MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
79  
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
20  
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
18  
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
18  
It's a thought -- I might consider it. – gowers Oct 4 '10 at 20:13
22  
Meta created tea.mathoverflow.net/discussion/1165/… – quid Oct 8 '11 at 14:27

209 Answers 209

Let $R$ be a ring with identity $e$, $A, B\in R$, $A\neq 0$, $B$ is invertible element. If $A\cdot B = A$ then $B = e$.

share|cite|improve this answer

Another common mistake. If $W = _P(e_1,\ldots, e_{n})$ is a vector space and $V$ is a subspace of $W$ of dimension $k$, then $V = _P(e_{i_1},\ldots, e_{i_k})$.

share|cite|improve this answer
1  
$W$ is a vector space over field $P$, $(e_1,\ldots, e_n)$ is a basis of $W$. $V$ is a subspace of $W$. – Mikhail Goltvanitsa 5 hours ago

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 $n \le 4$, it is true that $\alpha \ge 0$ (see this proof), but it's false for $n>5$ (see this comment). For the case $n=5$, I don't know.

share|cite|improve this answer

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).

share|cite|improve this answer

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

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

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

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

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

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

share|cite|improve this answer
3  
Correcting that misunderstanding is crucial to prove that an accumulation point of a net is always the limit of some subnet, which does not hold for sequences. – Alfonso Gracia-Saz Oct 19 '10 at 6:26

Most people that study Riemannian geometry for their first time make the following assumption at some point: "Let $(e_1,\dots,e_n)$ be a local orthonormal frame of $TM$ such that all Lie brackets $[e_i,e_j]$ vanish..."

This one is not so common (maybe special to me), but here we go: "$\mathbb{RP}^\infty$ and $\mathbb{CP}^\infty$ are Eilenberg-Mac Lane spaces, so $\mathbb{HP}^\infty$ is one, too."

share|cite|improve this answer

This one has bit me and some very good mathematicians I know.

Let $X,Y$ be Banach spaces, and let $E \subset X$ be a dense subspace. Suppose $T : E \to Y$ is a bounded linear operator. Then $T$ has a unique bounded extension $\tilde{T} : X \to Y$. (True, this is the well-known and elementary "BLT theorem".)

If $T$ is injective then so is $\tilde{T}$. (False! See this answer for a counterexample.)

share|cite|improve this answer

Some false beliefs in linear algebra:

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

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

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

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

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

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

share|cite|improve this answer
23  
Wow. I believed that second one until now. Which is ridiculous, of course, since the operator norm of a nilpotent matrix can't be zero or else it wouldn't be a norm! – Qiaochu Yuan Jun 6 '10 at 23:45
1  
The parethentical comment in 2nd bulleted point is worded as if, $\textit{in general},$ the operator norm were equal both to the spectral radius and the largest singular value (or, perhaps, that $\|A\|=\rho(A)$ and $\lambda_1(A)=s_1(A).$) But for a nilpotent matrix the spectral radius is 0, whereas the operator norm and the largest singular values aren't. – Victor Protsak Jun 10 '10 at 7:59
2  
Fair enough; I've reworded the parenthetical. – Terry Tao Jun 10 '10 at 16:35
3  
Yes, I meant right inverse, thanks. Getting a continuous right-inverse is actually a subtle question - the OMT only gets boundedness, which is not equivalent to continuity when one is not linear. I believe that the existence of a continuous right inverse may follow from a classical theorem of Bartle and Graves, but this is nontrivial. – Terry Tao Jun 10 '10 at 19:06

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.

share|cite|improve this answer

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

share|cite|improve this answer
3  
Hm, wonder how common that false belief actually is. It seems obviously implausible in the nonabelian case. – Todd Trimble Sep 6 '15 at 15:39

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

share|cite|improve this answer
11  
"If you want to prove that a certain infinite structure exists it is enough to show that there is no upper bound to the sizes of such structures." This is not necessarily false. In some important cases this does work! One uses the compactness theorem for such proofs. – Johannes Hahn May 7 '10 at 12:46
10  
Probability: There are two opposite errors. Both are common. Say we are flipping a fair coin repeatedly. (1) if there have been more heads than tails, then tails is "overdue" and thus more likely on the next flip. (2) if there have been more heads than tails, then heads is "hot" and thus more likely on the next flip. – Gerald Edgar May 7 '10 at 15:15
14  
I don't know anything about polytopes, but I'm having a hard time disbelieving this false result. Are we talking about finite polytopes here? – Tom Ellis May 9 '10 at 19:26
25  
Entirely finite, Tom. There are 4-dimensional polytopes with 33 vertices that cannot be presented with rational coordinates. Here is a reference arxiv.org/PS_cache/arxiv/pdf/0710/0710.4453v2.pdf – Gil Kalai May 9 '10 at 20:46
25  
You list the statement "Quantum computers can solve NP complete problems in polynomial time" as a false belief, but I don't believe you actually know this belief to be false. For example, the assertion that this belief is false implies $P\neq NP$. Perhaps the false belief that you intend to mention is: "It has been proved that Quantum computers can solve NP complete problems in polynomial time." – Joel David Hamkins May 17 '10 at 12:34

$\pi$ is equal to 22/7.

This was touched upon in the comments to a totally unrelated answer but I think this false belief is important enough to warrant its own answer (and as far as I could tell it does not have one yet, my apologies if I overlooked one.)

Of course, it's unlikely anyone on this site believes this, or ever believed it, which is why I think it's important to insist on this: it does not really resonate with us, we are unlikely to warn students against it, yet we probably see in front of us many students who have that false belief and then will move on to spread it around.

A Piece of Evidence

Let me offer as evidence this gem taken off the comments section of an unrelated (but quite thought-provoking) article on Psychology Today, of all places! When Less is More: The Case for Teaching Less Math in Schools (The title is a misnomer, it's a case for starting math later, but I think that with such a scheme you should be able to teach more math overall; anyway, read it for yourselves.)

Some years ago, my (now ex-) wife was involved in a "trivia night" fundraiser at her elementary school, and they wanted me on their "teacher team" to round out their knowledge. They had almost everything covered except some technology-related topics and I was an IT guy. In round four, my moment to shine arrived, as the category was "Math & Science" and one of the questions was, "give the first five digits of pi." I quickly said, "3.1415." The 9 teachers at the table ignored me and wrote down "22/7" on scrap paper and began to divide it out. I observed this quietly at first, assuming that 22/7ths gave the right answer for the first 5 digits, but it doesn't. It gives something like 3.1427. I said, "Whoops, that won't work." They ignored me and consulted among themselves, concluding that they had all done the division properly on 22/7ths out to five digits. I said, "That's not right, it's 3.1415." [...]

I'm cutting it off here because it's a long story: hilarity ensues when the non-teacher at the table stands up for the truth (when he finds out that the decimals of 22/7 were the expected answer!) The final decision of the judges:

"We've got a correction on the 'pi' question, apparently there's been confusion, but we will now be accepting 3.1415 as a correct answer as well" [as 3.1427].

The Moral of the Story

I used to dismiss out of hand this kind of confusion: who could be dumb enough to believe that $\pi$ is 22/7? (Many people apparently: in the portion of the story I cut was another gem - "I'm sorry, but I'm a civil engineer, and math is my job. Pi is 22/7ths.")

Now, I treat this very seriously, and depending on where you live, you should too. Damage wrought during the influential early years is very hard to undo, so that the contradictory facts "$\pi$ is irrational" and "$\pi$=22/7" can coexist in an undergraduate's mind. And when that person leaves school, guess which of the two beliefs will get discarded: the one implanted since childhood, or the one involving a notion (rational numbers) which is already getting fuzzy in the person's brain? I'm afraid it's no contest there, unless this confusion has been specifically addressed.

So if you have any future teachers in your classes (and even if you don't, cf. the civil engineer above), consider addressing this false belief at some point.

share|cite|improve this answer
6  
You know what the most funny part is? They can't divide, because actually 22/7 is between 3.1428 and 3.1429, but this way they got something closer to the real answer. – Zsbán Ambrus Nov 27 '10 at 19:36
6  
Well, doesn't the Bible say that π=3 ? – ACL Dec 1 '10 at 22:52
2  
ACL: I don't have the passage handy now, but as I recall, it does not outright say perimeter over diameter equals three; you can make the case that it's a matter of overenthusiastic rounding down of the value of the perimeter. It is a bit sloppy though, even for that time period... – Thierry Zell Dec 2 '10 at 2:00
3  
This one is depressing... – Andrés Caicedo Dec 7 '10 at 22:26
3  
This is from the AFP report on the recent computation of the first $5\times 10^{12}$ (decimal) digits of $\pi$: "Pi, the ratio of a circle's circumference to its diameter, starts with 3.14159 in a string whose digits are believed to never repeat or end." google.com/hostednews/afp/article/… – Andrés Caicedo Dec 7 '10 at 22:29

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

share|cite|improve this answer
1  
Some evidence. – Michael Albanese Dec 24 '15 at 10:21

A common false assumption is that that two non-orthogonal pure states of a quantum mechanical system may never be unambiguously distinguished by a measurement. (See http://arxiv.org/pdf/quant-ph/9807022.pdf)

Another false belief is that a quantum computer is similar to an analogue computer, in that large computations will necessarily fail because of accumulated error. (See, for example, http://arxiv.org/abs/quant-ph/9712048)

For that matter, another common false believe is that Bell Inequalities aren't violated, although it is mostly held by people who have never heard of Bell Inequalities.

share|cite|improve this answer

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.

share|cite|improve this answer

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 continous 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 disapointed 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?

share|cite|improve this answer
  • The category of commutative C*-algebras is equivalent to the opposite category of locally compact Hausdorff spaces.

It's actually not quite that simple! There is some discussion on math.SE.

share|cite|improve this answer
3  
Whoops! Thanks for posting this and particularly the link, which shows that the nLab got this one wrong as well. – Todd Trimble Nov 12 '14 at 13:03

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.

share|cite|improve this answer

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.
share|cite|improve this answer
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

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).

share|cite|improve this answer

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.

share|cite|improve this answer

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 $I$.

share|cite|improve this answer
1  
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 '15 at 6:07
1  
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 '15 at 8:05
1  
@SébastienPalcoux Yes, I think if $f'$ is continuous in $x_0$ then the statement is true. – Shamisen Aug 10 '15 at 15:16

"A real symmetric matrix is positive-definite iff all the leading principal minors are positive, and positive-semidefinite iff all the leading principal minors are nonnegative."

This paper collects some evidence that this belief is "common", and presents a counterexample (of size $3\times 3$. Exercise: find an example of size $2\times 2$).

(Related to, but not the same as this answer.)

share|cite|improve this answer
3  
$$\pmatrix{0&0\cr0&-1\cr}$$ – Gerry Myerson Jul 29 '15 at 3:22

I once very briefly thought that:

Given a vector space $V$ and a sub-space $U \subset V$ that $V-U$ is also a subspace.

I've heard this several times as a TA also.

share|cite|improve this answer
2  
I think this falls under $(x+y)^2=x^2+y^2$, – Thomas Rot Aug 10 '15 at 12:48
3  
It always fails... But I don't think this is a common held belief. – Thomas Rot Aug 10 '15 at 21:40

"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)

share|cite|improve this answer

Here's one from basic set theory. Let k be a cardinal and consider the operation "adding k", meaning

$l \mapsto k+l$

on cardinals. We know that this operation "stabilizes" to the identity after $k$, that is, for any $l>k$, we have $l+k = l$. Similarly, the "multiplying by $k$" operation,

$l \mapsto l * k$

stabilizes to the identity after $k$.

Everyone also knows that if $l$ is an infinite cardinal then $l^2$ is equipotent to $l$, and more generally $l^n$ is equipotent to $l$ for every natural number $n$. I.e. all the finite power functions stabilize to the identity at $\omega$.

Well, obviously "exponentiation by $\omega$" also stabilizes at some point, right? Like, $l^\omega$ is equal to $l$ for sufficiently large $l$? Look, we probably already have the stabilization point at $2^\omega$.

Right?

share|cite|improve this answer
2  
Why not? As an algebraist, my reaction already after "addition of k stabilizes" would be "if THAT holds, than WHATEVER". – Victor Protsak Jun 10 '10 at 6:45
2  
Victor, I held this belief for a good while when first learning set theory. I tried proving it a couple of times and failed, but I was in that stage just after I'd gotten the hang of basic cardinality arguments and they all seemed simple, so I figured it was just a matter of small details. – Pietro KC Jun 10 '10 at 9:01
6  
But it turns out that k^l is intimately linked with the cofinality of k, which is the length of the shortest unbounded sequence in k. For example, cof(omega) = omega, since sequences of length less than omega are finite, and thus bounded in omega. Similarly, cof(aleph_1) is aleph_1, since any countable sequence in aleph_1 is bounded. It's not immediately obvious that some cardinal k has cof(k) < k, but aleph_omega does! Anyway, the relevant theorem is that k^cof(k) > k, so there are arbitrarily large k s.t. k^omega > k. – Pietro KC Jun 10 '10 at 9:06

A Banach space $X$ is reflexive if it is isomorphic to its double dual ${X^*}^*$.

(Couldn't find this is the list…)

share|cite|improve this answer
3  
Even isometric fails. (Lindenstrauss & Tzafriri, in the '60s I believe.) – Hachino May 12 '15 at 8:19

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.

share|cite|improve this answer

"Some real numbers are not definable, by Cantor's diagonal argument."

There are subtleties involved in formalizing the statment "some real numbers are not definable", as explained in Joel's answer to this question. The statement can be seen to hold in some models and fail in other models. However, the claim that the statement follows from Cantor's diagonal argument is clearly false, yet seems to be fairly common.

The false reasoning typically proceeds in three steps:

  1. There are only countably many definitions of real numbers: $\varphi_0(x),\varphi_1(x),\ldots$ (this part is ok.)

  2. Consider the countably many real numbers so defined: $x_0,x_1,\ldots$ (this part is problematic for subtle reasons.)

  3. Use Cantor's diagonal argument to obtain a real number $y$ that is not in the sequence from step 2, and is therefore not definable.

For the moment, let us assume that step 2 succeeds in the way that one might naively think it would. Then we have defined a sequence $x_0,x_1,\ldots$ containing all definable real numbers. Therefore Cantor's diagonal argument in step 3 defines, from this sequence, a real number $y$ that is not in the sequence. So $y$ is both definable and not definable, and we obtain a contradiction outright! Clearly, something is wrong (and it turns out to be in step 2.)

share|cite|improve this answer

For $p$ prime and the chain of embeddings $\mathbb{Z}/p\mathbb{Z} \hookrightarrow \mathbb{Z}/p^2\mathbb{Z} \hookrightarrow \cdots$ given by multiplication by $p$, then $\bigcup_n \mathbb{Z}/p^n\mathbb{Z}$ is not the group of $p$-adic integers $\mathbb{Z}_p$, but its Pontryagin dual, the Prüfer $p$-group $\mathbb{Z}(p^{\infty})$.

share|cite|improve this answer
2  
Is that actually a common false belief? After all, $\mathbb{Z}_p$ is uncountable, as everyone realizes! – Todd Trimble Mar 5 '15 at 14:25
4  
It's hard for me to believe it's at all common. I wasn't the downvoter, but I think it would be better if answers were rooted either in instances that can be found in the literature, or widely encountered in one's experience as an instructor. – Todd Trimble Mar 5 '15 at 14:52

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

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site.

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.