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

Surprisingly, I didn't find this answer in the ten pages above, so...

If two measures agree on a class generating their sigma-algebras, they are equal, right?

(And there actually are two ways in which this is false. Even knowing the total measure and assuming that it is finite is not enough to make this true)

Answer 2

Many people believe that Cantor proved the uncountability of the real line using a diagonal argument. This paper does not that provide that proof; Cantor's stated purpose was to prove the existence of `uncountable infinities' without using the theory of irrational numbers.

Answer 3

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

Same flavor: for $n \le 5$, it is true that $\alpha \ge 0$ (see this proof), but it's false for $n>5$ (see this comment).

Answer 4

Given a bundle $ E \to X$, let $\mathcal{E}$ denote its sheaf of sections.

False belief: Given a map $f: Y \to X$, the inverse image sheaf $f^{-1}\mathcal{E}$ is the sheaf of sections of the pullback bundle $f^* E \to X$.

This is true if $E \to X$ is a local homeomorphism (e.g. a covering space), or if $f: U \hookrightarrow X$ is the inclusion of an open subset, but not for general maps and bundles.

For instance, taking $x^{-1}\mathcal{E}= \mathcal{E}_x$ for $x: 1 \to X$ the inclusion of a point and $\mathcal{E}$ the sheaf of smooth functions on a manifold will demonstrate that it is false.

For vector bundles (or sheaves of modules over the structure sheaf of a ringed space in general), the correct statement is obtained by using the pullback functor $$f^*\mathcal{V} = \mathcal{O}_Y \otimes_{f^{-1}\mathcal{O}_X} f^{-1} \mathcal{V}$$ which is the inverse image followed by extension of scalars.

One issue which leads to this false belief is that texts on sheaves often use $f^*$ in place of $f^{-1}$ for the inverse image functor, rather than reserving the former for sheaves of modules over ringed spaces.

Answer 5

(Cauchy and/or ordinary) product of two summable families. Until recently, I thought that, in a topological ring (i.e. a ring $R$ with topology $\tau$ such that, the maps $x\mapsto -x;\ (x,y)\mapsto x+y;\ (x,y)\mapsto x.y$ are continuous), products of two summable families were summable. In the following contexts, were my (false) beliefs

• $(a_i)_{i\in I},\ (b_j)_{j\in J}$ supposed summable and then $(a_ib_j)_{(i,j)\in I\times J}$ is summable
• Same situation with $I=J=\mathbb{N}$ and $c_n=\sum_{p+q=n}a_pb_q$ (Cauchy product).

But, I found this question and discussion (which proved me that this belief was false in general), returned to Bourbaki General Topology Chapter III, § 6, and there were Exercises 4-5 which proved me that this question was very delicate. Then I could debunk it.

Late addition: See also discussions and the beautiful answer by Robert Furber here.

Answer 6

It is a common mistake to believe that epimorphisms are either identical to surjections or that they are a better concept. Unfortunately this is rarely the case; epimorphisms can be very mysterious and have unexpected behavior

Answer 7

"A closed unit-ball (in a Banach space) is compact!"

It only true in finite-dimensional spaces.

Answer 8

These are 2 instances which i have seen to happen with my friends. If $A$ and $B$ are 2 matrices, then they believe that $(A+B)^{2}=A^{2}+ 2 \cdot A \cdot B +B^{2}$.

Another mistake is if one i asked to solve this equation, $ \displaystyle\frac{\sqrt{x}}{2}=-1$, people generally square both the sides and do get $x$ as $4$.

Answer 9

I have checked the existing answers, but I think this one is not given yet. Sorry, if I missed it.

Although the incompleteness theory of Gödel is generally correctly understood, the consequence of it has multiple false beliefs:

• Due to the incompleteness theory it is not possible to make an AI. Humans will always be be superior to the AI. This assumes that human thinking is complete and will eventually find the answer on any question.

• Due to the incompleteness theory, it is not possible to formalize mathematics. This is refuted by many proof systems, which can formalize almost all mathematics.

As side note, I think this is partly fueled how logic is taught. It puts more emphasis on impossibilities (incompleteness theory), than possibilities (a proof system).

Answer 10

The GNS construction

Let $\phi$ be a state on a $C^*$-algebra $\mathcal{A}$, and put $N_{\phi}:=\{a\in\mathcal{A}\;|\;\phi(a^*a)=0\}$. Then $N_{\phi}$ is a norm-closed left ideal in $\mathcal{A}$. The sesquilinear form $\left<\cdot,\cdot\right>:\mathcal{A}/N_{\phi}\times\mathcal{A}/N_{\phi}\to\mathbb{C}$ defined by $\left<a+N_{\phi},b+N_{\phi}\right>:=\phi(b^*a)$ is a well-defined inner product on $\mathcal{A}/N_{\phi}$. The completion of $\mathcal{A}/N_{\phi}$ establishes a Hilbert space.

False belief: The completion is in the quotient norm.

Surprisingly, Wikipedia (as of April 27, 2018) presents a false statement "The quotient space of the A by the vector subspace I is an inner product space. The Cauchy completion of A/I in the quotient norm is a Hilbert space, which we label H."(https://en.wikipedia.org/wiki/Gelfand%E2%80%93Naimark%E2%80%93Segal_construction#The_GNS_construction) First of all, the quotient of a Banach space by its closed subspace is again a Banach space in the quotient norm, which is a very elementary fact in functional analysis. Thus A/I is already complete in the quotient norm, and hence there is no need to complete it in the quotient norm!

The correct completion is, of course, in the norm induced by the inner product, and this norm is not equivalent to the quotient norm in general. In fact, let $\mathcal{H}$ be a separable infinite-dimensional Hilbert space and $\{\xi_n\}_{n=1}^{\infty}$ be an orthonormal basis for $\mathcal{H}$. The linear functional $\phi:\mathbb{B}(\mathcal{H})\to\mathbb{C}$ defined by $\phi(a):=\sum_{n=1}^{\infty}\frac{1}{2^n}\left<a\xi_n,\xi_n\right>$ is a state on $\mathbb{B}(\mathcal{H})$, and $N_{\phi}=\{0\}$. Let $\xi_k\otimes\xi_k$ be the canonical rank-one operator, and put $p_n:=\sum_{k=1}^n\xi_k\otimes\xi_k$. Then $(p_n)_{n=1}^{\infty}$ is a Cauchy sequence in $\mathbb{B}(\mathcal{H})/N_{\phi}$ in the norm induced by the inner product defined at the beginning, but it is NOT a Cauchy sequence in the quotient norm.

I have never seen a remark which clearly states the distinction between the norm induced by the inner product and the quotient norm in the literature on $C^*$-algebras. Since a quotient space is involved, students are easily tempted to think that the completion is in the quotient norm. (Even the Wikipedia editor was confused!) Or, they may thoughtlessly assume that these two norms are the same. So it will be instructive to clearly state the distinction between these two norms when one teaches this subject to undergraduate students.

Answer 11

This is more sort of a convention issue than an outright false belief (connected to the usual $\emptyset$ vs $\{\emptyset\}$ stuff), but I find it funny. I guess a fair share of mathematicians believe that: \begin{equation} \bigcap\emptyset=\emptyset\label{eq}\end{equation} while retaining the standard definition for intersection: $$\bigcap S:=\{x\ \text{such that}\ \forall Y(Y\in S\implies x\in Y)\}$$ according to which in fact: $$\bigcap\emptyset=V$$ where $V$ is the universal class. The condition in round brackets is of course vacuously true. So in a way - this is what I find funny - the former is the worst possible tentative solution of an equation ever.

Answer 12

If $R$ is a commutative ring with $char(R)=p$ ($p$ is prime), then

$$\varphi:R \to R$$ $$x\mapsto x^p$$

is an automorphism.

Which is false of course.

(If $R$ is a field then see comments).

Answer 13

Here are two beliefs. I think everybody will agree that one of them, at least, is false. I adhere to the second one.

Belief 1. There is no simple generalization of the Hodge Theorem to noncompact manifolds.

Belief 2. The most naive statement which would, if true, generalize the Hodge Theorem to noncompact manifolds is this.

The inclusion of the complex of coclosed harmonic forms into the de Rham complex of a riemannian manifold is a quasi-isomorphism.

This statement happens to be true.

Here is a reference:

http://www.iecl.univ-lorraine.fr/~Pierre-Yves.Gaillard/DIVERS/Hodgegaillard/ (Wayback Machine)

The simplest example is that of the real line with its standard metric. In degree zero the complex of coclosed harmonic forms is $\mathbb C\oplus\mathbb Cx$, and in degree one it is $\mathbb Cdx$, which gives the right cohomology.

Here is the (trivial) algebra background.

Let $A$ be a module over some unnamed ring, and let $d,\delta$ be two endomorphisms of $A$ satisfying $d^2=0=\delta^2$. Put $\Delta:=d\delta+\delta d$. Assume $A=\Delta A+A_{d,\delta}$ where $A_{d,\delta}$ stands for $\ker d\cap\ker\delta$. Write $A_{\delta,\Delta}$ for $\ker\Delta\cap\ker\delta$.

We claim that the natural map $$H(A_{\delta,\Delta},d)\to H(A,d)$$ between homology modules is bijective.

Injectivity. Assume $\delta da=0$ form some $a$ in $A$. We must find an $x$ in $A_{\delta,\Delta}$ such that $dx=da$. We have $a=\Delta b+c$ for some $b\in A$ and some $c\in A_{d,\delta}$. One easily checks that $x:=\delta db+c$ does the trick.

Surjectivity. Let $a$ be in $\ker d$. We must find $x\in A$, $y\in A_{d,\delta}$ such that $a=dx+y$. We have $a=\Delta b+c$ for some $b\in A$ and some $c\in A_{d,\delta}$. One easily checks that $x:=\delta b$, $y:=\delta db+c$ works.

Answer 14

I searched through the answers and hope that I haven't missed this one:

The Minkowski sum of two closed and convex sets is closed and convex.

Answer 15

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

Answer 16

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

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

Answer 17

A few mistakes I remember:

• The quotient groups $\frac{G}{N}$ and $\frac{H}{K}$ are isomorphic if $G \thicksim H$ and $N\thicksim K$.
• A closed interval of a complete lattice is a complete sublattice.
• Two homeomorphic topologies on a set are the same.
• The set of all compatible uniformities of a topological group forms a complete lattice.
• The trace of the identity matrix is 1.

Answer 18

I'm seven years late to the game, but here is mine:

False belief: The irrational numbers, in their usual topology as a subset of $\mathbb{R}$, are not a complete metric space.

Answer 19

False belief: It is obvious how to prove that $\sin'=\cos$.

Not so much... if $\cos$ and $\sin$ are defined geometrically. You need to prove geometrically that $$\lim\limits_{x\to 0}\frac{\sin x}x=1$$ and a (non-circular) proof of that is not obvious (see here).

Personally I'm aware of that just today! (thanks to a remedial course given to my niece).

Answer 20

False belief: relativization is well-defined and the corresponding notation $C^A$ is unambiguous. Which is not quite true because $P=NP$ would not imply $P^A=NP^A$.

Answer 21

If we regard a ring $R$ (with identity) as a right module ($R_{R}$), then there is a ring isomorphism $\text{End}(R_{R}) \simeq R$, however the same does not happen if we regard $R$ as a left module!

The correct is $\text{End}(_{R}R) \simeq R^{\text{op}}$.

Answer 22

False belief: << Let $M$ be the von Neumann algebra generated by a $\rm{C}^{\star}$-algebra $\mathcal{A}$. >>

The false belief is to think that the above sentence makes sense. In fact, a von Neumann algebras and a $\rm{C}^{\star}$-algebra don't have the same status. A von Neumann algebra is an operator algebra by definition, i.e. it is defined inside $B(H)$ for some separable Hilbert space $H$. Now, some subalgebras of $B(H)$ are (separable) $\rm{C}^{\star}$-algebras, but a $\rm{C}^{\star}$-algebra can also be defined abstractly. It can next be represented and a given representation $H$ (defined for example by GNS construction for a given state), if it is faithful, induces an embedding in $B(H)$.
So to make sense, the sentence above should be modified as:

<< Let $M$ be the von Neumann algebra generated by $(\mathcal{A},\rho)$, a couple of $\rm{C}^{\star}$-algebra and state. >>
or
<< Let $M$ be the von Neumann algebra generated by a $\rm{C}^{\star}$-algebra $\mathcal{A}$ represented on $H$. >>

Then, $M = \pi_H(\mathcal{A})''$. We can use $M$ to characterize the representation $H$, for example, we can talk about a representation of type ${\rm I}$, ${\rm II}$ or ${\rm III}$ if $M$ is a von Neumann algebra of type ${\rm I}$, ${\rm II}$ or ${\rm III}$. There is a $\rm{C}^{\star}$-algebra with representations of every type, for example the Cuntz algebra.

Finally, there exists a universal representation for every $\rm{C}^{\star}$-algebra (i.e. the direct sum of the corresponding GNS representations of all states; it is faithful). The associated von Neumann algebra is called the enveloping von Neumann algebra (it can also be defined as the double dual); it contains all the operator-algebraic information about the given $\rm{C}^{\star}$-algebra.

Answer 23

"It is less risky to place 100 one-dollar bets on 100 independent coin flips than to place a single one-dollar bet on a single coin flip".

This is quite evidently false for any reasonable definition of "risky"; for example, in the former case the probability of losing more than a dollar is about 38%; in the latter case the probability is zero.

Nevertheless I am prepared to attest that at least two excellent mathematicians have told me that they believed this all their lives until it was pointed out to them that nothing like it can be true.

What is true is that it is less risky to place 100 one-dollar bets on a 100 independent coin flips than to place a single hundred dollar bet on a single coin flip.

A related fallacy (or the same fallacy in different language), which I have heard many times from both students and professional colleagues, is that "insurance works because the insurance company takes on many independent risks, thereby reducing their overall risk to nearly zero". This is of course false. The more independent risks the company takes on, the more risk it faces.

The correct statement is that insurance works because each risk is apportioned among a great many shareholders, each of whom is now effectively betting a very small amount on each of a great many independent coin flips. If there were only one shareholder, there would be no point in insurance unless either a) the single shareholder was pathologically risk-preferring or b) the premiums were sufficiently actuarially unfair for the company to earn supernormal profits, which for some reason did not get competed away.

Answer 24

The following analogue of this result is clearly false:

Falsehood. If $M$ is a module over a commutative ring $R$, then $M^\vee = \operatorname{Hom}_R(M,R)$ is at least as big as $M$ (e.g. in terms of cardinality or rank).

For example, if $R$ is a domain and $M$ is torsion, then $M^\vee = 0$. But what's much more surprising is that the following is still false:

False belief: If $M$ is a torsion-free module over a principal ideal domain $R$ (even $R = \mathbf Z$), then $|M^\vee| \geq |M|$ and/or $\operatorname{rk}(M^\vee) \geq \operatorname{rk}(M)$.

(Even assuming $M$ has no divisible elements doesn't help.)

Answer 25

"It cannot be shown without some form of AC that the union (or disjoint union) of countably many countable sets is countable. I have a countably infinite set X of countably infinite sets. Therefore, the union of X cannot be shown to be countable without Choice."

The fallacy is that in many cases of interest, it is possible to exhibit an explicit counting of every element of X. In such a case a counting of X by antidiagonals is easily constructed. The usual counting of the rationals is an example of this.

I think this may even be an example of a more general phenomenon of "people think AC is necessary for a certain construction, but in fact it turns out not to be necessary for the example they have in mind". For example, AC is necessary to find a maximal ideal in an arbitrary ring ... but it isn't if you're prepared to assume the ring is Noetherian.

Answer 26

If every collection of disjoint open sets in a topological space is at most countable, then the space is separable

Answer 27

Hopefully this isn't a repeat answer. False belief: a matrix is positive definite if its determinant is positive.

Answer 28

A set is compact iff it is closed and bounded.

Answer 29

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

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

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

Answer 30

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

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