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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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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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Maybe I am a bit late to the party, but here is something that I falsely believed in for a while:

False Belief: If $f$ is a continuously differentiable function with a horizontal asymptote, i.e. $\lim_{x\to +\infty}f(x)=L$, then $f'(x)\to 0$ as $x$ goes to infinity.

Of course, this is incorrect even for smooth functions; take $g$ to be a smooth $L^1$ function with oscillatory behavior so that $\lim_{x\to +\infty}g(x)$ does not exist and then set $f(x)=\int_{0}^xg(s)ds$.

I came to this misconception during an ODE course I took in my undergrad. In the study of the asymptotic behavior of solutions of an autonomous equation $$ \frac{dy}{dx}=F(y) $$ one argues that if a solution $y=y(x)$ tends to a limit $L$ as $x\to +\infty$, that limit must be a zero of the vector field. We then proceeds with determining which zeros of $F$ are asymptotically stable to find $L$. The point is that here not only $\lim_{x\to +\infty}y$, but also $\lim_{x\to +\infty}\frac{dy}{dx}$ exists due to the ODE. Hence:

Correct Statement: If $f$ is a continuously differentiable function with a horizontal asymptote, and if $f'$ admits a limit as $x$ goes to infinity, that limit must be zero.

This readily follows from the mean value theorem.

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    $\begingroup$ A related old answer. I have found that people usually need some prompting to think of an $L^1$ function that does not converge to $0$ at infinity, even though all one needs is a sequence of spikes of increasing height, but sufficiently rapidly decreasing width so that the sum of the areas under them converges. $\endgroup$ – Robert Furber Mar 3 at 17:02
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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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It is a common mistake to believe that epimorphisms are either identical to surjections or that they are a better concept. Unfortunately this is rarely the case; epimorphisms can be very mysterious and have unexpected behavior

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    $\begingroup$ Can you give a concrete example why it is a "mistake"? $\endgroup$ – Willie Wong Apr 25 at 13:15
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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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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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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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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 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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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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