# 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

Yet another common false belief is :

"For non constant periodic functions $$f: \mathbb{R} \to \mathbb{R}$$ and $$g: \mathbb{R} \to \mathbb{R}$$ with smallest positive periods $$p_1 , p_2$$ respectively, the sum $$f+g$$ is periodic if and only if $$\frac{p_1}{p_2}$$ is rational."

One side of the statement above is true (the latter implies the former, but the former does not necessarily imply the latter. (But it's true for continuous functions.)

As an exercise one may like to prove the following : (Source : Miklos Schweitzer competition)

Given any two positive real numbers $$p_1,p_2$$, there exists functions $$f_1 : \mathbb{R} \to \mathbb{R}$$ with smallest positive period $$=p_1$$, and $$f_2 : \mathbb{R} \to \mathbb{R}$$ with smallest positive period $$=p_2$$, such that $$f_1+f_2$$ is also periodic.

One more interesting false belief is :

"If $$f: \mathbb{R} \to \mathbb{R}$$ is a continuous function taking unequal values at points $$x_1,x_2 \in \mathbb{R}$$ with $$x_1 < x_2$$ then there is some sub-interval of $$(x_1,x_2)$$ on which $$f$$ is either strictly increasing or strictly decreasing."

This is wrong. Among the most familiar counterexamples is the Devil's staircase.

I believe this false belief is often due to the habit of seeing continuous functions wearing the glasses provided by the Intermediate Value Property these functions have.

Addendum : Some nice examples from wikipdia, here : https://en.wikipedia.org/wiki/Normally_distributed_and_uncorrelated_does_not_imply_independent

• I think this false belief is mostly due to the underlying "$f$ is continuous" hypothesis that people often carry around – Maxime Ramzi Jun 12 '19 at 9:51
• @Max the condition of continuity can be relaxed. In my opinion this false belief often arises due to the negligence of the subtle role played by continuity in proving the statement for continuous functions. – Aditya Guha Roy Jun 12 '19 at 11:09

"The set A = {a, b} has two elements..."

It's quite simple to notice that a can be the same as b, but after 5 years of university there were people still believing it...

• I'm not sure there is a false belief here, as much as awkward writing. Depending on context, I might very well write "The set $\{a,b \}$ (where $a$ and $b$ might be equal)..." if this issue mattered. – David E Speyer May 6 '10 at 11:16
• There are many situations where one needs to speak of a set of two numbers that may or may not be equal. E.g.: "Let x<sub>1</sub>, x<sub>2</sub> &isin; ℝ. Then among all the open intervals containing the set {x<sub>1</sub>, x<sub>2</sub>}, none of them is contained in all the others." If one is addressing mathematicians, there is no need to specify that x<sub>1</sub> might be equal to x<sub>2</sub>. – Daniel Asimov Jun 17 '10 at 23:34
• E.g. if you said something like "for all a,b, (in some given universe) the set {a,b} has two elements", then I would agree. – roy smith Dec 1 '10 at 19:35
• Single-letter symbols are usually assumed to be variables, if the context doesn't determine otherwise, even in the absence of quantifiers. (You can put in an implicit universal quantifier to close up all sentences.) – Toby Bartels Apr 4 '11 at 9:41
• Here is a related but slightly less obvious situation. The ordered pair $(a,b)$ is generally defined in set theory to be $\{\{a\},\{a,b\}\}$. This is generally thought of as a set with two elements. But what if $a=b$? – Bruce Blackadar Sep 26 '14 at 4:12

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.

• More to the point, I think, is that the paper proves that the power set of any set has greater cardinality than the set itself. This is the first proof that there is no greatest cardinality. (The uncountability of the real line easily follows, even if Cantor does not mention it because he has bigger fish to fry.) – John Stillwell May 31 '10 at 5:12
• Just to fill in some history here: if I remember right, Cantor first proved the uncountability of the reals by other arguments, then later (as you reference) found the diagonal argument, as a proof of the more general statement about power sets. – Peter LeFanu Lumsdaine Sep 27 '10 at 3:01
• The link in the answer goes to the wrong page - it should go to page 75, not page 72. – David Roberts Jun 13 '12 at 6:41
• And it looks like a diagonal argument to me. – David Roberts Jun 13 '12 at 6:43

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).

• +1. I always found it wrong that classes in logic put so much emphasis on negative results. (And I wish they had prepared me better for proof assistants... though I guess one semester does not suffice for the ones that exist today.) – darij grinberg Sep 5 '15 at 22:17
• There's arguably too much fascination with incompleteness and not enough with completeness, which is more of a cornerstone of model theory. – Todd Trimble Sep 6 '15 at 1:50

Taylor's Formula and displacement operator: I (too often) see in papers (mathematical physics but a recent paper (a) by mathematicians also) the statement

False belief 1 : a) Let $D=\frac{d}{dx}$ be the derivation operator. Then, for all $f\in C^\infty(\mathbb{R})$, $$e^{tD}[f](x)=f(x+t)$$

which is false (take any $\phi\in C^\infty(\mathbb{R})$ with compact support, for instance).

False belief 2 : b) In the same vein, for formal power series (`our object is formal then we do not have to ensure convergences''). Let $S(x)\in \mathbb{R}[[x]]$ ($x$ is a formal variable) then for $t\in \mathbb{R}$, one has $$e^{tD}[S](x)=S(x+t)$$

which is false as we must have $t$ in the domain of convergence of $S$.

Remarks (i) The function $f\in C^\infty(\mathbb{R})$ is analytic over $\mathbb{R}$ iff $$(\forall x\in \mathbb{R})(\exists R>0)(\forall t\in ]-R,R[) (\sum_{n\geq 0}\frac{t^n}{n!}D^n[f](x)=f(x+t))\qquad (1)$$ (ii) Even if $f\in C^\omega(\mathbb{R})$, it can happen that the left hand side of eq. (1) do not converge otherwise $f$ would be the restriction of an entire function (which e.g. $\frac{1}{1+x^2}$ is not, for example).

(iii) Even if the LHS of (1) converges for all $x,t\in \mathbb{R}$, $f$ need not be analytic. Consider the following function (classic in theory of distributions)
$$f(x)=0\mbox{ if } x\notin ]-1,1[\mbox{ and } f(x)=e^{\frac{1}{1-x^2}} \mbox{ if } x\in ]-1,1[$$ (iv) In the (b) case $S=\sum_{n\geq 0}n!\, x^n$ for example cannot be displaced.

• What is the correct statement? – ಠ_ಠ Jul 13 '17 at 7:33
• @ಠ_ಠ: it is true for real analytic functions defined on the entire real number line. That is one possible choice of correct statement. – Ben McKay Jul 13 '17 at 8:07
• Maybe you could point out the error in my reasoning: for a Lie group $G$, $G$ has a canonical left action on $C^\infty(G, \mathbb{R})$ by $(g.f)(x) =f(g^{-1}.x)$. Since $D=\frac{d}{dx}$ is a left-invariant vector field on $(\mathbb{R}, +)$, then $(e^{tD}.f) (x)=f(e^{-tD}.x)=f(x-t)$. – ಠ_ಠ Jul 13 '17 at 8:35
• I think we probably mean different things with our notation. For me, $\exp: \mathfrak{g} \to G$ always denotes the exponential map of the Lie group, which always exists. But it looks to me like you mean by this notation that you are integrating the Lie algebra action rather than the Lie algebra. – ಠ_ಠ Jul 13 '17 at 21:02
• @ಠ_ಠ I think I see what you meant. In the (Fréchet) space $C^\infty(\mathbb{R})$, the evolution equation $$y(0)=f ; y′(t)=D[y(t)]$$ has a (unique) solution $y(t)[x]=f(x+t)$ which we could (legitimately ?) note $y(t)=e^{tD}[f]$, but one must keep in mind that, in this case $e^{tD}$ cannot be developed without caution. – Duchamp Gérard H. E. Jul 16 '17 at 14:17

Here is a short list of some false beliefs I had when I was studying mathematics, I suppose they may be common but I have never checked:

• I was in the last year of high school and studying university-level math in advance. I remember trying for a week to prove that a continuous injective map from an open subset of $\mathbb{R}^2$ to $\mathbb{R}^2$ that preserves "being aligned" (I mean that maps aligned triples to aligned triples) must be the restriction of an affine map (over $\mathbb{R}$). That is disproved by restrictions of projective transformations... Which I knew of but I was not able to see they contradicted my belief. When my teacher told me "What about projective transformations?"... I felt dumb.
• I was in the 1st year of PhD studies. My advisor, Adrien Douady, had an idea to build polynomial Julia sets with positive Lebesgue measure. Julia sets are fractals, often with complicated topological structures at every scale. Surely that must be the source of measure? So as an exercise, I tried for a week to prove that Jordan curves are necessarily of Lebesgue measure 0. I told Adrien about my attempts. He gave me a counter-example. I felt dumb.
• Learning that there are closed subsets of the interval with positive Lebesgue measure but no interior did not surprise me as much, as the construction is very simple, but still that's a bit counterintuitive.
• When you zoom on the Mandelbrot set, you see all that round components with smooth boundaries. They look so round. Surely they must be circles, for otherwise the difference would be visible. Well... they are not (except one). Guess how I felt when I learned.
• Frankly, when learning the first time about complex numbers, did anybody here expect that, adding the square root of -1 to the reals would add the roots of all other polynomials?
• I was giving a lecture to math teachers about sensitivity to initial condition (call it chaos) and showing strange attractors on the computer, one told me that by the very presence of chaos, what we see may be quite far from the actual behaviour of the equation, save reality. It turns out hyperbolic systems are stable, so I believe this is still representative (it does not prove it but it is an encouraging hint).
• ... Chaos in deterministic systems. I won't develop on that.
• Surely before hearing of set theory and Cantor's argument, you will believe that all sets are countable. Then after learning that this is not the case, you will think that $\mathbb{R}^2$ must be bigger than $\mathbb{R}$, right?
• You have a $C^\infty$ function on the right half plane, all of whose derivatives have a continuous extension to the boundary line. Surely, it must be easy to extend it to a $C^\infty$ function of the whole plane, isn't it? Well... You can but I would not call it easy.
• Short statements have short proofs. Disproved by Fermat's last theorem (among others).
• I was quite disappointed to learn that there cannot be a finite non-commutative field (division algebra).

I have a few other examples, that I would not term "common false beliefs" but rather "fun and surprising math facts". Is there already a MO question about that?

Here's another howler some people commit: If $m$, $n$ are integers such that $m$ divides $n^2$ then $m$ divides $n$.

It's true sometimes, for example if $m$ is prime (or more generally squarefree, i.e. a product of distinct primes). But in general all one can conclude is that there exists integers $p$, $q$, $r$ with $p$ squarefree such that $m = p q^2$ and $n = p q r$

The usual counterexample is that $8$ divides $4^2$ but not $4$ ;-)

• An even more trivial counterexample is that 4 divides 2^2 but not 2 :-P – Peter LeFanu Lumsdaine Feb 23 '11 at 9:40

A somewhat common belief among students starting out in cryptography:

Breaking RSA requires factoring the modulus.

Although it is not quite known to be a "false" belief, there is no known reduction showing that breaking RSA implies finding the prime factors of the modulus. This in contrast to e.g. Rabin's cryptosystem, and various cryptographic schemes built on other hard problems, whose security provably relies on the underlying hard problems.

• I would put this on the same level as the "false belief" that NP-complete problems really are hard. Nobody knows for sure if it actually is false or not, but given the current state of affairs it is a somewhat reasonable conjecture. And the very fact that nobody knows how to do it means that in practical applications NP-complete problems really are hard to solve and breaking RSA really does require factorisation of the modulus. – Johannes Hahn Mar 16 '18 at 22:56
• Unless you say what the term "breaking RSA" is supposed to mean then it's impossible to evaluate whether that is a false belief or not. For example, say $m = pq$ is an RSA modulus with two different primes $p$ and $q$ and $e$ and $d$ are encryption and decryption exponents for that modulus (so $e$ an$d$ are positive integers such that $ed \equiv 1 \bmod \varphi(m)$). In practice $e > 1$ and $d > 1$, and also in practice $p$ and $q$ are both odd. If "Breaking RSA" means somehow determining $\varphi(m)$ from $m$ and $e$ then you can factor $m$ from knowing $\varphi(m)$ and $m$. If (cont.) ... – KConrad Feb 25 '19 at 11:33
• ... "Breaking RSA" means somehow determining $d$ from $m$ and $e$, then you know $ed - 1$, which is a positive multiple of $\varphi(m)$, and there is a probabilistic algorithm that with over 50% probability of success each time will lead to a nontrivial factor of $m$ starting with a random choice of an integer from 1 to $m-1$. See Theorem 5.6 of kconrad.math.uconn.edu/blurbs/ugradnumthy/RSAnotes.pdf. If "Breaking RSA" means "a wizard tells you how to decode each message without explaining how it is done" then that's not math and thus can't be judged as being a false belief in math. – KConrad Feb 25 '19 at 11:36
• @KConrad This answer is referring to the fact that, unlike for the mentioned Rabin variant, the RSA problem is not known to be as hard as factoring: given an oracle that decrypts ciphertexts, there is no efficient way known to use this oracle to recover the factorization of the modulus. This in contrast to many other schemes and protocols, for which such an oracle would immediately yield the solution for the hard problem it is presumed to rely on. (And many would disagree with you that such oracle reductions are "not math"...) – TMM Apr 2 '19 at 23:46
• @JohannesHahn I'd say the difference is that in cryptography, there do exist plenty reductions for other schemes proving that breaking these schemes implies solving the underlying well-studied hard problems. The whole hardness-hierarchy at least gives us some faith that some problems are probably really hard, but the RSA decoding problem is not connected to factoring in this hierarchy, let alone to any NP-hard problems. – TMM Apr 4 '19 at 1:24

For a bounded subset of a metric space the diameter is two times the radius!

Let $S\subset X$ be bounded. The definitions are:

$\mathrm{diameter}(S):=\sup\{d(x,y)\,|\,x,y\in S\}$

$\mathrm{radius}(S):=\inf\{r>0\,|\,\exists x\in X:\,S\subset B(x,r)\}$

where $B(x,r)$ denotes the open ball of radius $r$ around $x$.

• Hםw do you define the radius of an arbitrary bounded subset? – Mark Apr 11 '11 at 15:34
• Disproved nicely by Reuleaux triangles. – darij grinberg Apr 12 '11 at 8:10
• Disproved nicely by a two-point metric space. – Tom Goodwillie Apr 17 '11 at 1:36
• An equilateral triangle in the Euclidean plane also does the job (diameter $1$ and radius $1/\sqrt{3}$): $2/\sqrt{3} > 1$. – Jean Van Schaftingen Nov 6 '13 at 15:07

For awhile, I used to think:
If $depth\ M\ge depth\ N$ then $depth\ M_p\ge depth\ N_p$; for any prime ideal $p$ and finite R-modules $M$ and $N$ (Which is not true).

(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.

Uncorrelatedness implies independence

This statement is indeed false. Suppose $$X \sim U[-1, 1]$$. Then $$Cov(X, X^2) = EX^3 - EXEX^2 = 0$$, but $$X$$ and $$X^2$$ are clearly not independent. However, that mistake is quite popular...

I already thought that the following two sets were one. $$M(\color{blue}{\Bbb R},2n)\qquad \text{and}\qquad M(\color{red}{\Bbb C},n).$$

Here are some various examples (I hope that some of them weren't already mentioned):
1. If a space $X$ have two different norms $\| \cdot \|_i, i=1,2$ such that $\| \cdot \|_1 \leq \| \cdot \|_2$ then the completion with respect to $\| \cdot \|_1$ is contained in the completion with respect to $\| \cdot \|_2$.
2. If $M_1,M_2$ are isomorphic modules and $N_1,N_2$ are isomorphic submodules then $M_1/N_1$ and $M_2/N_2$ are isomorphic.
3. If $A,B$ are subsets of topological spaces $X,Y$ (resp.) and $A,B$ are homeomorphic then the closures $\overline{A}$ and $\overline{B}$ are also homeomorphic.
4. The standard construction of adjoining unit to the Banach algebra $A$ yields nothing new if $A$ already was unital.
5. The phrase "a function is almost everywhere continuous" means the same as: "the function is almost everywhere equal to the continuous function".
6. Suppose you are trying to prove that some function space $F$ is complete (say that functions are defined on $X$ and real valued): you take a Cauchy sequence $\{f_n\}_n$ and prove that for each point $x \in X$ the sequence $\{f_n(x)\}_n$ is Cauchy. Then form the completeness of $\mathbb{R}$ you obtain a function $f$. The false belief is that it is now enough to show that $f$ belong to $F$.
7. If you have an ascending family $\{A_i\}_i$ then to obtain it's union $\bigcup_{i}A_i$ it is enough to take some countable subfamily
8. A convergent net $\{x_i\}_i$ in a metric space is bounded and the set $\{x_i\}_i \cup \{x\}$ is compact (where $x$ is the limit).
9. If $D$ is an open dense subset of a topological space $X$ then $card \; D= card \; X$

Some undergraduate common false beliefs that I found

(1) If $H$ is a subgroup of $\mathbb{Z}$ and $H$ and $\mathbb{Z}$ are isomorphic, then $H = \mathbb{Z}$;

(2) In a metric space every two open balls are homeomorphic;

(3) For $p \in [1, \infty]$, $L^p(X, \mathfrak{M}, \mu) = \left\{ f \in \mathbb{C}^X : \int_X |f|^p \, d \mu < \infty \right\}$ is a $\mathbb{C}$-normed vector space, with the norm $\lVert f \rVert_p = (\int_X |f|^p \, d \mu)^{1/p}$.

Belief (1) is very naive, for every nontrivial subgroup of $\mathbb{Z}$ is of the form $n \mathbb{Z}$, all of them isomorphic with $\mathbb{Z}$. For (2) people tend to think of normed vector spaces and forgets the discrete metric spaces. For (3) some people just forget that one have to consider the quotient space, where the classes $[f]=[g]$ iff $f=g$ $\mu$-almost everywhere.

Belief (1) is very naive, because every nontrivial subgroup of $\mathbb{Z}$ is of the form $n \mathbb{Z}$, all of them isomorphic to $\mathbb{Z}$. For (2) people tend to think of normed vector spaces and they forget the discrete metric spaces. For (3) some people just forget that one have to consider the quotient space, where the classes $[f]=[g]$ iff $f=g$ $\ \mu$-almost everywhere.

• Well, in (1) I think I can replace $\mathbb{Z}$ by an arbitrary group $G$, because $\mathbb{Z}$ do not come in mind so quickly. – Gustavo Jan 8 '16 at 4:03

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).

This example is similar to this earlier answer.

If $k$ is a field, then $k[x] \otimes_k k[y] \cong k[x,y]$. Therefore also $k[[x]] \otimes_k k[[y]] \cong k[[x,y]]$, right?

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/

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.

This may not count as a false belief, but it is an amusing misconception I had. In college I had a numerical analysis professor who had both a strong accent and messy handwriting, so it was hard to know exactly what he was talking about sometimes.

I was not yet familiar with the Greek letter $\xi$, and that was the variable he always used to represent the error of a computation, but with his handwriting it just looked like a purposeful scribble. So he would say,"Here we have the calculated value and then of course with some error" (scribble).

I thought he was just being dismissive about the error and trying to represent it in a pejorative way.

To be fair, the letter $\xi$ is not one of the easier ones to draw by hand.

• One of my high school teachers was known for exclaiming "I'm a genius!" in reference to certain multivariate polynomials. – Dan Brumleve Jan 13 '18 at 17:46
• (Well, I spent way too long trying to figure this out. In case anyone else comes across this, to save you some time, the joke is that the high school teacher was discussing homogeneous (multivariate) polynomials. With appropriate pronunciation, such as an Australian accent, "homogeneous" could sound like "I'm a genius.") – Zach Teitler Dec 14 '18 at 9:24

True: Given a graded algebra $A$, there is a notion of a "homogeneous" ideal of $A$. It is a property that connects an ideal of $I$ with the grading and is often necessary to require. For example, if $I$ is a homogeneous ideal of $A$, then the algebra $A / I$ is graded again. If $I$ is not homogeneous, then it is not graded in general (since the projections of different graded components of $A$ onto $A / I$ might have nonzero intersection).

False: Given a filtered algebra $A$, there is a notion of a "filtered" ideal of $A$.

There is no such notion. We can require $I$ to be generated by $I\cap A_n$ for some $n$, or actually to lie inside $A_n$ for some $n$, but in most cases none of these is actually needed. (Correct me if I am wrong.) Formulations like "Let $I$ be an ideal compatible with (or respecting) the filtration" are cargo cult.

But: Given a filtered algebra $A$ and a generating set $G$ of an ideal $I$ of $A$, it is an important question whether $I\cap A_n$ is generated by $G\cap A_n$ for every $n\in \mathbb N$. This is not always satisfied, often nontrivial (in many cases it can be proved by using the diamond lemma to show that every element of $A_n$ has a unique "remainder" modulo $I$ in a certain sense, and this remainder can be obtained by repeated subtraction multiples of elements of $G\cap A_n$) and used tacitly in various texts.

• Good point, but "cargo cult"? – Tom Goodwillie Mar 15 '11 at 14:32
• What I mean is: People use these formulations as a protective charm against a danger they don't see but intuitively feel is there, although closer inspection shows that it is pure superstition. – darij grinberg Mar 15 '11 at 17:26

A set is compact iff it is closed and bounded.

• This is perhaps a common false belief among undergraduates, but one that is dispelled by just a superficial acquaintance with functional analysis. – Todd Trimble Dec 9 '13 at 2:45
• @ Todd Trimble: true, but then also the belief about $sin$ suggested by the OP is only common among people who have not completed a course in complex analysis. – Delio Mugnolo Dec 13 '13 at 8:34
• I thought "bounded" is only defined on metric spaces, and this is true on metric spaces. Is that wrong? – Akiva Weinberger Sep 1 '15 at 2:48
• I have seen analysis textbooks take this as a definition. I hope they realize that they are contributing to future confusion in their readers once they move on to topology or even metric spaces. @AkivaWeinberger, The Heine-Borel theorem stated in this way makes sense for arbitrary metric spaces, but it is only true for complete metric spaces for which balls are totally bounded. The correct statement of H-B for general metric spaces is "a metric space is compact iff it is complete and totally bounded". – Mario Carneiro Oct 20 '15 at 21:27
• @AkivaWeinberger: Yes, it is wrong. The closed unit ball of an normed vector space is compact if and only if the space is finite dimensional. – ACL Apr 21 '16 at 13:43

Anytime I wanted to write an answer to this question, I doubted maybe it is not as common as worthy of mentioning here. In fact, I am also not sure how common is the false belief that I observed today in a PDE class. I didn't observe that in many years of teaching calculus, but today four or five students in a small PDE class when calculating a definite integral by parts only applied the limits of the integral to the "second" integral, that is:

$$\int_{a}^b{f(x) g'(x) dx}=f(x) g(x) - \int_{a}^b{f'(x) g(x) dx}$$

Haven't I observed well enough in my calculus classes?

• Definitely integrals are numbers and $f(x)g(x)$ is a function of variable $x$. Formula as written is something very strange. – Fedor Petrov Apr 20 '16 at 18:58
• @FedorPetrov More strange is that most students don't see such a very strange something :) – Amir Asghari Apr 20 '16 at 19:50

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.

• Could you write that more carefully? You mean the false belief to be that the irrationals, as a topological space, can't be complete for some metric. What you wrote can be easily confused with saying the irrational numbers are not complete for the usual metric coming from $\mathbf R$, which is true rather than false. Consider the "false belief" that $(-1,1)$ with its topology from $\mathbf R$ can't be made into a complete metric space for some metric. Certainly it's not complete for the usual metric, but it is if we use $\tan(\pi x/2)$ to identify $(-1,1)$ with $\mathbf R$ topologically. – KConrad Jul 15 '17 at 3:26
• "are not completely metrizable" is the wording you want. – Andrés E. Caicedo Jul 15 '17 at 3:28
• @KConrad and Andres: The wording is part of what made that false belief so believable! At the time I didn't think about the fact that there could be multiple metrics, much less that the completeness of those metrics wasn't a topological property. I only realized my mistake when I was introduced to the ideas contained in your two comments. (That, and picturing the irrationals as complete is HARD!) – Pace Nielsen Jul 15 '17 at 4:15
• Pace, you probably realize by now that continued fractions make the picture a lot easier (whereby the space of irrationals between $0$ and $1$ can be identified with a product space $\mathbb{N}^\mathbb{N}$). – Todd Trimble Jul 23 '17 at 21:07
• @DuchampGérardH.E. A topological subspace of a completely metrizable topological space is completely metrizable if and only if it is a $G_\delta$, that is a countable intersection of open sets. One can use Baire's category theorem to show that $\mathbb{Q}$ is not a $G_\delta$. All this can be found at: en.wikipedia.org/wiki/G%CE%B4_set – Michael Greinecker Aug 11 '17 at 14:41

I didn't notice this in the long list. A student beginning to learn group theory may believe that the converse of Lagrange's Theorem is true, because it is true for subgroups of prime power. They may also believe that a Sylow subgroup is normal because it has a special name. A counter example to both is $A_{4}$ of order $12$ which has no subgroup of order $6$ and whose four different Sylow $3$-subgroups are all conjugates of one another.

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.

• Of course, this is well known to people who are used to this sort of things, but I have found that most of my non-set-theorist friends (and me) believed this. – Fernando Martin Oct 16 at 5:38
• It is especially tricky since it's not that easy to come up with cardinals having countable cofinality, and most familiar uncountable sets do have this property. – Fernando Martin Oct 16 at 5:40
• Yes. The cofinality of $\mathbb{R}$ with its usual ordering is $ℵ_0$, since $\mathbb{N}$ is cofinal in $\mathbb{R}$. But the cofinality of its cardinality $c$ has cofinality strictly greater than $ℵ_0$ (the usual ordering of $\mathbb{R}$ is not order isomorphic to $c$, so that the cofinality depends on the order). Question: Is there really an uncountable cardinal with countable cofinality? reference? – Sebastien Palcoux Oct 16 at 11:12
• But $|X^\mathbb{N}|\geqslant 2^{\mathbb{N}}$ that may be more than $|X|$ if continuum hypothesis is not true. – Fedor Petrov Oct 16 at 12:26
• @SebastienPalcoux I believe the standard example is $\aleph_\omega$ – Denis Nardin Oct 16 at 12:51

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 $4$.

• What "people"? Non-mathematicians? – Todd Trimble May 4 '11 at 0:03
• @Todd: No i was talking of high school students. – crskhr May 4 '11 at 4:08
• @S.C.:if squarring both sides will not give the solution then how can second problem be solved? – Styles Oct 25 '17 at 16:08

"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.

• If "Noetherian" is defined by the ascending chain condition or by requiring all ideals to be finitely generated, then in order to deduce the existence of maximal ideals, you still need a weak form of the axiom of choice. The usual argument uses the axiom of dependent choice. (Of course, if you define "Noetherian" to mean that every set of ideals has a maximal element, then deducing the existence of maximal ideals is a choiceless triviality.) A good reference is "Six impossible rings" by Wilfrid Hodges (J. Algebra 31 (1974) 218-244). – Andreas Blass Oct 22 '10 at 15:29
• Thanks Andreas! I had a feeling there was a technicality somewhere there, but couldn't remember what it was. As a philosophical point I personally think that of course in the absence of AC you want to define Noetherian so that my original statement is true, but admittedly that's a harder sell than my countable-sets example. – Karol Nov 16 '10 at 21:06
• @AndreasBlass's reference, clickably: Hodges - 6 impossible rings. – LSpice Feb 5 '19 at 1:09

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

• Is this really a common(!) false belief? – Martin Brandenburg Oct 3 '11 at 7:23

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 $2$.

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.
• A closed interval of a complete lattice does form a lattice that is complete, right? So that the mistake is that sups and infs in the interval (particularly the sup and inf over the empty set) are not necessarily computed as they would be in the ambient complete lattice; is that what you have in mind? – Todd Trimble Sep 6 '15 at 1:47
• Yes‌‌‌‌‌‌‌‌‌‌‌‌. – Minimus Heximus Sep 6 '15 at 1:55