@Astyx consider a finite extension of a local field as a f.d. vector space over your field and show the topology generated by the unique extension of the absolute value to your extension is locally compact
can someone take a look into this question which has ID 3987120 ? Although a nice person showed up and gave some hint, and while I commend his effort. It doesnt address the problem as the intended method. Perhaps someone out there with some experience could help me with ideas on how to solve it using euclidean geometry?
@Thorgott wat CP^2 covers itself. There should be many examples, like the Whitehead manifold
It seems annoying to prove, but something like the following: The Whitehead manifold is contractible, so if it covers a compact manifold $M$, $M = K(\pi_1(M), 1)$. Finite dimensionality implies $\pi_1(M)$ is torsion-free; in particular it must contain a copy of $\Bbb Z$
Pick the generator, this guy is a self-homeomorphism of the Whitehead manifold which pushes everything to infinity eventually (by proper discontinuity). This should mean the Whitehead manifold is simply connected at infinity.
That would be a contradiction
The point is universal covers of compact manifolds cannot have complicated space of ends; they have periodic tilings
@EdwardEvans Thanks, gonna listen. Glad you liked Nachtterror btw
@BalarkaSen i'm intrigued by the last name. Apparently Larry Guth is the son of Alan Guth, who i recognize as one of the people who came up with the idea of cosmic inflation
(which is something so far out of my expertise that i can't even pretend to talk about it)
@user586228 you are given the numbers for $>=1$, $=2$, $>=2$ passes. from this you can figure out the numbers for $=0$, $=1$, $=2$, $=3$. assuming independence, you can get that $pmc$ is the same as $=3$. this cuts out two options. now you need to compute $p+m+c$. draw a venn diagram with 3 circles corresponding to $p,m,c$. you want to add the 'areas' of $p,m,c$. note that the area is one times the chances of $=1$ plus two times the chances of $=2$ plus threes times the chances of $=3$.
Directional derivative of a function $f$ along a curve $\Gamma$ at a point $\mathbf x$ do not necessarily equal to the directional derivative of $f$ along the tangent vector of $\Gamma$ at the point $\mathbf x$. This statement is correct, right?
@copper.hat Could anyone tell me two things...(1) Why are we assuming independence?(2)Why is pcm=p3(3) Why does it cut down two options?4)Why is the last line holding true,""note that area is one times the....".
In the proof of weierstrass preparation, we have that $f(z_1,0,\cdots,0) \neq 0$. This gives us an $\epsilon_1$ sized disc around $z_1$ in $\Bbb{C}$ where $f(z_1,0,\cdots,0)$ is nonzero. Then we go on to let $a_1(z_2,\cdots,z_n),\cdots,a_d(z_2,\cdots,z_n)$ denote the zeroes of $f(z_1,z_2,\cdots,z_n)$ inside the $\epsilon_1$ sized disc. Why should this zero set be finite?
so if you fix $(z_2,...,z_n)$, that's an analytic function of one variable and those have discrete zero sets, so if you pass to a slightly smaller closed disc, it will be finite
analytic functions of more than one variable can have infinitely many zeroes, but those of one variable can't (in a disk, that is)
I'm reading through a proof on the uniqueness theorem of Laurent series in my lecture notes and came across this sentence:
The point $c$ ($\in \mathbb{C})$ is a limit point of zero points of the (analytic) function $f$ and therefore there exists a circle $K$ around $c$ where $f=0$.
Under wh...
So how does having $k$ perfect and $f''(x) = 0$ (possibly not necessary condition for this question) imply that $f'(x) = \Sigma_{i \textit{ } even, 0 \leq i \leq s -1} a_{i+1} x^i$ when $f(x) = \Sigma_{i = 0}^s a_i x^i$?
For stopping times this is super intuitive. Define a random time as the infimum of some random set, for example this first time that a random walk hits 0
I'm reading through a proof on the uniqueness theorem of Laurent series in my lecture notes and came across this sentence:
The point $c$ ($\in \mathbb{C})$ is a limit point of zero points of the (analytic) function $f$ and therefore there exists a circle $K$ around $c$ where $f=0$.
Under wh...
