me, some once they understand that 0.9999... = 1 think that 0.8888888...= 0.9 $\endgroup$
    – Gil Kalai
    May 5, 2010 at 18:44
  • 15
    $\begingroup$ That's great Gil! And according to this copiously referenced article, en.wikipedia.org/wiki/Parity_of_zero, many don't think that 0 is an even number $\endgroup$ May 5, 2010 at 23:37
  • 42
    $\begingroup$ @Victor: Many don't think that 0 is even a number. :) $\endgroup$ May 7, 2010 at 12:35
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    $\begingroup$ Those that think 0 is a number sometimes don't think that -1 is a number, and of the latter, some don't think that i is a number. I'm not sure if I think that j and k are numbers. $\endgroup$
    – Tom Ellis
    May 9, 2010 at 19:16
  • 35
    $\begingroup$ Another argument that I have sometimes found effective is to get people to agree that 0.33333... = 1/3 and then multiply both sides by 3. But sometimes this just causes people to believe that 0.3333... is not quite equal to 1/3. $\endgroup$ Mar 24, 2011 at 14:18

This is perhaps a misunderstood definition rather than a false belief, but:

"A subnet of a net $( x_\alpha )_{\alpha \in A}$ takes the form

$( x_\alpha )_{\alpha \in B}$ for some subset $B$ of $A$."

In truth, subnets are allowed to contain repetitions, and can be indexed by sets much larger than the original net. (In particular, there are subnets of sequences that are not subsequences.)

This false belief, incidentally, reinforces the false belief noted in a different answer, namely that compactness implies sequential compactness.

A precise Counterexample: The sequence $\sin(nx)$ is a sequence in the compact topological space $[-1, 1]^{\mathbb{R}}$ with product topology. So this net has a convergence subnet. But it is well known that the above sequence has no subsequence which is pointwise convergent (See the last page of the book of Walter Rudin's Principles of mathematical Analysis). So in this example the convergent subnet cannot be counted as a subsequence.

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    $\begingroup$ Correcting that misunderstanding is crucial to prove that an accumulation point of a net is always the limit of some subnet, which does not hold for sequences. $\endgroup$ Oct 19, 2010 at 6:26
  • $\begingroup$ I cannot find this (counter)example in Rudin's book (I have the 3rd edition). $\endgroup$ Dec 25, 2016 at 6:32
  • $\begingroup$ It's Chapter 11, Exercise 16 on page 334 just before the bibliography in my copy of the 3rd edition of Rudin's Principles of Mathematical Analysis. $\endgroup$ Jan 3, 2017 at 22:57
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    $\begingroup$ I was cured of this one when I learned about compact spaces that have no convergent sequences other than eventually constant ones (why not take a sequence with distinct elements, pass to a convergent subnet, and there you have it ...?). $\endgroup$ Aug 22, 2017 at 1:47
  • $\begingroup$ Also the converse is false. The ordinal $\omega_1$ with the order topology (generated by the open intervals $(x,y)$ with $x\in\omega_1\cup\{-\infty\}$ and $y\in\omega_1\cup\{+\infty\}$) is obviously not compact, but it is sequentially compact: from any sequence $(x_k)$ one can either extract a monotone decreasing or a monotone increasing subsequence $(x_{k_\ell})$, whose limit is $\bigcap x_{k_\ell}$ or $\bigcup x_{k_\ell}$ respectively. Notice that in the second case the limit is a countable ordinal and thus does belong to $\omega_1$. $\endgroup$
    – Mizar
    Sep 16, 2017 at 14:57

"It is impossible in principle to well-order the reals in a definable manner."

To be more precise, the belief I am talking about is the belief that well-orderings of the reals are provably chaotic in some sense and certainly not definable. For example, the belief would be that we can prove in ZFC that no well-ordering of the reals arises in the projective hierarchy (that is, definable in the real field, using a definition quantifying over reals and integers).

This belief is relatively common, but false, if the axioms of set theory are themselves consistent, since Goedel proved that in the constructible universe $L$, there is a definable well-ordering of the reals having complexity $\Delta^1_2$, which means it can be obtained from a Borel subset of $R^3$ by a few projections and complements. See this answer for a sketch of the definition of the well-order.

The idea nevertheless has a truth at its core, which is that although it is consistent that there is a definable well-ordering of the reals (or the universe), it is also consistent that there is no such definable well-ordering. Thus, there is no definable relation that we can prove is a well-ordering of the reals (although we also cannot prove that none is).

  • $\begingroup$ Your 'always' kept me wondering for a while! $\endgroup$ May 4, 2010 at 23:35
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    $\begingroup$ I think that you are taking an imprecise statement and making it precise in such a way that it is wrong. $\endgroup$ May 5, 2010 at 1:00
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    $\begingroup$ Marcos, the false belief I was aiming at was the belief that it is impossible in principle for a well ordering of the reals to be definable. That is indeed a false belief, since it IS possible in principle for there to be such a definable well-ordering, as there is one in Goedel's universe L. (Note: the constructible universe L has nothing at all to do with constructivism or constructive proofs, in the sense of your comment; Goedel used only classical logic.) And my remark was not aimed particularly at undergrads or even grads, but rather at research mathematicians holding that false view. $\endgroup$ May 5, 2010 at 1:29
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    $\begingroup$ Marcos, I have edited. One paradox with the equation "constructive"="not using AC" is that in the constructible universe $L$, everything is constructible (in the sense of Goedel), but AC holds, and there are constructible well-orderings! $\endgroup$ May 5, 2010 at 12:50
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    $\begingroup$ The best part of this answer is how the constructible universe subverts all the intuition you learn about AC from doing non-model theory. Experience would lead one to think that "AC = nonconstructive" in "the usual model of the real numbers", not realizing that there is no usual model. Your (my, everyone's) mental image of the reals is a sort of "lazy evaluation" (to use a programming term) of the model we would really like but haven't even specified fully. As you show in your answer, once given the facts we wouldn't even know which model that would be. $\endgroup$
    – Ryan Reich
    Oct 20, 2010 at 11:21

Occasionally seen on this site: if a polynomial $P:\mathbb{Q}\rightarrow\mathbb{Q}$ is injective, so must be its extension to $\mathbb{R}$.

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    $\begingroup$ +1 Yaakov. $P(x) = x^3 - 5x$ is a counterexample. $\endgroup$
    – Todd Trimble
    Mar 31, 2011 at 21:47
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    $\begingroup$ That's another funny one, because you would be very unlikely to see the same mistake with $\mathbb{C}$. So why are people liklely to make it with $\mathbb{R}$? $\endgroup$ Apr 7, 2011 at 0:30
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    $\begingroup$ @Tierry: continuity, of course. $\mathbb{R}$ is far from dense in $\mathbb{C}$ but $\mathbb{Q}$ is dense in $\mathbb{R}$. $\endgroup$
    – Ryan Reich
    Apr 22, 2011 at 18:06
  • $\begingroup$ @ToddTrimble: Could you explain, how this is a counterexample? Taking the polynomial $P(X)=X^3-aX$ and assuming $P(X)=P(Y)$, I get: \begin{equation} (X-Y)\left((X-Y)^2-3XY-a\right)=0. \end{equation} Now I'm clueless how to show the second factor can't vanish. (I used a general $a$ because of interest, what properties besides $a>0$ would be necessary.) $\endgroup$ Apr 19, 2022 at 16:45
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    $\begingroup$ @SamuelAdrianAntz Here's what I have. For any given rational $q$, suppose $r$ is a rational root of $x^3 - 5x -q$. After a long division, we conclude $x^3 - 5x - q = (x-r)(x^2 + rx + (r^2 - 5))$. The discriminant of the quadratic factor is $D = r^2 - 4(r^2-5) = 20 - 3r^2$, and another rational root of $x^3 - 5x - q$ exists only if $D$ is a rational square. So it suffices to show that there are no rational solutions $(r, s)$ to $3r^2 + s^2 = 20$. For this, show there are no integer solutions to $3r^2 + s^2 = 5t^2$ (except $(r,s,t)=(0,0,0)$). This boils down to simple modular arithmetic. $\endgroup$
    – Todd Trimble
    Apr 20, 2022 at 2:39
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