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Jan
25
comment Fibonacci numbers expressed as squares of lower Fibonacci numbers
en.wikipedia.org/wiki/Fibonacci_number
Dec
13
awarded  Critic
Oct
17
answered Eigenvalues of the “Laplacian” on [0,2$\pi]\subset\mathbb{R}$
Oct
13
comment Finite automorphism groups of $\mathbb{P}^1$
I think the third groupe of your line of isomorphism is in fact $SO(3,\mathbb{R})$, not $SO(3,\mathbb{C})$ and hence your problem is solved (you can check that $SO(3,\mathbb{C})$ doesn't have the good dimension as a real manifold if you're not covinced...).
Oct
6
revised a set containing every limit points but not closed
added 34 characters in body
Oct
6
answered a set containing every limit points but not closed
Oct
6
awarded  Commentator
Oct
6
comment Why are locally compact groups Weil complete?
(I wrote as if the group was commutative, but that doesn't change anything if it is not, this will work in an arbitrary uniform space)
Oct
6
comment Why are locally compact groups Weil complete?
Exactly as you would do with a sequence on a metric space : You have $(u_i)$ a cauchy net and $(u_j)$ a subnet converging to $u$. Take $V$ a neighborhood of $u$, and $V'$ a neighborhood of $0$ such that if $(u-x)$ and $(x-a)$ are in $V'$ then $a$ is in $V$. write the cauchy propertie for $V'$, then write that there is therm of $(u_j)$ which is $V'$ close to $u$ for $j$ big enough to be converne bu the previous cauchy propertie you wrote, you can deduce from those two thing, that for any $i$ concerne bu the cauchy properties, $u_i$ is in $V$.
Oct
5
answered Why are locally compact groups Weil complete?
Oct
4
comment $A = B^2$ for which matrix $A$?
by Starting to compute the Jourdan reduction of your matrix. you juste have to determine whether or not a matrix of the form $\lambda * I_n +N$ with $N$ a reduced nilpotent matrix has a square root or not. it seem that when $\lambda$ is non zero it's always the case. So in the end you have to find a way to know when a reduced nilpotent matrix has a square root or not, this seem to be a purely combinatorics question, and the answer will only depend on the dimension. (so the general answer will depend on the dimmension of the jordan block of the 0 eigenvalue)
Oct
4
comment What if every function in $C(X)$ has finite spectrum?
Well, if you have a net $(u_i)$ which does not converge to a value $x$. Then by (the negation of the) definition there exist a neighborhood $V$ of $u$ such that for every $i \in I$ there exist $j \in J$ such that $u_j$ is not in $V$. Hence the set of $i$ such that $u_i$ is not in $V$ is cofininal in $I$ and it gives us a sub net (the increasing function you're talking about is just the inclusion) of $u_i$ such that no "sub-sub-net" converge to $x$. which leads to a contradiction if the space is compact and if every converging subnet of $(u_i)$ converge to $u$.
Oct
3
awarded  Scholar
Oct
3
accepted Convex hull of an open set…
Oct
3
comment What if every function in $C(X)$ has finite spectrum?
and the argument still hold for $C_0(X)$ on a localy compact space as $C_0(X)$ separate the point of the alexandrov compactification of $X$. ($C_0(X)$ is the set of function on the compactification which vanish at infinity... )
Oct
3
comment What if every function in $C(X)$ has finite spectrum?
Yes you are right, in my mind they were positive but i forgot to write it ^^
Oct
3
comment What if every function in $C(X)$ has finite spectrum?
Yes, but if $(u_i)$ was not converging to $u$, then you can construct a sub net of $u_i$ which stay away from $u$, and hence extract a "sub-sub-net" which converge to something diferent than u, which leads to a contradiction. that a classical fact in topologie : in a compact set, if every convergent extration of a sequence/net converge to the same value, then the sequence/net is convergente.
Oct
3
awarded  Teacher
Oct
3
answered What if every function in $C(X)$ has finite spectrum?
Oct
3
answered What if every function in $C(X)$ has finite spectrum?