@Thorgott lets say we are interested into maps into $\mathbb{R}^n$, and we take a subset of it
If $A\subseteq \mathbb{R}^n$, I am wondering about this version of path-connectedness that for every $x,y\in A$ there exists a differentiable map $p:[0, 1]\to A$ such that $p(0) = x$ and $p(1) = y$
Is there a good graduate or undergraduate level textbook that introduces fiber bundles and sections? I'm looking for a book that either has an entire chapter dedicated to that topic or the whole book is about that topic. I've read short sections on the topic but want to go more in depth.
To prove the above theorem, Rudin uses the topological definition of continuity. He shows that for an open subset $V$ of $X$, $f(V^c)$ is closed in $Y$. Then he says "since $f$ is one-to-one and onto, $f(V)$ is the complement of $f(V^c)$." Why do we need the function to be one-to-one? I thought $f(V)^c=f(V^c)$ is equivalent to $f$ being surjective only.
No, for this you need bijectivity. For example, $f(x) = x^2$ is a surjective function from $[-1,1]$ to $[0,1]$. Take $V = [0,1] \subset [-1,1]$. We have $f(V) = [0, 1]$ and $f(V^c) = (0, 1]$
@VladimirLysikov interesting, thank you. Right, for equality to hold we need the function to be bijective, to only show $f(V)^c\subset f(V^c)$ we need it to be surjective.
so I can say what the map is, it's obtained by twisting the map $\Gamma(X,\mathcal{F}\otimes\mathcal{L}_{\alpha}^{\otimes n})\rightarrow\mathcal{F}\otimes\mathcal{L}_{\alpha}^{\otimes n}$ (here, the abelian group is interpreted as a constant sheaf)
this map is induced by literal restriction of global sections to open subsets
let's focus on ample line bundles rather than ample families, the condition we want is rather than an epimorphism $\mathcal{O}_X^k\otimes\mathcal{L}_{\alpha}^{-\otimes n}\rightarrow\mathcal{F}$ obtained by twisting a map $\mathcal{O}_X^k\rightarrow\mathcal{F}\otimes\mathcal{L}_{\alpha}^n$ classifying a bunch of global sections
in other words, $\mathcal{F}\otimes\mathcal{L}_{\alpha}^n$ is generated by global sections
the condition that $\Gamma(X,\mathcal{F}\otimes\mathcal{L}_{\alpha}^{\otimes n})\rightarrow\mathcal{F}\otimes\mathcal{L}_{\alpha}^n$ seems much stronger for it demands that every section comes from a global section
so it should be possible to cook up a counter-example over projective space
I'm also a bit confused that $\mathcal{F}$ is not required to be of finite type or sth
anyway, I can point to the issue, but I'm not the right person to resolve it
If we have something like $$1 = \lim_{z \to 0} \frac{\sin^2(z)}{z^2} = \lim_{z \to 0} \frac{(1 - \cos(z))(1 + \cos(z))}{z^2}$$ and we know $1 + \cos(z) \to 2$, we still can't just pull the limit apart and conclude $$\lim_{z \to 0} \frac{1 - \cos(z)}{z^2} = \frac 12,$$ right? We would need to know both limits exist. But that's exactly what my book did.
mo: the notation ln^k x is nonstandard, but might you recognize that as a series you know, or at least something very close to a series you know, with ln x plugged into it?
mo: whenever you're asked to find the sum of a series, that's a sign that you're expected to recognize something 'special' about the series. it is not generally possible to write down a simpler expression for the sum of a series than the series itself, so if a resource suggests that this is possible, that is telling you something.
and your toolkit of series you "actually know how to sum" is probably relatively small (maybe limited to: geometric series, series you can recognize as 'telescoping' [a vague category], and taylor series of functions you know, perhaps with other functions plugged into them)
well, i don't internally think of it as 'cancelling,' but if that is how you think of "if cosh(t) = a + b + [series], then [series] = cosh(t) - a - b," then yes
@Thorgott well there do exist such subjects like geometric measure theory where "differentiable components" of this sort could be studied, and where sets are not exactly well-behaved
although I don't know about it to be confident in saying that this really would be of any use there, they seem to be concerned in generalizing things much more past differentiability
Call a number despicable if it cannot be written in the form p + n³ where p is prime and n is an integer. Are all despicable numbers cubes?
Has anyone got any ideas for this problem appreciate any help
I tried finding despicable cubes
Let us consider writing y³ in this form. Then, we either have y³ = p + n³ or y³ = p - n³. Rearranging these equations, we get one of: p = y³+n² = (y+n)(y² -ny+n²) p = y³ -n³ = (y - n)(y²+ny+n²)
If p is prime, one of these factors must be 1. In equation (1), we know y + n >= 1+1 = 2 and y² - ny + n² = (y - n)² + ny > 1 unless n = y = 1. However, if n = y = 1, then p = 2 would be prime. Therefore this expression is never prime. In equation (2), y² + ny + n² > 1, so this can only be prime if n = y - 1, then p 1 (y² + y(y - 1) + (y - 1)2) = 3y² - 3y + 1. If this factor is composite, then there is no way to write y³ in the form p ± n³ with p prime. Then all that remains is to find such a y.
I used trial and error to find some values of y such as 6 but I'm not sure where to go from here
Find the complex roots of the equation: $z^* + z |z| = 24 - 12j $ where $z^* = r \cos\theta - j \sin\theta, \quad z = r (\cos\theta + j \sin\theta), \quad |z| = r$
@AlessandroCodenotti if $X\cup_f Y$ is an adjunction space with $f:A\to Y$, can you call this adjunction space by $A$? How does one refer to the set $A$ here?
I wrote a question here titled "Adjunction space of compact $F$-spaces by a $P$-set is an $F$-space" but I'm not sure if this is correct nomenclature
In remark 4.31 in Rudin's PMA, he constructs a function which is monotonic on $(a,b)$ and discontinuous at every of point of a countable subset $E$ of $(a,b)$. Let $E=\{x_n\}_{n=1}^\infty$. Let $(c_n)$ be a sequence of positive numbers such that $\sum c_n$ converges. Define $$f(x)=\sum_{x_n<x}c_n\qquad (a<x<b).$$Here we sum over those $n$ for which $x_n<x$.
Now, I think it is pretty clear that $f(x-)=f(x)$ ($f(x-)$ denotes the left-hand limit at $x$), but I don't see why, if we change the summation to $x_n\leq x$, that $f(x+)=f(x)$. Is this obvious to you?
