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

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

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

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

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

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

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

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Let $M_1$ be a finitely generated module over a PID and let $M_2$ be a submodule.

We may pick $L_i$ and $T_i$ submodules of $M_i$ such that $L_i$ is free, $T_i$ is torsion, $M_i = L_i \oplus T_i$, $L_2\subseteq L_1$ and $T_2\subseteq T_1$.

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If we regard a ring $R$ (with identity) as a right module ($R_{R}$), then there is a ring isomorphism $\text{End}(R_{R}) \simeq R$, however the same does not happen if we regard $R$ as a left module!

The correct is $\text{End}(_{R}R) \simeq R^{\text{op}}$.

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  • $\begingroup$ Here is a discussion about the condition for Morita equivalence between rings, which is related to this subtle detail: math.stackexchange.com/questions/3566579/… $\endgroup$ – user144185 Mar 27 at 12:19
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I don’t know how common is the following false belief, but I had it for several years, so maybe some other people also have it. I apologize to those to whom I shared this false belief. I hope this post will help.

Kaplansky’s 6th conjecture (here, 1975) states that if $H$ is a finite dimensional semisimple Hopf algebra and $V$ an irreducible representation of $H$, then $\dim (V)$ divides $\dim (H)$. This conjecture is open over the complex field $\mathbb{C}$, but false in positive characteristic. So we assume to be over $\mathbb{C}$.

For the group case, this property was proved by Frobenius, that is why a finite dimensional semisimple Hopf algebra (over $\mathbb{C}$) with this property is called of Frobenius type.

A finite dimensional Hopf algebra (over $\mathbb{C}$) is called a finite quantum group (or Kac algebra) if it has a $*$-structure. And then it is also semisimple. It is an open problem whether such a $*$-structure always exists.

False belief: George Kac proved Kaplansky’s 6th conjecture for the finite quantum groups.

This false belief was pointed out to me by Pavel Etingof after this talk I gave for Harvard University, and where I mentioned it. Fortunately, that does not affect the content of the talk.

What I had in mind is Theorem 2 in the following paper:
G. I. Kac, Certain arithmetic properties of ring groups., Funct. Anal. Appl., 6 (1972), pp. 158–160.

In modern language, Theorem 2 proves the following: let $H$ be a finite quantum group, and let $\mathcal{C} = Corep(H)$ be the fusion category of complex corepresentations of $H$. For every simple object $X$ of the Drinfeld center $Z(\mathcal{C})$ which contains the trivial object of $\mathcal{C}$ under the forgetful functor, $FPdim(X)$ divides $FPdim(\mathcal{C}) = \dim(H)$ (the quotients are called the formal codegrees).

Note that these $X$ correspond to the irreducible representations of the Grothendieck ring $K(\mathcal{C})$ of $\mathcal{C}$ (see Theorem 2.13 here). In particular, for $G$ a finite group, $\mathcal{C} = Corep(G) = Vec(G)$, and $Irr(K(\mathcal{C})) = Irr(G)$. That is why Theorem 2 implies Kaplansky’s 6th conjecture in the group case (i.e. covers Frobenius theorem). But it is not clear for a finite quantum group in general. It could be relevant to search in this direction, in particular to check whether for every object $Y$ of $Irr(H)$ there exists an $X$ as above such that $\dim(Y)$ divides $FPdim(X)$, because this would prove that $H$ is a Frobenius type.

Note that Theorem 2 (as stated above) holds more generally for every (complex) fusion category $\mathcal{C}$. The case $\mathcal{C} = Rep(G)$, with $G$ a finite group, recovers the fact that the size of each conjugacy class of $G$ divides $|G|$. Finally, according to Pavel, the theorem holds more generally without the assumption ‘which contains the trivial object’ (I don’t have the exact reference for that, so if you know it, please put it in comment).

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The assumption that a cubic surface expressed as a foliation of Weierstrass curves cannot be rational, because a general Weierstrass curve is not rational.

I've seen this false assumption more than once on sci.math over the years. But there are simple counterexamples, such as:

$ (x + y) (x^2 + y^2) = z^2 $

On defining $ u = x/y $ and $ v = z/y $ one obtains $ y (u + 1) (u^2 + 1) = v^2 $, and hence x, y, z as rational functions of u, v.

I'd love to have a reference to a procedure for calculating the geometric genus and algebraic genus of surfaces like this, because they are rational if and only if both these quantities are zero, and for other cubic surfaces that interest me it would save a lot of fruitless hacking around trying to find a rational solution that probably doesn't exist! Are there any symbolic algebra packages that can do this?

I mean for example is $ x y (x y + 1) (x + y) = z^2 $ rational? I'm almost sure it isn't; but how can one be sure?

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  • $\begingroup$ Algebraic genus in software: Seems to be available in Singular and Maple and may be available in Mathematica by now (from mathematica.stackexchange.com/a/5453/16237 ). (Aside: I don't know anything about the software mentioned.) $\endgroup$ – Eric Towers Aug 13 '17 at 2:06
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Here's one that I think will surprise many.

False belief. Let $E$ be an elliptic curve over an algebraically closed field $k$ of characteristic $p > 0$. Then $\operatorname{End}^\circ(E)$ is strictly larger than $\mathbb Q$.

While this is true for all elliptic curves defined over finite fields, most elliptic curves whose field of definition is transcendental over $\mathbb F_p$ have $\operatorname{End}^\circ(E) \cong \mathbb Q$. The extra automorphism on elliptic curves over a finite field comes from the geometric Frobenius. For varieties over larger fields, this is not a thing.

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Let $M \subset B(H)$ be a von Neumann algebra, $p \in B(H)$ a projection and $q=I-p$.

False belief: If $pM=Mp$ then $M=pMp \oplus qMq$.
(I think it is a quite common careless mistake)

Counter-example: diagonal embedding of $\mathbb{C}$ into $M_2(\mathbb{C})$.

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False belief: << Let $M$ be the von Neumann algebra generated by a $\rm{C}^{\star}$-algebra $\mathcal{A}$. >>

The false belief is to think that the above sentence makes sense. In fact, a von Neumann algebras and a $\rm{C}^{\star}$-algebra don't have the same status. A von Neumann algebra is an operator algebra by definition, i.e. it is defined inside $B(H)$ for some separable Hilbert space $H$. Now, some subalgebras of $B(H)$ are (separable) $\rm{C}^{\star}$-algebras, but a $\rm{C}^{\star}$-algebra can also be defined abstractly. It can next be representated and a given representation $H$ (defined for example by GNS construction for a given state), if it is faithful, induces an embedding in $B(H)$.
So to make sense, the sentence above should be modified as:

<< Let $M$ be the von Neumann algebra generated by $(\mathcal{A},\rho)$, a couple of $\rm{C}^{\star}$-algebra and state. >>
or
<< Let $M$ be the von Neumann algebra generated by a $\rm{C}^{\star}$-algebra $\mathcal{A}$ represented on $H$. >>

Then, $M = \pi_H(\mathcal{A})''$. We can use $M$ to characterize the representation $H$, for example, we can talk about a representation of type ${\rm I}$, ${\rm II}$ or ${\rm III}$ if $M$ is a von Neumann algebra of type ${\rm I}$, ${\rm II}$ or ${\rm III}$. There is a $\rm{C}^{\star}$-algebra with representations of every type, for example the Cuntz algebra.

Finally, there exists a universal representation for every $\rm{C}^{\star}$-algebra (i.e. the direct sum of the corresponding GNS representations of all states; it is faithful). The associated von Neumann algebra is called the enveloping von Neumann algebra (it can also be defined as the double dual); it contains all the operator-algebraic information about the given $\rm{C}^{\star}$-algebra.

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  • $\begingroup$ So there is no abstract version of the notion of a von Neumann algebra? Like, say, isomorphism classes of "usual" von Neumann algebras, or something like that? $\endgroup$ – მამუკა ჯიბლაძე Jan 12 '18 at 21:53
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    $\begingroup$ @მამუკაჯიბლაძე: A von Neumann algebra can be defined abstractly as a (non-necessarily separable) $\rm{C}^{\star}$-algebra that have a predual; but it is not the usual definition, some authors call this abstract version a $\rm{W}^{\star}$-algebra, see the last paragraph of en.wikipedia.org/wiki/Von_Neumann_algebra#Definitions $\endgroup$ – Sebastien Palcoux Jan 12 '18 at 22:23
  • $\begingroup$ @SebastienPalcoux If one takes this abstract definition (or something equivalent to it), how does one recognise the concrete von Neumann algebras, i.e. what condition on a continuous *-homomorphism $\rho: A\to B(H)$ is equivalent to $\rho(A)$ being a von-Neumann-algebra? I'd guess it is something like "$\rho$ is still continuous if one chooses certain other natural topologies on $A$ and $B(H)$ instead of the norm topologies". Is that the case? $\endgroup$ – Johannes Hahn Mar 16 '18 at 23:24
  • $\begingroup$ @JohannesHahn: A $\rm{C}^{\star}$-algebra (resp. von Neumann algebra) can be defined concretely as a $\star$-subalgebra of $B(H)$ closed by the operator norm topology (resp. the weak operator topology). The problem in your question is that these topologies are operator topologies, and $A$ is abstract. You could be satisfied by the following paragraph on the predual. $\endgroup$ – Sebastien Palcoux Mar 17 '18 at 8:27
  • $\begingroup$ @SebastienPalcoux That paragraph is part of the reason why I'm asking. I checked wikipedia first of course. I don't see if or how it answers my question. Since the predual is intrinsic, the ultraweak topology can be defined intrinsically as well. So it makes sense to say "$\rho$ is continuous w.r.t. the ultraweak topologies on $A$ and $B(H)$". That's what makes me think that a characterisation like what I'm asking is even possible. But I don't see if it's true. $\endgroup$ – Johannes Hahn Mar 17 '18 at 14:35
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The derived subgroup of a finite group equals to the set of all its commutators

or equivalently

A product of two commutators in a finite group is always a commutator

This mistake is very widespread, probably because counterexamples to it tend to be quite large. The smallest group, for which it is not true has order $96$.

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    $\begingroup$ What is your evidence that this is a commonly held belief? $\endgroup$ – Geoff Robinson Aug 28 at 12:06
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False belief: a subgroup isomorphic to a quotient is a retract.

Formally: Let $H,N$ be subgroups of $G$ with $N$ normal and $H \simeq G/N$, then $H$ is a retract of $G$.

It is false, because otherwise $C_2$ would be a retract of $C_4$, but it is not.

In fact, $H$ is a retract of $G$ if and only if $G$ is isomorphic to $H \ltimes N$ (semidirect product).

This false belief caused this post.

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Let $R$ be a ring with identity $e$, $A, B\in R$, $A\neq 0$, $B$ is invertible element. If $A\cdot B = A$ then $B = e$.

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  • $\begingroup$ I think, it is closely related to the following false "deduction": because invertible element cannot be at the same time zero divisor, therefore sum of any unit and zero divisor is not invertible. Ok, maybe it isn't popular, but I've got this belief at my first algebra course, until I discovered counterexample $1+X$ in $R[X]/X^2$. This is almost exactly the thing you mentioned, just put $B:=X, A:=X+1$. $\endgroup$ – Przemek Nov 5 at 20:09
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This is more sort of a convention issue than an outright false belief (connected to the usual $\emptyset$ vs $\{\emptyset\}$ stuff), but I find it funny. I guess a fair share of mathematicians believe that: \begin{equation} \bigcap\emptyset=\emptyset\label{eq}\end{equation} while retaining the standard definition for intersection: $$\bigcap S:=\{x\ \text{such that}\ \forall Y(Y\in S\implies x\in Y)\}$$ according to which in fact: $$\bigcap\emptyset=V$$ where $V$ is the universal class. The condition in round brackets is of course vacuously true. So in a way - this is what I find funny - the former is the worst possible tentative solution of an equation ever.

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Way late to the party...

"$ \mathrm{polymod}\ p$ and $\mathrm{mod}\ p$ are the same thing."

And it's cousin: "$\forall{x}, f(x) \cong g(x) \pmod{q} \implies f(x) = g(x)$"

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    $\begingroup$ What does polymod mean? $\endgroup$ – darij grinberg Oct 20 '10 at 11:47
  • $\begingroup$ Either the cousin needs a bit more detail if it is to be false, it is quite naive! $\endgroup$ – Mariano Suárez-Álvarez Oct 20 '10 at 18:25
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    $\begingroup$ Probably I understand what this means: if $f(x)=0\pmod 2$ for all $x$, then $f=0$ over $\mathbb F_2$. This is similar to my second example: mathoverflow.net/questions/23478/… $\endgroup$ – zhoraster Oct 20 '10 at 18:33
  • $\begingroup$ Consequently, there are only $4$ polynomials over $\mathbb F_2$ Isn't this convenient? :-) $\endgroup$ – zhoraster Oct 20 '10 at 18:40
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    $\begingroup$ $\mathrm{polymod}$ is "polynomial mod". Two polynomials are congruent $\mathrm{polymod} p$ iff the coefficients each power of the variable are congruent $\pmod{p}$. The equivalence classes are sets of polynomials where each coefficient ranges over an equivalence class $\pmod{p}$. For the cousin, there are many local/globals but they all seem to require additional conditions (q.v. Hensel lifting). I think the set from which $x$ was chosen was left unspecified because this "imprecise mental abbreviation" pops up at various levels of sophistication each with a different such set. $\endgroup$ – Anonymous Oct 23 '10 at 15:22
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A common false assumption is that that two non-orthogonal pure states of a quantum mechanical system may never be unambiguously distinguished by a measurement. (See http://arxiv.org/pdf/quant-ph/9807022.pdf)

Another false belief is that a quantum computer is similar to an analogue computer, in that large computations will necessarily fail because of accumulated error. (See, for example, http://arxiv.org/abs/quant-ph/9712048)

For that matter, another common false believe is that Bell Inequalities aren't violated, although it is mostly held by people who have never heard of Bell Inequalities.

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    $\begingroup$ I'm not sure how you can believe that something you have never heard of isn't violated. $\endgroup$ – Geoff Robinson Apr 10 '16 at 17:55
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Another common mistake. If $W = _P(e_1,\ldots, e_{n})$ is a vector space and $V$ is a subspace of $W$ of dimension $k$, then $V = _P(e_{i_1},\ldots, e_{i_k})$.

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  • $\begingroup$ What does that little subscript $p$ on the equals sign mean? $\endgroup$ – Gerry Myerson Feb 11 '16 at 21:36
  • $\begingroup$ $V$ is a vector space over field $P$. $\endgroup$ – Mikhail Goltvanitsa Feb 12 '16 at 6:52
  • $\begingroup$ So, what does "$V$ is a vector space over field $P$ $(e_1,\dots,e_n)$" mean? $\endgroup$ – Gerry Myerson Feb 12 '16 at 8:37
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    $\begingroup$ $W$ is a vector space over field $P$, $(e_1,\ldots, e_n)$ is a basis of $W$. $V$ is a subspace of $W$. $\endgroup$ – Mikhail Goltvanitsa Feb 12 '16 at 8:40
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I don't know how common this is, but it occurs as a corollary of a theorem in the fine, and widely used, text by Shafarevich on algebraic geometry: namely, if $f \colon X \longrightarrow Y$ is a surjective algebraic map of varieties, then 1) for all $y \in Y$, the fiber over $y$ has dimension $≥ \dim(X)-\dim(Y)$; 2) on some non empty open set in $Y$ the dimension of the fibers equals $\dim(X)-\dim(Y)$; 3) for all $r$, the set of $y \in Y$ such that the fiber over $y$ has dimension $≥ r$, is closed in $Y$.

The first two are true, but the third is false. Upper semicontinuity of fiber dimension is true on the source, not the target. For the conclusion as stated to hold, one can add properness to the hypothesis on the map. I think this is not at all widely believed by experts, but for some reason it persists in the text, hence may be believed by students.

Since I have myself written notes in which blatantly false statements occur, I do not think for a moment that Shafarevich himself believed this false statement. But such things do slip by, and may mislead beginners. In fact I believed it for some time until enlightened by a friend.

In keeping with the OP's desire to know the psychological reason for the error, it seems for some reason common in my experience for people to assume unconsciously that maps are proper.

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From Keith Devlin

"Multiplication is not the same as repeated addition", as put forward in Devlin's MAA column.

I'm not really sure how I feel about this one; I might be one of the unfortunate souls who are still prey to that delusion.

Caution

In case you missed it, the column ended up spilling a lot of electronic ink (as evidenced in this follow-up column), so I don't believe it would be wise to start yet a new one on MO. Thanks in advance!

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  • $\begingroup$ I followed your link, and I cannot even tell what is wrong about attaching helium balloons to both sides of a balance to model substraction on both sides of an equation. $\endgroup$ – user11235 Apr 10 '11 at 20:32
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    $\begingroup$ The more I think about this "error", the less I am convinced. It's like saying that you cannot say that $\binom n k$ is the number of $k$-element sets in an $n$-element set because then you will be unable to generalize to complex values of $n$. Or you cannot define the chromatic polynomial as the function counting the colourings and then plug in $-1$ to get the acyclic orientations of the graph. Also, I think it is perfectly understandable what it means to add something halfways. $\endgroup$ – user11235 Apr 10 '11 at 20:50
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    $\begingroup$ It's not a "false belief". It's a false heuristic. And it's actually here: mathoverflow.net/questions/2358/most-harmful-heuristic $\endgroup$ – darij grinberg Apr 10 '11 at 21:17
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    $\begingroup$ When I taught elementary teachers the course on arithmetic, they all had been taught that multiplication is repeated addition, but I myself thought it was the cardinality of the cartesian product. We enjoyed discussing this difference in point of view. $\endgroup$ – roy smith May 9 '11 at 3:06
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    $\begingroup$ The "repeated addition" characterization has an advantage over the "cardinality of the Cartesian product" characterization (which possibly in some contexts could be considered a disadvantage). That is that it's not self-evident that it's commutative, and so one has a useful exercise for certain kinds of students: figure out why it's commutative. $\endgroup$ – Michael Hardy May 20 '11 at 2:28
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For $p$ prime and the chain of embeddings $\mathbb{Z}/p\mathbb{Z} \hookrightarrow \mathbb{Z}/p^2\mathbb{Z} \hookrightarrow \cdots$ given by multiplication by $p$, then $\bigcup_n \mathbb{Z}/p^n\mathbb{Z}$ is not the group of $p$-adic integers $\mathbb{Z}_p$, but its Pontryagin dual, the Prüfer $p$-group $\mathbb{Z}(p^{\infty})$.

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    $\begingroup$ Is that actually a common false belief? After all, $\mathbb{Z}_p$ is uncountable, as everyone realizes! $\endgroup$ – Todd Trimble Mar 5 '15 at 14:25
  • $\begingroup$ "$\mathbb{Z}_p$ is countable" is also a false belief for people who didn't really read the definition of $\mathbb{Z}_p$, but I don't know how much it is common. $\endgroup$ – Sebastien Palcoux Mar 5 '15 at 14:34
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    $\begingroup$ It's hard for me to believe it's at all common. I wasn't the downvoter, but I think it would be better if answers were rooted either in instances that can be found in the literature, or widely encountered in one's experience as an instructor. $\endgroup$ – Todd Trimble Mar 5 '15 at 14:52
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In algebraic topology, I thought for a while:

  • "For $k \geq 2$, $H_k$ is the abelianization of $\pi_k$." False. True for $k = 1$. Also for all $k$ up to $n-1$ if the space is $(n-1)$-connected for $n \geq 2$ (vacuously, since this says the first $n-1$ homotopy groups are trivial and for these, the Hurewicz homomorphism is the isomorphism, $\pi_k \cong H_k$). See the Hurewicz theorem for more.
  • "Generically, all the $\pi_k$ are nonabelian." False. For $k \geq 2$, $\pi_k$ is abelian.

Edit: I had a third error in thinking when I first posted this, mangling the above into something further from true. Which I suppose makes the first version of this post meta-appropriate for this thread (but I've fixed it anyway). Thankfully, user Michael gently pointed out my mangling.

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  • $\begingroup$ First bullet: did you mean "True for $n=1$"? $\endgroup$ – Michael Jan 15 '19 at 23:06
  • $\begingroup$ @Michael : It's not always true for $n=1$, $\pi_1$ can be abelian, e.g. the fundamental group of the circle. For $n > 1$, $H_n \cong \pi_n$. It's easy to imagine "$\pi_n$s are (usually) nonabelian monsters and their associated homology groups are friendly abelian groups", but this difference *only* happens for $n=1$. $\endgroup$ – Eric Towers Jan 15 '19 at 23:31
  • $\begingroup$ I think you are confusing a few things here. Compare $H_2$ of the 2-dimensional torus with its $\pi_2$, for example. $\endgroup$ – Michael Jan 15 '19 at 23:34
  • $\begingroup$ @Michael : After actually looking up what I was talking about, I find that I have mashed together (at least) two errors to make another. Yay? $\endgroup$ – Eric Towers Jan 16 '19 at 4:42
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    $\begingroup$ @Michael : I think I've disentangled my mangling. I may still have a fumble-thought in the first bullet that I'm just not seeing. $\endgroup$ – Eric Towers Jan 16 '19 at 5:11
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Two commons believes :

Undergrad belief :

  1. The godel incompleteness theorem said that there is true things which are not provable, this is certainly false and I think that this create some spiritual pseudo mathematical reflection!

I think that the right way to imagine godel incompleteness theorem is by analogy to linear independence or transcendentality

Graduate belief:

Limit in analysis is a special case of limit in category theory

This is not true since $2\mathbb{N}$ is cofinal in $\mathbb{N}$

(This breaked my heart when I realize this thing!!!)

If there is another way to see why limit in analysis is a special case of category theory I will be very very happy so if you have a way react !

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  • $\begingroup$ Regarding Godel incompleteness theorem: why you think it is false? The Godel theorem constructs a program $P$ such that $P$ does not halt (in reality) but $PA$ does not prove "P does not halt", so this is an unprovable statement that is true. Regarding your second question, see: mathoverflow.net/questions/9951/… $\endgroup$ – kp9r4d Sep 16 at 7:37
  • $\begingroup$ @kp9r4d The point is that Godel create a statement P for a theory T with certain conditions such that $\{ P \} \cup T $ *and* $\{ \neg P \} \cup T $ are coherent this can be think as $P$ is logically independent of the theory $\endgroup$ – Anonyme Sep 16 at 8:01
  • $\begingroup$ But (if $PA$ is consistent) one of statement $P$ or $\neg P$ should be true in the standard model of arithmetic and the second one is false, right? And we can even say which one is true. $\endgroup$ – kp9r4d Sep 16 at 8:08
  • $\begingroup$ @kp9r4d you right we need to put caution on the difference of true In general (relatively to a theory $T$) and true relatively to a special model the point is that respectively to the model $\mathbb{N}$ for arithmetic we can find a true statement which is not provable but this statement is not true in general(in any model) ! $\endgroup$ – Anonyme Sep 16 at 8:11
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    $\begingroup$ Sure, but it seems to me to say "this is certainly false" is too harsh, usually when one talk about the falsity or truth of an arithmetic statement, one means exactly its falsity/truth in the standard model. Therefore, I do not see anything criminal in the statement "if $PA$ consistent, there is an arithmetic statement that is true but not provable" $\endgroup$ – kp9r4d Sep 16 at 8:17
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I once very briefly thought that:

Given a vector space $V$ and a sub-space $U \subset V$ that $V-U$ is also a subspace.

I've heard this several times as a TA also.

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  • $\begingroup$ Why the downvote! I heard this from more than one student in introductory linear algebra classes and when marking. $\endgroup$ – Benjamin May 12 '15 at 22:21
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    $\begingroup$ I think this falls under $(x+y)^2=x^2+y^2$, $\endgroup$ – Thomas Rot Aug 10 '15 at 12:48
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    $\begingroup$ It always fails... But I don't think this is a common held belief. $\endgroup$ – Thomas Rot Aug 10 '15 at 21:40
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    $\begingroup$ @ThomasRot But it always fails, while $(x+y)^2=x^2+y^2$ sometimes holds, especially in characteristic 2. $\endgroup$ – ACL Apr 21 '16 at 6:37
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    $\begingroup$ I meant that $V-U$ cannot be a subspace since it doesn't contain 0. On the other hand, in any commutative ring where $1+1=0$, then the formula $(x+ y )^2=x^2+y^2$ holds. $\endgroup$ – ACL Apr 21 '16 at 10:02
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I'm not sure how common it is but I've certainly been able to trick a few people into answering the following question wrong:

Given $n$ identical and independently distributed random variables, $X_k$, what is the limiting distribution of their sum, $S_n = \sum_{k=0}^{n-1} X_k $, as $n \to \infty$?

Most (?) people's answer is the Normal distribution when in actuality the sum is drawn from a Levy-stable distribution. I've cheated a little by making some extra assumptions on the random variables but I think the question is still valid.

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  • $\begingroup$ I don't understand your third paragraph. Are you saying that under the assumptions in the 2nd paragraph, the limiting distribution (rescaling if necessary) is always Levy-stable? $\endgroup$ – Yemon Choi Apr 12 '11 at 1:28
  • $\begingroup$ @Yemon, Yes, this is what I was implying. Perhaps I was a little too cavalier? Certainly the sum of (well enough behaved) i.i.d. r.v.'s with power law tails converge to a Levy-Stable distribution... $\endgroup$ – dorkusmonkey Apr 12 '11 at 23:53
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    $\begingroup$ Generally such a limiting distribution doesn't exist. Perhaps you need to divide your sum by the square root of $n$? $\endgroup$ – John Bentin Dec 29 '11 at 13:56
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I had the false belief that recursive functions are always decidable in ZFC.

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When I was a kid (8th grade), I solved a bunch of math problems in an exam using the ``well-known identity'' that $(x+y)^2=x^2+y^2$, which I was sure I had been taught the year before. It was of course way before I heard about characteristic two and I didn't get a good grade that day!

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    $\begingroup$ Quoth the question, "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)". $\endgroup$ – JBL Dec 1 '10 at 23:39
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    $\begingroup$ Also, this is of course just a special case of the more general “law of universal linearity”, which iirc was mentioned in earlier answers… $\endgroup$ – Peter LeFanu Lumsdaine Dec 2 '10 at 0:40
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I don't know if this is what you are looking for, but I keep hearing that "a differentiable function is one that is locally linear", not one whose local variation can be approximated linearly. No one stops to think about e.g, $x^2$, and the fact that its graph does not look like a line at any value of $x$.

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    $\begingroup$ I would say this is more a heuristic than a false statement; as such, it would be more appropriate as an answer to mathoverflow.net/questions/2358/most-harmful-heuristic (although I do not think anyone interprets it the way you apparently do). $\endgroup$ – Qiaochu Yuan May 5 '10 at 4:53
  • $\begingroup$ Yes, I did not read the question very carefully. I realize it is not a good comment, and, yes, it is more of a abd heuristic than anything else. $\endgroup$ – Herb May 25 '10 at 23:59
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    $\begingroup$ it is also a comment on the imprecision of the words locally, infinitesimally,.... This once led Oort-Steenbrink to give some careful restatements of results previously called as "local Torelli theorems"... $\endgroup$ – roy smith Apr 14 '11 at 19:02
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