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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 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 R must be the whole of 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
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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
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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
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It's a thought -- I might consider it. –  gowers Oct 4 '10 at 20:13
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Meta created tea.mathoverflow.net/discussion/1165/… –  quid Oct 8 '11 at 14:27

199 Answers 199

Multiplication of differential forms is inherently anti-commutative. Thus, if $x$ and $y$ are coordinates on a surface, then $dx \wedge dy$ makes sense but $(dx)^2+(dy)^2$ is either nonsense or, if it means anything, is $0$.

I'm not sure why I believed this, but I did for several years. I tried my best to avoid creating this impression in my students, but I think it still happened in some of them, simply because the curriculum spends a lot of time on integration and Stokes theorem and very little time on metrics, curvature, etc.

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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 at 6:07
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A counter-example is necessarily with $f'$ discontinuous in $x_0$, right? For example $f(x)=x^2 sin (1/x)+x/2$ and $x_0 = 0$. –  Sébastien Palcoux Aug 10 at 8:05
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@SébastienPalcoux Yes, I think if $f'$ is continuous in $x_0$ then the statement is true. –  Shamisen Aug 10 at 15:16

When $H$ and $K$ are subgroups of $G$ then $HK$ is subgroup of $G$...

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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 at 3:22

$\mathbb{R}^2$ has a unique complex structure; it's just $\mathbb{C}$ right?

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As a sequel of this famous answer on $\dim(U+V+W)$, the following inequality is not true $\forall n \ge 4$:
$$\dim(\sum_{i = 1}^{n} U_i) \le \sum_{r=1}^{n} (-1)^{r+1} \sum_{i_1 < i_2 < \dots < i_r} \dim(\bigcap_{s=1}^{r}U_{i_s})$$
Darij Grinberg has found a counter-example (see this post).

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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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"Let $E$ be a complete locally convex topological vector space (or a complete topological vector space or a complete topological group) and let $F$ be closed vector subspace (or a closed subgroup). Then the quotient $E/F$ is complete."

This just has to be true. One can almost see the proof. And in fact it is true for Banach spaces. So it has to be true for locally convex spaces as well.

Another one with completions:

"Every topological group is a dense subgroup of a complete topological group." True for abelian groups but false in general (take the homeomorphism group of $[0,1]$ with the compact open topology)

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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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Why not? As an algebraist, my reaction already after "addition of k stabilizes" would be "if THAT holds, than WHATEVER". –  Victor Protsak Jun 10 '10 at 6:45
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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. –  Pietro KC Jun 10 '10 at 9:01
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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. –  Pietro KC Jun 10 '10 at 9:06

A Banach space $X$ is reflexive if it is isomorphic to its double dual ${X^*}^*$.

(Couldn't find this is the list…)

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Even isometric fails. (Lindenstrauss & Tzafriri, in the '60s I believe.) –  Hachino May 12 at 8:19

This might not be common, but it gave me a headache once. I'll delete if it gets heavily downvoted.

I once had to think really hard about a contradiction in the great scheme of things that followed from my unwitting assumption that if $f$ was a function from a semigroup to a semigroup, then if its kernel was a congruence, $f$ had to be a homomorphism. I encountered a function whose kernel clearly was a congruence but which clearly wasn't a homomorphism, and it took about an hour's walk in a park for my vague notions and incoherent thought to produce the necessary realization.

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"Some real numbers are not definable, by Cantor's diagonal argument."

There are subtleties involved in formalizing the statment "some real numbers are not definable", as explained in Joel's answer to this question. The statement can be seen to hold in some models and fail in other models. However, the claim that the statement follows from Cantor's diagonal argument is clearly false, yet seems to be fairly common.

The false reasoning typically proceeds in three steps:

  1. There are only countably many definitions of real numbers: $\varphi_0(x),\varphi_1(x),\ldots$ (this part is ok.)

  2. Consider the countably many real numbers so defined: $x_0,x_1,\ldots$ (this part is problematic for subtle reasons.)

  3. Use Cantor's diagonal argument to obtain a real number $y$ that is not in the sequence from step 2, and is therefore not definable.

For the moment, let us assume that step 2 succeeds in the way that one might naively think it would. Then we have defined a sequence $x_0,x_1,\ldots$ containing all definable real numbers. Therefore Cantor's diagonal argument in step 3 defines, from this sequence, a real number $y$ that is not in the sequence. So $y$ is both definable and not definable, and we obtain a contradiction outright! Clearly, something is wrong (and it turns out to be in step 2.)

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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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Is that actually a common false belief? After all, $\mathbb{Z}_p$ is uncountable, as everyone realizes! –  Todd Trimble Mar 5 at 14:25
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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. –  Todd Trimble Mar 5 at 14:52

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

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Although the original intent of this question seems to have long since evaporated, I can't help asking: is this really a "common false belief"? –  Yemon Choi Feb 17 at 1:24

The closure of the open ball of radius $r$ in a metric space, is the closed ball of radius $r$ in that metric space.

In a somewhat related spirit: the boundary of a subset of (say) Euclidean space has empty interior, and furthermore has Lebesgue measure zero. (This false belief is closely related to Gowers' example of the belief that there are no non-trivial open dense sets.)

More generally, point set topology and measure theory abound with all sorts of false beliefs that only tend to be expunged once one plays with the canonical counterexamples (Cantor sets, bullet-ridden squares, space-filling curves, the long line, $\sin\left(\dfrac{1}{x}\right)$ and its variants, etc.).

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I remember being assigned as an exercise to find a counterexample to the first statement, but I can't remember where. Rudin? –  Qiaochu Yuan Jun 6 '10 at 23:39
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What about a space with 2 points a distance 1 apart, and the open/closed ball having radius 1? I don't remember seeing this before, though. –  Peter Samuelson Jun 6 '10 at 23:53
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@Terry Really good examples. We can count on you to do anything but waste our time with a post, Terry. I hope you keep finding the time to post here and lend your support! –  The Mathemagician Jun 7 '10 at 0:11
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These seem to be more "interesting mistakes" than "false beliefs", especially the last part. –  Victor Protsak Jun 10 '10 at 6:02
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Peter: actually the simplest counterexample is the open/closed ball of radius $0$, empty set vs a singleton. –  Pietro Majer Aug 1 '11 at 15:43

Some false beliefs in linear algebra:

  • If two operators or matrices A, B commute, then they are simultaneously diagonalisable. (Of course, this overlooks the obvious necessary condition that each of A, B must first be individually diagonalisable. Part of the problem is that this is not an issue in the Hermitian case, which is usually the case one is most frequently exposed to.)

  • The operator norm of a matrix is the same as the magnitude of the most extreme eigenvalue. (Again, true in the Hermitian or normal case, but in the general case one has to either replace "operator norm" with "spectral radius", or else replace "eigenvalue" with "singular value".)

  • The singular values of a matrix are the absolute values of the eigenvalues of the matrix. (Closely related to the previous false belief.)

  • If a matrix has distinct eigenvalues, then one can find an orthonormal eigenbasis. (The orthonormality is only possible when the matrix is, well, normal.)

  • A matrix is diagonalisable if and only if it has distinct eigenvalues. (Only the "if" part is true. The identity matrix and zero matrix are blatant counterexamples, but this false belief is remarkably persistent nonetheless.)

  • If L: X -> Y is a bounded linear transformation that is surjective (i.e. Lu=f is always solvable for any data f in Y), and X and Y are Banach spaces then it has a bounded linear right inverse. (This is subtle. Zorn's lemma gives a linear right inverse; the open mapping theorem gives a bounded right inverse. But getting a right inverse that is simultaneously bounded and linear is not always possible!)

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Wow. I believed that second one until now. Which is ridiculous, of course, since the operator norm of a nilpotent matrix can't be zero or else it wouldn't be a norm! –  Qiaochu Yuan Jun 6 '10 at 23:45
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The parethentical comment in 2nd bulleted point is worded as if, $\textit{in general},$ the operator norm were equal both to the spectral radius and the largest singular value (or, perhaps, that $\|A\|=\rho(A)$ and $\lambda_1(A)=s_1(A).$) But for a nilpotent matrix the spectral radius is 0, whereas the operator norm and the largest singular values aren't. –  Victor Protsak Jun 10 '10 at 7:59
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Fair enough; I've reworded the parenthetical. –  Terry Tao Jun 10 '10 at 16:35
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Yes, I meant right inverse, thanks. Getting a continuous right-inverse is actually a subtle question - the OMT only gets boundedness, which is not equivalent to continuity when one is not linear. I believe that the existence of a continuous right inverse may follow from a classical theorem of Bartle and Graves, but this is nontrivial. –  Terry Tao Jun 10 '10 at 19:06

In ${\mathbb F}_p^\times$, the non-squares are the opposite of the squares. In other words, $a$ is square iff $-a$ is not a square.

This is a confusion with the facts that the kernel of $x\mapsto x^2$ is $\{1,-1\}$ and the subgroup of squares has index $2$.

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The ring $\mathbb{C}[x]$ has countable dimension over $\mathbb{C}$; therefore its field of fractions $\mathbb{C}(x)$ also has countable dimension over $\mathbb{C}$.

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The uncountably many elements $1/(x-a)$ for all $a \in \mathbb C$ are linearly independent. –  darij grinberg Nov 29 '14 at 22:47

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

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

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

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

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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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Here are a few more: (Everything between quotation marks is a false belief.)

Basic logic: Among students: "If A implies B then B implies A" (or "if A implies B then not A implies not B").

Even among mature mathematicians a frequent false belief is to forget that the conclusion of a theorem need not hold once the conditions of the theorem fail. Another common frequent belief is to assume that once the conditions fail then the conclusion must fail too.

Calculus: "The derivative of a differentiable function is continuous."

"An infinite series whose general term tend to 0 is convergent."

Geometry: "The circle is the only figure which has the same width in all directions." (Feynmann regarded this mistake as one reason for the space shuttle challenger disaster).

Polytopes: Often people believe that "given a convex polytope P you can slightly move the vertices to rational positions keeping the structure of the polytope unchanged."

(From Udi de Shalit): Some people believe that "if you hold a cube along a main diagonal, the remaining vertices all lie on a plane." Some even say that their number is 4.

Algebra (Also from Udi) "I have encountered many misconceptions about solvability by radicals. Some people think that 'the solution of an irreducible equation of degree 5 and higher, say over Q, is never expressible by radicals'. Some amateur mathematicians even say that 'equations of degree 5 and higher have no solutions'."

Probability: "If you play the casino patiently and carefully you will win in the long run" (and "you do not believe that?, this is my own experience on the matter!" and "Indeed when I am calm and patient I win, but when I lose my temper I lose big time".)

"an event which may occur has positive probability": (not true for infinite probability spaces)

Various places: "If you want to prove that a certain infinite structure exists it is enough to show that there is no upper bound to the sizes of such structures."

Combinatorics: "This is a finite problem, surely you can solve it with a computer."

"Hall marriage theorem is very nice and I am surprised no combinatorialist bothered to extend it to a matching built from triples instead of pairs." (It is unlikely that a general characteriztion when a hypergraph built from triples has a perfect matching (of triangles) will be found.)

Computer science: "It is known that quantum computers can solve NP complete problems in polynomial time."

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"If you want to prove that a certain infinite structure exists it is enough to show that there is no upper bound to the sizes of such structures." This is not necessarily false. In some important cases this does work! One uses the compactness theorem for such proofs. –  Johannes Hahn May 7 '10 at 12:46
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Probability: There are two opposite errors. Both are common. Say we are flipping a fair coin repeatedly. (1) if there have been more heads than tails, then tails is "overdue" and thus more likely on the next flip. (2) if there have been more heads than tails, then heads is "hot" and thus more likely on the next flip. –  Gerald Edgar May 7 '10 at 15:15
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I don't know anything about polytopes, but I'm having a hard time disbelieving this false result. Are we talking about finite polytopes here? –  Tom Ellis May 9 '10 at 19:26
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Entirely finite, Tom. There are 4-dimensional polytopes with 33 vertices that cannot be presented with rational coordinates. Here is a reference arxiv.org/PS_cache/arxiv/pdf/0710/0710.4453v2.pdf –  Gil Kalai May 9 '10 at 20:46
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You list the statement "Quantum computers can solve NP complete problems in polynomial time" as a false belief, but I don't believe you actually know this belief to be false. For example, the assertion that this belief is false implies $P\neq NP$. Perhaps the false belief that you intend to mention is: "It has been proved that Quantum computers can solve NP complete problems in polynomial time." –  Joel David Hamkins May 17 '10 at 12:34

The false belif could be:

A sequence is a net. Then a subnet of a sequence is a subsequence

Counterexample: The sequence $sin(nx)$is a sequence in the compact topological space $[-1, 1]^{\mathbb{R}}$. So this net has a convergence subnet. But it is well known that the above sequence has no a subsequence which is point wise convegent(See the last page of the book of Walter Rudin Principle of mathematical Analysis). So in this example the convergent subnet can not be counted as a subsequence.

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This seems essentially a dupe as it is contained explictly in mathoverflow.net/a/40091 –  quid Nov 12 '14 at 13:31
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@quid Thank you. I did not read it already. however this one introduce an explicte counter example. –  Ali Taghavi Nov 12 '14 at 13:59
  • The category of commutative C*-algebras is equivalent to the opposite category of locally compact Hausdorff spaces.

Let me know in case that this has been mentioned before; I haven't been able to find it. There is some discussion on math.SE.

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Whoops! Thanks for posting this and particularly the link, which shows that the nLab got this one wrong as well. –  Todd Trimble Nov 12 '14 at 13:03

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 quaternion 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 quternion 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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This is not a common false belief except among people who do not understand the definition of an algebra over a field –  Yemon Choi Nov 12 '14 at 23:43
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Moreover, surely the quaternions are a real vector space, not a complex vector space –  Yemon Choi Nov 13 '14 at 1:34
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@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! –  Todd Trimble Nov 13 '14 at 11:58

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 http://dominiczypen.wordpress.com/2014/10/13/accumulation-without-converging-subsequence/

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Here's my list of false beliefs ;-):

  • If $U$ is a subspace of a Banach space $V$, then $U$ is a direct summand of $V$.
  • If $M/L, L/K$ are normal field extensions, then the same is true for $M/K$.
  • Submodules/groups/algebras of finitely generated modules/groups/algebras are finitely generated.
  • The Krull dimension of a subring is at most the Krull dimension of the ring.
  • The Krull dimension of a noetherian domain is finite.
  • If $A \otimes B = 0$, then either $A=0$ or $B=0$.
  • If $f$ is a smooth function with $df=0$, then $f$ is constant.
  • If $X,Y$ are sets such that $P(X), P(Y)$ are equipotent, then $X,Y$ are equipotent.
  • Every short exact sequence of the form $0 \to A \to A \oplus B \to B \to 0$ splits.
  • $R[[x,y]] = R[[x]][[y]]$ as topological rings.
  • $R[x]^* = R^*$, even if $R$ is not a domain.
  • Every presheaf on a site has an associated sheaf. (Hint: the index category of the usual colimit has to be essentially small!)
  • (Co)limits may be computed in full subcategories. For example, $Spec(\prod_i R_i) = \coprod_i Spec(R_i)$ as schemes because $Spec$ is an antiequivalence.
  • Every finite CW-complex is compact, thus every CW-complex is locally compact.
  • The smash product of pointed spaces is associative (this is even false for CW complexes when you don't use the compactly-generated product!), products commute with quotients, and so on: Topologists often assume that everything behaves well, but sometimes it does not.
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+1: you had me at "Here's my list of false beliefs". –  Pete L. Clark May 5 '10 at 23:41
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$A \to A \oplus B$ does not have to be the inclusion; likewise $A \oplus B$ does not have to be the projection. Thus the error here is: Two chain complexes, which are isomorphic "pointwise", don't have to be isomorphic. This occurs sometimes. –  Martin Brandenburg May 6 '10 at 16:12
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Your fifth example reminds me of an even more plausible false belief I once held: if $A \otimes A = 0$, then $A = 0$. –  Reid Barton May 11 '10 at 2:12
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@Regenbogen: Take the abelian group $\mathbb{Q}/\mathbb{Z}$. –  Steve D May 15 '10 at 13:44
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$f$ is just locally constant ;-) –  Martin Brandenburg May 19 '10 at 8:39

The product of two symmetric matrices is symmetric!

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I don't know how common this mistake is, but I think it's worth mentioning. I used to think that existence of non-measurable sets is guaranteed by the axiom of choice only.

In the presence of AC, there cannot be a $\sigma$-additive measure on $\mathcal{P}(\mathbb{R})$ that extends the usual Lebesgue measure.

It is true that we cannot extend the Lebesgue measure in a translation-invariant way by various Vitali set constructions. On the other hand, if you do not insist that the extension is translation-invariant, it might be possible to do this relative to a real-valued measurable cardinal assumption.

Theorem (Ulam): If there exists a cardinal $\kappa$ such that there exists an atomless $\kappa$-additive probability measure on $\mathcal{P}(\kappa)$, then there exists a $\sigma$-additive measure on $\mathcal{P}(\mathbb{R})$ extending the Lebesgue measure.

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

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