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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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I have to say this is proving to be one of the more useful CW big-list questions on the site... – Qiaochu Yuan May 6 '10 at 0:55
The answers below are truly informative. Big thanks for your question. I have always loved your post here in MO and wordpress. – Unknown May 22 '10 at 9:04
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. – Suvrit Sep 20 '10 at 12:39
It's a thought -- I might consider it. – gowers Oct 4 '10 at 20:13
Meta created… – user9072 Oct 8 '11 at 14:27

218 Answers 218

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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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? – Richard Kent Mar 6 '11 at 17:52
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. – Dan Ramras Mar 6 '11 at 19:57
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! – Richard Kent Mar 6 '11 at 20:01
Yes, precisely. It's an odd little example, but helpful when people forget to include the proper conditions... – Dan Ramras Mar 6 '11 at 21:06
Maybe even more amazing wrong belief in this field: $\dim(E/G)\le\dim E$ (there are counterexamples by A.N. Kolmogorov) – mikhail skopenkov Jun 9 '11 at 14:52

I don't know how common this is, but I've noticed it half an hour ago in some notes I had written: If $J$ is a finitely generated right ideal of a not necessarily commutative ring $R$, and $n$ is natural, then $J^n$ is finitely generated, isn't it?

No, it isn't. For an example, try $R=\mathbb Z\left\langle X_1,X_2,X_3,...\right\rangle $ (ring of noncommutative polynomials) and $J=X_1R$.

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Omg, I will have to be careful about that. Thanks Darij ;). – Martin Brandenburg Apr 12 '11 at 8:45

A degree $k$ map $S^n\to S^n$ induces multiplication by $k$ on all the homotopy groups $\pi_m(S^n)$.

(Not sure if this is a common error, but I believed it implicitly for a while and it confused me about some things. If you unravel what degree $k$ means and what multiplication by $k$ in $\pi_m$ means, there's no reason at all to expect this to be true, and indeed it is false in general. It is true in the stable range, since $S^n$ looks like $\Omega S^{n+1}$ in the stable range, "degree k" can be defined in terms of the H-space structure on $\Omega S^{n+1}$, and an Eckmann-Hilton argument applies.)

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If $n$ is even and $x \in \pi_{2n-1}(S^n)$ and $f$ a degree $k$ map and $H$ the Hopf invariant, then $H(f_* (x)) = k^2 H(x)$. A related misbelief: if $M$ is a framed manifold and $N\to $M a finite cover, of degree $d$. Then the framed bordism classes satisfy $[N]=d [M]$. Completely wrong. – Johannes Ebert Apr 14 '11 at 9:04

A random $k$-coloring of the vertices of a graph $G$ is more likely to be proper than a random $(k-1)$-coloring of the same graph.

(A vertex coloring is proper if no two adjacent vertices are colored identically. In this case, random means uniform among all colorings, or equivalently, that each vertex is i.i.d. colored uniformly from the space of colors.)

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...wait, what's the truth then? – Harry Altman May 10 '11 at 0:06
It sounds plausible. – Michael Hardy May 10 '11 at 0:34
For some graphs $G$ and integers $k$, the opposite. The easiest example is the complete bipartite graph $K_{n,n}$ with $k=3$. The probability a $2$-coloring is proper is about $(1/4)^n$ while the same for a $3$-coloring is about $(2/9)^n$, where I've ignored minor terms like constants. The actual probabilities cross at $n=10$, so as an explicit example, a random $2$-coloring of $K_{10,10}$ is more likely to be proper than a random $3$-coloring. – aorq May 10 '11 at 0:37
This seems like a good example of a counterintuitive statement, but to call it a common false belief would mean that there are lots of people who think it's true. The question would probably never have occurred to me it I hadn't seen it here. The false belief that Euclid's proof of the infinitude of primes, on the other hand, actually gets asserted in print by mathematicians---in some cases good ones. – Michael Hardy May 10 '11 at 15:36

False statement: If $A$ and $B$ are subsets of $\mathbb{R}^d$, then their Hausdorff dimension $\dim_H$ satisfies

$$\dim_H(A \times B) = \dim_H(A) + \dim_H(B). $$

EDIT: To answer Benoit's question, I do not know about a simple counterexample for $d = 1$, but here is the usual one (taken from Falconer's "The Geometry of Fractal Sets"):

Let $(m_i)$ be a sequence of rapidly increasing integers (say $m_{i+1} > m_i^i$). Let $A \subset [0,1]$ denote the numbers with a zero in the $r^{th}$ decimal place if $m_j + 1 \leq r \leq m_{j+1}$ and $j$ is odd. Let $B \subset [0,1]$ denote the numbers with a zero in the $r^{th}$ decimal place if $m_{j} + 1 \leq r \leq m_{j+1}$ and $j$ is even. Then $\dim_H(A) = \dim_B(A) = 0$. To see this, you can cover $A$, for example, by $10^k$ covers of length $10^{- m_{2j}}$, where $k = (m_1 - m_0) + (m_3 - m_2) + \dots + (m_{2j - 1} - m_{2j - 2})$.

Furthermore, if $\mathcal{H}^1$ denotes the Hausdorff $1$-dimensional (metric) outer measure of $E$, then the result follows by showing $\mathcal{H}^1(A \times B) > 0$. This is accomplished by considering $u \in [0,1]$ and writing $u = x + y$, where $x \in A$ and $y \in B$. Let $proj$ denote orthogonal projection from the plane to $L$, the line $y = x$. Then $proj(x,y)$ is the point of $L$ with distance $2^{-1/2}(x+y)$ from the origin. Thus, $proj( A \times B)$ is a subinterval of $L$ of length $2^{-1/2}$. Finally, it follows:

$$ \mathcal{H}^1(A \times B) \geq \mathcal{H}^1(proj(A \times B)) = 2^{-1/2} > 0. $$

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Well, it's disappointing that this fails, although it hadn't occurred to me to conjecture it. – Toby Bartels Apr 4 '11 at 9:53
Actually, the situation is worse than I say: there exist sets $A, B \subset \mathbb{R}$ with $dim_H(A \times B )= 1$, and yet $\dim_h(A) = \dim_H(B) = 0$. – JavaMan Apr 5 '11 at 6:22
By the way, is there a simple counter-example with $A=B$? – Benoît Kloeckner May 9 '11 at 7:51
Nice, I did not know that, though Hausdorff dimension is part of my mathematical life! But the sets I study (Julia sets in complex dimension one) usually are uniform enough that this does not occurr, I guess. Here's what happens, morally, in the example given here: the scales epsilon at which you have good covers of A and the scales at which you have good covers of B are disjoint. The products of these good covers are extremely distorted : they are thin rectangles, instead of squares. – Arnaud Chéritat Oct 18 '15 at 13:25

A possible false belief is that "a maximal Abelian subgroup of a compact connected Lie group is a maximal torus". Think of the $\mathbf Z_2\times\mathbf Z_2$-subgroup of $SO(3)$ given by diagonal matrices with $\pm1$ entries.

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Fu... I just "proved" that again as an exercise a few days ago. – Johannes Hahn Mar 6 '13 at 0:02

A common trap which sometimes I see people fall is that a Hermitian matrix $M$ is negative definite if and only if its leading principal minors are negative.

What is true is the Sylvester's criterion, which says that $M$ is positive definite if and only if its principal minors are positive. Thus, the true statement is that $M$ is negative definite if and only if the principal minors of $-M$ are positive.

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That Darboux functions are continuous is certainly a widely held belief among students, at least in France where it is induced by the way continuity is taught in high school.

I remember having gone through all the five "stages of grief" when shaken from this false belief with the $sin(1/x)$ example : denial, anger ( "then the definition of continuity must be wrong ! Let's change it !), bargaining ("Ok, but a Darboux function must surely be continuous except at exceptional points. Let's prove that..."), depression (when shown a nowhere continuous Darboux function), acceptance ("Hey guys, you really think the intermediate value theoem has a converse ? C'mon, you're smarter than that...")

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Let $(X,\tau)$ be a topological space. The false belief is: "Every sequence $(x_n)$ in $X$ with an accumulation point $a\in X$ has a subsequence that converges to $a$". I subscribed to this intuitively until I stumbled over a counterexample, see

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Some things from pseudo-Riemannian geometry are a bit hard to swallow for students who have had previous exposure to Riemannian geometry. Aside from the usual ones arising from sign issues (like, in a two dimensional Lorentzian manifold with positive scalar curvature, time-like geodesics will not have conjugate points), an example is that in Riemannian manifolds, connectedness + geodesic completeness implies geodesic connectedness (every two points is connected by a geodesic). This is not true for Lorentzian manifolds, and the usual example is the pseudo-sphere.

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I just realized yesterday that, given $A \to C, B \to C$ in an abelian category, the kernel of $A \oplus B \to C$ is not the direct sum of the kernels of $A \to C, B \to C$.

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A common false belief is that all Gödel sentences are true because they say of themselves they are unprovable. See Peter Milne's "On Goedel Sentences and What They Say", Philosophia Mathematica (III) 15 (2007), 193–226. doi:10.1093/philmat/nkm015

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Two very common errors I see in (bad) statistics textbooks are

(i) zero 3rd moment implies symmetry (though generally stated in terms of "skewness", where skewness has just been defined as a scaled third moment)

(ii) the median lies between the mean and the mode

(I have seen a bunch of related errors as well.)

Another one I often see is some form of claim that the t-statistic goes to the t-distribution (with the usual degrees of freedom) in large samples from non-normal distributions.

Even if we take as given that the samples are drawn under conditions where the central limit theorem holds, this is not the case. I have even seen (flawed) informal arguments given for it.

What does happen is (given some form of the CLT applies) Slutzky's theorem implies that the t-statistic goes to a standard normal as the sample size goes to infinity, and of course the t-distribution also goes to the same thing in the limit - but so, for example, would a t-distribution with only half the degrees of freedom - and countless other things would as well.

The first two errors are readily demonstrated to be false by simple counterexample, and to convince people that they don't have the third usually only requires pointing out that the numerator and denominator of the t-statistic won't be independent if the distribution is non-normal, or any of several other issues, and they usually realize quite quickly that you can't just hand-wave this folk-theorem into existence.

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In the statistics text at the college where I teach, (ii) is universal among the examples given, so I formulated the conjecture; but when I tried to prove it and thought about what the mode really is, I realised how badly behaved that can be and found immediate counterexamples. (Then this gets me wondering why anybody would bother using the mode as a statistic for anything, since it's pretty much meaningless, but never mind.) – Toby Bartels Apr 4 '11 at 9:24
Toby: sure, you use the mode for cases when the domain of the measurement is not an ordered set but just a set without structure and so the median wouldn't make sense. – Zsbán Ambrus Apr 7 '11 at 12:01

Here's one I was reminded recently during lunch in the common room.

A maximal abelian subalgebra of a semisimple Lie algebra is a Cartan subalgebra.

This is true for compact real forms of semisimple Lie algebras, but fails in general. The missing condition is that the subalgebra should equal its normaliser.

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For example, the usual proof of Schur's theorem on commutative algebras of matrices produces abelian Lie subalgebras of $\mathfrak{sl}_n$ much larger than the rank of $\mathfrak{sl}_n$. – Mariano Suárez-Alvarez Aug 4 '10 at 23:41
Yes, my favourite example (being a physicist) is the stabiliser in $\mathfrak{so}(1,n)$ of a nonzero zero-norm vector in $\mathbb{R}^{1,n}$ for $n>4$. The stabiliser contains an abelian ideal (infinitesimal null rotations) of dimension $n-1$. – José Figueroa-O'Farrill Aug 5 '10 at 0:02
I meant to write $n>3$ above. – José Figueroa-O'Farrill Aug 5 '10 at 0:03
``The missing condition is that the subalgebra should equal its normaliser'', or that the subalgebra consists of semisimple elements, no? (That provides another perspective on why it's true for compact real forms.) – L Spice Dec 12 '13 at 23:26

Consider the following well-known result: Let $(E,\leq)$ be an ordered set. Then the following are equivalent: (i) Every nonempty subset of $E$ has a maximal element. (ii) Every increasing sequence in $E$ is stationary.

It is immediate that (i) implies (ii). To prove the converse, one assumes that (i) is false and then "constructs step by step" a strictly increasing sequence.

The common mistake (which I have seen in textbooks) is to describe the latter construction as a proof by induction. In fact, the construction uses the axiom of choice (or at least the dependent choice axiom).

(As a special case, I don't think ZF can prove that every PID is a UFD.)

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It’s not exactly wrong to call it a proof by induction. In ZFC, the proof of dependent choice — or of just about any instance of it, eg the one here — works by combining induction and choice. So I’d agree it’s wrong to sweep the choice under the carpet; but if you’re not explicitly invoking DC, then you will be using induction as well. – Peter LeFanu Lumsdaine Dec 1 '10 at 15:34
Peter, let's state DC as follows: "If $(p_n:X_{n+1}\to X_n)$ is an $\mathbb{N}$-projective system of nonempty sets with all $p_n$ surjective , then projlim($X_n$) is nonempty." Proof from AC: put $X:=\coprod_{n\geq0}X_n$ and $X^+=\coprod_{n>0}X_n$ with obvious map $p:X^+\to X$. Then $p$ is onto, so has a section $s$ (family of sections of all $p_n$'s). Given $x_0\in X_0$, sequence $(s^n(x_0))$ is an element of projlim($X_n$). I agree that we do need induction to define $s^n$. But iteration of a map is such a basic tool that I don't agree to call any proof using it a "proof by induction". – Laurent Moret-Bailly Dec 7 '10 at 11:49

Draw the graph of a continuous function $f$ (from $\mathbb{R}$ to $\mathbb{R}$). Now draw two dashed curves: one which everywhere a distance $\epsilon$ above the graph of $f$ and one which is everywhere a distance $\epsilon$ below the graph of $f$. Then the open $\epsilon$-ball around $f$ (with respect to the uniform norm) is all functions which fit strictly between the two dashed curves.

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Surely this is true if you are talking about the closed ball, and only just barely false for the open ball (and if we were talking about functions from $[a,b]$ to $\mathbb{R}$ it would be true)? Or else I am one of those with the false belief... – Nate Eldredge Oct 10 '10 at 18:26
You are right, I should have specified open ball, thanks. I think it is just barely false for the open ball. Honestly, I held this false belief until a couple of days ago, and I haven't thought much about correcting my belief. Probably the real open epsilon ball is the union of all functions that fit between dashed curves a distance strictly less than epsilon away from f? At any rate, I think the above picture is the right way to think about it most of the time. But it gives results such as $tan^{-1}$ being in the open ball of radious pi/2 centered at 0 if you interpret it literally. – user4977 Oct 10 '10 at 19:24
Hmm, very nice (once clarified to the open ball)! Easily dispelled as soon as you question it, but I could easily imagine using it without thinking and missing the alternation of quantifiers that’s going on under the surface. – Peter LeFanu Lumsdaine Dec 1 '10 at 15:30

"If a field $K$ has characteristic 0 and $G$ is a group, then all $KG$-modules are completely reducible."

True for finite groups but very false in general.

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1- A very common mistake that 1st year students (but not even a single mathematician) think that it is true is "a transitive and symmetric relation on a set is reflexive". But as the empty set is a transitive and symmetric relation but not reflexive on any non-empty set. Of course there lots of non-trivial examples also.

2- Another common mistake is that the expression "countable union of countable sets is again countable" is independent of axiom of choice (AC). Many people make the proof of this statement without mentioning axiom of choice. Indeed, in his holly book Algebra, Lang proves this statement just by taking an ordering from each countable set and continues without the mentioning AC.

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For big-list questions, it's usually best to post independent answers as separate answers. – Nate Eldredge Dec 2 '10 at 15:13
+1 for #2. Baby Rudin is another offender. And many authors use so-called "diagonalization tricks" for proving compactness theorems like Arzela-Ascoli and Prohorov, which typically reduce to the compactness of $[0,1]^\mathbb{N}$. – Nate Eldredge Dec 2 '10 at 15:21
Isn't the more right statement that a transitive and irreflexive relation is assymetric? – Zsbán Ambrus Dec 4 '10 at 22:46

Duality reverses inclusions of vector spaces.

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That's funny, because I don't imagine this kind of idea would occur to someone who has just learned the definition of a dual space. That would be a strangely sophisticated mistake to make. – Thierry Zell Apr 7 '11 at 0:21
And, once you learn that this is wrong, you can make the opposite mistake. See my comments here on how surprised I was that duality DOES reverse the inclusions between Soboloev spaces. – David Speyer Apr 11 '11 at 12:07
The mistake is somehow in the wording. The dual of the inclusion morphism is reversed, it's just not an inclusion anymore. – Turion Sep 4 '15 at 15:42

I have heard the following a few times :

"If $f$ is holomorphic on a region $\Omega$ and not one-to-one, then $f'$ must vanish somewhere in $\Omega$."

$f(z)=e^z$ of course is a counterexample.

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is it true though if the image is simply-connected? – KotelKanim Apr 17 '11 at 11:33
thanks. that's great. I never thought about it before, but it just sounded right... – KotelKanim Apr 19 '11 at 7:46
No true. Take $f(z)=z^3-3z$ and restrict it to the complement of $\lbrace 1,-1\rbrace$ so that $f'(z)$ is never $0$. It maps this domain onto $\mathbb C$. – Tom Goodwillie May 4 '11 at 0:16
@TomGoodwillie: what if both domain and image are simply-connected? – Michael Dec 3 '13 at 0:45

This is more of a false philosophy than a clear mistake, but nevertheless it is very common:

A compact topological space must be "small" in some sense: it should be second countable or separable or have cardinality $ \le 2^{\aleph_0}$, etc.

This is all true for compact metric spaces, but in the general case, Tychonoff's theorem gives plenty of examples of compact spaces which are "huge" in the above sense.

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More or less along the same lines, one is usually lead to think that "every topological space is Hausdorff". – Marco Golla May 4 '11 at 14:43

This is a common error made by mature mathematicians in many books and papers in analysis, especially in differential equations: If $X$ is a closed subspace of a Banach space $Y$, then the $Y^*$ (the dual of $Y$) is isomorphic to a subspace of $X^*$ (the dual of $X$). It is false (of course) since Euclidian space $\mathbb R$ is a subspace of $\mathbb R^2$, yet the dual of $\mathbb R^2=\mathbb R^2$ is not isomorphic to a subspace of the dual of $\mathbb R=\mathbb R$. I guess, sometimes they really, really want it to be true. Cheers Boris

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I'll take your word for it, but since that statement is false even without introducing norms and topology, it staggers me that people could even believe that. They might say it without thinking, I guess – Yemon Choi Oct 20 '10 at 2:58
I would also be shocked if this really gets believed often! It seems to be a sort of “mis-dualisation”: they dualise “$X$ is a subobject of $Y$” to “$Y^*$ is a subobject of $X^*$”, where the correct dual is “$X^*$ is a quotient of $Y^*$”. – Peter LeFanu Lumsdaine Dec 1 '10 at 15:19

Teaching introduction to analysis, I had students using the "fact" that if $f: [a,b] \rightarrow \mathbb{R}$ is continuous, then $[a,b]$ can be divided to subintervals $[a,c_1],[c_1,c_2],...,[c_n,b]$ such that $f$ is monotone on every subinterval. For instance you can use this "fact" to "prove" the (true) fact that $f$ must be bounded on $[a,b]$. Also, some students used the same "fact", but with countably many subintervals. I found this mistake hard to explain to students, because constructing a counterexample (such as the Weierstrass function) is impossible at the knowledge level of an introduction course.

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Why not $x \sin(1/x)$ as example? – user9072 Jan 2 '14 at 17:33
It is in the case of finitely many subintervals, but not in the case of countably many subintervals. – Izhar Oppenheim Jan 2 '14 at 19:17
You can surely discuss fractal shapes without needing to go into the details of a technical counterexample. The point seems to be that it is hard to imagine that "increasing at a point" and "increasing in a neighborhood of a point" are not the same for continuous functions. You can give easy examples showing that indeed they disagree, locally, and fractals suggest that you can make the disagreement happen everywhere. You can revisit this later, once more technology has been set in place. – Andrés E. Caicedo Jan 2 '14 at 23:44
While technically it is true one can do it with countably many for the function I gave (if one includes degenerate intervals) I would be surprised if not at least some (or rather most) of the confusion of the students could be addressed by the example (possibly continuing with discussion along the lines suggested by @AndresCaicedo). – user9072 Jan 5 '14 at 16:50

Many students believe that every abelian subgroup is a normal subgroup.

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"A real symmetric matrix is positive-definite iff all the leading principal minors are positive, and positive-semidefinite iff all the leading principal minors are nonnegative."

This paper collects some evidence that this belief is "common", and presents a counterexample (of size $3\times 3$. Exercise: find an example of size $2\times 2$).

(Related to, but not the same as this answer.)

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$$\pmatrix{0&0\cr0&-1\cr}$$ – Gerry Myerson Jul 29 '15 at 3:22

Yet another one:

Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be differentiable. If $f'(x_0) > 0$, then there exists an interval $I$ containing $x_0$ such that $f$ is increasing in $I$.

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I sort of find it hard to believe that amongst the nearly 200 answers on this thread (and just over 20 deleted ones), no one has posted this. – Asaf Karagila Aug 10 '15 at 6:07
A counter-example is necessarily with $f'$ discontinuous in $x_0$, right? For example $f(x)=x^2 sin (1/x)+x/2$ and $x_0 = 0$. – Sebastien Palcoux Aug 10 '15 at 8:05
@SébastienPalcoux Yes, I think if $f'$ is continuous in $x_0$ then the statement is true. – Shamisen Aug 10 '15 at 15:16

Common false belief: a space that is locally homeomorphic to $\mathbb{R}^n$ must be Hausdorff. More generally, many people forget that the usual definition of a manifold contains the Hausdorff and paracompact conditions.

There are of course examples that show that forgetting this assumption leads to unexpected result, and they are in fact much wilder than I knew a few weeks ago. Notably, among examples of (Hausdorff) non-paracompact "manifolds" are the well-known long line, but also the Prüfer manifold constructed from a closed half-plane by attaching to it a half plane at each boundary point.

Added: Let me give a particular case of this false belief to illustrate what kind of weird things can happen that most people would not realize when they are sloppy with the paracompact hypothesis: there exists a path-connected, locally contractible, simply-connected space that admits non-trivial locally trivial bundles with fiber $[0,1]$. Indeed, the first octant in the product of two long line is not homeomorphic to a product a long ray with an interval, but has a natural bundle structure over a long ray.

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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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Here's a mistake I've seen from students taking a first course in linear analysis. For a vector $g$ in a Hilbert space $H$, it is true that $\langle f,g\rangle=0$ for every $f\in H$ implies $g=0$. This leads us to the mistaken:

“Let $(g_n)$ be a sequence in $H$. If, for every $f\in H$, $\langle f,g_n\rangle\to0$, then $g_n\to 0$.”

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You wrote: "Here's a mistake I've seen from students taking a first course in linear analysis." Then you wrote: "For a vector $g$ in a Hilbert space, $\langle f,g\rangle$ for every $f \in H$ implies $g = 0$." At this point the reader could be wondering what that is a mistake. – Michael Hardy Dec 1 '10 at 22:35
....sorry; I meant "$\langle f,g \rangle = 0$ for every[....]" – Michael Hardy Dec 1 '10 at 22:36
@Michael: all answers are CW; so if we think some wording needs clarifying, we can do it ourselves! – Peter LeFanu Lumsdaine Dec 2 '10 at 0:43
Thanks for the edits - that's a lot clearer.. – Ollie Margetts Dec 2 '10 at 13:49

I saw many students using the "fact" that for a subset $S$ of a group one has $SS^{-1}=\{e\}$

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This is an interesting example, because it addresses the mistakes that come from the all-too frequent confusion with notations. But we need our shortcuts, our $f^{-1}(x)$ versus $x^{-1}$, etc. Obtaining concise notations while avoiding confusion: a tricky proposition! – Thierry Zell Apr 14 '11 at 15:50

protected by François G. Dorais Oct 15 '13 at 2:34

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