@Chris'ssistheartist I presume we're talking about very clever integration bits. Out of curiousity, do you have a source of interesting problems (and perhaps solutions) outside of forums like here or AoPS or Integrals&Series?
@mixedmath I'm working on it at the moment, it's going to be a collection of integrals, series and limits major part being created by me, around 500 problems.
@mixedmath Thank you. I think I'll need some support at some time. :-)
OK
Th hardest part is indeed the way you present the content to the readers, not the mathematics, not at all. That makes the difference between an ordinary book and a special book that a reader might like.
@mixedmath I often browse through the math integral answers in Brilliant and Quora ... but I guess the AoPS, I&S and M.Se are unrivaled in terms of the enormous variation and flavors of integrals asked .. :)
anyway, $\Bbb R^2 - (0, 0)$ can be shown to be path-connected by joining points by straightlines and joining them by paths when it makes sense, and deforming them around the origin a bit if they hit the origin.
Then I found that since it is a lone passing through the origin then one point will be negative and other will be positive... So using these facts I worked out an example
@Rememberme I'd suggest if you want to read topology .. for a gentler introduction you can read Kumaresan's Topology of Metric Spaces book .. which has very good exposition and a good number of exercises :)
I think it can safely be said that Math.StackExchange (and MathOverflow and perhaps some of the sister sites on the SE network) are the best resources for (English-speaking) people with objective math questions on the web. There are other math sites, such as those mentioned in Useful Mathematical...
@Rememberme the book contains topics on connectedness as well .. why topologist's sine curve is not path connected has two proofs in that book as far as I can recall ..
@BalarkaSen every field homomorphism $A\to B$ can be extended to algebraic closures $\bar{A}\to\bar{B}$ by transfinite induction (I'm assuming everything's separable so I don't have to think about that).
Fine..... (I dont have anything else to do then :p)
@Balarka ... last question (I wont disturb you anymore) You asked for the example of a set which is not path connected right, $\Bbb{R} \setminus {x}$ is not path connected right.....
@anon mortal flaw with my analogy : k-isoms between separable closures are paths, so k-auts of k^alg should be loops. precisely these form Gal(\bar k/k), so that becomes an analog for loopspace. no notion of homotopy there. something's amiss.
i am not talking about path connectedness anymore, @Remember
paths in L between points $*_1 : L \hookrightarrow \bar k$ and $*_2 : L \hookrightarrow \bar k$ are then k-auts $\sigma : \bar k \to \bar k$ such that the obvious diagram consisting of $*_1$, $*_2$ and $\sigma$ commutes.
i am trying to develop a path-lifting lemma right now
i guess i have kind of a "loop lifting lemma"
although my proof is ad-hoc. I'd want to mimic the topological proof, but that'd require developing the notion of an open cover -- that sounds hard, as covers are topological.
hmm, the problem with Gal(k^alg/k) is still not resolved. an element of Gal(k^alg/k) is indeed an aut k^alg --> k^alg which is absorbed by the standard basepoint k --> k^alg. :s
ok. let $A \stackrel{f}{\to} B \stackrel{g}{\to} C$ be a short exact sequence in $\mathsf{AbGrp}$. if there is a homomorphism $s : C \to B$ such that $gs = \text{id}_C$, then $B \cong A \oplus C$.
this is called the splitting lemma in $\mathsf{AbGrp}$
$s$, if exists, is called a section (of the short exact sequence), and the sequence is said to split.
I found this theorem pretty surprising when I first got to know this.
it doesn't hold in $\mathsf{Grp}$, unfortunately, but something of a partial generalization works. that same criterion, except that $B$ is a semidirect product of $A$ and $C$.
that's powerful too : semidirect products have nice presentations
@Tobias I am on the chapter called conectedness and compactness....(In munkres) Though this is not a theorem I want to think about this: A set X being connected can I say anything about its complement ?
@Rememberme its complement could be literally anything. let Y be any topological space, then consider the disjoint union of X and Y. in here, X's complement is Y... which is anything we wanted it to be.
@LucioD Basically, I can't say anything that Crostul didn't already say. You have a family of sets you want to be the closed sets, so you check whether the family has the required properties (closed under finite unions and arbitrary intersections).
@robjohn Excuse me. Suppose I want to evaluate $\displaystyle\lim_{n\to\infty}\left(\sqrt{\frac{9^n}{n^2} + \frac{3^n}{5n}+2}-\frac{3^n}{n}\right)$. Is it valid if I make the change of variable $y=\dfrac{3^n}{n}$ ?
oops, that was supposed to be $-\dfrac{3^n}{n}$ there
@Cristopher Add a couple of spaces [I just did that here]. Otherwise the software inserts a zero-width joiner after 80 characters, and that tends to break some latex commands.
