# 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.

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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/… – quid Oct 8 '11 at 14:27

Many students believe that every abelian subgroup is a normal subgroup.

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"Let $E$ be a complete locally convex topological vector space (or a complete topological vector space or a complete topological group) and let $F$ be closed vector subspace (or a closed subgroup). Then the quotient $E/F$ is complete."

This just has to be true. One can almost see the proof. And in fact it is true for Banach spaces. So it has to be true for locally convex spaces as well.

Another one with completions:

"Every topological group is a dense subgroup of a complete topological group." True for abelian groups but false in general (take the homeomorphism group of $[0,1]$ with the compact open topology)

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Yet another one:

Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be differentiable. If $f'(x_0) > 0$, then there exists an interval $I$ containing $x_0$ such that $f$ is increasing in $I$.

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I sort of find it hard to believe that amongst the nearly 200 answers on this thread (and just over 20 deleted ones), no one has posted this. – Asaf Karagila Aug 10 at 6:07
A counter-example is necessarily with $f'$ discontinuous in $x_0$, right? For example $f(x)=x^2 sin (1/x)+x/2$ and $x_0 = 0$. – Sébastien Palcoux Aug 10 at 8:05
@SébastienPalcoux Yes, I think if $f'$ is continuous in $x_0$ then the statement is true. – Shamisen Aug 10 at 15:16

When I was studying Banach spaces, I was confused with the following: We know that, in any Banach Space $V$, the closed unit ball is compact in the topology generated by the norm if, and only if, the dimension of $V$ is finite. But thinking about $\mathbb R$ as a vector space over $\mathbb Q$, we have an infinite-dimensional vector space which is complete in the norm (given by the modulus) but the closed unit ball is, of course, compact in topology generated by the norm.

I took some time to discover that my mistake was that I thought about $\mathbb R$ over $\mathbb Q$ as a Banach space. In fact, this vector space is a complete metric space (in the sense of Cauchy sequences), but I realized later that the word Banach space is reserved only for vector spaces defined over the fields $\mathbb R$ or $\mathbb C$.

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You can define Banach spaces over any complete field. For example, one can define p-adic Banach spaces. But Q isn't complete with respect to any of its norms. – Qiaochu Yuan May 4 '10 at 22:14
In fact the Theorem I mentioned in my answer, is based on the Riesz lemma and this lemma is not valid if the scalar fields is not complete. – Leandro May 4 '10 at 22:34
@QiaochuYuan ...unless you use the trivial norm. – Mario Carneiro Oct 20 at 21:16

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

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

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@Michael: all answers are CW; so if we think some wording needs clarifying, we can do it ourselves! – Peter LeFanu Lumsdaine Dec 2 '10 at 0:43

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 $x$.

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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

I don't know how common this mistake is, but I think it's worth mentioning. I used to think that existence of non-measurable sets is guaranteed by the axiom of choice only.

In the presence of AC, there cannot be a $\sigma$-additive measure on $\mathcal{P}(\mathbb{R})$ that extends the usual Lebesgue measure.

It is true that we cannot extend the Lebesgue measure in a translation-invariant way by various Vitali set constructions. On the other hand, if you do not insist that the extension is translation-invariant, it might be possible to do this relative to a real-valued measurable cardinal assumption.

Theorem (Ulam): If there exists a cardinal $\kappa$ such that there exists an atomless $\kappa$-additive probability measure on $\mathcal{P}(\kappa)$, then there exists a $\sigma$-additive measure on $\mathcal{P}(\mathbb{R})$ extending the Lebesgue measure.

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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.

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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

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 ;-)

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

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.

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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

If $\alpha>0$ is not an integer, the set of functions $f:[a,b]\rightarrow\mathbb R$ such that $$\sup_{y\ne x}\frac{|f(y)-f(x)|}{|y-x|^\alpha}<+\infty$$ is ${\mathcal C}^\alpha([a,b])$.

False for $\alpha>1$, because this set contains only constant functions.

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"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...

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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 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
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
In a context where one is discussing real analysis, $e$ and $\pi$ are generally taken to be the famous constants. But this is hardly universal; in other contexts, they may have very different meanings. – Toby Bartels Dec 15 '13 at 5:16

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:

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.

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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. The simplest way to compute the exponential $e^A$ of a complex square matrix $A$ is to use the Jordan decomposition.

Belief 2. It's simpler and more efficient to use the following fact.

Let $f(z)$ be the minimal polynomial of $A$, let $g(z)$ be $f(z)$ times the singular part of $e^z/f(z)$, and observe $e^A=g(A)$.

(By abuse of notation $z$ is at the same time an indeterminate and a complex variable.) (The problems of computing the exponential of $A$ and that of computing the Jordan decomposition of $A$ have the same difficulty level. But, to solve one of them, there is no need to refer to the other.) Here are two references

Jordan decomposition is often mentioned in relation with matrix exponentials. I'm convinced (rightly or wrongly) that the association of these notions in this context is purely irrational. I think somebody once made this association by accident, and then many people repeated it mechanically.

Here is another attempt to describe the situation.

Put $B:=\mathbb C[A]$. This is a Banach algebra, and also a $\mathbb C[X]$-algebra ($X$ being an indeterminate). Let $$\mu=\prod_{s\in S}\ (X-s)^{m(s)}$$ be the minimal polynomial of $A$, and identify $B$ to $\mathbb C[X]/(\mu)$. The Chinese Remainder Theorem says that the canonical $\mathbb C[X]$-algebra morphism $$\Phi:B\to C:=\prod_{s\in S}\ \mathbb C[X]/(X-s)^{m(s)}$$ is bijective. Computing exponentials in $C$ is trivial, so the only missing piece in our puzzle is the explicit inversion of $\Phi$. Fix $s$ in $S$ and let $e_s$ be the element of $C$ which has a one at the $s$ place and zeros elsewhere. It suffices to compute $\Phi^{-1}(e_s)$. This element will be of the form $$f=g\ \frac{\mu}{(X-s)^{m(s)}}\mbox{ mod }\mu$$ with $f,g\in\mathbb C[X]$, the only requirement being $$g\equiv\frac{(X-s)^{m(s)}}{\mu}\mbox{ mod }(X-s)^{m(s)}$$ (the congruence taking place in the ring of rational fractions defined at $s$). So $g$ is given by Taylor's Formula.

This can be summarized as follows:

There is a unique polynomial $E$ such that $\deg E<\deg\mu$ and $e^A=E(A)$. Moreover $E$ can be uniquely written as $$E=\sum_{s\in S}\\ E_s\\ \frac{\mu}{(X-s)^{m(s)}}$$ with (for all $s$) $\deg E_s < m(s)$ and $$E_s\equiv e^s\ e^{X-s}\\ \frac{(X-s)^{m(s)}}{\mu}\mbox{ mod }(X-s)^{m(s)},$$ the congruence taking place in $\mathbb C[[X-s]]$.

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"$\emptyset$ is a basis for {0}" is an immediate consequence of the definitions. There is no false belief in this point. – Johannes Hahn May 12 '10 at 15:14
Dear Johannes, please reread my post. – Pierre-Yves Gaillard May 12 '10 at 15:18
Your opinions are normative statements: "one should" and "it is better". It is naive to suppose that there is one best method that one should use to compute the matrix exponential. – Robin Chapman May 15 '10 at 14:07
I don't think the OP wants examples of normative statements. As I read it, the question is about conceptual errors regarding non-normative mathematical statements. – Qiaochu Yuan May 17 '10 at 6:19

A set is compact iff it is closed and bounded.

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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

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 $card \; D= card \; X$

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Multiplication of differential forms is inherently anti-commutative. Thus, if $x$ and $y$ are coordinates on a surface, then $dx \wedge dy$ makes sense but $(dx)^2+(dy)^2$ is either nonsense or, if it means anything, is $0$.

I'm not sure why I believed this, but I did for several years. I tried my best to avoid creating this impression in my students, but I think it still happened in some of them, simply because the curriculum spends a lot of time on integration and Stokes theorem and very little time on metrics, curvature, etc.

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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).

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$\mathbb{R}^2$ has a unique complex manifold structure; it's just $\mathbb{C}$ right?

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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 $4$.

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@Todd: No i was talking of high school students. – S.C. May 4 '11 at 4:08

If a matrix $A$ is self-adjoint/skew-self-adjoint with respect to a symmetric bilinear form, then it is diagonalizable.

True for matrices over $\mathbb{R}$, with respect to a positive definite inner product.

False over other fields. For example, over $\mathbb{C}$, $\left( \begin{smallmatrix} 1 & i \\ i & -1 \end{smallmatrix} \right)$ and $\left( \begin{smallmatrix} 0 & 1 & i \\ -1 & 0 & 0 \\ -i & 0 & 0 \end{smallmatrix} \right)$ are nilpotent, but self-adjoint and skew self-adjoint respectively with respect to the standard inner product.

False for other nondegenerate symmetric bilinear forms: $\left( \begin{smallmatrix} 1 & 1 \\ -1 & -1 \end{smallmatrix} \right)$ and $\left( \begin{smallmatrix} 0 & -1 & -1 \\ 1 & 0 & 0 \\ -1 & 0 & 0 \end{smallmatrix} \right)$ are nilpotent, but self-adjoint and skew self-adjoint respectively with respect to $\left( \begin{smallmatrix} 1 & 0 \\ 0 & -1 \end{smallmatrix} \right)$ and $\left( \begin{smallmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{smallmatrix} \right)$.

You can exponentiate the skew-self-adjoint matrices to get examples of matrices preserving a nondegenerate symmetric bilinear form, with Jordan blocks of the form $\left( \begin{smallmatrix} 1 & 1 \\ 0 & 1 \end{smallmatrix} \right)$.

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You seem to have a different definition of "the standard inner product on $\mathbb{C}^n$" than I do. I think that phrase normally refers to the familiar positive definite sesquilinear form, with respect to which self-adjoint matrices are indeed diagonalizable. – Mark Meckes Jan 28 '11 at 16:46
Of course it's not bilinear -- an "inner product" on a complex vector space is defined to be sesquilinear, not bilinear -- I've spent a lot of time trying to get my linear algebra students to remember that. The failure of such a form to generalize to other fields is indeed sad, but I think the richness of Hilbert space theory helps to make up for that disappointment. :) – Mark Meckes Jan 28 '11 at 21:24

If $a$ is a real zero of a cubic polynomial with rational coefficients then $a$ can be written as a combination of cube roots of rational numbers.

More generally if $a$ is a real zero of an irreducible polynomial with rational coefficients that is solvable by radicals then students expect the following:

1. Any expression inside a radical evaluates to a real number
2. Any sub-expression of the expression for $a$ evaluates to an algebraic number of order less than or equal to the order of $a$

Of course the problem is that from Cardan's solution to the cubic we can have negative rational numbers inside a square root. Let $c$ = $4*(-1 + \sqrt{-3})$.

$a$ = $\frac{\sqrt[3]{c}}{4} + \frac{1}{\sqrt[3]{c}}$

$f(x) = 4x^3 - 3x + \frac{1}{2}$.

So while $a$ is an algebraic number of degree three, it can not be written as combination of cube roots of rational numbers. Indeed, it is counter-intuitive that $\sqrt[3]{c}$ has degree 6 over the rational numbers yet we can use this number and simple arithmetic to produce an algebraic number of degree 3.

Also $a$ = $\sin(50^{\circ})$. For many values of $\theta$, $\sin \theta$ is a radical number. See also radical values for sine and cosine

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This has a name: en.wikipedia.org/wiki/Casus_irreducibilis – Qiaochu Yuan Apr 6 '11 at 21:21

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

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

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

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

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

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

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Although the original intent of this question seems to have long since evaporated, I can't help asking: is this really a "common false belief"? – Yemon Choi Feb 17 at 1:24

When $H$ and $K$ are subgroups of $G$ then $HK$ is subgroup of $G$...

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Hm, wonder how common that false belief actually is. It seems obviously implausible in the nonabelian case. – Todd Trimble Sep 6 at 15:39

False Belief: "The suspension spectrum map from spaces to (edit: symmetric) spectra preserves smash-products"

The facts that one denotes the smash product of spectra and the smash product of a space with a spectrum (levelwise) with the same $\wedge$ and tends to leave away the $\Sigma^\infty$ when one embeds a space into spectra are also not helpful in getting used to the harsh reality that the above is wrong.

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I don't see that this qualifies as a false belief. In order for the question of whether it is true or false to even be meaningful, you have to first commit yourself to one of the many different notions of spectrum, not to mention smash product of spectra. – Tom Goodwillie Oct 5 '10 at 0:35
True. I meant symmetric spectra with the smash product coming from their description as modules over the symmetric sequence of spheres. – Peter Arndt Oct 5 '10 at 10:52

Coordinates on a manifold do not have an immediate metric meaning. Until becoming familiar with differential geometry one tends to think they do. (Einstein wrote that he took seven years to free himself from this idea.)

For example, linear control theory is for the most part metric with variables in $R^n$. When moving away from linear control theory, variables are represented as coordinates on a manifold. Nevertheless, much of the literature tends to either abandon metric notions altogether, or to keep using an Euclidean metric though it is no longer very useful.

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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 $2$.

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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.

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## protected by François G. Dorais♦Oct 15 '13 at 2:34

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