Let $S^{d-1} \subseteq \mathbb{R}^d$ denote the $d$-dimensional sphere. For a point $x \in S^{d-1}$, let $A_x = \{y \in S^{d-1}: (x,y) \geq p \}$, where $(x,y)$ is the euclidean inner product. For my application it is enough to assume that, say, $p=3/4$. Question: What is the minimal $m=m(d)$ such that there are $m$ points $x_1,...,x_m$ for which $\bigcup_{i=1}^{m}{A_{x_i}} = S^{d-1}$?
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It growth exponentially in $d$, is this enough information or you need more precise bounds? – Fedor Petrov Feb 28 '16 at 14:14
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I need an upper bound; is it known to be exponential? – Slava Feb 28 '16 at 14:15
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Of course it is. Your set is a ball of some radius $d$ on the sphere (considered as a metric space). Choose maximal number $N$ of points with all mutual distances at least $d$. Then balls centered in these points cover the sphere (else $N$ is not maximal), while balls of radius $d/2$ are disjoint, thus $N$ does not exceed area of a sphere divided by area of $d/2$-ball. This growth exponentially in $d$. – Fedor Petrov Feb 28 '16 at 14:20
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(I think all d but the last one should be rotated of 180 degrees to get p ) – Pietro Majer Feb 28 '16 at 16:11