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

For vector spaces, $\dim (U + V) = \dim U + \dim V - \dim (U \cap V)$, so $$ \dim(U +V + W) = \dim U + \dim V + \dim W - \dim (U \cap V) - \dim (U \cap W) - \dim (V \cap W) + \dim(U \cap V \cap W), $$ right?

Answer 2

Everyone knows that for any two square matrices $A$ and $B$ (with coefficients in a commutative ring) that $$\operatorname{tr}(AB) = \operatorname{tr}(BA).$$

I once thought that this implied (via induction) that the trace of a product of any finite number of matrices was independent of the order they are multiplied.

Answer 3

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

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

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

Answer 4

Many students believe that 1 plus the product of the first $n$ primes is always a prime number. They have misunderstood the contradiction in Euclid's proof that there are infinitely many primes. (By the way, $2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13 + 1$ is not prime and there are many other such examples.)

Much later edit: As pointed out elsewhere in this thread, Euclid's proof is not by contradiction; that is another widespread false belief.

Much much later edit: Euclid's proof is not not by contradiction. This is another very widespread false belief. It depends on personal opinion and interpretation what a proof by contradiction is and whether Euclid's proof belongs to this category. In fact, if the derivation of an absurdity or the contradiction of an assumption is a proof by contradiction, then Euclid's proof is a proof by contradiction. Euclid says (Elements Book 9 Proposition 20): The very thing (is) absurd. Thus, G is not the same as one of A, B, C. And it was assumed (to be) prime.

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Nb. The above edits were not added by the OP of this answer.

Edit on 24 July 2017: Euclid's proof was not by contradiction, but contains a small lemma in the middle of it that is proved by contradiction. The proof shows that if $S$ is any finite set of primes (not assumed to be the set of all primes) then the prime factors of $1+\prod S$ are not in $S$, so there is at least one more prime than those in $S.$ The proof that $\prod$ and $1+\prod$ have no common factors is the part that is by contradiction. All of this is shown in the following paper: M. Hardy and C. Woodgold, "Prime simplicity", Mathematical Intelligencer 31 (2009), 44–52.

Answer 5

Here's my list of false beliefs 😉

• If $U$ is a subspace of a Banach space $V$, then $U$ is a direct summand of $V$.
• If $M/L$ and $L/K$ are normal field extensions, then the same is true for $M/K$.
• Submodules / subgroups / subalgebras of finitely generated modules / groups / algebras are finitely generated.
• For a subring $S \subseteq R$ of a commutative ring the Krull dimension satisfies $\dim(S) \leq \dim(R)$.
• The Krull dimension of a noetherian integral domain is finite.
• If $A \otimes B = 0$ for abelian groups $A,B$, then either $A=0$ or $B=0$.
• If $f$ is a smooth function with $df=0$, then $f$ is constant.
• If $X,Y$ are sets such that $P(X), P(Y)$ are equipotent, then $X,Y$ are equipotent.
• Every short exact sequence of the form $0 \to A \xrightarrow{f} A \oplus B \xrightarrow{g} B \to 0$ splits.
• $R[x]^{\times} = R^{\times}$ for any commutative ring $R$.
• Every presheaf on a site has an associated sheaf.
• (Co)limits may be computed in full subcategories. For example, $\mathrm{Spec}(\prod_i R_i) = \coprod_i \mathrm{Spec}(R_i)$ as schemes because $\mathrm{Spec}$ is an anti-equivalence between commutative rings and affine schemes.
• Every finite CW-complex is compact, thus every CW-complex is locally compact.
• The smash product of pointed spaces is associative, products of topological spaces commute with quotients, and so on.

Answer 6

I don't know if this is common or not, but I spent a very long time believing that a group $G$ with a normal subgroup $N$ is always a semidirect product of $N$ and $G/N$. I don't think I was ever shown an example in a class where this isn't true.

Answer 7

These are actually metamathematical (false) beliefs that many intelligent people have while they are learning mathematics, but usually abandon when their mistake is pointed out, and I am almost certain to draw fire for saying it from those who haven't, together with the reasons for them:

The results must be stated in complete and utter generality.

Easy examples are left as an exercise to the reader.

It is more important to be correct than to be understood.

(Applicable to talks as well as papers.)

Reasons: 1. Von Neumann is in the audience. 2. This is just a generalization of Lemma 1.2.3 in volume X of Bourbaki. 3. The results are impressive and speak for themselves.

Answer 8

a student, this afternoon: "this set is open, hence it is not closed: this is why [...]"

Answer 9

Some false beliefs in linear algebra:

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

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

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

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

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

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

Answer 10

I once thought that if $A$, $B$, $C$, and $D$ were $n$-by-$n$ matrices, then the determinant of the block matrix $\pmatrix{A & B \\\ C & D}$ would be $\det(A) \det(D) - \det(B) \det(C)$.

Answer 11

Here are two things that I have mistakenly believed at various points in my "adult mathematical life":

For a field $k$, we have an equality of formal Laurent series fields $k((x,y)) = k((x))((y))$.

Note that the first one is the fraction field of the formal power series ring $k[[x,y]]$. For instance, for a sequence $\{a_n\}$ of elements of $k$, $\sum_{n=1}^{\infty} a_n x^{-n} y^n$ lies in the second field but not necessarily in the first. [Originally I had $a_n = 1$ for all $n$; quite a while after my original post, AS pointed out that that this actually does lie in the smaller field!]

I think this is a plausible mistaken belief, since e.g. the analogous statements for polynomial rings, fields of rational functions and rings of formal power series are true and very frequently used. No one ever warned me that formal Laurent series behave differently!

[Added later: I just found the following passage on p. 149 of Lam's Introduction to Quadratic Forms over Fields: "...bigger field $\mathbb{R}((x))((y))$. (This is an iterated Laurent series field, not to be confused with $\mathbb{R}((x,y))$, which is usually taken to mean the quotient field of the power series ring $\mathbb{R}[[x,y]]$.)" If only all math books were written by T.-Y. Lam...]

Note that, even more than KConrad's example of $\mathbb{Q}_p^{\operatorname{unr}}$ versus the fraction field of the Witt vector ring $W(\overline{\mathbb{F}_p})$, conflating these two fields is very likely to screw you up, since they are in fact very different (and, in particular, not elementarily equivalent). For instance, the field $\mathbb{C}((x))((y))$ has absolute Galois group isomorphic to $\hat{\mathbb{Z}}^2$ -- hence every finite extension is abelian -- whereas the field $\mathbb{C}((x,y))$ is Hilbertian so has e.g. finite Galois extensions with Galois group $S_n$ for all $n$ (and conjecturally provably every finite group arises as a Galois group!). In my early work on the period-index problem I actually reached a contradiction via this mistake and remained there for several days until Cathy O'Neil set me straight.

Every finite index subgroup of a profinite group is open.

This I believed as a postdoc, even while explicitly contemplating what is probably the easiest counterexample, the "Bernoulli group" $\mathbb{B} = \prod_{i=1}^{\infty} \mathbb{Z}/2\mathbb{Z}$. Indeed, note that there are uncountably many index $2$ subgroups -- because they correspond to elements of the dual space of $\mathbb{B}$ viewed as a $\mathbb{F}_2$-vector space, whereas an open subgroup has to project surjectively onto all but finitely many factors, so there are certainly only countably many such (of any and all indices). Thanks to Hugo Chapdelaine for setting me straight, patiently and persistently. It took me a while to get it.

Again, I blame the standard expositions for not being more explicit about this. If you are a serious student of profinite groups, you will know that the property that every finite index subgroup is open is a very important one, called strongly complete and that recently it was proven that each topologically finitely generated profinite group is strongly complete. (This also comes up as a distinction between the two different kinds of "profinite completion": in the category of groups, or in the category of topological groups.)

Moreover, this point is usually sloughed over in discussions of local class field theory, in which they make a point of the theorem that every finite index open subgroup of $K^{\times}$ is the image of the norm of a finite abelian extension, but the obvious question of whether this includes every finite index subgroup is typically not addressed. In fact the answer is "yes" in characteristic zero (indeed $p$-adic fields have topologically finitely generated absolute Galois groups) and "no" in positive characteristic (indeed Laurent series fields do not, not that they usually tell you that either). I want to single out J. Milne's class field theory notes for being very clear and informative on this point. It is certainly the exception here.

Answer 12

"Any subspace of a separable topological space is separable, too." Sounds natural.

Answer 13

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Answer 14

From the Markov property of the random walk $(X_n)$ we have

$$P(X_4>0 \ |\ X_3>0, X_2>0) = P(X_4>0\ |\ X_3>0).$$

To paraphrase Kai Lai Chung in his book "Green, Brown, and Probability",

"The Markov property means that the past has no after-effect on the future when the present is known; but beware, big mistakes have been made through misunderstanding the exact meaning of the words "when the present is known"."

Answer 15

$$2^{\aleph_0} = \aleph_1$$

This is a pet peeve of mine, I'm always surprised at the number of people who think that $\aleph_1$ is defined as $2^{\aleph_0}$ or $|\mathbb{R}|$.

Answer 16

I think, there are different types of false beliefs. The first kind are statements which are quite natural to believe, but a moment of thought shows the contradiction. Of this type is the sin-example in the opening post or a favorite of mine (also occurred to me):

• The underlying additive group of the field with $p^n$ elements is $\mathbb{Z}/p^n\mathbb{Z}$.

The other type is also quite natural to believe, but one has really to think to construct a counter example:

• Every contractible manifold is homeomorphic to $\mathbb{R}^n$.
• Every manifold is homotopy equivalent to a compact one.
• Quotients commute with products in topological spaces.
• Every connected component of a topological space is open and closed. Or related to this:
• To give a continuous action of a topological group $G$ on a discrete space $X$ is the same as to give an action of the group of connected components of $G$ on $X$.

Answer 17

I used to believe that a continuous algebra homomorphism from $k[[x_1,\dots, x_m]]$ to $k[[y_1,\dots,y_n]]$, with $m > n$, could not be injective. Konstantin Ardakov set me straight on this.

Answer 18

Here are two group theory errors I've seen professionals make in public.

1) Believing that if $G_1 \subset G_2 \subset \cdots$ is an ascending union of groups such that $G_i$ is free, then $\bigcup_{i=1}^{\infty} G_i$ is free. Probably the vague idea they have is that any relation has to live in some $G_i$, so there are no nontrivial relations.

2) Consider a group $G$ acting on a vector space $V$ (over $\mathbb{C}$, say). Assume that $G$ acts as the identity on a subspace $W$ and that the induced action of $G$ on $V/W$ is trivial. Then I've seen people conclude that the action of $G$ on $V$ is trivial. Of course, this is true if $G$ is finite since then all short exact sequences of $G$-modules split, but it is trivial to construct counterexamples for infinite $G$.

Answer 19

The field of $p$-adic numbers has characteristic $p$.

Answer 20

"Either you can prove the statement, or you can find a counterexample."

This statement is usually applied to universal statements, those having the form $\forall x\ \varphi(x)$, where the concept of counterexample makes sense, but the general sentiment is the belief that every statement in mathematics is either provable or refutable.

The belief is false, because of the independence phenomenon.

Answer 21

I remember from my first analysis class thinking that if $\mathbb{Q}\subset E\subset\mathbb{R}$ with $E$ open, then $E$ would have to be all of $\mathbb{R}$ (at least more or less, maybe up to countably many points). And once we started measure theory I remember arguing with a friend over it for a good two hours.

Answer 22

If $f(x,y)$ is a polynomial with real coefficients, then the image of $f$ is a closed subset of $\mathbb{R}$. Note. Problem A1 on the 1969 Putnam exam asked to describe all possible images of $f$. I was told that the writers of this problem did not realize its subtlety.

Answer 23

In order to show that a polynomial $P \in F[x_1,\ldots,x_n]$ vanishes, it suffices to show that $P(x_1,\ldots,x_n) = 0$ for all $x_1,\ldots,x_n \in F$. True in infinite fields, but very false for small finite fields.

Closely related: if two polynomials $P$, $Q$ agree at all points, then their coefficients agree. Again, true in infinite fields, but false for finite fields.

(This is ultimately caused by a conflation of the concept of a polynomial as a formal algebraic expression, and the concept of a polynomial as a function. Once one learns enough algebraic geometry to be comfortable with concepts such as "the $F$-points $V(F)$ of a variety $V$" then this confusion goes away, though.)

Answer 24

"If any two of the $3$ random variables $X,Y,Z$ are independent, all three are mutually independent." In fact, they may be dependent; the simplest example is probably $(X, Y, Z)$ chosen uniformly from $\{(0, 0, 0), (1, 1, 0), (1, 0, 1), (0, 1, 1)\}$.

Answer 25

"Automorphisms of the symmetric group $S_n$ are inner (that is, each one is of the form $x \to axa^{-1}$ for some $a \in S_n$)" is a popular misconception, false for nontrivial reasons when $n=6$. That is both an easy mistake to make and important conceptually as an early hint of the complexities and special combinatorics that arise in finite group theory. Many people make it through a first class in group theory without understanding that something different happens for $S_6$ and in doing so have missed an important piece of the the big picture, as far as finite groups are concerned.

It is easy to implicitly or explicitly acquire this belief, because:

1. those really are all the automorphisms for $n$ other than 6, and

2. the inner automorphisms are used so often, for all values of $n$ (or $n>2$) without distinguishing any specific case as unusual.

3. $S_n$ behaves in many ways as a family of similar groups rather than a list of individual groups with their own diverse features. A typical proof might show some property of $S_n$ by induction on $n$, starting from a small value such as $n=1$ for basic properties, or $n=3$ to assure noncommutativity. Apart from the classification of symmetric group automorphisms itself (exposure to which would be an explicit articulation and correction of the false belief), these arguments never start as high as $n=7$ and I don't know of any that distinguish $n=6$ or some equivalent case as a lone nontrivial exception. So it is easy to get the idea of more uniformity in the $S_n$ than really exists.

In essence, there are no obvious clues in the environment that $n=6$ might be special, and a number of indicators that no special case should exist at all.

Answer 26

There are a couple of false beliefs regarding the $I$-adic completion functor, where $I$ is an ideal in a commutative ring $A$.

The first is that the completion of an $A$-module $M$ is complete, or in other words, that the completion functor is idempotent. This is true if $I$ is finitely generated (in particular when $A$ is Noetherian), but false in general. I find this quite unexpected - you take a module, "complete" it, and the result is not complete...

Another issue is the exactness of the completion functor. The completion functor is exact on the category of finitely generated modules over a Noetherian ring, but when you consider arbitrary modules, it is neither left-exact (this is easy to see) nor it is right-exact (this probably less known), even when $I$ is finitely generated and the modules in question are finitely presented.

Answer 27

Some people have trouble understanding that (and why is) 0.999... = 1

Answer 28

This is perhaps a misunderstood definition rather than a false belief, but:

"A subnet of a net $( x_\alpha )_{\alpha \in A}$ takes the form

$( x_\alpha )_{\alpha \in B}$ for some subset $B$ of $A$."

In truth, subnets are allowed to contain repetitions, and can be indexed by sets much larger than the original net. (In particular, there are subnets of sequences that are not subsequences.)

This false belief, incidentally, reinforces the false belief noted in a different answer, namely that compactness implies sequential compactness.

A precise Counterexample: The sequence $\sin(nx)$ is a sequence in the compact topological space $[-1, 1]^{\mathbb{R}}$ with product topology. So this net has a convergence subnet. But it is well known that the above sequence has no subsequence which is pointwise convergent (See the last page of the book of Walter Rudin's Principles of mathematical Analysis). So in this example the convergent subnet cannot be counted as a subsequence.

Answer 29

"It is impossible in principle to well-order the reals in a definable manner."

To be more precise, the belief I am talking about is the belief that well-orderings of the reals are provably chaotic in some sense and certainly not definable. For example, the belief would be that we can prove in ZFC that no well-ordering of the reals arises in the projective hierarchy (that is, definable in the real field, using a definition quantifying over reals and integers).

This belief is relatively common, but false, if the axioms of set theory are themselves consistent, since Goedel proved that in the constructible universe $L$, there is a definable well-ordering of the reals having complexity $\Delta^1_2$, which means it can be obtained from a Borel subset of $R^3$ by a few projections and complements. See this answer for a sketch of the definition of the well-order.

The idea nevertheless has a truth at its core, which is that although it is consistent that there is a definable well-ordering of the reals (or the universe), it is also consistent that there is no such definable well-ordering. Thus, there is no definable relation that we can prove is a well-ordering of the reals (although we also cannot prove that none is).

Answer 30

Occasionally seen on this site: if a polynomial $P:\mathbb{Q}\rightarrow\mathbb{Q}$ is injective, so must be its extension to $\mathbb{R}$.