bio | website | mathnet.ru/php/… |
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location | ||
age | ||
visits | member for | 3 years, 9 months |
seen | 7 hours ago | |
stats | profile views | 1,643 |
Aug 5 |
comment |
A fixed point problem
Oh, I missed this. |
Aug 5 |
comment |
A fixed point problem
@dimo Something is wrong, we do not understand each other. You consider a map from the closed interval $[0,1]$ into the plane $\mathbb R^2$ acting by formula $t\mapsto (tr,1-t)$ -- is this correct? And $A$ is the image of this map. If this is correct (and $[0,1]$ and $\mathbb R^2$ have usual topology), then this map is a homeomorphism between $[0,1]$ and $A$. (By the way, in this case it doesn't matter whether $r$ belongs to $\mathbb Q$ or to $\mathbb R$.) |
Aug 5 |
comment |
A fixed point problem
Of course, this works only if I understand you correctly: the topology on $\mathbb R^2$ must be the usual topology of direct product of two $\mathbb R$, and each of those $\mathbb R$ is endowed with the usual topology of $\mathbb R$ (i.e. intervals $(a,b)$ form a base of open sets). |
Aug 5 |
comment |
A fixed point problem
This is a general theorem: every bijective and continuous map from a compact space to a Hausdorff space is a homeomorphism, see R.Engelking, General topology, 3.1.13, see also Wikipedia en.wikipedia.org/wiki/Compact_space#cite_note-10 (the properties of Hausdorff spaces). |
Aug 3 |
comment |
What are the qualities of a good (math) teacher?
I saw math teachers who are unable to build a logical chain carefully, without gaps and/or mistakes. I think, Jose had in mind that a teacher must have enough qualification for doing this in case when he sees that his pupils don't understand the logic of the exposition. Not that every time he must explain each millimeter of the proof. Am I right, Jose? :) |
Jul 31 |
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A single paper everyone should read?
I wonder if there is a progress since that time in the directions that Dyson mentioned as what should be done. |
Jul 29 |
comment |
A fixed point problem
I don't understand the problem. $A$ is homeomorphic to the interval $[0,1]$. So everything follows from the Brouwer fixed-point theorem (it's variant for the dimension 1). The statement of the problem looks strange. |
Jul 21 |
awarded | Revival |
Jul 21 |
answered | Hahn-Banach restricted to a pre-dual |
Jul 21 |
accepted | “Uniformly (co)well-powered” categories? |
Jul 21 |
comment |
“Uniformly (co)well-powered” categories?
Andreas, which reading do you recommend? |
Jul 20 |
revised |
“Uniformly (co)well-powered” categories?
deleted 466 characters in body |
Jul 20 |
comment |
“Uniformly (co)well-powered” categories?
Yes, I found this result about well-ordering of the class of all sets in A.Levy's book. Andreas, I think you should put what you wrote in form of answer, and I will close this question. I'll make corrections in the question so that your answer will be natural. Or I do not know what to do with this. Eric, I still do not understand what you wrote. That is interesting, please, give me a reference. |
Jul 20 |
comment |
“Uniformly (co)well-powered” categories?
@Eric and Andreas, I need a reference. This is the first time I hear this. The class of all sets can be well-ordered? And this trick with sending X to the set of all skeleta... Is there a text to look how people do this? |
Jul 20 |
comment |
“Uniformly (co)well-powered” categories?
To apply the axiom of choice (i.e. to choose this map $X\mapsto S_X$) we must already have a map which to every $X$ assigns a set which contains some $S$ as element. But we do not have such a map. Or you mean something else? |
Jul 20 |
comment |
“Uniformly (co)well-powered” categories?
@Eric Wolfsey: this must be a misunderstanding... I don't see how this follows from global choice. |
Jul 20 |
asked | “Uniformly (co)well-powered” categories? |
Jul 19 |
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Tangent vectors on the algebra of trigonometric polynomials
@Qiaochu Yuan: Yes, I do not see the difference. To moderators: I flagged the previous comment by chance, excuse me, that was not intentionally! |
Jul 19 |
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Tangent vectors on the algebra of trigonometric polynomials
For me it is not obvious that ${\sf Trig}(G)$ coincides with the algebra of pokynomials on the complexification of $G$. Why is this so? |
Jul 18 |
comment |
Tangent vectors on the algebra of trigonometric polynomials
Yes, this is simple... However you should have some experience in algebraic groups to feel free in this field. Can you recommend a textbook for reading and references? |