Re math.stackexchange.com/a/93406/13425, If you are going to hide everything with a big-O, why bother with such methods? Why not bound each term by $\log k$ and be happy? (Just like the second answer does...)
@Srivatsan the end result hides the method, but bounding each term by the biggest, might overestimate the sum. Of course, once you know that the term is $O(k\log(k))$ the other proof looks sufficient.
@robjohn I'm not saying the method is without value. But if you're using a stronger method and that would yield a stronger bound, it makes sense to note that in the answer, no?
@Srivatsan I guess using the Riemann sums is a natural reflex. You might not even think of the other trick if you're not doing these things on a regular basis.
@RajeshD In the stereotypical image, Russians are like Texans in that they like to exaggerate about the importance or size of things from their homeland.
@ZhenLin As mentioned in that paper, the cubic has to have all its roots real (casus irreducibilis) for trisection to apply. You can't duplicate the cube with an angle trisector, for instance.
@Matt: your answer was good. Bill D seems to like to downplay others' answers and push his own as more general or more understandable. It happened to me a lot on sci.math.
@robjohn Thanks. I was so unsure about it that I deleted it right after hitting the post button. Then I thought, never mind: if it's wrong, I'll learn something and I can still delete it : )
Were the results of Nash embedding theorems been conjectured by anyone before he came up with ? @all sorry for reposting... i thought you have missed it
@ZhenLin oh, you mean that for $G$ being a group we can consider a category $\mathscr G$ such that $\mathrm{Ob}(\mathscr G) = \{G\}$ and $f\in\mathrm{Hom}_\mathscr G(G,G)$ if and only if $f:G\to G$ is an isomorphism, don't you?
@robjohn Nice, thanks. I think I don't understand why I need the Lebesgue number lemma. If I have a finite subcover consisting of balls of radii $\delta_1, \dots , \delta_n$ around points $x_1, \dots , x_n$ and I then choose the smallest of these $\delta_i$ and call it $\delta$ then surely $B(x_i, \delta) \subset B(x_i, \delta_i)$.
@ZhenLin didn't get it, sorry. I've just started to read about categories and need some time to digest new info. E.g. I know that $*$ is a terminal object and one example can be a singleton in $\mathrm{Set}$ but that's it
@Zhen what is the motivation (or the sense) of considering opposite categories? Maybe I should tell you that I'm reading PhD thesis of one student where he uses category theory a lot - and he gives quite comprehensive but very brief description of CT
@Ilya: Some interesting concrete categories turn out to be opposites, but otherwise it's an abstraction to let us treat covariant and contravariant functors on the same footing.
I don't think I'll have time at the current moment to read about this in a book (where would be more examples and more motivation I suppose) so I'm trying to break through the material he gave.
@ZhenLin I guess, this is one of the sentences non-mathematicians associate to mathematicians: even more basic then [complicated maths] is [very complicated math]
@Zhen: unfortunately, I don't know what is contravariant representable functor, so I'm about to give up and borrow the book. CT seems to be something to make things more comfortable and it does not work for me now (maybe I have a lack of particular examples). As a main reference this student advises Mac Lane's book 'Categories for working mathematicians'. Have you heard about it?
@robjohn I thought that if the distance between two points was less than $\delta$ then the function values at those points were less than $\varepsilon$.
@ZhenLin ok, 300p is not so much. I know, it's weird to count the pages in the book - but I have some time restrictions. Language seems to be very good, so thanks a lot for the reference, Zhen
@Srivatsan It just seems verrry hopeful that the thing would be a cube root of some power of itself. That doesn't seem to be implied by the cyclicity or anything.
@DylanMoreland Let's find out. The only thing we know is $a^{p} = a$ or $a^{p+1} = a^2$. So $a^{(p+1)/3} = a^{2/3} = a / a^{1/3}$, so you can write $a^{1/3}$ as $a^{(2-p)/3}$, right?
@DylanMoreland Ok, let's go fraction-free. $a^{2-p} = a$, so $a^{(2-p)/3}$ is a cube-root of $a$.
That's pretty much the content of my earlier comment. The only thing is $a^{2-p}$ might seem a bit magical, although it's not if you think about it a bit...
I typically pride myself on making sure that I've nailed something down before asserting it here, and twice this week I've had to be saved from myself.
If $a,b,c,d$ are in $R=\mathbb{C}[t]$ and $ad-bc \ne 0$, $L= R(a,b)+R(c,d)$ in $R^{2}$. I want to show that $\dim_{\mathbb{C}}R^{2}/L = \deg(ad-bc)$.
In a previous theorem it was shown that : $\dim_{\mathbb{C}}R /tR = \deg(t)$. So I think of : $\dim_{\mathbb{C}}R^{2}/tR$ and I believe this ...
@Matt Some users wrote a browser extension to add some stuff to chat. That thing is currently broken, it makes broken and unwanted requests to the server on behalf of the user who has installed it.
@Zhen: Awodey considers the category of finite sets with arrows being rectangular matrices [quote starts] $F = (n_{ij})_{i,j\in\mathbb N}$ where $i=|A|$ and $j = |B|$ [quote ends]
@Srivatsan For any two points x,y with distance less than delta you need to argue that they lie inside the same delta_i ball of the finite subcover. For that you can either use the Lebegue number lemma or then take half of the delta then they also lie inside the same ball.
If $a,b,c,d$ are in $R=\mathbb{C}[t]$ and $ad-bc \ne 0$, $L= R(a,b)+R(c,d)$ in $R^{2}$. I want to show that $\dim_{\mathbb{C}}R^{2}/L = \deg(ad-bc)$.
In a previous theorem it was shown that : $\dim_{\mathbb{C}}R /tR = \deg(t)$. So I think of : $\dim_{\mathbb{C}}R^{2}/tR$ and I believe this ...
