Ah, it works! You more than deserve my encouragements. People use their real names because there is practically no risk of seeing malicious people taking advantage of this frank display. You see, captchas can be faked, but very few people in the world (and probably no criminal) can explain as well as you why anon-degenerate curve of degree d can be embedded only in projective space of dimension at most d :-)
Also, since no money is involved on this site, there is no incentive for criminals to try and interfere. The positive side is that real names create a friendly atmosphere: I prefer to be addressed as Georges rather than, say, Donaldduckfan...
Oh, fantastic: thanks for the change. What part (if any) of your name corresponds to the first name in English, like John or Peter? In other words, what is the friendly address I can use when communicating with you?
Things are happening fast here! Rereading the thread I realize that my posts look a bit surrealistic because there were many interventions by other users between mine. And I see it is happening right now! Anyway, thanks for your kind words.
He claimed: for any finitely generated graded ring $S$ which is also generated by $S_1$, then $X:=\operatorname{Proj} S\cong \operatorname{Proj}(\oplus_{n\geq 0}\Gamma(X,\mathcal{O}_X(n)))$.
I failed to prove it. There is a natural graded ring morphism from $S$ to $\oplus_{n\geq 0}\Gamma(X,\mathcal{O}_X(n))$, but I cannot prove that this morphism induces a regular map (i.e. morphism) of their Proj.
Dear Yuchen, thanks for your explanation of Chinese etiquette: I think i prefer Yuchen to jerry. I'd rather not discuss mathematics on chat, since the pressure of answering immediately would be too stressful. Also I would prefer to have the option of not answering if, as is very usual, I have no idea for solving the problem! So, may I suggest that you go on asking your future questions on the main site?
OK, I will consider to post it on the main site. I am a little hesitate to ask a question, since I want to work out it by myself, and I doubt the validity of this statement.
Also, I would like to take my leave now. Is the etiquette here to warn that you are suspending the communication or do you just discreetly quit the chat room?
I am not sure whether I should post this on meta, since that thread is emotional enough so I am not sure whether it is a good idea to bump it.
But to me it seems that the problems with that particular user are more about not following the rules written down by room owner than about not following network-wide rules.
If that is the case, proper action seems to be ask the room owner to do something about it. (I suppose he has possibility to ban user from a room, if that would be necessary..)
I thought I'd ask in chat, since a room owner is here now. @robjohn
@Chris'swisesister I wanted to say, however angry you may become with the site, do not delete your account, whenever I am in mood to solve some integrals, I go to your user page and try to solve the integrals from there. :-| :-)
I've started my journey in the study of mathematics with mathematics for the nonmathematician - I couldn't simply open a textbook and study it for no reason. This book was very important for giving me motivation to study stuff today. The problem with textobooks is that they are kinda cold, there are no dreams, no speculations, no fantasies, only that black, cold text. - I kinda need a little of that dreams and stuff to proceed. — Gustavo BandeiraMay 17 at 16:58
@MartinSleziak Sorry, it was almost 2 AM and I was only here for a short bit. anon and I are room owners (I made anon an owner when I was made a moderator since I knew that my time would be divided). However, it may be that I am not here as much, but my observance has been that skullpatrol's behavior has mellowed recently. If that is not the case, then someone needs to notify me or anon (don't bother Mark Gravell). I see that mixedmath has added himself as a room owner, so he is also fair game.
My questions was whether contacting room owner would not be a more appropriate course of action than flagging in situations similar to the one described in the meta-thread.
He writes there: I have seen passionate letters (e.g. click here) written by integration theorists urging that the Kurzweil-Henstock integral replace the Riemann integral in all courses starting with calculus. Their argument seems to be that almost no calculus student completely understands what a tagged partition is, so their understanding of a tagging which is d-fine with respect to a gauge function d will be about the same.
I don't quite buy it at the calculus level but it might be interesting to teach this integral in an undergraduate real analysis course.
Basically, we enumerate the rationals of $[0,1]$ as $$\{r_1,r_2,\dots\}$$ and define the gauge $\delta$ to be $\delta(x)=1$ when $x$ is irrational and $$\delta(r_k)=\frac{\epsilon}{2^{k+1}}$$
This makes any generalized sum less than $\epsilon$.
I think I am being won over by this integral. Might as well study it.
@robjohn I have read about a interesting sequence of functions such that $$\operatorname{l.i.m.}\limits_{n\to\infty}f_n=0$$ but the pointwise limit of the $f_n$ fails to exist!
Take $[0,1]$. $I_1=\chi_{[0,1]}$. Split it in half. $I_2$ is the indicator on the first half, $I_3$ on the second. Split in fourths. $I_4$ is the indicator on the first fourth, ($[0,1/4]$) and then up to $I_5,I_6,I_7$ in each fourth, in order. Then split in eighths, and so on.
**PROP** Let $S$ be compact and let $\langle f_n\rangle$ be a sequence of *continuous* functions defined over $S$. Then $f_n\to f$ uniformly if and only if $(i)$ $f$ is continuous on $S$. $(ii)$ For each $ \epsilon >0$ there exists a $\delta >0$ and $m>0$ such that $n>m$ and $|f(x)-f_k(x)|<\delta\implies |f_{n+k}(x)_f(x)|<\epsilon$ for each $x\in S$ and each $k=1,2,\dots$.
