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

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

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

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

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

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

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

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    $\begingroup$ Although it's not an Eilenberg-MacLane space, $\mathbb{HP}^\infty$ is a classifying space (specifically $B(SU(2))$), just as $\mathbb{CP}^\infty = B(U(1))$ and $\mathbb{RP}^\infty = B(\mathbb{Z}/(2))$. $\endgroup$ Mar 3 '20 at 15:55
  • $\begingroup$ @RobertFurber That's right. And because $BO(1)$ and $BU(1)$ are Eilenberg-Mac Lane spaces, you can immediately classify real and complex line bundles by a single cohomology class. But not quaternionic line bundles. $\endgroup$ Mar 4 '20 at 19:46
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(1) All Lebesgue-null sets are countable, or are strongly measure zero. (2) The following,verbatim, was a Q in American Mathematical Monthly : " A student asserted that any uncountable real set has a closed uncountable subset. Is this true ?" .

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People seem to believe that conventional computation (for example, running a chaotic irreversible cellular automaton) can be as efficient as one wants simply with good engineering, but this is not the case. Landauer's principle states that erasing a bit of information always takes $\ln(2)\cdot k\cdot T$ energy where $k$ is Boltzmann's constant ($k=1.38065\cdot 10^{-23}$ Joules/Kelvin) and $T$ is the temperature. Landauer's principle is a consequence of the second law of thermodynamics since if Landauer's principle were violated, then entropy would decrease. Landauer's principle means that conventional irreversible computation always must take $\ln(2)\cdot k\cdot T$ energy per bit erased (and one can erase data just by running it through AND and OR gates, so every irreversible gate must take a minimum amount of energy by Landauer's principle). However, Landauer's principle does not apply to reversible computation since reversible computers are not allowed to erase data.

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    $\begingroup$ Okay, I see what you are aiming at with this last edit. The idea that ordinary computation can be made arbitrarily efficient is a reasonably common false belief about our physical world (and may even have a somewhat solid mathematical interpretation). I withdraw my initial objection. $\endgroup$
    – S. Carnahan
    Aug 11 '17 at 14:04
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The GNS construction

Let $\phi$ be a state on a $C^*$-algebra $\mathcal{A}$, and put $N_{\phi}:=\{a\in\mathcal{A}\;|\;\phi(a^*a)=0\}$. Then $N_{\phi}$ is a norm-closed left ideal in $\mathcal{A}$. The sesquilinear form $\left<\cdot,\cdot\right>:\mathcal{A}/N_{\phi}\times\mathcal{A}/N_{\phi}\to\mathbb{C}$ defined by $\left<a+N_{\phi},b+N_{\phi}\right>:=\phi(b^*a)$ is a well-defined inner product on $\mathcal{A}/N_{\phi}$. The completion of $\mathcal{A}/N_{\phi}$ establishes a Hilbert space.

False belief: The completion is in the quotient norm.

Surprisingly, Wikipedia (as of April 27, 2018) presents a false statement "The quotient space of the A by the vector subspace I is an inner product space. The Cauchy completion of A/I in the quotient norm is a Hilbert space, which we label H."(https://en.wikipedia.org/wiki/Gelfand%E2%80%93Naimark%E2%80%93Segal_construction#The_GNS_construction) First of all, the quotient of a Banach space by its closed subspace is again a Banach space in the quotient norm, which is a very elementary fact in functional analysis. Thus A/I is already complete in the quotient norm, and hence there is no need to complete it in the quotient norm!

The correct completion is, of course, in the norm induced by the inner product, and this norm is not equivalent to the quotient norm in general. In fact, let $\mathcal{H}$ be a separable infinite-dimensional Hilbert space and $\{\xi_n\}_{n=1}^{\infty}$ be an orthonormal basis for $\mathcal{H}$. The linear functional $\phi:\mathbb{B}(\mathcal{H})\to\mathbb{C}$ defined by $\phi(a):=\sum_{n=1}^{\infty}\frac{1}{2^n}\left<a\xi_n,\xi_n\right>$ is a state on $\mathbb{B}(\mathcal{H})$, and $N_{\phi}=\{0\}$. Let $\xi_k\otimes\xi_k$ be the canonical rank-one operator, and put $p_n:=\sum_{k=1}^n\xi_k\otimes\xi_k$. Then $(p_n)_{n=1}^{\infty}$ is a Cauchy sequence in $\mathbb{B}(\mathcal{H})/N_{\phi}$ in the norm induced by the inner product defined at the beginning, but it is NOT a Cauchy sequence in the quotient norm.

I have never seen a remark which clearly states the distinction between the norm induced by the inner product and the quotient norm in the literature on $C^*$-algebras. Since a quotient space is involved, students are easily tempted to think that the completion is in the quotient norm. (Even the Wikipedia editor was confused!) Or, they may thoughtlessly assume that these two norms are the same. So it will be instructive to clearly state the distinction between these two norms when one teaches this subject to undergraduate students.

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    $\begingroup$ I think it is surprising to be surprised by a wrong statement on Wikipedia. Fortunately it need not remain wrong! $\endgroup$
    – LSpice
    Apr 27 '18 at 22:31
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    $\begingroup$ @LSpice: Well, the surprising thing is that this false statement appeared on May 6, 2004 and has remained since then, and nobody has corrected the error for 14 years! See the editing history. $\endgroup$ Apr 27 '18 at 22:55
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    $\begingroup$ I read the "quotient norm" statement as saying that it's the quotient of the seminorm induced by the inner product. Then it's correct, no? $\endgroup$ Apr 28 '18 at 12:24
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    $\begingroup$ @MasayoshiKaneda: I agree that the Wikipedia entry has been ambiguous at that point, and I've clarified it. $\endgroup$ Apr 29 '18 at 7:14
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    $\begingroup$ @Tobias Fritz: Good job! I also modified the sentence before the one you clarified. $\endgroup$ May 1 '18 at 10:51
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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? $\endgroup$ Aug 28 '20 at 12:06
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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 '20 at 12:19
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Here's one that I think will surprise some number theorists:

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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The following analogue of this result is clearly false:

Falsehood. If $M$ is a module over a commutative ring $R$, then $M^\vee = \operatorname{Hom}_R(M,R)$ is at least as big as $M$ (e.g. in terms of cardinality or rank).

For example, if $R$ is a domain and $M$ is torsion, then $M^\vee = 0$. But what's much more surprising is that the following is still false:

False belief: If $M$ is a torsion-free module over a principal ideal domain $R$ (even $R = \mathbf Z$), then $|M^\vee| \geq |M|$ and/or $\operatorname{rk}(M^\vee) \geq \operatorname{rk}(M)$.

(Even assuming $M$ has no divisble elements doesn't help.)

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    $\begingroup$ One might add that what is being overlooked here is that the "correct dualizer" is an injective cogenerator, rather than a projective generator such as $R$. What makes it work for vector spaces is that a field is an injective cogenerator over itself. By the way, what are the rings for which injective hull of $R$ works? It seems to always work for torsion frees but not for all - even over integers. Presumably $R$ must be local for that? $\endgroup$ Feb 7 at 6:37
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Fans: (related to the one of polytopes written above) all convex cones are rational, i.e. one would expect that a line would eventually hit a point in the lattice. It is obviously not true, just take the one-dimensional cone generated by $(1,\sqrt{2})$. A similar one was thinking that if I rotate the cone a bit, I can always make it rational.

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    $\begingroup$ reminds me of the curious fact that some circles in the plane, too, have no points in $\mathbb Q^2$. (proven simply by cardinality!) $\endgroup$ Oct 4 '10 at 19:21
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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 '20 at 20:09
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Assume that $a,b\in \mathbb{R}\setminus \{0\}$ which satisfy $a^{3}= 2b^{3}$.

Then $a-2b$ is a non zero nilpotent element of group ring $\mathbb{Z}_{3} \mathbb{R}$, that is $(a-2b)^{3}=0$.

This would be a counterexample to the zero divisor Kaplansky conjecture

The false lies in an obvious abuse in the definition of the group ring multiplication.

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    $\begingroup$ This does not seem like a common false belief. $\endgroup$
    – Yemon Choi
    May 15 '16 at 11:47
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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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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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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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If an Abelian category $\mathcal{A}$ is a full subcategory of an Abelian category $\mathcal{B}$, then for all objects $M,N$ of $\mathcal{A}$, we have an injection $$\operatorname{Ext}^i_{\mathcal{A}}(M,N) \hookrightarrow \operatorname{Ext}^i_{\mathcal{B}}(M,N).$$

As an example, let $G$ be the free group on $2$ letters, $A$ its abelianization, $\mathcal{B} = G-mod$, $\mathcal{A}=A-mod$, and $M=N=\mathbb{Z}$ with the trivial action. Then $\operatorname{Ext}^i_{\mathcal{A}}(M,N) \cong \mathbb{Z}$, while $\operatorname{Ext}^i_{\mathcal{B}}(M,N) \cong 0$.

(This example comes from the topological fact that a torus has nontrivial $H^2$, while a punctured torus has trivial $H^2$. In algebra, it's related to the idea that group homology $H_1$ is space of generators for a group while $H_2$ is a space of relations.)

This false belief came up a context where $\mathcal{B}$ was the category of all Galois representations while $\mathcal{A}$ was a certain subcategory. See the comments to Status of the conjectured vanishing of Bloch-Kato H^2.

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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 https://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, https://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. $\endgroup$ Apr 10 '16 at 17:55
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Initially when I started studying sequences I believed that:

Consider $(x_n)$ and $(y_n)$ are two convergent real sequences having limits $x$ and $y$ as limits respectively. If $ x_n < y_n \quad \forall n\in \mathbb{N}$ then $x < y$

which turned out to be false.

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    $\begingroup$ but the question is "less interested in very elementary false statement like $(x+y)^2=x^2+y^2$ " $\endgroup$ Aug 16 at 9:21
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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? $\endgroup$ Feb 11 '16 at 21:36
  • $\begingroup$ $V$ is a vector space over field $P$. $\endgroup$ Feb 12 '16 at 6:52
  • $\begingroup$ So, what does "$V$ is a vector space over field $P$ $(e_1,\dots,e_n)$" mean? $\endgroup$ 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 $W$. $\endgroup$ 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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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? $\endgroup$ 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! $\endgroup$ 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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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 $\endgroup$ 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. $\endgroup$ 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. $\endgroup$ 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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I cannot believe this example was not yet given (but, if my belief is false, I will happily delete this answer):

It is very common among "lay people" (who do not understand what it means for lines to be parallel) to believe that "in some kind of geometry" (frequently described as non-euclidean) parallel lines can intersect.

One finds many instances (my guess, the count is in hundreds of thousands) of this false believe just by searching the internet. Here is a random example, the article "How Looking At A Basketball Disproves Something Everybody Learns In High School Geometry" from "Business Insider", 2014. The article concludes with

And voila! We’ve successfully disproven “parallel lines never intersect” using just a basketball.

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    $\begingroup$ i.redd.it/r7etb5kayl961.jpg $\endgroup$ Feb 9 at 4:39
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    $\begingroup$ I feel like this is a confusion of terminology and not mathematics. The definition of parallel being used in the Business Insider article is clearly not 'lines that do not meet and are a constant distance apart' or whatever the official definition of parallel is. It is closer to 'distinct lines that go in the same direction'. Insofar as there is a false belief, it is that there is a coherent notion of 'same direction'! But this is not essential to the belief that lines that go in the 'same direction' can meet in some kind of geometry, which is a belief that is closer to true than false. $\endgroup$
    – Chan Bae
    Feb 9 at 13:37
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    $\begingroup$ @ChanBae Mathematics starts by establishing common terminology and axioms. This is what Greeks realized over 2000 years ago. Sadly, this understanding was lost with changes in math education in the last century. The thing is, math is part language and part science. You cannot separate the two and claim that inability to understand definitions is just a matter of terminological disagreements. $\endgroup$ Feb 10 at 1:37
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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 $n=1$. $\endgroup$ 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? $\endgroup$ 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. $\endgroup$ Jan 16 '19 at 5:11
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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... $\endgroup$ 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 $n$? $\endgroup$ Dec 29 '11 at 13:56
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