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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, 2010 at 0:55
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    $\begingroup$ The answers below are truly informative. Big thanks for your question. I have always loved your post here in MO and wordpress. $\endgroup$
    – Unknown
    May 22, 2010 at 9:04
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    $\begingroup$ wouldn't it be great to compile all the nice examples (and some of the most relevant discussion / comments) presented below into a little writeup? that would make for a highly educative and entertaining read. $\endgroup$
    – Suvrit
    Sep 20, 2010 at 12:39
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    $\begingroup$ It's a thought -- I might consider it. $\endgroup$
    – gowers
    Oct 4, 2010 at 20:13
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    $\begingroup$ Meta created tea.mathoverflow.net/discussion/1165/… $\endgroup$
    – user9072
    Oct 8, 2011 at 14:27

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This is (I think) a fairly common misconception about maths that arises in connection with quantum mechanics. Given a Hermitian operator A acting on a finite dimensional Hilbert space H, the eigenvectors of A span H. It's easy to think that the infinite dimensional case is "basically the same", or that any "nice" operator that physicists might want to consider has a spanning eigenspace. However, neither the position nor the momentum operator acting on $L^2(\mathbb{R})$ have any eigenvectors at all, and these are certainly important physical operators! Based on an admittedly fairly small sample size, it seems that it's not uncommon to simultaneously believe that Heisenberg's uncertainty relation holds and that the position and momentum operators possess eigenvectors.

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    $\begingroup$ Yeah, for some reason many physicists are taught exactly no functional analysis... In fact, I know of no "quantum mechanics for physicists" books which use much more than a beginning undergrad level of analysis. Though admittedly these details are not so important for doing simple calculations, though they can be important in doing more sophisticated calculations, or understanding, e.g., why field theory works the way it does... $\endgroup$
    – jeremy
    Jun 1, 2010 at 23:33
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    $\begingroup$ Reciprocally, many mathematicians are taught no quantum mecha... make it, no physics at all! This is shocking, since the biggest impetus to the development of PDEs and functional analysis was given by what? You guessed it, physics. $\endgroup$ Jun 10, 2010 at 6:56
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I got 2 well earned downvotes for a false belief I claimed proudly, it is time to balance that by exposing it here:

Let $(P,\le)$ be any poset, and let $\le^*$ be an order on $P$ extending $\le$. Any Endomorphism of $\le^*$ also is an endomorphism of $\le$

($f:P\to P$ endomorphism of $\le$ meaning $x\le y \implies f(x)\le f(y)$).

Of course this is a particular case of a very general fallacy: by extending $\le$ into $\le^*$ one weakens both the conclusion and the premise of the implication, so that there is no general relation between orders that extend one another.

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  • $\begingroup$ More generally, $\operatorname{End}(X)$ is often not functorial in $X$! $\endgroup$ Jun 19, 2020 at 3:54
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To my knowledge, noone has proven that the scheme of pairs of matrices (A,B) satisfying the equations AB=BA is reduced. But whenever I mention this to people someone says "Surely that's known to be reduced!"

(Similar-sounding problem: consider matrices M with $M^2=0$. They must be nilpotent, hence have all eigenvalues zero, hence $Tr(M)=0$. But that linear equation can't be derived from the original homogeneous quadratic equations. Hence this scheme is not reduced.)

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  • $\begingroup$ I remember Etingof thinking it was open as recently as 2006 in his lectures on Calogero-Moser systems. Also, it's fairly easy to check that in small cases the commutator scheme is reduced, though getting it in general seems to be beyond current technology. $\endgroup$ Jun 6, 2010 at 21:13
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    $\begingroup$ Sadly, people rely on technology so much nowadays that it gets increasingly unlikely that it will $\textit{ever}$ be proved. $\endgroup$ Jun 10, 2010 at 6:40
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A stunning, ignorance-based false belief I have witnessed while observing a class of a math education colleague is that there is no general formula for the n-th Fibonacci number. I wonder if this false belief comes from conflating the (difficult) lack of formulas for prime numbers with something that is just over the horizon of someone whose interests never stretch beyond high-school math.

Behind a number of the elementary false beliefs listed here there is a widespread tendency among people to give up too easily (maybe when having to read at least to page 2 in a book), or to nourish an ego that allows to conclude that something is impossible if they cannot do it themselves.

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    $\begingroup$ I hope at least your colleague had it right! There is another one along these lines: there is no formula for the sequence $1,0,1,0,1,0,... .$ Your second paragraph is right on target, but I also think that the specific beliefs you and I mentioned have a lot to do with a very limited understanding of what is a "formula". $\endgroup$ Jun 10, 2010 at 7:02
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    $\begingroup$ Perhaps the people who believe this are using the meta-reasoning that the sequence would not be interesting as an example of recursion if it could be solved exactly. Since it is a popular example of recursion, then... $\endgroup$
    – Ryan Reich
    Oct 5, 2011 at 17:01
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    $\begingroup$ When, as an undergrad, I couldn't solve a problem given to me by the advisor, and asserted that it's "unsolvable", the advisor replied that "solvability of a problem is a function of two arguments: the problem and the solver." $\endgroup$
    – Michael
    Dec 3, 2013 at 1:14
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    $\begingroup$ @Michael (The constant function is still a function. :P ) $\endgroup$ Sep 1, 2015 at 0:03
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It always confused me as an undergraduate that $\mathbb{Q}\subset\mathbb{R}$ has an open neighborhood $N_\epsilon \supset\mathbb{Q}$ of arbitrarily small measure $\epsilon$, because $\mathbb{Q}$ is dense in $\mathbb{R}$.

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    $\begingroup$ (This answer is very similar to Owen Sizemore's answer that is currently on the first page.) $\endgroup$ Jul 24, 2017 at 14:41
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$\mathbb{R}^2$ has a unique complex manifold structure; it's just $\mathbb{C}$ right?

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

So, what is the error?

The error is the following:

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

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    $\begingroup$ This is not a common false belief except among people who do not understand the definition of an algebra over a field $\endgroup$
    – Yemon Choi
    Nov 12, 2014 at 23:43
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    $\begingroup$ Moreover, surely the quaternions are a real vector space, not a complex vector space $\endgroup$
    – Yemon Choi
    Nov 13, 2014 at 1:34
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    $\begingroup$ @YemonChoi the field of complex number is a subring of the ring of quaternions. So quaternions is a complex vec. space.More generaly if a ring R contains a field F then R is a F-vector space,Ok? $\endgroup$ Nov 13, 2014 at 5:31
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    $\begingroup$ @YemonChoi I think this example is perhaps interesting unless a participant do not read it carefully. please think again to the main motivation and aim of the question of "common false..." $\endgroup$ Nov 13, 2014 at 5:44
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    $\begingroup$ @AliTaghavi You're right that $R$-multiplication induces an $F$-module ($F$-vector space) structure via the evident composite $F \times R \to R \times R \to R$. To be fair to both you and Yemon: a very common slip even among professionals is in knowing that for commutative algebras an $F$-algebra is tantamount to a homomorphism $F \to R$, but temporarily forgetting this doesn't apply in the noncommutative setting (except of course when $F$ is central in $R$) -- not rising to the level of false belief so much as a temporary slip-up. I've made that slip myself! $\endgroup$
    – Todd Trimble
    Nov 13, 2014 at 11:58
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There are cases that people know that a certain naive mathematical thought is incorrect but largely overestimate the amount by which it is incorrect. I remember hearing on the radio somebody explaining: "We make five experiments where the probability for success in every experiment is 10%. Now, a naive person will think that the probability that at least one of the experiment succeed is five times ten, 50%. But this is incorrect! the probability for success is not much larger than the 10% we started with."

Of course, the truth is much closer to 50% than to 10%.

(Let me also mention that there are various common false beliefs about mathematical terms: NP stands for "not polynomial" [in fact it stands for "Nondeterministic Polynomial" time]; the word "Killing" in Killing form is an adjective [in fact it is based on the name of the mathematician "Wilhelm Killing"] etc.)

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    $\begingroup$ And the Killing field has nothing to do with Pol Pot. $\endgroup$ May 5, 2010 at 14:40
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    $\begingroup$ Unfortunately I often slip up in class and say that the Killing vector field $T$ kills the metric term (well, I use the verb kills when a differential operator hits something and makes it zero, because, you know, bad terms are always "the enemy"). I'm not sure how much damage I did to the students' impressions... $\endgroup$ May 5, 2010 at 17:19
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    $\begingroup$ "Kills" is one of those terms I hear mathematicians use surprisingly often. The other one is "this guy." I never really understood the prevalence of either. $\endgroup$ May 6, 2010 at 7:38
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    $\begingroup$ "Guy" is a pretty standard English colloquialism for "person"; combine this with humans' tendency to anthropomorphize and this usage is understandable. (Though we shouldn't anthropomorphize mathematical objects, because they hate that.) $\endgroup$ May 6, 2010 at 14:51
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    $\begingroup$ In the only lecture I saw by David Goss he started with "guy", quickly went to something like "uncanny fellow" and then stayed with "sucker" for most of the talk. I don't know what those poor Drinfeld modules had done to him the day before :-) $\endgroup$ May 19, 2010 at 12:24
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Here's a little factoid: (The Mean-value theorem for functions taking values in $\mathbb{R} ^n$.) If $\alpha : [a,b]\rightarrow \mathbb{R}^n$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then there exists a $c\in (a,b)$ such that $\frac{\alpha (b)-\alpha (a)}{b-a}=\alpha '(c)$

A counterexample is the helix $(\cos (t),\sin (t), t)$ with $a=0$, $b=2\pi$.

Another common misunderstanding (although not mathematical) is about the meaning of the word factoid. In fact, the common mistaken definition of the word factoid is factoidal.

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    $\begingroup$ On the other hand, perhaps the most useful corollary of the mean value theorem is the "mean value inequality": that $|\alpha(b) - \alpha(a)| \le (b-a) \sup_{t \in [a,b]} |\alpha'(t)|$. If you look carefully, most applications of the MVT in calculus are really using this "MVI". The MVI remains true for absolutely continuous functions taking values in any Banach space, and so is probably the right generalization to keep in mind. $\endgroup$ May 6, 2010 at 14:37
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    $\begingroup$ According to at least one dictionary, there are two different definitions of factoid: (1) an insignificant or trivial fact, and (2) something fictitious or unsubstantiated that is presented as fact, devised especially to gain publicity and accepted because of constant repetition. I am not convinced that the multi-d mean value “theorem” fits either definition. $\endgroup$ May 8, 2010 at 19:09
  • $\begingroup$ Related to the M-V Thm is the following fact. If $f:I=(a,b)\rightarrow{\mathbb R}$ is differentiable (not necessarily ${\mathcal C}^1$), then $f'(I)$ is connected (i.e. is an interval). This is false when $f:I=(a,b)\rightarrow{\mathbb R}^n$ is differentiable, and $n\ge2$. $\endgroup$ Oct 20, 2010 at 10:48
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By googling one sees that each of the following statements has a significant number of believers:

(1) the vector space {0} has no basis,

(2) the empty set is a basis of {0} by convention,

(3) the statements "{0} has no basis" and "the empty set is a basis of {0}" are equivalent,

(4) the statements "{0} has no basis" and "the empty set is a basis of {0}" are NOT equivalent,

(5) the statement "the empty set is a basis of {0}" is an immediate consequence of the definitions of the terms involved.

I think that we'll all agree that the 5 beliefs are not ALL true. My personal religion is to believe in (4) and (5). I don't think I'll ever understand the arguments in favor of (1), (2) or (3).

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    $\begingroup$ I feel like there are a lot of areas in mathematics in which the empty set is interpreted in a certain way (for example, the empty product is one, the empty sum is zero, the empty set has one map into any non-empty set, etc). Given each of these particular situations locally, I might agree that it is a convention in each case. However, given the ubiquity of such "conventions," one might think that there is a uniform description of what the empty set really "means" in these contexts. If this becomes the case, then I might argue for (5), which would follow from this conception of the empty set. $\endgroup$ Jul 7, 2010 at 23:48
  • $\begingroup$ Given that the free space on the empty set is the zero space (high-fallutin general nonsense-maximizing proof: free-ification is a left adjoint => it is cocontinuous => takes initials to initials + the initial vector space is the zero space and the initial set is the empty set), and that for free spaces $F(X)$, $X$ is a basis, I would definitely say (4) and (5). $\endgroup$ Jul 22, 2010 at 13:39
  • $\begingroup$ I think one can chase the controversy here down a little further, to the statement: "the sum of the empty set is 0". I think most people who accept this then accept (5). $\endgroup$ Sep 27, 2010 at 2:59
  • $\begingroup$ I don't see how anybody could use language such that (3) is true and (4) is false. After that, it is up to how the terms are defined, but (of course!) I agree that (5) is the way to go here. $\endgroup$ Apr 4, 2011 at 9:13
  • $\begingroup$ At least, $\{0\}$ is a vector space. I have seen "a vector space has at least two elements" from a professional mathematician. $\endgroup$
    – user11235
    Apr 10, 2011 at 21:28
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There is a bijection between the set of [true: prime!] ideals of $S^{-1}R$ and the set of [true: prime!] ideals of $R$ which do not intersect $S$.

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    $\begingroup$ Yes! In a review of a text on commutative algebra I have suggested to extend the prime ideal correspondence in localizations to some ideal correspondence, because I wasn't aware that we have to actually use the prime ideal condition somewhere ... $\endgroup$ Apr 12, 2011 at 8:41
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The cost of multiplying two $n$-digit numbers is of order $n^2$ (because each digit of the first number has to be multiplied with each digit of the second number).


A lot of information is found on http://en.wikipedia.org/wiki/Multiplication_algorithm .

The first faster (and easily understandable) algorithm was http://en.wikipedia.org/wiki/Karatsuba_algorithm with complexity $n^{log_2 3} \sim n^{1.585}$.

Basic idea: To multiply $x_1x_2$ and $y_1y_2$ where all letters refer to $n/2$-digit parts of $n$-digit numbers, calculate $x_1 \cdot y_1$, $x_2\cdot y_2$ and $(x_1+x_2)\cdot(y_1+y_2)$ and note that this is sufficient to calculate the result with three such products instead of four.

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    $\begingroup$ It would be better if these misconceptions would come with explanations how things really are... $\endgroup$ Apr 10, 2011 at 18:28
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    $\begingroup$ Along these lines: there is a widespread misapprehension that multiplication is the same thing as a multiplication algorithm (whichever one the speaker learned in elementary school). $\endgroup$ Apr 10, 2011 at 19:25
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    $\begingroup$ At least it's better than people thinking multiplication is constant-time. :P $\endgroup$ Apr 10, 2011 at 19:35
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I confess that I didn't carefully comb through all the answers, although I've read through this thread a few times in the past. So maybe these are repeats.

  • "The category of compact Hausdorff spaces is complete but not cocomplete; for example, it doesn't have all coproducts."

  • "The category of torsion abelian groups is cocomplete but not complete; for example, it doesn't have all products."

One of my professors in graduate school (quite a well-known and strong mathematician actually) insisted on the first, and quite a few people here at MO have mistakenly believed the second before the error was pointed out.

The moral of the story: sometimes categorical limits/colimits aren't computed the way you might first think of, e.g., colimits of compact Hausdorff spaces aren't always computed as colimits in $\mathrm{Top}$, and limits of torsion abelian groups aren't always computed as limits in $\mathrm{Ab}$.

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  • $\begingroup$ Left adjoints preserves colimits, therefore a colimit taken in CHaus is "just" the Stone-Čech compactification of the colimit taken in Top. $\endgroup$
    – user20948
    Feb 8, 2021 at 21:07
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This one has bit me and some very good mathematicians I know.

Let $X,Y$ be Banach spaces, and let $E \subset X$ be a dense subspace. Suppose $T : E \to Y$ is a bounded linear operator. Then $T$ has a unique bounded extension $\tilde{T} : X \to Y$. (True, this is the well-known and elementary "BLT theorem".)

If $T$ is injective then so is $\tilde{T}$. (False! See this answer for a counterexample.)

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Let $V$ be a vector space. Then the intersection of $n$ hyperplanes (i.e. subspaces of codimension 1) is a subspace of codimension at most $n$.

So, naturally, the intersection of countably many hyperplanes is a subspace of countable codimension. Hence if $V$ is of uncountable dimension, this intersection is non-trivial.

Except this is of course wrong. For example, consider the vector space $V:=\mathbb{K}^{\mathbb{Z}}:=\{ f:\mathbb{Z}\to\mathbb{K} \}$ of uncountable dimension. The kernels of the projections $\pi_i:V\to\mathbb{K},\ f\mapsto f(i)$ are hyperplanes. Their intersection is the trivial subspace of $V$, and thus has uncountable codimension.

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    $\begingroup$ Is it clear that $\mathbb{K}^{\mathbb{Z}}$ is of uncountable dimension for all $\mathbb{K}$ ? (for uncountable $\mathbb{K}$ (like $\mathbb{R}$ or $\mathbb{C}$) one can consider the functions $\frac{1}{X-a}$ with $a\in \mathbb{K}^\times$ which are independant, identifying $\mathbb{K}^{\mathbb{N}}$ with $\mathbb{K}[[X]]$) $\endgroup$ Oct 29, 2017 at 20:59
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    $\begingroup$ OK, you're right, I found the "Erdos-Kaplansky theorem" in Bourbaki (Algebra II § 7 ex. 3) which explains that $dim(\mathbb{K}^J)$ ($J$ infinite) is $card(\mathbb{K})^{card(J)}$. $\endgroup$ Oct 30, 2017 at 7:47
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False belief: ${\cal P}(\omega)$ has only countable chains with respect to $\subseteq$.

It seems mind-boggling to me that you can start with $\emptyset$, and "add stuff" uncountably many times until you reach $\mathbb{N}$ itself! I learnt this today in a comment by Andreas Blass that he wrote referring to this question.

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    $\begingroup$ Well, if "adding stuff ... many times" is meant in an iterative sense, i.e., is parametrized by an ordinal, then your original intuition would be right. $\endgroup$
    – Todd Trimble
    Oct 8, 2017 at 2:34
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    $\begingroup$ On a similar line: the Hilbert space $\ell_2$ has a family of closed linear subspaces, parametrized on the unit interval, and increasing by inclusion. This looks a bit paradoxical. Should we add orthogonal vectors uncountably many times as $t$ increases from $0$ to $1$? No: just think to $L^2[0,t]$ as subspaces of $L^2[0,1]$ for $0\le t\le 1$ $\endgroup$ Oct 22, 2017 at 14:21
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"Every 1-dimensional knot in $R^n, n\ge 4,$ is trivial." This is true for tame knots and false for wild knots. See here.

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In geometric combinatorics, there is a widespread belief that polytopes of equal volume are not scissor congruent (as in Hilbert's third problem) only because their dihedral angles are incomparable. The standard example is a cube and a regular tetrahedron, where dihedral angles are in $\Bbb Q\cdot \pi$ for the cube, and $\notin \Bbb Q\cdot \pi\ $ for the regular tetrahedron. In fact, things are rather more complicated, and having similar dihedral angles doesn't always help. For example, the regular tetrahedron is never scissor congruent to a union of several smaller regular tetrahedra (even though the dihedral angles are obviously identical). This is a very special case of a general result due to Sydler (1944).

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    $\begingroup$ Oh, I previously thought that eight regular tetrahedra with side length 1 fit in one with side length 2. That might be a more common false belief. $\endgroup$
    – Junyan Xu
    May 5, 2012 at 7:45
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Sequence $\{a_n\}$ has a limit $A$ in $\mathbb{R}$ and a limit $B$ in $\mathbb{Q}_p$. Then $A$ is rational iff $B$ is rational.

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    $\begingroup$ Or: if a sequence has a rational limit in Q_p and in Q_r, then they're the same. $\endgroup$ May 5, 2010 at 4:16
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    $\begingroup$ But if a rational sequence has a limit in all Q_p, including Q_\infty ... $\endgroup$ May 5, 2010 at 12:17
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    $\begingroup$ It reminds me of this fake proof that $\pi$ is an irrational number. $\endgroup$
    – Watson
    Jan 10, 2020 at 9:22
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The following false belief enjoyed a certain success in the '70. (See R.S.Palais, Critical point theory and the minimax principle for an account.)

A second countable, Hausdorff, Banach manifold is paracompact.

Regular is necessary, otherwise there are counterexamples!

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The distinction between convergence and uniform convergence. It even got Cauchy in its time.

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

A better, but still false, version:

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

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

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

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

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

  • It is common to imagine rotation in an n-dimensional space, as a rotation through an "axis". this is of course true only in 3D, In higher dimensions there is no "axis".

  • In calculus, I had some troubles with the following wrong idea. A curve in a plane parametrized by a smooth function is "smooth" in the intuitive sense (having no corners). the curve that is defined by $(t^2,t^2)$ for $t\ge0$ and $(-t^2,t^2)$ for $t<0$ is the graph of the absolute value function with a "corner" at the origin, though the coordinate functions are smooth. the "non-regularity" of the parametrization resolves the conflict.

  • When first encountering the concept of a spectrum of a ring, the belief that a continuous function between the spectra of two rings must come from a ring homomorphism between the rings.

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    $\begingroup$ What do you mean by "In higher dimensions there is no "axis" but a n-2 dimensional subspace instead" ? Whenever n is even, there are rotations without real eigenvectors. $\endgroup$ Apr 14, 2011 at 13:12
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    $\begingroup$ Unfortunately, "smooth" is a word which means whatever its utterer does not want to specify. Differentiable, C^infty, continuous, everything is mixed. $\endgroup$ Apr 14, 2011 at 15:12
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    $\begingroup$ +1 for the discrete $\neq$ totally disconnected example. $\endgroup$
    – Jim Conant
    May 4, 2011 at 15:12
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    $\begingroup$ It's still possible to have an example where the parametric definition is a smooth (as in $ C^{\infty} $) function of time but you get a corner, right? $\endgroup$ May 9, 2011 at 14:46
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    $\begingroup$ Discrete $\ne$ totally disconnected is a good one that I thought of today and just had to check to see if it was posted already. It adds to the confusion that every finite subset of a totally disconnected space must have the discrete topology, and that in most topological spaces encountered "in nature," the connected components are open sets. $\endgroup$ Oct 20, 2011 at 14:30
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"The universal cover of $SL_2(R)$ is a universal central extension" (which I believed until recently...)

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It took me a bit too long to realize that these two beliefs are contradictory:

  • Period 3 $\Rightarrow$ chaos: if a continuous self-map on the interval has a period-3 orbit, then it has orbits of all periods.
  • The black dots on each horizontal slice of this picture above $x=a$ show the location of the periodic points of the logistic map $f_a(y) = ay(1-y)$: Bifurcation diagram for the logistic map

You can clearly see a 3-cycle in the light area towards the right; yet we know that if there is a 3-cycle in that slice then there must be a cycle of any period in that slice... so where are they?

(The other cycles are there of course, but they are repelling and hence are not visible. You can see artifacts from these repelling cycles near the period-doubling bifurcations in this picture)

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Piggybacking on one of Pierre's answers, I once had to teach beginning linear algebra from a textbook wherein the authors at one point stated words to the effect that the the trivial vector space {0} has no basis, or that the notion of basis for the trivial vector space makes no sense. It is bad enough as a student to generate one's own false beliefs without having textbooks presenting falsehoods as facts.

My personal belief is that the authors of this text actually know better, but they don't believe that their students can handle the truth, or perhaps that it is too much work or too time-consuming on the part of the instructor to explain such points. Whatever their motivation was, I cannot countenance such rationalizations. I told the students that the textbook was just plain wrong.

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    $\begingroup$ Bjorn Poonen once gave a lecture at MIT about the empty set; it really opened my eyes. If someone wrote a textbook or something on the matter I think everyone would be a lot less confused. $\endgroup$ Jul 7, 2010 at 23:56
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    $\begingroup$ I can combine Qiaochu's and Victor's remarks in this memory I have of a coffee break conversation between two colleagues, who were arguing on whether it made sense to say that the 1-element group acts on the empty set. I wisely decided to stay out of the controversy... $\endgroup$ Aug 31, 2010 at 2:24
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    $\begingroup$ Thierry: of course it makes sense. But the action is not transitive. $\endgroup$
    – ACL
    Dec 1, 2010 at 22:53
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    $\begingroup$ @kow: I disagree. That is the "wrong" definition of transitivity for empty G-sets. See the discussion at qchu.wordpress.com/2010/12/03/empty-sets . $\endgroup$ Dec 16, 2010 at 23:08
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    $\begingroup$ I once taught abstract algebra from a book that adopted the artificial convention that the domain of a map of sets must be nonempty. I eventually figured out that the reason was in order to be able to say that every one-to-one map has a left inverse. And I have many times taught topology from a book that adopts the artificial convention that when speaking of the product of two spaces we require both spaces to be nonempty. I eventually figured out that the reason was in order to be able to say that $X\times Y$ is compact if and only if both $X$ and $Y$ are compact. $\endgroup$ Mar 14, 2012 at 22:01
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I'm pretty sure I've heard both of the following multiple times:

  1. Transfinite induction requires the axiom of choice. False, though many applications of transfinite induction require axiom of choice (either in the form of the well-ordering theorem, or directly (though using transfinite induction together with choice directly is essentially the same as just using Zorn's Lemma)).

  2. Transfinite induction requires the axiom of foundation. I guess some people get transfinite induction mixed up with epsilon-induction?

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Inversion is an automorphism of a group. ('Cause it, like, preserves the conjugacy classes and all that...)

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" Every open dense subset of $\mathbb{R}^n$ has full Lebesgue measure. "

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  • $\begingroup$ (See Craig's answer below) $\endgroup$ Dec 9, 2013 at 12:13
  • $\begingroup$ Well, although similar to Craig's answer on an open neighborhood of $\mathbb{Q}$ with arbitrarily small measure, I find this formulation much more appealing. $\endgroup$
    – Dirk
    Jan 2, 2014 at 17:18
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Here's one from basic set theory. Let k be a cardinal and consider the operation "adding k", meaning

$l \mapsto k+l$

on cardinals. We know that this operation "stabilizes" to the identity after $k$, that is, for any $l>k$, we have $l+k = l$. Similarly, the "multiplying by $k$" operation,

$l \mapsto l * k$

stabilizes to the identity after $k$.

Everyone also knows that if $l$ is an infinite cardinal then $l^2$ is equipotent to $l$, and more generally $l^n$ is equipotent to $l$ for every natural number $n$. I.e. all the finite power functions stabilize to the identity at $\omega$.

Well, obviously "exponentiation by $\omega$" also stabilizes at some point, right? Like, $l^\omega$ is equal to $l$ for sufficiently large $l$? Look, we probably already have the stabilization point at $2^\omega$.

Right?

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    $\begingroup$ Why not? As an algebraist, my reaction already after "addition of k stabilizes" would be "if THAT holds, than WHATEVER". $\endgroup$ Jun 10, 2010 at 6:45
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    $\begingroup$ Victor, I held this belief for a good while when first learning set theory. I tried proving it a couple of times and failed, but I was in that stage just after I'd gotten the hang of basic cardinality arguments and they all seemed simple, so I figured it was just a matter of small details. $\endgroup$
    – Pietro
    Jun 10, 2010 at 9:01
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    $\begingroup$ But it turns out that k^l is intimately linked with the cofinality of k, which is the length of the shortest unbounded sequence in k. For example, cof(omega) = omega, since sequences of length less than omega are finite, and thus bounded in omega. Similarly, cof(aleph_1) is aleph_1, since any countable sequence in aleph_1 is bounded. It's not immediately obvious that some cardinal k has cof(k) < k, but aleph_omega does! Anyway, the relevant theorem is that k^cof(k) > k, so there are arbitrarily large k s.t. k^omega > k. $\endgroup$
    – Pietro
    Jun 10, 2010 at 9:06
  • $\begingroup$ Actually, you can find that belief proclaimed here at MO, until someone points out the mistake. $\endgroup$
    – Todd Trimble
    Sep 6, 2015 at 2:09
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