# 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... 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. 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. Sep 20 '10 at 12:39
• It's a thought -- I might consider it. Oct 4 '10 at 20:13
• Meta created tea.mathoverflow.net/discussion/1165/…
– user9072
Oct 8 '11 at 14:27

"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. 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>. 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. 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.) 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 $a=b$? Sep 26 '14 at 4:12

I saw many students using the "fact" that for a subset $S$ of a group one has $SS^{-1}=\{e\}$

• This is an interesting example, because it addresses the mistakes that come from the all-too frequent confusion with notations. But we need our shortcuts, our $f^{-1}(x)$ versus $x^{-1}$, etc. Obtaining concise notations while avoiding confusion: a tricky proposition! Apr 14 '11 at 15:50

Probably my fault for not paying enough attention in analysis, but:

Any continuous function on the interval that has derivative equal to zero almost everywhere is constant.

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. 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. 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?

"This algebraic variety is a $C^\infty$-smooth manifold, therefore it must be non-singular". This sounds obvious (and in fact it is true over $\mathbb{C}$) however it is false in general (for instance over $\mathbb{R}$). See the discussion here for many details.

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. 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.) ... 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. 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

Here is a false belief I've been carrying in my head for 5 years and only woken up from recently. When one defines symmetric and anti-symmetric tensors on a vector space $$V$$, one usually starts with an action of the symmetric group on tensors, which is usually defined as:

$$S_n \curvearrowright V^{\otimes n}, \quad \sigma (v_1 \otimes \ldots \otimes v_n) = v_{\sigma(1)} \otimes \ldots \otimes v_{\sigma(n)}.$$

Now, if $$\tau \in S_n$$ is another permutation, we have

$$\tau(\sigma (v_1 \otimes \ldots \otimes v_n)) = \tau(v_{\sigma(1)} \otimes \ldots \otimes v_{\sigma(n)}) = v_{\tau(\sigma(1))} \otimes \ldots \otimes v_{\tau(\sigma(n))},$$

right? Wrong! The action above is a right action, so

$$\tau(v_{\sigma(1)} \otimes \ldots \otimes v_{\sigma(n)}) = v_{\sigma(\tau(1))} \otimes \ldots \otimes v_{\sigma(\tau(n))} = (\sigma \cdot \tau)(v_1 \otimes \ldots \otimes v_n).$$

The reason for that is that the defining formula for the action doesn't directly tell you where $$v_i$$ goes. Rather, it shows who comes to the $$i$$-th place.

• This is something I have reteach myself every time I teach it to others. Happily I remember that I need to do so. Feb 7 at 19:24
• Probably I will now demonstrate another false belief, but how is this possible? A $\tau$ is presented with an $v_{\sigma(1)}\otimes...\otimes v_{\sigma(n)}$ to act on. It is not told what $\sigma$ was, how can it guess? In other words, this $v_{\sigma(1)}\otimes...\otimes v_{\sigma(n)}$ is just some (rank 1) tensor, and $\tau$ is just supposed to permute its entries, without knowing anything about any previous actions on it, no? Feb 9 at 6:05
• @მამუკა ჯიბლაძე, one's mind gets tricked because in the original tensor, for each $v_i$, its subscript number coincides with its place within the tensor. And the action of $S_n$ is unfortunately formulated in terms of the subscripts. If you look at the definition of the action, it actually says: $\sigma$ takes a decomposable tensor and sends its $i$-th factor to the $\sigma^{-1}(i)$-th place. This inverse explains why the action is actually from the right. Feb 9 at 8:15
• Thank you! I overlooked your statement that the action is from the right, and was thinking about the left action that permutes the vectors according to $\sigma$. It is true that for different spaces $V_1$, ..., $V_n$ the simplest things available are canonical isomorphisms $V_1\otimes...\otimes V_n\to V_{\sigma(1)}\otimes...\otimes V_{\sigma(n)}$ and this gives that right action when the spaces coincide. When they do coincide there also is the left action that I had in mind but maybe it can also be defined abstractly? Say, in any symmetric monoidal category? Feb 9 at 9:45
• The same already happens with the permutation action of $S_n$ on $k^n$: the action $\sigma \cdot (x_1,\ldots,x_n) = (x_{\sigma(1)},\ldots,x_{\sigma(n)})$ is a right action. The better definition is $(\sigma x)_i = x_{\sigma(i)}$; the confusion comes from viewing coordinates as maps into versus maps out of your space. Feb 10 at 1:38

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.

• Would we get an equivalence between PMI and WOP, if we replaced axiom 2 with $\forall n\in N\left(n\ne0\Longleftrightarrow\exists m\in N\left(S\!\left(m)=n\right)\right)\right)$ Jul 29 at 19:32
• Jul 29 at 19:59
• @VladimirReshetnikov: Good night! I've just corrected the typo that you mentioned in your first comment. Thanks for the links you have shared with me... Jul 30 at 4:38
• @VladimirReshetnikov: Answer to the question in your second comment: Yes. Öhman touches upon this issue on the antepenultimate page of his paper: «One way of actually making the w.o.p. and the p.m.i. equivalent relative to the other axioms is to supplant axiom $\forall n \in N, \, S(n) \neq 0$ with a slightly different version of it, namely: (2') "$\forall n \in N, \, S(n) \neq 0$ and $0$ is the only element that is not a successor". Axiom (2') is obviously stronger than axiom (2). In a sense, this added strength makes up for the weakness of the w.o.p. in relation to the p.m.i.» Jul 30 at 5:05

Every matrix is the sum of a symmetric and an antisymmetric matrix. Hence:

If $V$ is a vector space and $k$ is a number, then the $k$-th tensor product of $V$ with itself decomposes as a direct sum into symmetric and antisymmetric tensors: $$\underbrace{V \otimes ... \otimes V}_{k\text{ times}} = \Lambda^kV \oplus \mathrm{Sym}^kV$$

Recall (in the finite-dimensional case) the dimensions: $$\dim \Lambda^k V = \binom{n}{k} \quad\text{ and }\quad \dim\mathrm{Sym}^kV = \binom{n+k-1}{k}$$

Looking at $k=1$ shows that we have non-trivial intersection.

Looking at $n=k=3$ shows that the sum is not exhausting.

• Is this a common false belief? Oct 18 '15 at 2:52
• @JimConant: I believed this once. Of course, if you count dimensions it's obviously false. But if $V$ is an infinite-dimensional Hilbert space it sure seems natural to decompose the full tensor product into its Bosonic and Fermionic parts, and you might not think right away to ask whether it works in finite dimensions. That's my excuse, anyway! Mar 4 '16 at 4:22

It’s tempting to believe that a knot and its mirror-image, if tied in succession on a rope, can cancel. View https://youtu.be/lwWeRMmXIoU to see John Conway’s lucid explanation of why this is false.

I would be interested in knowing whether Lord Kelvin was under this misapprehension when he proposed his vortex theory of the atom. (Knots don’t cancel but vortices can.)

I once went to a conference on recreational mathematics in which a speaker claimed that a knot and its mirror-image can be cancelled; I was kind enough not to ask for a demonstration. (There is a way to pretend to make them cancel, but I don’t know the details of the trick.)

• Although a knot and its mirror image do cancel in the concordance group! Jan 25 at 18:56

Heisenberg-Weyl and enveloping algebras

I have heard (and have read) the following belief even among distinguished mathematical physicists.

Let $$HW_\mathbb{C}$$ be the (associative with unit) algebra generated by two elements $$\{a,a^\dagger\}$$ subjected to the relation $$[a,a^\dagger]=1$$. It is easy to check that $$\mathfrak{g}=span_\mathbb{C}\{a,a^\dagger,1_{HW_\mathbb{C}}\}$$ is a Lie subalgebra.

False belief The algebra $$HW_\mathbb{C}$$ is the universal enveloping algebra of $$\mathfrak{g}$$ i.e. $$HW_\mathbb{C}=U(\mathfrak{g})\ .$$

One way to see that this is false at once is to observe that any enveloping algebra $$U(\mathfrak{g})$$ possesses (at least) a character $$\varepsilon$$, but there is none on $$HW_\mathbb{C}$$ (one would have indeed $$1=\varepsilon([a,a^\dagger])=0$$).

Late edit (After Darij's post) Indeed, the enveloping algebra of $$\mathfrak{g}=span_\mathbb{C}\{a,a^\dagger,1_{HW_\mathbb{C}}\}$$ is the algebra (associative with unit) generated by $$\{a,a^\dagger,e\}$$ ($$e$$ is a clone of $$1_{HW_\mathbb{C}}$$) subjected to the relations $$[a,a^\dagger]=e ; [a,e]=[a^\dagger,e]=0\qquad \mbox{[Rel 1]}$$ and the canonical arrow $$U(\mathfrak{g})\to HW_\mathbb{C}$$ transforms an expression in $$a,a^\dagger,e$$, already reduced by [Rel 1] into its image in $$HW_\mathbb{C}$$. For relations [Rel 1] one can take, for instance, the normal form: all $$a^\dagger$$ (creations) on the left, all $$a$$ (annihilations) on the right and $$e$$ anywhere (as they are central).

Combinatorially, in the reduced expressions of $$U(\mathfrak{g})$$, the power of e counts the number of times Wick commutations have been applied. I'd include this in my post if it were not out of place.

• Yeah. There is a surjection $U\left(\mathfrak{g}\right) \to HW_{\mathbb{C}}$, though; its kernel is the ideal generated by $1_{HW_{\mathbb{C}}} - 1$. Thus, $HW_{\mathbb{C}}$ can be viewed as a "reduced" $U\left(\mathfrak{g}\right)$ (similarly to how the free product $A * B$ of two algebras $A$ and $B$ can be viewed as a "reduced" tensor algebra $T\left(A \otimes B\right)$). This is one of the things Pavel Etingof taught me; before that, I thought Weyl/Clifford algebras and universal enveloping algebras were similar but un-relatable constructions. Oct 3 '17 at 6:34
• I suppose a simple way to convince oneself this is not right is simply to note that the unit' in $\mathfrak g$ is not special in any way (and in particular, does not become the unit of the universal envelope), as darij does. Mar 21 at 20:36
• @PedroTamaroff Of course yes. Units have no meaning in the world of Lie algebras. Note that, beyond the (philosophical) question of the meaning, the canonical arrow $U(\mathfrak{g})\to HW_\mathbb{C}$ has a combinatorial interest as the reduction counts something. Mar 22 at 7:29
• @darijgrinberg [$𝐴$ and $𝐵$ can be viewed as a "reduced" tensor algebra $𝑇(𝐴\otimes 𝐵)$]---> didn't you mean $𝑇(𝐴\oplus 𝐵)$]? Mar 22 at 8:05
• Well, as a second thought, both are possible. The $T(A\oplus B)$ version allows noncommutative grading and embedding proofs though. Mar 22 at 8:23

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

Belief (1) is very naive, for every nontrivial subgroup of $$\mathbb{Z}$$ is of the form $$n \mathbb{Z}$$, all of them isomorphic to $$\mathbb{Z}$$. For (2) people tend to think of normed vector spaces and 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. Jan 8 '16 at 4:03

"Frac" and "Const" commute: A differential ring is simply a pair $$(R,\partial)$$ where $$R$$ is a ring and $$\partial$$ is a derivation in $$R$$ (i.e. $$\partial\in End_{\mathbb{Z}}(R)$$ follows identically Liebniz rule $$\partial(ab)=\partial(a)b+a\partial(b)$$). The constants of $$R$$, $$\ker(\partial)$$ form a subring $$Const(R)\subset R$$ called subring of constants.

If $$R$$ is a commutative domain (i.e. without zero divisors), it is standard to consider the field of fractions $$Frac(R)$$ and to extend $$\partial$$ by the formula of calculus $$\partial(1/g)=-\partial(g)/g^2$$.

False belief: $$Const(Frac(R))=Frac(Const(R))$$, in other words, every constant in $$Frac(R)$$ is of the form $$\alpha/\beta$$, where $$\alpha,\beta\in Const(R)$$.

Until yesterday morning, I postponed to prove this (false) lemma (thinking it was an easy exercise). Then, preparing a talk, I could not prove this and searched for a counterexample. Finally I found a simple one in MSE (see below).

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.) 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. 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. Jun 13 '12 at 6:41
• And it looks like a diagonal argument to me. Jun 13 '12 at 6:43

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. Apr 12 '11 at 8:10
• Disproved nicely by a two-point metric space. Apr 17 '11 at 1:36
• An equilateral triangle in the Euclidean plane also does the job (diameter $1$ and radius $1/\sqrt{3}$): $2/\sqrt{3} > 1$. Nov 6 '13 at 15:07

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. 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.") Dec 14 '18 at 9:24

(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...

For some reason, I believed until recently that every compact operator on a Hilbert space must have eigenvalues. It's obviously true for finite-rank operators and Hermitian operators, so I must have subconsciously generalized.

This was despite being well aware of (say) the Volterra operator, say; I simply had two contradictory ideas in my head.

• The example I had heard of for this was the operator on $\ell^2$ you get by multiplying pointwise by $\frac{1}{n}$ after doing a unilateral right shift. I realize now that this is the Volterra operator in disguise. Jun 2 at 2:31
• Yes, your example was what I first thought of as well, and only afterwards did someone bring up the Volterra operator. Jun 2 at 19:30

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 $card \; D= card \; X$

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.) 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. Sep 6 '15 at 1:50

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).

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.

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.

• "Normal because it has a special name" belief looks pretty weird. All people have special names, do they consider all of them normal? Mar 21 at 20:49

I always thought that the GCD of two elements existed if and only if the LCM of those two elements existed, because of my experience with GCDs and LCMs in the ring $$\mathbb{Z}$$. Only recently did I discover that this isn't true in other commutative rings.

• For example, ...? Nov 5 '20 at 21:36
• See link.springer.com/article/10.1007/BF02837870 There are two elements of the integral domain $\mathbb{Z}[\sqrt{-3}]$ that have an LCM but not a GCD. (Theorem 2 gives a nice relationship between LCM's and GCD's in general.) Nov 5 '20 at 23:23
• Thanks – but, at 35 euros, I think I'll pass. For free, there are examples at math.stackexchange.com/questions/1449521/… (where it clarifies that if there's an LCM, then there's a GCD; only the converse fails). Nov 6 '20 at 6:31

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"? 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. Mar 15 '11 at 17:26

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.