# Examples of common false beliefs in mathematics

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 tea.mathoverflow.net/discussion/1165/… – user9072 Oct 8 '11 at 14:27

## 218 Answers

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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And the Killing field has nothing to do with Pol Pot. – Nate Eldredge May 5 '10 at 14:40
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... – Willie Wong May 5 '10 at 17:19
"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. – Qiaochu Yuan May 6 '10 at 7:38
"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.) – Nate Eldredge May 6 '10 at 14:51
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 :-) – Peter Arndt May 19 '10 at 12:24

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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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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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. – David Corwin Jul 7 '10 at 23:48
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). – G. Rodrigues Jul 22 '10 at 13:39
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). – Peter LeFanu Lumsdaine Sep 27 '10 at 2:59
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. – Toby Bartels Apr 4 '11 at 9:13
At least, $\{0\}$ is a vector space. I have seen "a vector space has at least two elements" from a professional mathematician. – user11235 Apr 10 '11 at 21:28

A common belief of students in real analysis is that if $$\lim_{x\to x_0}f(x,y_0),\qquad\lim_{y\to y_0}f(x_0,y)$$ exist and are both equal to $l$, then the function has limit $l$ in $(x_0,y_0)$. It is easly to show counter-examples. More difficult is to show that also the belief $$\lim_{t\to 0}f(x_0+ht,y_0+kt)=l,\quad\forall\;(h,k)\neq(0,0)\quad\Rightarrow\quad\lim_{(x,y)\to(x_0,y_0)}f(x,y)=l$$ is false. For completeness's sake (presumably anybody who ever taught calculus has seen it, but it's easily forgotten) the standard counterexample is $$f(x,y)=\frac{xy^2}{x^2+y^4}$$ at $(0,0$).

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That counterexample has the advantage of being well-behaved away from $(0,0)$, but the (related) disadvantages of being easily forgotten and requiring a bit of thought to come up with. This can make things look trickier than they are. For this reason, I prefer brain-dead counterexamples like $f(x,y)=1$ if $y=x^2 \neq 0$, $f(x,y)=0$ otherwise. – Chris Eagle Jan 12 '11 at 17:11
@Chris As you know, this is not a "real function" to the minds of calculus students. – Ryan Reich Jan 2 '14 at 3:04
Can I try to generate a simpler counterexample? Consider $f(x,y)=\begin{cases}1,&x^2+y^2=1\\0,&x^2+y^2\ne1\end{cases}$. Then it's not hard to show that all straight-line limits to $(x_0,y_0)$ exist for all $x_0,y_0$, and are equal to $0$, but clearly the limit doesn't exist on the unit circle. EDIT: Didn't see Eagle's comment. – Akiva Weinberger Sep 1 '15 at 0:07

In a finite abelian $p$-group, every cyclic subgroup is contained in a cyclic direct summand.

Added for Gowers: Maybe one reason why people fall into this error goes something like this: First you learn linear algebra, so you know about vector spaces, bases for same, splittings of same. Then you run into elementary abelian $p$-groups and recognize this as a special case of vector spaces. Then you learn the pleasant fact that all finite abelian $p$-groups are direct sums of cyclic $p$-groups, and a corresponding uniqueness statement. You notice that all of the cyclic subgroups of order $p^2$ in $\mathbb Z/p^2\times \mathbb Z/p$ are summands, and if you have a certain sort of inquiring mind then you also notice that not every subgroup of order $p$ is a summand: one of them is contained in a copy of $\mathbb Z/p^2$, in fact in all of those copies of it. Having learned so much, both positive and negative, from the example of $\mathbb Z/p^2\times \mathbb Z/p$, you may think that it shows all the interesting basic features of the general case and overlook the fact that in $\mathbb Z/p^3\times \mathbb Z/p$ there is a $\mathbb Z/p^2$ not contained in any $\mathbb Z/p^3$.

In any case, reputable people sometimes make this blunder; it happened to somebody here at MO just the other day.

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Can you sketch the "proof" that makes this plausible? – gowers Jul 7 '10 at 16:40
Finite abelian $p$-groups are direct sum of cyclic subgroups so they look a bit like vector spaces. Therefore, you expect them to behave the same way, i.e. every subspace should have a complement. In other words, take a minimal generating set for your subgroup and complete it to a minimal generating set for the whole group. This fails since your generating set for the subgroup might be depended modulo the Fratinni subgroup of the whole group. (A set is a minimal generating set for a finite $p$-group iff it is abasis for the group modulo the Fratinni subgroup). – Yiftach Barnea Jul 7 '10 at 17:58
Is there an easily stated classification of the ways one can place a subgroup inside a finite abelian p-group (up to automorphisms of the larger group)? – T.. Jul 7 '10 at 22:31
I once worked out a classification of the ways one can place an element inside a finitely generated abelian group (up to automorphisms of the larger group), but I don't recall how it went exactly. – Tom Goodwillie Jul 8 '10 at 0:24
This is related to a somewhat subtle issue of characterizing inclusions between the closures of the conjugacy classes of matrices. Suppose $A$ is a nilpotent $n\times n$ matrix of type $\lambda$ (i.e. with Jordan blocks of sizes $\lambda_1\geq \lambda_2\geq\ldots$ adding up to $n$) and $B$ is ... $\mu.$ Can $B$ be obtained as a limit of the conjugates of $A$? This is clearly possible if $\lambda$ is componentwise greater or equal than $\mu$, but the necessary and sufficient condition is given by the dominance order, en.wikipedia.org/wiki/Dominance_order. – Victor Protsak Jul 9 '10 at 4:05

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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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. – Qiaochu Yuan Jul 7 '10 at 23:56
For most of the history of civilization, zero was very controversial... – Victor Protsak Jul 9 '10 at 4:12
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... – Thierry Zell Aug 31 '10 at 2:24
Thierry: of course it makes sense. But the action is not transitive. – ACL Dec 1 '10 at 22:53
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. – Tom Goodwillie Mar 14 '12 at 22:01

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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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 ... – Martin Brandenburg Apr 12 '11 at 8:41

In group theory, if $G_1 \cong G_2$ and $H_1 \cong H_2$, then

$G_1 / H_1 \cong G_2 / H_2$.

For example, $\mathbb{Z} / 2\mathbb{Z} \not \cong \mathbb{Z} / \mathbb{Z}$. The point is that the inclusion of $H_j$ into $G_j$ is needed in order to define the quocient.

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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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Is it because $\{x_0,x_1,\dots\}$ isn't necessarily a set? – Akiva Weinberger Sep 1 '15 at 0:17
@columbus8myhw It's related to that, although it's possible for $\{x_0, x_1, \ldots\}$ to be a set for unrelated reasons. (For example, given a pointwise definable model of set theory, we can consider $\{x_0, x_1, \ldots\}$ from the outside, and see that it happens to equal a set in the model, namely $\mathbb{R}$ itself.) But in this case $(x_0,x_1,\ldots)$ will fail to be a sequence (of the model) and the argument will still fail (in the model). – Trevor Wilson Sep 1 '15 at 18:30
That being said, understanding why $\{x_0,x_1,\ldots\}$ isn't necessarily a set is similar to understanding why the argument isn't valid. – Trevor Wilson Sep 1 '15 at 18:31

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
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
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
Actually, you can find that belief proclaimed here at MO, until someone points out the mistake. – Todd Trimble Sep 6 '15 at 2:09

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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"The universal cover of $SL_2(R)$ is a universal central extension" (which I believed until recently...)

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

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(See Craig's answer below) – Pietro Majer Dec 9 '13 at 12:13
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. – Dirk Jan 2 '14 at 17:18

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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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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Or: if a sequence has a rational limit in Q_p and in Q_r, then they're the same. – Qiaochu Yuan May 5 '10 at 4:16
But if a rational sequence has a limit in all Q_p, including Q_\infty ... – Gerald Edgar May 5 '10 at 12:17

"A 'random' number field has large class number"

I've heard this belief quite a few times. Usually random means taking a not-too-small degree (7?) and then somehow taking integer coefficients (around 10,000?).

But in fact class number tend to be much smaller than one expects. Usually they are logarithmic in the size of the discriminant.

The main reasons for the belief are the common examples of fields given in undergraduate and early graduate courses - imaginary quadratic fields and cyclotomic fields. In more advanced courses students see abelian extensions and CM-fields, which also have special arithmetic properties that make their class groups somewhat larger. In the courses I have taken the actual size of 'random' number fields was not addressed, and, say, the Cohen-Lenstra heuristics were not mentioned.

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Complex variables: "An entire function that is onto and locally one-to-one is globally one-to-one."

Counterexample: $f(z) := \int_0^z \exp(\zeta^2)\,d\zeta$

I'll leave the proof that this is indeed a counterexample as a pleasant exercise.

(I believe this example is due to Lawrence Zalcman.)

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Let's see if you TeX code can be improved: $$f(z) := \int_0^z \exp(\zeta^2)\,d\zeta$$ (The backslash in \exp not only should prevent italicization but should also result in proper spacing in things like "a \exp b", and the space before d\zeta seems appropriate.) – Michael Hardy Jul 8 '10 at 15:19
@MichaelHardy, if we're going to {\TeX}pick, then surely it should be something like ${\mathrm d}\zeta$ (rather than $d\zeta$), since the $\mathrm d$ is an operator (rather than a variable)? – L Spice Dec 12 '13 at 23:20
@LSpice : I understand the case for that usage; in particular, it allow the use of $d$ as a variable, so that one can write $\dfrac{\mathrm{d}f}{\mathrm{d}d}$, etc. However, the usage with the $d$ italicized as if it were a variable is standard although not universal. – Michael Hardy Dec 13 '13 at 0:58

The fundamental group of the Klein bottle is $D_\infty$, the infinite dihedral group (which is $\mathbb Z \rtimes \mathbb Z_2$).

I believed this for some time, and I seem to recall some others having the same confusion.

The group that has been mistaken for $D_\infty$ is in fact $\mathbb Z \rtimes\mathbb Z$, which can also be written with the presentation $x^2y^2=1$. The former abelianizes to $\mathbb Z_2\oplus \mathbb Z_2$, the latter to $\mathbb Z\oplus \mathbb Z_2$.

A 2-dimensional Lie group is a product of circles and lines, in particular it is abelian.

I don't know if anyone else suffered this one. The mistake is (a) in forgetting that the classification of surfaces doesn't apply since homeomorphic Lie groups are not necessarily isomorphic (e.g., the (bijective, orientation preserving) affine transformations $x\mapsto ax+b$, where $a>0, b\in \mathbb R$ are homeomorphic to $\mathbb R^2$, though not isomorphic) and (b) that Lie groups aren't necessarily connected, in particular $\mathbb R^2$ cross any finite non-abelian group is non-abelian.

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Count me in for the 2nd fallacy. – Michael Dec 3 '13 at 0:41

(*) "Let $(I,\leq)$ be a directed ordered set, and $E=(f_{ij}:E_i\to E_j)_{i\geq j}$ be an inverse system of nonempty sets with surjective transition maps. Then the inverse limit $\varprojlim_I\,E$ is nonempty."

This is true if $I=\mathbb{N}$ ("dependent choices"), and hence more generally if $I$ has a countable cofinal subset. But surprisingly (to me), those are the only sets $I$ for which (*) holds for every system $E$. (This is proved somewhere in Bourbaki's exercises, for instance).

Of course, other useful cases where (*) holds are when the $E_i$'s are finite, or more generally compact spaces with continuous transition maps.

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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 $f_a(y) = ay(1-y)$:

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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Here's a relevant back-and-forth about the use of the term 'chaos' in the AMS Notices (in response to Freeman Dyson's 'Birds and Frogs'): ams.org/notices/200906/rtx090600688p.pdf ams.org/notices/200910/rtx091001232p.pdf – Mike Hall Aug 8 '11 at 0:17

Here are mistakes I find surprisingly sharp people make about the weak$^{*}$ topology on the dual of $X,$ where $X$ is a Banach space.

-It is metrizable if $X$ is separable.

-It is locally compact by Banach-Alaoglu.

-The statement $X$ is weak$^{*}$ dense in the double dual of $X$ proves that the unit ball of $X$ is weak$^{*}$ dense in the unit ball of the double dual of $X.$

The first two are in fact never true if $X$ is infinite dimensional. While both statements in the third claim are true, the second one is significantly stronger, but a lot of people believe you can get it from the first by just "rescaling the elements" to have norm $\leq 1.$ (Although the proof of the statements in the third claim is not hard). The difficulty is that if $X$ is infinite dimensional then for any $\phi$ in the dual of $X,$ there exists a net $\phi_{i}$ in the dual of $X$ with $\|\phi_{i}\|\to \infty$ and $\phi_{i}\to \phi$ weak$^{*},$ so this rescaling trick cannot be uniformly applied. Really these all boil down to the following false belief:

-The dual of $X$ has a non-empty norm bounded weak$^{*}$ open set.

Again when $X$ is infinite dimensional this always fails.

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Minor nitpick: Consider a locally compact Hausdorff space $T$. The $*$ topology on the dual of the $C^*$ algebra $C_0(T)$ is metrizable, if and only if $X$ is second countable. That is a theorem in Choquet's book on functional analysis. So your claim, that the first statement is never true in infinite dimensional situations, is false. Take e.g. $T$ being a circle. – Marc Palm Oct 6 '11 at 13:38
I think $M(T)$ is not metrizable in the weak$^\ast$ topology, and in fact my claim that this fails for every infinite dimensional Banach space i also think is true. The rough outline of the proof I saw was this: 1. If $X^\ast$ is weak$^\ast$ metrizable, then a first countabliity at the origin argument implies that $X^\ast$ has a translation invariant metric given the weak$^\ast$ topology. 2. One can characterize completeness topologically for translation-invariant metrics, and see directly that if $X^\ast$ had a translation-invariant metric given the weak$^\ast$ topology it would be complete. – Benjamin Hayes Oct 12 '11 at 3:42
$X^{∗}$ in the weak∗ topology is a countable union of $\{\phi\in X^{*}:\|\phi\|\leq N\}$, which have empty weak∗ interior. Hence, if the weak∗ topology were metrizable, we get a contradiction to the Baire Category Theorem. Are you sure you don't mean the weak∗ topology on the state space of $C_{0}(X)? – Benjamin Hayes Oct 12 '11 at 3:47 Okay, excuse my false claim, I was overlooking that this holds for the subset$M^+(T)$of positive Radon measure, and does not generalize to the complex linear span. – Marc Palm Oct 16 '11 at 10:24 The cost of multiplying two$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)$and note that this is sufficient to calculate the result with three such products instead of four. - It would be better if these misconceptions would come with explanations how things really are... – darij grinberg Apr 10 '11 at 18:28 Along these lines: there is a widespread misapprehension that multiplication is the same thing as a multiplication algorithm (whichever one the speaker learned in elementary school). – Thierry Zell Apr 10 '11 at 19:25 At least it's better than people thinking multiplication is constant-time. :P – Harry Altman Apr 10 '11 at 19:35 Nice, thanks for the explanation! – darij grinberg Apr 10 '11 at 20:19 should be$(x_1+x_2)(y_1+y_2)$. – Junyan Xu May 6 '12 at 4:09 False belief: A function being continuous in some open interval implies that it is also differentiable on some point in that interval: Counterexample: The Weierstrass function is an example of a function that is continuous everywhere but differentiable nowhere:$f(x) = \sum_{n=0}^\infty a^n \cos(b^n \pi x)$Where$a \in (0, 1)$,$b$is a positive odd integer, and$ab > 1 + \frac{3\pi}{2}$. The function has fractal-like behavior, which leads to it not being differentiable. This notion is rather disheartening to most calculus students, though. - Related: if f is continuous on the interval I, there must be an interval J in I on which f is monotone. Easily believed by the beginner. – Thierry Zell Aug 31 '10 at 2:34 Did you mean "differentiable on some point in that interval:" ? – Rasmus Bentmann Sep 18 '13 at 19:13 Haha, figures that I edit a 3 year old post to introduce an even worse typo. Yes, that is what I meant. – Jon Paprocki Sep 18 '13 at 22:53 I guess you don't want commonly held beliefs of students that for every real number there is a next real number, or that convergent sequences are eventually constant. A version I saw in a book asked whether points on a line "touch." Understanding the topology of a line is a challenge for many students, although presumably not for most mathematicians. Here is a more esoteric belief that I have even seen in some books: "The Banach-Tarski Paradox says that a ball the size of a pea can be cut into 5 pieces and reassembled to make a ball the size of the sun." As a consequence of the Banach-Tarski paradox, a ball the size of a pea can be partitioned (not really "cut") into a finite number of pieces which can be reassembled into a ball the size of the sun, but a simple outer measure argument implies that the number of pieces must be very large (I roughly estimate at least$10^{30}$). The number 5 probably comes from the fact that the basic Banach-Tarski paradox is that a ball of radius 1 can be partitioned into 5 pieces which can be reassembled into two disjoint balls of radius 1. (It can almost, but not quite, be done with four pieces; one of the five pieces can be taken to be a single point.) - Since points do not touch, this was an objection to the set theoretic view of the geometric continuum as a set of points, for example by Veronese. A decent account of this can be found in Debates about infinity in mathematics around 1890: The Cantor-Veronese controversy, its origins and its outcome, by Detlef Laugwitz. – Andrés E. Caicedo Dec 2 '13 at 23:22 – Andrés E. Caicedo Jan 2 '14 at 4:03 Convergent sequences are eventually constant! With the discrete topology/metric/norm, that is. – Akiva Weinberger Sep 1 '15 at 0:19 Of course I meant sequences in$\mathbb{R}$with the usual topology. Hopefully by the time students study general metric spaces or topological spaces they understand the topology of$\mathbb{R}$. – Bruce Blackadar Jan 7 at 19:35 An incredibly common false belief is: For a (say smooth, projective) algebraic variety$X$the$K_X$-negative part of the cone$NE(X)$is locally polyhedral. A right statement of the theorem of the cone is$\overline{NE(X)} = \overline{NE(X)}_{K_X \geq 0} + \sum_{i} \mathbb{R}[C_i]$for a denumerable set$\{ C_i \}$of rational curves, which accumulate at most on the hyperplane$K_X = 0$. At a first glance this seems to imply that$\overline{NE(X)}_{K_X < 0}$is locally poyhedral, but this is not true. It depends on the shape of the intersection$\overline{NE(X)} \cap \{ K_X = 0 \}$. For instance if this latter intersection is round, and there is only one curve$C_i$, the half-cone$\overline{NE(X)}_{K_X < 0}$is actually a circular cone! Definitely not polyhedral in any sense. I believe this behaviour can happen even with varieties birational to abelian varieties. The strange thing about this false belief is that it is held true by many competent mathematicians (and indeed I don't believe that many undergraduates meet the theorem of the cone!). - You meant: I believe this behaviour can happen even with (varieties birationally isomorphic to) abelian varieties. Nice example although perhaps too technical for MO. – VA. May 5 '10 at 3:27 Incredibly common? The number of people who can even understand the statement, let alone believe it, isn't all that large... – Victor Protsak May 5 '10 at 6:57 Yes, but among those, almost all believe that the wrong version is true. – Andrea Ferretti May 5 '10 at 10:13 And about 50% of the large community who cannot understand the point will believe that the right version is true! Rather high percentage... – Wadim Zudilin May 5 '10 at 11:41 I'm not sure to what extent this is a "false belief", and to what extent people are just being sloppy with the terminology "locally polyhedral". But I agree, it's disturbing to hear experts happily making this false statement, without any further comment. <i>Mea culpa:</i> An old version of the wikipedia article entitled "Cone of curves" contained this false statement. If one looks through the article history, it's not hard to see who is to blame... – user5117 May 6 '10 at 7:24 "the quadratic variation of a Brownian motion between$0$and$T$is equal to$T$" this is only true that if$\mathcal{D}^N$is a nested sequence of partitions of$[0,T]$(with mesh size going to$0$) then the quadratic variation of a Brownian motion along these partitions converges towards$T$, almost surely. If we define the quadratic variation of a continuous function$f$as we would like to, $$Q(f,[0,T]) = \sup_{0=t_0<\ldots, t_n=T } \sum |f(t_k)-f(t_{k+1})|^2,$$ then the Brownian paths have almost surely infinite quadratic variation. This was something I had never noticed until I read the wonderful book "Brownian motion" by Peter Morters and Yuval Peres. - The key here is that quadratic variation is defined as a limit in probability, not a limit almost surely. – nullUser Jul 8 '13 at 15:46 As a student, I thought (for quite a while) that our textbook had stated that tensoring commutes with taking homology groups. It wasn't until calculating the homology groups of the real projective plane over rings Z and Z/2Z that I realized my mistake. - Before reading about it, I really thought that if$f \colon [0,1] \times [0,1] \to [0,1]$is a function with the following properties: 1. for any$x \in [0,1]$the function$f_x\colon [0,1] \to [0,1]$defined by$f_x(y)=f(x,y)$is Lebesgue measurable, and also the function$f^y \colon [0,1]\to[0,1]$defined by$f^y(x)=f(x,y)$is Lebesgue measurable, for all$y \in [0,1]$; 2. both$\varphi(x)=\int_0^1 f_x d\mu$and$\psi(y)=\int_0^1 f_y d\mu$are Lebesgue measurable. Then the two iterated integrals $$\int_0^1\varphi(x)dx \mbox{ and } \int_0^1\psi(y)dy$$ should be equal. This is false (see Rudin's "Real and Complex Analysis", pag. 167), at least if you assume the continuum hypothesis. - I really like this example from Rudin's book. Do you know if there exist such an example that does not use the continuum hypothesis (or if it's even possible to find one)? – Kalim Jul 28 '10 at 13:39 I don't know, but this could be a good questions for MO! – Ricky Jul 28 '10 at 14:28 For others reading, the hypothesis left off here is that one must assume$f$is measurable with respect to the product$\mathcal{B}[0,1] \times \mathcal{B}[0,1]$. – nullUser Jul 8 '13 at 15:39 As is well known, if$V$is a vector space and$S, T \subset V$are subspaces, then$S \cup T$is a subspace iff$S \subset T$or viceversa. However,$S \cup T \cup U\$ can be a subspace even if no two spaces are contained in each other (think finite fields...)

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But only finite fields... – darij grinberg Oct 19 '10 at 8:42

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

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