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1d
comment Strong contraction of hyperbolic space
No, I don't think so.
2d
comment Strong contraction of hyperbolic space
I found a more quantitative solution in the book Géométrie et théorie des groupes, les groupes hyperboliques de Gromov.
2d
revised Strong contraction of hyperbolic space
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2d
comment Strong contraction of hyperbolic space
I agree, and there probably exists a proof avoiding ultralimits. But such a proof would be, I think, a laborious manipulation of inequalities. It is why I used ultralimits: thinking of hyperbolic spaces as approximations of real trees seems to be more natural here.
2d
comment Strong contraction of hyperbolic space
$n$ is just the index of the sequence; it corresponds also to a lower bound on the diameter of the projection of $B_n$ on $\gamma_n$, by definition. Then, $Y$ is a real tree because it is a $0$-hyperbolic geodesic space. Of course, you have to choose $y_n$ carefully, but I just gave a hint.
2d
revised Strong contraction of hyperbolic space
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answered Strong contraction of hyperbolic space
Nov
14
accepted Explicit norm on $\mathcal{C}^0(\mathbb{R},\mathbb{R})$
Nov
14
asked Finding free subgroups thanks to Lie algebras
Nov
13
comment Mathematical Science Writers without PhD
Related: mathoverflow.net/questions/44244/… and mathoverflow.net/questions/132413/…
Nov
13
awarded  Nice Question
Nov
11
comment Set of orthogonal matrix over $\Bbb{R}$: Closed, convex, open?
Absolutely, it works.
Nov
11
comment Set of orthogonal matrix over $\Bbb{R}$: Closed, convex, open?
If $O(n)$ were open, there would exist a open neighborhood around $I_n$, so that any sequence converging to $I_n$ would be eventually in $O(n)$.
Nov
11
answered Set of orthogonal matrix over $\Bbb{R}$: Closed, convex, open?
Nov
10
revised Von Dyck's theorem (group theory)
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Nov
10
revised Von Dyck's theorem (group theory)
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Nov
10
revised Von Dyck's theorem (group theory)
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Nov
10
answered Von Dyck's theorem (group theory)
Nov
8
comment How can we proof that $2^{\sqrt 2}$ is irrational?
A proof can be found on Mark Sapir's blog: marksapir.wordpress.com/2013/08/01/…
Nov
7
answered If the normalizer of a subgroup in a group is equal to the subgroup then is the subgroup abelian?