Suppose I have a chain complex of chains $C_n$. Then one can obtain the homology groups of this complex. Now if I choose any abelian group $G$ and I consider the cochain group $C_n^*=Hom(C_n,G)$ then I can obtain the cohomology groups. Now the question is: If I form the cocohomology group by cons...
Suppose that for $n \geq 1$, $X_n$ is uniformly distributed on {1, 2, ..., n}. how to show $\lim\limits_{n\to\infty}P(\frac{X_n}{n}\leq y)= y$ for $y\in(0, 1)$.
sure, but most of the time when there is cancellation from double negation it's coming from an expression that's an early hold out from old or middle english, where there is never any cancellation of double negatives
The problem ask to use it, the idea I have is $\lim\limits_{n\to\infty}P\left(\frac{X_n}{n}\leq y\right)=\lim\limits_{n\to\infty}P\left(X_n\leq ny\right)=\lim\limits_{n\to\infty}\frac{\lfloor{ny}\rfloor}{n}=y$
@Akiva I have no idea what partitions of unity are, I just know they're a thing that exist and are important if you want to translate integration to manifolds, so you tend to ask that the topology is second countable
oh yeah so if you want the higher dim analogue to be like, simply connected, complete, and with everywhere negative sectional curvature you still have a diffeomorphism with $\mathbf{R}^{n}$.
