@iwriteonbananas: Well, anything becomes a line bundle once you specify transition functions. You in addition need to show that the notion of homological orientation agrees with the notion of "section of that line bundle". In fact, since that's the hard part, I'm telling you to do too much work anyway. Can we start by putting a metric on M?
@iwriteonbananas: My hands are too cold to type much right now. The point, roughly, is that $H_n(M,M-p)$ is in bijection with orientation classes of orthonormal bases at a point.
In particular, find something that N sqcup N bounds (not N sqcup \bar N). Then take the "boundary sum" of this and M x [0,1]. If you care, I leave it to you to interpret this appropriately.
@MikeMiller Oh, true. Take a nonorientable manifold and # with N x [0, 1], remove a tubular nbhd of a path from N x {0} to N x {1} going through the nonorientable summand. N $\sqcup$ N bounds this fellow. Connected sum with M x [0, 1], and it bounds M $\sqcup$ N and \bar M $\sqcup$ N.
Actually, isn't that enough? Now you just use that # is cobordant to $\sqcup$
Does anyone know when we're checking for existence and uniqueness are we supposed to only check whether that function and it's partial derivative w.r.t $y$ are continuous or do we also have to check whether they're both bounded?
@MikeMiller The resulting group is called the unoriented cobordism group of a point, right? Because you can replace every manifold $M$ by a map $M \to pt$.
Do you remember some examples of $\Omega_n$ for $n \leq 5$ off the top of yor head? :P
I don't think I've ever seen someone care about $\Omega_n(X)$ for any $X$ other than the point. They care about it in the abstract (It's a cohomology theory!) and in the specific of the point. But I've never seen it for other $X$.
@Huy when we're checking for existence and uniqueness are we supposed to only check whether that function and it's partial derivative w.r.t $y$ are continuous or do we also have to check whether they're both bounded?
@Samurai If I got your point, you have a positive quantity in denominator $$a^2\ln^2 x +1,$$ right? Well, from this point you're done (after making a little observation).
if I have two whole numbers x and y and x is chosen from one range and y from a non-overlapping range and I give you x+ y.. what can you tell me about x and y?
and note that the resulting $0$ can be views simply as $\frac{1}{\infty}$. In these conditions how do you relate more infinities arising from different ways and without estabilishing a relation between the producing sources?
I think it could be turned into an interesting post (not this one - it's dead - but a modified version of it). What mathematics can be done in an internally consistent way if you reject the law of the excluded middle? Can you still develop calculus? algebraic geometry?
That is a good question only if you care and only if you can't, after a chunk of work, cannot find the answer. KCd might hate you no matter what you do anyway.
Induced from $H^l(X; R) \otimes H^k(X; R) \to H^{k+l}(X \times X; R) \to H^{k+l}(X; R)$, the second map being inclusion as a diagonal, the first cross product.
Note that cross product means the map induced from $C^k(X) \otimes C^l(X) \to C^{k+l}(X \times X)$, the E-Z map
Not the geometric thing in Hatcher
Otherwise it's circular.
Yes.
That's in Qiaochu's blog, if I interpret the broken latex correctly.
Guys, from Engeling - "Every Tychonoff space X has a compactification (Y,c), such that weight(X) = weigh(Y)" - any ideas on the proof of this? I think I get it for the case of m=weight(X) $\geq $ $\mathbb{N}$ , as then I could embed X in I^m with some fuction f, and I would have weight(f(X)) = m, weight(I^m), so any dense set containing f(X) will have weight m. But I don't know how it would work for m finite.
@MikeMiller Nope, can't find a reference. Alexander, Whitney, Lefshetz, Cech, Eilenberg, Zilberg, and all sorts of people but no mention of who discovered which definition.
@Jake1234 Probably the only person in this room who has a shot at answering that is Daniel Fischer. If you post it on main, Brian Scott will make short work of it.
I have a Hausdorff space $X$ and $E$ is a compact subspace, i want to prove that for all $x\in X$, if $x\notin E$ then there exists a neighborhood of $V$ of $x$ and an other $U$ for $E$ , such that $V\cap U=\emptyset.$
I let $y\in E$ then $x\neq y$, as $X$ is a Hausdorff space there exist a neighborhood $V$ for $x$ and $W$ for $y$ such that $V\cap W=\emptyset$
@MikeMiller Do you have a jstor account? Can you tell me if this define cup in an abstract way? It seems to start with simplicial sets, so I'd guess YES
@Vrouvrou For every element y of E take a neightborhood U_y of it that is disjunct with some neightborhood of U_x of x, union of those neightborhoods U_y covers E, take a finite subcovering - then the finite intersection the appropriate V_i will be disjunct with the covering of E.
"I had a class of 10 students. I had just taught the Birthday paradox, and told them that, of course, because we only have 10 students, it would be smart for me to take a bet that none of them had the same birthday. But one student wanted to take the bet with me, even though I reminded him of the probabilities involved. He won. Keeping in mind that he had no special knowledge of everyone else's birthdays, why was it a bad idea for me to take the bet?"
@Jake1234: i say let $y\in E$ then $x\neq y$ as X is Hausdorff there exists $V\in\mathcal{V}_x$ and $U_y\in \mathcal{V}_y$ we can choose $V,U$ opens so $E=\cup_{y\in E} U_y\cap E$ as $E$ is compact $E=\cup_{i=1}^n U_{y_i}\cap E$
@Vrouvrou The intersection of V_i from 1 to n (Where V_i contains x, and is disjunct with U_y_i and contain- the ones you wrote at the end of your last post) - it is a finite intersection of open sets, so it's an open set, let's call it A - it contains x, and it is disjunct with all the sets U_y_i. The finite union of U_y_i covers E in X, and so it's an open set containing E, let's call it B. Now A and B are disjunct, A contains x, B contains E.
@Jake1234 It's not the only one I met, but they seem to me so rare, or I simply had a very bad luck not to meet more.
@Jake1234 He is very OK to do mathematics with. I don't know him, but my inuition tells me that I'm safe even if I gave him the solution to the Riemann hypothesis.
I see, I thought you meant that it's best to do mathematics alone for efficiency of learning new stuff etc.
The perspective of "do mathematics alone, because others might end up annoying you" is not something that ever occured to me, but I'm just a bad student, so there's no risk of someone 'stealing' an idea from me. I must say I've met a large amount of very kind people at university.
I'd estimate it as one of the best places with relatively lot of people, where there's a relatively large chance I'd get my wallet back if I lost it or something etc. Compared to economics etc. departments.
3.5.2. Engelking
Every Tychonoff space $X$ has a compactification $(Y,c)$, such that $w(X) = w(Y)$.
Any ideas on the proof of this?
I think I get it for the case of $m=w(X) \geq \mathbb{N}$ , as then I could embed $X$ in $I^m$ with some fuction $f$, and I would have $w(f(X)) = m$ (homeomorp...
3.5.2. Engelking
Every Tychonoff space $X$ has a compactification $(Y,c)$, such that $w(X) = w(Y)$.
Any ideas on the proof of this?
I think I get it for the case of $m=w(X) \geq \mathbb{N}$ , as then I could embed $X$ in $I^m$ with some fuction $f$, and I would have $w(f(X)) = m$ (homeomorp...
