@TedShifrin We both know that if a smooth manifold has a nonvanishing vector field, then $\chi(M)=0$. I seemed to remember the converse was true but forgot how to prove it. Turns out it's the very last proposition of Bredon... guess I never learned in the first place :P
it was a multipart question, first to simplify the negation, and then, as a follow-up, it asked if, in case "(2)" (the second negation asked about, reproduced above), the negation is true as written.
"why or why not"
$\neg \exists x\in\mathbb{R} (\forall y\in\mathbb{R} ( x > y))$ seems to be false when spoken verbally, yet its simplification $\forall x\in\mathbb{R} (\exists y\in\mathbb{R} (x\leq y))$ seems to be true?
$\neg \exists x\in\mathbb{R} (\forall y\in\mathbb{R} ( x > y))$ says "there does not exist an $x$ in $\mathbb{R}$ for all $y$ in $\mathbb{R}$ such that $x>y$", which is true because there does not exist a single real number that is greater than than all real numbers.
@Chris'ssis @robjohn @N3buchadnezzar @rehband @r9m $c_{k, n}$ be the number of integer partitions of $k$ in exactly two parts, each of size exactly $n$.
Lesson 1 : Don't underestimate number theory.
Lesson 2 : Don't underestimate me, in particular.
Lesson 3 : Next time, @Chris'ssis, you might want to post something a wee bit harder. =p
@BalarkaSen I can understand a partition of $k$ into exactly two parts, but how does each of those have size exactly $n$? If I ignore the one term with the $c_{k,n}$, the rest of your derivation looks good.
@robjohn Like, for example, in a free partition $2n = (2n - 1) + 1$ is allowed. But when the sizes of the parts are restricted to $n$, $2n = n + n$ is the only possible partition.
$c_{k, n} = k - 1$ when $k \leq n$. $c_{k, n} = 2n - k + 1$ when $2n \geq k > n$
@robjohn I was kidding, though, don't take it too literally.
@BalarkaSen Generally the partitions ones are interested in are, say, the ones in which the number of terms is bounded (above or below), or the size of the terms, etc...
@DanielFischer (If you care, the answer is that for your favorite positive $\varepsilon$, there's an $f \in O\left(x^{4+\varepsilon}\right)$ such that $|\text{Aut}(G \times G)| = f\left(\left|\text{Aut}(G)\right|\right)$ for any abelian group $G$. I've no idea about how things play in general.)
@TedShifrin Pretty sure it's working. Also, about yesterday's group theory question: you said consider $GL_2$. Would upper- (or lower-) triangular matrices work? They contain the center, but I'm pretty sure they're not normal in $GL_2$.
I can conjugate an upper triangular matrix to one with no 0s. It's just picking a different basis for the transformation, one that isn't particularly "nice", right?
@r9m So she used integration to get the sums? I prefer manipulating the sums if possible, but sometimes it is necessary to use an integral for these sums.
@robjohn of all the ways I have derived the integral representation of the terms .. Kneser's proof seems to be the key idea (atleast that's how I did it .. manipulating the integrals/series and connecting it to Kneser's proof :) .. )
@robjohn can you delete the binomial identity above .. she asked me not to show it .. :P
@Khallil ya .. :) but nowadays I have a feeling they are just making minor changes to the fan made colored versions of the manga and putting it into the anime frame with background music and special sound effects :P
I feel like it is deviating form the original Naruto anime style =(
Oh, in that case, I know what you mean! I think the anime's been doing some great stuff. I love canonical filler, especially the Sannin-cave scene, @r9m.
Hey guys, I've got a small (and basic) mathematics question
I'm reading Reed & Simon's book Functional Analysis and have reached the chapter on topological spaces. I noticed they talk about norms inducing a topology, but I don't quite understand how this 'happens'.
I'm wondering whether someone could put me on the right track - it's probably simple.
These kinds of mathematical structures are defined to have certain properties. They are talking about these properties arising from some construction or study. Thus they form the strucutre.
For example @Danu Objects in physics are defined to have properties like mass. Now if you show that a thing has properties of and object then it is an object see where I am going with this?
Hmm. I have not that much under my belt to tell you that and also I would need to see the text. Best bet is to ask a question on the Main quoting the extract and explaining that you do not understand. People will help you if you show effort at trying tp understand. (mathematics main)
It's a little annoying because I find this particular chapter of the book is rather lacking - e.g. they never even defined convergence in the topological space setting
so there's not a clean cut paragraph that I can quote
The issue I'm running into is rather because they don't explain :)
@Danu Nobody else is replying here atm so by the time someone here who is able to answer answers it would have been answered on the main. So go post the question and carry on with your reading. Then when answers come up you can go back and think.
@Khallil Then still a whole month is plenty to lose your mental sharpness. Don't throw yourself into a snake pit without wall climbing skills before you start.
@BalarkaSen The 30-day limit on name changes makes it difficult to keep track of the days of week. I was in sync yesterday, not anymore. Considering a switch to month-based names.
@Alizter Solve in reals $$\frac{1}{\left\lfloor{a}\right\rfloor}+\frac{1}{\left\lfloor{2a}\right\rfloor}=a-\left\lfloor{a}\right\rfloor+\frac{1}{3}$$
For a sequence x_n and a in R, then a is a subsequential limit of x_n if and only if for all e>0 , there exists infinitely many n in N with |x_n - a| < e . Does this have antyhing to do with the sequence x_n being frequently on the set {a} ? Does the consequent of this implication have anything to do with a being a supremum/infimum of any set ?
I'm trying to discover what "for all e>0 , there exists infinitely many n in N with |x_n - a| < e" really means, that is, equivalent statements to it .
