# 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.

• 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

Here's a little factoid: (The Mean-value theorem for functions taking values in $\mathbb{R} ^n$.) If $\alpha : [a,b]\rightarrow \mathbb{R}^n$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then there exists a $c\in (a,b)$ such that $\frac{\alpha (b)-\alpha (a)}{b-a}=\alpha '(c)$

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

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

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

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).

• 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

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

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 $I$.

• 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

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

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

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

• Is it clear that $\mathbb{K}^{\mathbb{Z}}$ is of uncountable dimension for all $\mathbb{K}$ ? (for uncountable $\mathbb{K}$ (like $\mathbb{R}$ or $\mathbb{C}$) one can consider the functions $\frac{1}{X-a}$ with $a\in \mathbb{K}^\times$ which are independant, identifying $\mathbb{K}^{\mathbb{N}}$ with $\mathbb{K}[[X]]$) – Duchamp Gérard H. E. Oct 29 '17 at 20:59
• OK, you're right, I found the "Erdos-Kaplansky theorem" in Bourbaki (Algebra II § 7 ex. 3) which explains that $dim(\mathbb{K}^J)$ ($J$ infinite) is $card(\mathbb{K})^{card(J)}$. – Duchamp Gérard H. E. Oct 30 '17 at 7:47

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

So, what is the error?

The error is the following:

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

• 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
• Moreover, surely the quaternions are a real vector space, not a complex vector space – Yemon Choi Nov 13 '14 at 1:34
• @YemonChoi the field of complex number is a subring of the ring of quaternions. So quaternions is a complex vec. space.More generaly if a ring R contains a field F then R is a F-vector space,Ok? – Ali Taghavi Nov 13 '14 at 5:31
• @YemonChoi I think this example is perhaps interesting unless a participant do not read it carefully. please think again to the main motivation and aim of the question of "common false..." – Ali Taghavi Nov 13 '14 at 5:44
• @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

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!

A stunning, ignorance-based false belief I have witnessed while observing a class of a math education colleague is that there is no general formula for the n-th Fibonacci number. I wonder if this false belief comes from conflating the (difficult) lack of formulas for prime numbers with something that is just over the horizon of someone whose interests never stretch beyond high-school math.

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

• I hope at least your colleague had it right! There is another one along these lines: there is no formula for the sequence $1,0,1,0,1,0,... .$ Your second paragraph is right on target, but I also think that the specific beliefs you and I mentioned have a lot to do with a very limited understanding of what is a "formula". – Victor Protsak Jun 10 '10 at 7:02
• Perhaps the people who believe this are using the meta-reasoning that the sequence would not be interesting as an example of recursion if it could be solved exactly. Since it is a popular example of recursion, then... – Ryan Reich Oct 5 '11 at 17:01
• When, as an undergrad, I couldn't solve a problem given to me by the advisor, and asserted that it's "unsolvable", the advisor replied that "solvability of a problem is a function of two arguments: the problem and the solver." – Michael Dec 3 '13 at 1:14
• @Michael (The constant function is still a function. :P ) – Akiva Weinberger Sep 1 '15 at 0:03

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 $S$.

• 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
• Many students have the false belief that if a topological space is totally disconnected, then it must be discrete (related to examples already given). The rationals are a simple counter-example of course.

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

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

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

• What do you mean by "In higher dimensions there is no "axis" but a n-2 dimensional subspace instead" ? Whenever n is even, there are rotations without real eigenvectors. – Johannes Hahn Apr 14 '11 at 13:12
• Unfortunately, "smooth" is a word which means whatever its utterer does not want to specify. Differentiable, C^infty, continuous, everything is mixed. – darij grinberg Apr 14 '11 at 15:12
• @Zsban: thanks for noticing. I edited the post. now I believe it is correct (though less impressive perhaps...). – KotelKanim May 3 '11 at 15:25
• +1 for the discrete $\neq$ totally disconnected example. – Jim Conant May 4 '11 at 15:12
• Discrete $\ne$ totally disconnected is a good one that I thought of today and just had to check to see if it was posted already. It adds to the confusion that every finite subset of a totally disconnected space must have the discrete topology, and that in most topological spaces encountered "in nature," the connected components are open sets. – Timothy Chow Oct 20 '11 at 14:30

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.)

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

• Oh, I previously thought that eight regular tetrahedra with side length 1 fit in one with side length 2. That might be a more common false belief. – Junyan Xu May 5 '12 at 7:45

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.)

• 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

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 $(0,0$).

• 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

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.

• 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

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

• (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 $\mathrm{Ab}$.

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?

• 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 got 2 well earned downvotes for a false belief I claimed proudly, it is time to balance that by exposing it here:

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

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

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

I had in mind that a $0$-sphere is only one point, but it is false, it is a collection of two points: $$\mathbb{S}^0 = \{ x \in \mathbb{R} \ \ | \ \ \|x\|=1 \} = \{-1,1\}$$

• Exact. Moreover, if it were connected, its suspension $\mathbb S^1$ would be simply connected. – ACL Apr 21 '16 at 6:23

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.

• 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
• It reminds me of this fake proof that $\pi$ is an irrational number. – Watson Jan 10 at 9:22

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.

• 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

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?

False belief: ${\cal P}(\omega)$ has only countable chains with respect to $\subseteq$.

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

• Well, if "adding stuff ... many times" is meant in an iterative sense, i.e., is parametrized by an ordinal, then your original intuition would be right. – Todd Trimble Oct 8 '17 at 2:34
• On a similar line: the Hilbert space $\ell_2$ has a family of closed linear subspaces, parametrized on the unit interval, and increasing by inclusion. This looks a bit paradoxical. Should we add orthogonal vectors uncountably many times as $t$ increases from $0$ to $1$? No: just think to $L^2[0,t]$ as subspaces of $L^2[0,1]$ for $0\le t\le 1$ – Pietro Majer Oct 22 '17 at 14:21

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)

"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.

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...)

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

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.

• 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? – Autumn 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! – Autumn 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

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

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:

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

• 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 $\dim_h(A) = \dim_H(B) = 0$. – David 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