@tb Well, I don't have a bibtex compiler installed in my head yet. I only have plain LaTeX compiler, and some set theory compiler. Also I have installed that emacs parenthesis highlighting for Lisp writing.
It's the contents of that reference I'd think you'd find intriguing, though. :-)
@AsafKaragila How bizarre. However, at some point in the near future (a few more answers of yours) I will be surprised if I'm not surprised by mindboggling conclusions in the absence of choice.
@HenningMakholm In this case I don't see how applying a canned theorem let's me make progress skill-wise. I needed to work out the details of that proof.
@HenningMakholm well, if you go for canned theorems, you could as well go for the canned theorem "continuous with compact support implies uniformly continuous".
Write down the difference quotient defining the derivative: $$H'(x) = \lim_{h \to 0} \frac{H(x+h) - H(x)}{h}$$ plug in $H = f+g$ distribute and use that $\lim (a_n + b_n) = \lim a_n + \lim b_n$.
There's no $n$ in your sequences. Note that you're assuming that $f$ and $g$ are differentiable at $x$ otherwise the right hand side of $H' = f' + g'$ doesn't make sense.
Again, for $\lim(a_n + b_n) = \lim a_n + \lim b_n$ to hold we really need the two guys on the right to exist. I always show my students $a_n = n$ and $b_n = -n$ and stuff like that.
@N3buchadnezzar Any nice object can typically be broken down into bizarre sub-objects. What one usually expects is that when you assemble an object from nice parts, then the resulting object will also be nice.
[Even that is false in many cases, but that is another issue.]
For example, imagine taking a mirror and smashing it; let's say you ended up with two weird-looking pieces. You did start with something nice, but smashing destroyed the niceness. And just because the two pieces join together to form the original single nice mirror, we do not expect that the pieces themselves are so nice.
And the master smiled and started sipping his tea again.
@Srivatsan I'm not sure if it must be open, I'm not a big fan, honestly. But I thought that I'd like to see how it works out and I'd like to see a really good informed answer to that.
I tried searching the various obvious Greek spellings of Alaoglu but never found more than a thousand results. That makes me think it's an unusual or mutated surname...
@ZhenLin Given that A.'s parents were of Greek nationality and he was born and raised in Canada, no one can tell for sure how to pronounce it from just reading the name.
Special props to Mariano for keeping his ranking constant between years :)
The reputation league pages have recently (?) changes to showing the logged-in user's ranking at the top of the table. I keep going "woo! I'm number 1!" and then "oops, it's just .."
Oh, I didn't mean to spoil your 2012! Anyway, djvu things are usually OCR'ed, so quickly searchable and that can be tremendously useful, especially with a book like Hartshorne.
Is this really mathematics? I had a quick glance at that paper and it didn't appear too mathematical to me, but admittedly, I didn't understand much of it.
@DylanMoreland Which djvu viewer supports things like bookmarks, hyperlinks etc.
I am using the viewer from lizardtech, several times I've seen djvu file, which should be bookmarked (according to descrition) but I did not see bookmarks there.
@MartinSleziak I think this is something worth consideridering. On the other hand, it might be better to have as few (faq)'s as possible. It could be worth considering extending your "why should we accept answers" to "why should we accept answers and how to accept answers" or something to that effect.
tb: I have already tagged that questions as faq-proposed and wrote a comment there. But if you think that is not good thing to do, feel free to edit/comment.
Some SE linkage indicate that he's about 13-14 years old. He probably didn't understand too much of my answer, and I want to improve it. I'm not sure how though, I tried to write more but I ended up writing additional 700+ words (the current answer is about 500), so I decided to save it on my hard drive and not post the edit.
I should think more closely about it, what and how I want to write.
I don't care writing 700 words, but I don't want to write 2000. No one will read a 2000 words monologue.
I think the first sentence In a very "raw" sense the symbol 2 is just a shorthand for 1+1. There is really not much to prove there. is understandable for 13-years old guy.
I discussed it here with Henning and Zhen yesterday, I should probably add some more about what is a mathematical proof (and what is a formal proof), and why do the symbols are mostly irrelevant while the fact that 1+1=2 says something about the axioms instead.
I don't understand that either. I find that rather annoying. What I find even more astounding is that people upvote that stuff. Recently an answer got 4 upvotes and the link didn't even remotely answer the question posed.
I think the problem with that is where to draw the line. I don't see a clear-cut criterion to decide on that.
That is: I don't like questions that can be answered by the most obvious Google search. On the other hand, many questions reveal some specific confusion that can be much better addressed by someone answering.
This doesn't answer your questsion, but according to this the only $n$ for which $\cos{\frac{2\pi}{n}}$ is algebraic of degree 2 are $n = 5,8,12$ while it is rational for $n=1,2,3,4,6$.
@AsafKaragila "In what class were you given this question?" - In none. I am just reading a textbook on set theory and computation and sometimes I stumble across the problems I do not know how to solve or material I do not fully understand.
@AsafKaragila It is, actually. However in this part of the book they do not deal with cardinals arithmetics yet. I thought I had to use Cantor-Berstein because the question was introduced in the same chapter under the C-B theorem.
There is also the same question about the square (the square on the plane is divided into 2 sets, prove that at least one of them is equinumerous to the whole square).
@Daniil This can be reduced back to the line segment.
Since there is a bijection between the line and the square, this gives a bijection from the partition of the square to a partition of the line, which we already solved.
@AsafKaragila yeah, true. But I still can not understand how to solve this problem without cardinal arithmetics. (In fact, the only thing about cardinal arithmetics they covered so far in this book is that if A is infinite and B is countable then $|A \cup B| = A$)
It is possible to have very strange sets which are not with bijection with any finite set (thus infinite) but still have no proper subset which is equinumerous to them.
@Daniil Interesting. Seems like a very hard question, if I knew all the contents covered before this question I'd probably be of more help. I'll keep thinking about it.
@Daniil I am a grad student... I also research in a field which has been somewhat low on activity over the past 30 years, so it's easy to get around there.
I was wondering whether covering systems of integers could be put to more general uses other than proving that there exist infinitely many numbers NOT of some type
does there exist some literature on this with EXAMPLES?
Hmmm. Well, if I remember correctly, it's a general fact that if $\alpha$ is a linear operator and $p$ is a polynomial such that $p(\alpha) = 0$, then any root of $p$ is an eigenvalue, and vice-versa. But this is most meaningful in the algebraically closed case.
