Timeline for Examples of common false beliefs in mathematics
Current License: CC BY-SA 4.0
55 events
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| Sep 8 at 18:05 | comment | added | მამუკა ჯიბლაძე | Maybe then for more clarity it would be better to write "If $U$ is a Banach subspace of a Banach space $V$..." But even then, it is still not entirely apparent that all this happens in the category of Banach spaces: no categories are mentioned. | |
| Sep 8 at 11:04 | comment | added | Martin Brandenburg | @მამუკაჯიბლაძე I am not tacitly applying forgetful functors. When "direct summand" refers to Banach spaces, I refer to the notion of "direct sum" for Banach spaces - not for their underlying vector spaces. I do not agree with "any Banach space is a vector space". | |
| Mar 28 at 13:23 | comment | added | მამუკა ჯიბლაძე | You probably need to add something to the first bullet, no? Because any subspace of any vector space is a direct summand, and any Banach space is a vector space... | |
| Jan 27, 2022 at 8:11 | comment | added | François Brunault | Another false belief I once had is $\mathrm{dim}(R[x]) = \mathrm{dim}(R) + 1$ for every commutative ring (Krull dimension). This is ok if $R$ is noetherian. | |
| May 21, 2021 at 23:53 | comment | added | Martin Brandenburg | PS: I now deleted that example. We have $R[[x,y]]=R[[x]][[y]]$ as you explained; what is wrong is $R[[x,y]] = U(R[[x]])[[y]]$ when $U$ is the forgetful functor $\mathbf{TopRing} \to \mathbf{Ring}$. I made the (common) mistake to not notate a forgetful functor. | |
| May 21, 2021 at 23:48 | history | edited | Martin Brandenburg | CC BY-SA 4.0 |
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| Mar 21, 2021 at 14:34 | comment | added | Martin Brandenburg | @PeterScholze I have to admit that I don't remember the exact reason (the post is like 11 years ago), but I guess that you are right that $R$ is usually seen as discrete when forming $R[[x]]$. See also en.wikipedia.org/wiki/… for a discussion about this issue. Perhaps I should just delete that example. | |
| Mar 19, 2021 at 21:47 | comment | added | Peter Scholze | Let me also embarrass myself: Why are $R[[x,y]]$ and $R[[x]][[y]]$ different topological rings? To me they are both $\prod_{\mathbb N\times \mathbb N} R$. Or are you implicitly making $R[[x]]$ discrete when forming $(R[[x]])[[y]]$? | |
| Sep 16, 2020 at 8:28 | comment | added | wlad | @user541686 Take $f:\mathbb R_{\neq 0} \to \mathbb R, x \mapsto \begin{cases} 1, & x > 0 \\ 0,& x < 0 \end{cases}$ | |
| Apr 16, 2019 at 20:09 | comment | added | user137767 | @Mehrdad take a space consisting two points, and a function that 0 on one point and 1 on the other. | |
| Dec 28, 2017 at 9:10 | comment | added | user541686 | @MartinBrandenburg: I don't understand the locally-constant hint. Would you mind giving an actual counterexample? | |
| Dec 21, 2017 at 16:19 | comment | added | Somatic Custard | The fact that the inclusion $\mathbb{Z} \subset \mathbb{Q}$ does not preserve dimension can be expressed by saying that $\mathbb{Q}$ is zero-dimensional, but not hereditarily zero dimensional. These are studied in the book Zero-Dimensional Commutative Rings, edited by David Dobbs. | |
| Apr 12, 2017 at 11:16 | comment | added | Martin Brandenburg | @FawzyHegab: It depends on the choice of the model of set theory whether this is true or not. | |
| Mar 8, 2017 at 21:12 | comment | added | FNH | Could you give an example for two sets whose power sets are equipotent whereas they are not equipotent? | |
| Jan 14, 2017 at 3:19 | comment | added | Will Sawin | @user1 but there could be maximal ideals of many different heights! | |
| Feb 4, 2016 at 14:34 | comment | added | Mohan | @user1 I do not have a reference, but it was mentioned with reference here earlier by others (e. g. Graham Leuschke) too. The proof, while not trivial, can be worked out and I would be happy to post one somewhere (how?) if you so desire. | |
| Feb 4, 2016 at 8:18 | comment | added | user 1 | @mohan do you have a reference? isnt this contradiction with Martin's statement? | |
| Feb 4, 2016 at 8:03 | comment | added | user 1 | @Martin: the statement "The Krull dimension of a noetherian domain is finite." is my false belief today :) . doesn't this implied by KRULL'S PRINCIPAL IDEAL THEOREM? I mean if $R$ be noetherian ring, height of every maximal ideal is finite. and $\dim R$ is $sup$ of these heights. | |
| Oct 16, 2014 at 20:29 | comment | added | Michael | @MartinBrandenburg: Oh! I was thinking in terms of function rings of algebraic varieties. | |
| Oct 16, 2014 at 20:14 | comment | added | Martin Brandenburg | @Michael: $\mathbb{Z} \subseteq \mathbb{Q}$ | |
| Oct 16, 2014 at 18:28 | comment | added | Michael | Wait a sec, what would be a counterexample for "The Krull dimension of a subring is at most the Krull dimension of the ring"? | |
| Oct 16, 2014 at 15:14 | history | edited | Martin Brandenburg | CC BY-SA 3.0 |
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| Oct 16, 2014 at 13:29 | comment | added | Todd Trimble♦ | I think the "topologists assume" sentence in the last bullet is unfair; it implies topologists are making mistakes. Certainly competent topologists are not making such rookie mistakes, and are well aware of the standard counterexamples. | |
| Jan 2, 2014 at 16:31 | comment | added | Mohan | As a positive result, if $0\to A\to A\oplus B\to B\to 0$ is an exact sequence of finitely generated modules over a commutative Noetherian ring, then the exact sequence does split. | |
| Sep 18, 2013 at 19:05 | comment | added | Michael | Hey, I currently share a half of these allegedly false beliefs! | |
| Nov 26, 2012 at 11:10 | comment | added | Zhen Lin | @Harry: Very late addendum, but for an explanation of why sheafifying over large sites really is problematic even when you assume universes, see Waterhouse (1975) - Basically bounded functors and flat sheaves. The point is that the result of sheafification depends on the choice of universe when you use universes to construct them, i.e. it is no longer intrinsic. | |
| Apr 26, 2012 at 9:42 | comment | added | Neil Epstein | Amazingly enough, the splitting belief IS true if you add the innocuous-looking condition that $A$ and $B$ are finitely generated modules over a commutative Noetherian ring. (Theorem 1 from T. Miyata, Note on direct summands of modules, J. Math. Kyoto Univ. 7 (1967) 65-69) | |
| Jan 28, 2012 at 16:40 | comment | added | Todd Trimble♦ | Olivier, you might want to check out Easton's theorem in forcing: en.wikipedia.org/wiki/Easton%27s_theorem | |
| Jun 29, 2011 at 11:47 | comment | added | Ostap Chervak | Generalised continuum hypotesis implies this statement, while Martin Axiom + negation of continuum hypothesis provide a counterexample $P(\aleph_1)=P(\aleph_0)$ hence this misbelief is in fact independent of ZFC | |
| May 3, 2011 at 15:44 | comment | added | Olivier Bégassat | I would like to know more about $\mathcal{P}(X)$ equipotent to $\mathcal{P}(Y)$ not implying $X$ and $Y$ being equipotent. Is there no proof with the axiom of choice? It seems the gen. continuum hypothesis should imply it. Can you point me to some reference? | |
| Feb 24, 2011 at 17:17 | history | edited | Martin Brandenburg | CC BY-SA 2.5 |
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| Jul 3, 2010 at 9:04 | history | edited | Martin Brandenburg | CC BY-SA 2.5 |
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| Jun 10, 2010 at 10:44 | comment | added | kakaz | I saw doctorate thesis defence when one of reviewers, prominent one, claimed that: "If f is a smooth function with df=0 , then f is constant." is true, and then work has serious flaw. f in this work was topological invariant with df=0 but clearly there was different topological charges here, not only one ( and it was shown in the work).... | |
| May 19, 2010 at 8:39 | comment | added | Martin Brandenburg | $f$ is just locally constant ;-) | |
| May 19, 2010 at 8:07 | comment | added | Lennart Meier | I may be stupid, but what is a non-constant smooth function with df = 0 everywhere? | |
| May 16, 2010 at 17:41 | comment | added | Harry Gindi | @Martin: No, that doesn't really matter as long as we keep track of relative size differences. There's no pathology there. | |
| May 16, 2010 at 15:18 | comment | added | Martin Brandenburg | @Harry: No, it's a real problem because often you want to stay in the fixed universe; otherwise mathematics becomes pathological. For example, you could claim that every continuous functor has a left adjoint since the solution set is satisfied if we make the universe large enough for the solution set condition. But then we are not talking anymore about the same categories and functors between them! | |
| May 16, 2010 at 6:55 | comment | added | Harry Gindi | The point about presheaves and associated sheaves is one of those unimportant size issues that can be rectified by using universes and is a technical point that depends on a specific choice of set-theoretic formalism (For this reason, I suspect that Grothendieck ignores this issue in SGA4). I don't know if it really warrants inclusion on this list, since the rest of the list is so good. | |
| May 15, 2010 at 13:44 | comment | added | Steve D | @Regenbogen: Take the abelian group $\mathbb{Q}/\mathbb{Z}$. | |
| May 15, 2010 at 10:17 | comment | added | Regenbogen | @Reid Barton: Could you please provide a counterexample? | |
| May 11, 2010 at 2:12 | comment | added | Reid Barton | Your fifth example reminds me of an even more plausible false belief I once held: if $A \otimes A = 0$, then $A = 0$. | |
| May 10, 2010 at 23:38 | history | edited | Martin Brandenburg | CC BY-SA 2.5 |
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| May 9, 2010 at 20:41 | comment | added | Martin Brandenburg | The left side is quasi compact, but the right side only when $R_i = 0$ for almost all $i$. The difference can be made precise if $R_i$ are fields. Then $Spec(\prod_i R_i)$ is the Stone-Cech-compactification of the discrete space $\coprod_i Spec(R_i)$. | |
| May 9, 2010 at 19:09 | comment | added | babubba | Is the one on Spec false or true-but-not-because-of-the-obvious-thing? | |
| May 9, 2010 at 9:53 | history | edited | Martin Brandenburg | CC BY-SA 2.5 |
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| May 8, 2010 at 2:49 | comment | added | Peter LeFanu Lumsdaine | Ooh, very nice --- a classic "check your implicit assumptions" example. Good point! | |
| May 7, 2010 at 11:11 | comment | added | Olivier | I once made this very mistake, and it invalidates one of the main theorems of a published article I once quoted. A good reason in my opinion to specify what are the arrows when writing a sequence or a diagram: they are usually what you think they are, but hey, let's check. | |
| May 6, 2010 at 16:12 | comment | added | Martin Brandenburg | $A \to A \oplus B$ does not have to be the inclusion; likewise $A \oplus B$ does not have to be the projection. Thus the error here is: Two chain complexes, which are isomorphic "pointwise", don't have to be isomorphic. This occurs sometimes. | |
| May 6, 2010 at 15:48 | comment | added | Peter LeFanu Lumsdaine | I'm sure you'll have me kicking myself in a moment... but how does a short exact sequence of the form 0 --> A --> A + B --> B --> 0 fail to split? In any Abelian (indeed, additive is enough) category, since A + B is a biproduct, there's a paired map (0,1_B): B --> A + B, and a copaired map [1_A,0]: A + B --> A, which split each half of the sequence... don't they? Or were you thinking of a context for this example that's wider than Abelian categories? | |
| May 6, 2010 at 0:21 | history | edited | Martin Brandenburg | CC BY-SA 2.5 |
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| May 5, 2010 at 23:47 | history | edited | Martin Brandenburg | CC BY-SA 2.5 |
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| May 5, 2010 at 23:41 | comment | added | Pete L. Clark | +1: you had me at "Here's my list of false beliefs". | |
| May 5, 2010 at 23:40 | history | edited | Martin Brandenburg | CC BY-SA 2.5 |
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| May 5, 2010 at 23:33 | history | edited | Martin Brandenburg | CC BY-SA 2.5 |
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| May 5, 2010 at 23:21 | history | answered | Martin Brandenburg | CC BY-SA 2.5 |