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olivier dot begassat dot cours at gmail dot com


15h
comment Borel Measures: Atoms (Summary)
It relies on the underlying set being numerable to prove that the equivalence class is in $\Sigma$, and some form of countable choice to represent the equivalence class as a countable intersection of measurble sets.
16h
comment Borel Measures: Atoms (Summary)
You could try something along the lines of considering the equivalence classes under the equivalence relation $x\sim y$ iff for any $A\in\Sigma$, $x\in A\Longleftrightarrow y\in A$. If this works then the equivalence classes are the smallest measurable sets, and define the smallest measurable partition.
17h
comment Solution verification: $G$ and $G/H$ contain elements of same order
The correct formula is $$|G/H|=\frac{|G|}{|H|}$$ at least for finite groups, and there are non-isomorphic groups of the same order, so your argument is incorrect on two fronts.
2d
comment Sam Harris' theory of probability on the Second Coming of Christ
@GrumpyParsnip I don't think dei could produce a rock that dei couldn't lift deiself, but dei may ask deis deity friends to make one for deis.
2d
comment Sam Harris' theory of probability on the Second Coming of Christ
I watched that interview a few days ago, and was in total agreement with Sam Harris on his probability argument. I'm quite amused this question found its way to this forum, @SteveMcQueen are you per chance trying to settle some internet dispute?
Oct
26
comment If $A$ is connected, is at least one of the sets $\mathrm{Int}A$ and $\mathrm{Bd}A$ connected?
@DanielFischer You are right. I'll correct that.
Oct
26
comment Topology on $C_{compact}^{\infty}(R)$
Are you equipping the $C^\infty(K)$, $K$ compact, with the topology defined by the family of semi-norms $$\|f\|_m=\sup_{x\in K}|f^{(m)}(x)|\;?$$
Oct
26
comment Topology on $C_{compact}^{\infty}(R)$
Equipped with what topology?
Oct
26
comment Topology on $C_{compact}^{\infty}(R)$
What is a "good semi-norm"?
Sep
22
comment $G$ is the semidirect product of $U_n$ & $D_n$
@MarianoSuárez-Alvarez My comment was about the original formulation of the question, where all $GL_n(\Bbb R)$ was supposed to be the semidirect product of $U_n$ and $D_n$.
Sep
22
comment $G$ is the semidirect product of $U_n$ & $D_n$
$U_n$ isn't a normal subgroup. Geometrically, it's the group that stabilises a complete flag of subspaces, conjugation of such an element stabilises a different flag. And dimensionwise the semidirect product can't fill up all of $GL_n$... Also, why tag this question "finite groups"?
Sep
20
comment Definition of the triad homotopy groups
I streamlined the exposition, your original post was a little cluttered. I hope you agree with the changes. You can always edit the post to your liking.
Sep
19
comment The vector space $L(X,Y)$ of linear maps.
@WantTobeAbstract great!
Sep
19
comment The vector space $L(X,Y)$ of linear maps.
Could you try to clearly spell out your confusion? I don't understand what it is : "we want to show that $L(X,Y)$ is a linear map" just doesn't make any sense...
Sep
19
comment The vector space $L(X,Y)$ of linear maps.
$aS+bT$ is, a priori, just a symbol assembled from two scalars and two linear maps. We want this symbol to describe a linear map $R$, so we need to specify $Rx$ for any $x\in X$. We simply define $R(x)$, for any $x\in X$, by the formula $aS(x)+bT(x)$.
Sep
14
comment how to prove a uniformly convex Banach space is reflexive
You should look online for course notes on functional analysis.
Sep
12
comment Why is differential geometry called differential geometry?
The fundamental theorems in differential geometry are about what the differential of a function reveals about the function's local behaviour. Integral geometry (and geometric measure theory) isn't separate from differential geometry, but it addresses different questions. It is much more about the interplay of submanifolds and measure theory, and often the differentiability assumptions are very weak.
Sep
8
comment Extensions of $\mathbb{Z}_p$ by $\mathbb{Z}$ (Hilton & Stammbach III.1.2)
No problem! ${}$
Sep
8
comment Category of pointed manifolds
You don't capture the germ of a function by its differential, as witnessed by "flat functions", functions that admit a point where all its derivatives (up to any order) vanish. So I don't think you can construct a functor $G$ in the opposite direction such that $GF$ is naturally equivalent to the identity functor of $\mathbf{\mathrm{Diff}_*}$.
Aug
21
comment What is $X^{\omega}$ where $X$ is a set?
It should be the set of sequences $(x_n)_{n\in\Bbb N}$ of elements of $X$ indexed by the nonnegative integers.