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(i) a bounded entire function is constant;
(ii) sin(z) is a bounded function;




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186 Answers 186


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"A 'random' number field has large class number"




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A common belief of students in real analysis is that if $$ \lim_{x\to x_0}f(x,y_0),\qquad\lim_{y\to y_0}f(x_0,y) $$ exist and are both equal to $l$, then the function has limit $l$ in $(x_0,y_0)$. It is easly to show counter-examples. More difficult is to show that also the belief $$ \lim_{t\to 0}f(x_0+ht,y_0+kt)=l,\quad\forall\;(h,k)\neq(0,0)\quad\Rightarrow\quad\lim_{(x,y)\to(x_0,y_0)}f(x,y)=l $$ is false. For completeness's sake (presumably anybody who ever taught calculus has seen it, but it's easily forgotten) the standard counterexample is $$ f(x,y)=\frac{xy^2}{x^2+y^4} $$ at $(0,0$).

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That counterexample has the advantage of being well-behaved away from $(0,0)$, but the (related) disadvantages of being easily forgotten and requiring a bit of thought to come up with. This can make things look trickier than they are. For this reason, I prefer brain-dead counterexamples like $f(x,y)=1$ if $y=x^2 \neq 0$, $f(x,y)=0$ otherwise. –  Chris Eagle Jan 12 '11 at 17:11
@Chris As you know, this is not a "real function" to the minds of calculus students. –  Ryan Reich Jan 2 at 3:04

Piggybacking on one of Pierre's answers, I once had to teach beginning linear algebra from a textbook wherein the authors at one point stated words to the effect that the the trivial vector space {0} has no basis, or that the notion of basis for the trivial vector space makes no sense. It is bad enough as a student to generate one's own false beliefs without having textbooks presenting falsehoods as facts.

My personal belief is that the authors of this text actually know better, but they don't believe that their students can handle the truth, or perhaps that it is too much work or too time-consuming on the part of the instructor to explain such points. Whatever their motivation was, I cannot countenance such rationalizations. I told the students that the textbook was just plain wrong.

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Bjorn Poonen once gave a lecture at MIT about the empty set; it really opened my eyes. If someone wrote a textbook or something on the matter I think everyone would be a lot less confused. –  Qiaochu Yuan Jul 7 '10 at 23:56
For most of the history of civilization, zero was very controversial... –  Victor Protsak Jul 9 '10 at 4:12
I can combine Qiaochu's and Victor's remarks in this memory I have of a coffee break conversation between two colleagues, who were arguing on whether it made sense to say that the 1-element group acts on the empty set. I wisely decided to stay out of the controversy... –  Thierry Zell Aug 31 '10 at 2:24
Thierry: of course it makes sense. But the action is not transitive. –  ACL Dec 1 '10 at 22:53
@kow: I disagree. That is the "wrong" definition of transitivity for empty G-sets. See the discussion at qchu.wordpress.com/2010/12/03/empty-sets . –  Qiaochu Yuan Dec 16 '10 at 23:08

The distinction between convergence and uniform convergence. It even got Cauchy in its time.

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A second countable, Hausdorff, Banach manifold is paracompact.

Regular is necessary, otherwise there are counterexamples!

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(I believe this example is due to Lawrence Zalcman.)

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-It is metrizable if $X$ is separable.

-It is locally compact by Banach-Alaoglu.




Again when $X$ is infinite dimensional this always fails.

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Draw the graph of a continuous function $f$ (from $\mathbb{R}$ to $\mathbb{R}$). Now draw two dashed curves: one which everywhere a distance $\epsilon$ above the graph of $f$ and one which is everywhere a distance $\epsilon$ below the graph of $f$. Then the open $\epsilon$-ball around $f$ (with respect to the uniform norm) is all functions which fit strictly between the two dashed curves.

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Surely this is true if you are talking about the closed ball, and only just barely false for the open ball (and if we were talking about functions from $[a,b]$ to $\mathbb{R}$ it would be true)? Or else I am one of those with the false belief... –  Nate Eldredge Oct 10 '10 at 18:26

Inversion is an automorphism of a group. ('Cause it, like, preserves the conjugacy classes and all that...)

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A better, but still false, version:





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I don't know how common this is, but I've noticed it half an hour ago in some notes I had written: If $J$ is a finitely generated right ideal of a not necessarily commutative ring $R$, and $n$ is natural, then $J^n$ is finitely generated, isn't it?

No, it isn't. For an example, try $R=\mathbb Z\left\langle X_1,X_2,X_3,...\right\rangle $ (ring of noncommutative polynomials) and $J=X_1R$.

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This is more of a false philosophy than a clear mistake, but nevertheless it is very common:

A compact topological space must be "small" in some sense: it should be second countable or separable or have cardinality $ \le 2^{\aleph_0}$, etc.

This is all true for compact metric spaces, but in the general case, Tychonoff's theorem gives plenty of examples of compact spaces which are "huge" in the above sense.

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False statement: If $A$ and $B$ are subsets of $\mathbb{R}^d$, then their Hausdorff dimension $\dim_H$ satisfies

$$\dim_H(A \times B) = \dim_H(A) + \dim_H(B). $$

EDIT: To answer Benoit's question, I do not know about a simple counterexample for $d = 1$, but here is the usual one (taken from Falconer's "The Geometry of Fractal Sets"):

Let $(m_i)$ be a sequence of rapidly increasing integers (say $m_{i+1} > m_i^i$). Let $A \subset [0,1]$ denote the numbers with a zero in the $r^{th}$ decimal place if $m_j + 1 \leq r \leq m_{j+1}$ and $j$ is odd. Let $B \subset [0,1]$ denote the numbers with a zero in the $r^{th}$ decimal place if $m_{j} + 1 \leq r \leq m_{j+1}$ and $j$ is even. Then $\dim_H(A) = \dim_B(A) = 0$. To see this, you can cover $A$, for example, by $10^k$ covers of length $10^{- m_{2j}}$, where $k = (m_1 - m_0) + (m_3 - m_2) + \dots + (m_{2j - 1} - m_{2j - 2})$.

Furthermore, if $\mathcal{H}^1$ denotes the Hausdorff $1$-dimensional (metric) outer measure of $E$, then the result follows by showing $\mathcal{H}^1(A \times B) > 0$. This is accomplished by considering $u \in [0,1]$ and writing $u = x + y$, where $x \in A$ and $y \in B$. Let $proj$ denote orthogonal projection from the plane to $L$, the line $y = x$. Then $proj(x,y)$ is the point of $L$ with distance $2^{-1/2}(x+y)$ from the origin. Thus, $proj( A \times B)$ is a subinterval of $L$ of length $2^{-1/2}$. Finally, it follows:

$$ \mathcal{H}^1(A \times B) \geq \mathcal{H}^1(proj(A \times B)) = 2^{-1/2} > 0. $$

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Actually, the situation is worse than I say: there exist sets $A, B \subset \mathbb{R}$ with $dim_H(A \times B )= 1$, and yet $\dim_h(A) = \dim_H(B) = 0$. –  JavaMan Apr 5 '11 at 6:22


$G_1 / H_1 \cong G_2 / H_2$.


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An incredibly common false belief is:


A right statement of the theorem of the cone is





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l |--> k+l


l |--> l * k

stabilizes to the identity after k.




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