I didn't intend to dogpile you haha @Alizter. Just wanted to say that I haven't yet encountered anything fun about Lie stuff.
To make you happy, there is a geometric-topology conjecture that says that Z_p (p-adics) can't act by homeomorphisms on any finite-dimensional manifold.
It's unsolved till now, AFAIK.
In fact it can be proved that what I said above is essentially equivalent to say that if G is any topological group that acts faithfully on a compact finite-dimensional manifold by homeomorphisms, G must be Lie.
Of course, I dunno any of that stuff. Just understand the statement of the problem prof told me.
Heh. I've a question. An open set in the basis of an infinite product topology has some $n$ so that for every $i>n$, $U_i=X_i$, right? Otherwise we would have infinite number of open sets $U_i \neq X_i$, contradicting it being the product topology
@BalarkaSen Certainly not if the two points you identify are the tips of the cones. Then you get a contractible space. If you identify other points, it may work, you need to pick the right points, however. (I think if you pick the right points it works, but I haven't verified that.)
@Chris'ssis By "increasing in chunk of $n$ terms" I mean the following : let $u_k=\min\{a_k,\dots,a_{k+n-1}\}$, then $u_k$ is increasing. Do you get the idea ?
OK, we have that $$\lim_{n\to\infty} n\left(\left(n \int_0^{\pi/4}e^{n x} (\tan^{n-1}(x)+\tan^{n}(x)+\tan^{n+1}(x)) \ dx\right)^{1/n^2}-1\right)=\frac{\pi}{4}$$
I also wonder what happens when we have $$\lim_{n\to\infty} n\left(\left(n \int_0^{\pi/4}e^{n x} (\tan^{n-k}(x)+\tan^{n}(x)+\tan^{n+k}(x)) \ dx\right)^{1/n^2}-1\right)$$
@Chris'ssis I liked this book people.reed.edu/~jerry/211/vcalc.pdf. It is more of a deeper understanding, instead of practical tricks. However I really like it
Let $A,B\subset\mathbb{R},n\ge2$.
Let $f:A\to B$ (not necessarily continuous) such that $\forall a\in A,f^{-1}(a)$ is a tuple of $n$ elements.
I know that if $f$ in continuous, for $A=B=\mathbb{R}$ and $n=2$, such a function does not exist. Therefore I was wondering :
When does such a functio...
@DonLarynx Thanks, however that does not give me an explicit formula for the general case. I know that for the example I showed above ($[0,1]\to[0,1]$) such a function does exist, so I guess for higher $n$s it should exist too...?
@TedShifrin @DonLarynx For my comment : I don't know which one exactly, but question b) of an exercise I have tells me to find one for $A=B=[0,1]$ and $n=2$. So there has to be one. And I'm pretty sure it's not continuous.
Let $A,B\subset\mathbb{R},n\ge2$.
Let $f:A\to B$ (not necessarily continuous) such that $\forall a\in A,f^{-1}(a)$ is a tuple of $n$ elements.
I know that if $f$ in continuous, for $A=B=\mathbb{R}$ and $n=2$, such a function does not exist. Therefore I was wondering :
When does such a functio...
