Examples of common false beliefs in mathematics

Question

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.

Answer 1

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

Answer 2

The following seems not to be here yet.

Misconception.

$R[[x_1,x_2,x_3,\dotsc]]/(x_2,x_3,\dotsc)$ isomorphic to $R[[x_1]]$ ${}\hspace{118pt}$ (f)

Source of the misconception. A fallacy of type false generalization: for any $n\in\mathbb{N}$ it is true that

$R[[x_1,x_2,x_3,\dotsc,x_n]]/(x_2,x_3,\dotsc,x_n)\cong R[[x_1]]$ ${}\hspace{125pt}$ (t)

but to conclude from this that (f) was true by passing to the limit $n\to\infty$ is fallacious.

Reason for why the misconception is false. E.g. the formal power series $f:=x_2+x_3+\dotsm$ is an element of $R[[x_1,x_2,x_3,...]]$, but by the standard definition of $I:=(x_2,x_3,\dotsc)$, which after all means nothing more than the $R[[x_1,x_2,x_3,\dotsc]]$-module generated by the infinite set $\{x_i\colon i\in \omega,\ i\geq 2\}$, the ideal $I$ does not contain $f$. (Having coefficients from the huge power series ring $R[[x_1,x_2,x_3,\dotsc]]$ does not help.)

Reason for including the example. I saw this misconception in a dissertation. For obvious reasons, I won't give the source.

Further remarks. In the above, $R$ can be any commutative unital ring, and $R[[x_1,x_2,x_3,\dotsc]]$ as usual means the projective limit in the category of commutative unital rings of the diagram $\dotsm\twoheadrightarrow R[[x_1,x_2,x_3]]\twoheadrightarrow R[[x_1,x_2]]\twoheadrightarrow R[[x_1]]$ consisting of the canonical projections.

Answer 3

I would like to turn the attention of mathematical community to a false beliefs related to the direct limit topologies.

Many years ago in the theory of topological groups there was a false belief that for every space $X$ the free topological group carries the topology of direct limit of the sequence $F_n(X)$ of words of length $\le n$. This illusion was broken up by Fay, Ordman and Thomas who showed that even for the space of rational numbers the free topological group $F(\mathbb Q)$ is not a $k$-space.

The problems with direct limit topologies is that for the direct limit $X=lim X_n$ of an increasing sequence $(X_n)$ of topological spaces the topology on $X\times X$ does not coincide with the direct limit topology of the sequence $ (X_n\times X_n)$.

Now specialists in General Topology and Topological Algebra are conscious of pathological behaviour of direct limit topologies and are careful with this delicate topic.

On the other hand, I was quite surprised lerning that in Algebraic Geometry this misbelief still is alive. For example, in this paper posted to arxiv (maybe it is already published) in the very introduction (on page 3) it is written that for any topological space $X$ the Ran space (of all non-empty finite subsets of $X$, endowed with the topology of direct limit of the sequence $R_n(X)$ of sets of cardinality $\le n$ in $X$) is a topological semilattice. But this is not true in general, see Proposition 4 here.

So, some false beliefs that have died in some areas of mathematics can be still alive in others. By the way, this situation also explains why mathematicians should not neglect general topology.

Answer 4

Let $X$ be a "nice" path-connected topological space, say a connected manifold or CW-complex.

False belief: "A universal covering $Y\to X$ of $X$ is unique up to unique isomorphism" and therefore can be called "the" universal covering.

The isomorphism far from unique in general (there are as many as elements in "the" fundamental group). However uniqueness (and the universal property) holds in the category of coverings of pointed topological spaces. (In particular, for topological groups there's a canonical choice.)

Browsing I found several textbooks teaching the above "false belief" (I saw several too that are careful with this issue).

Answer 5

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

Answer 6

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.

Answer 7

Duality reverses inclusions of vector spaces.

Answer 8

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

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

Answer 9

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.

Answer 10

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

Answer 11

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

(Couldn't find this is the list…)

Answer 12

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?

Answer 13

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

Answer 14

If $X$ is uncountable, then $X^{\mathbb{N}}$ is in bijection with $X$.

König's theorem implies that $|X^{\mathbb{N}}|>|X|$ whenever the cardinality of $X$ has countable cofinality.

Answer 15

The following inexact belief can be spotted in many a textbook for undergraduates: the principle of mathematical induction and the well-ordering principle for $\mathbb{N}$ are equivalent.

Lars-Daniel Öhman wrote about this misbelief in his paper Are induction and well-ordering equivalent? (Math. Intelligencer, vol. 41 (2019), no. 3, pp. 33-40.). To make a long story short, one of the (main) points by Öhman in the said article is that if we define the natural numbers à la Peano, i.e. as a set $N$ endowed with a function $S \colon N \to N$ satisfying the following axioms:

1. $0\in N$,
2. $\forall n \in N \, (S(n) \neq 0)$,
3. $S$ is injective, and
4. (P.M.I.) if $M$ is a subset of $N$ such that $0 \in M$ and $S(m) \in M$ for every $m \in M$, then $M=N$.;

then, the P.M.I. is not equivalent to the well-ordering principle (every nonempty subset of $\mathbb{N}$ has a least element; heretofore, W.O.P.) relative to axioms 1 - 3.

One route that Öhman follows to evince that there are issues with the so-called equivalence of both principles is by illustrating that, in the popular proof of the implication W.O.P. $\Rightarrow$ P.M.I., the existence of an immediate predecessor for every $n \in \mathbb{N}$ is assumed: this is an assumption that can not be obtained as a consequence of 1, 2, 3, and W.B.O., "as evidenced by the existence of a model... in which this property [about immediate predecessors] does not hold [whereas 1, 2, 3, and W.B.O. do]"... In point of fact, the model he provides to exemplify the validity of the assertion between quotation marks is one in which the P.M.I. does not hold either.

Answer 16

"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.

Answer 17

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.

Answer 18

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.)

Answer 19

(*) "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.

Answer 20

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

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

Answer 21

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

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

Answer 22

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:

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

Answer 23

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

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

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

Answer 24

False belief: Any orthonormal basis of a subvectorspace $W\subset V$ of an inner product space $V$ can always be extended to an ONB of $V$.

Counterexample: Let $V$ be $\bigoplus_{i\ge 1} \mathbb{R}$ with the inner product given by $\langle a_*,b_*\rangle =\sum_{i\ge 1} a_ib_i$ and let $W$ be the subvectorspace of $V$ spanned by $e_1+e_i$ for $i\ge 2$. The given set is basis and we can apply Gram-Schmidt to obtain an ONB.

However $W^\perp = 0$ so there is no way to complete it. Related false belief: $(W^\perp)^\perp=W$. These beliefs are all true in finite dimensions, but false in general.

Answer 25

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...)

Answer 26

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

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

Answer 27

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

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

Answer 28

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.)

Answer 29

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.

Answer 30

True: The solution operator of the linear one-dimensional time-dependent ordinary differential equation (ODE) $x' = a(t) x$ is $$ \exp\Bigl(\int\limits_{t_0}^{t} a(s) \, ds\Bigr). $$ True: The solution operator of the linear multi-dimensional time-independent ODE $x' = A x$ is $$ \exp\,(A(t - t_0)). $$ A quite popular misconception, even among research mathematicians:

The solution operator $\Phi(t;t_0)$ of the linear multi-dimensional time-dependent ODE $x' = A(t) x$ is $$ \exp\Bigl(\int\limits_{t_0}^{t} A(s) \, ds \Bigr), $$

perhaps strengthened by Liouville's formula:

$$ \det{\Phi(t;t_0)} = \exp\Bigl(\int\limits_{t_0}^{t} \operatorname{tr}{A(s)} \, ds\Bigr). $$