I’ve a set of $d$-dimensional vectors $V = {+1, 0, -1}^d $. Then $P(V)$ constitutes the ability set of $V$. I now assemble a set of unit vectors $V_{sum}$ from the ability set $P(V)$ such that

$$

V_{sum} = left{frac{bar{v}}{|bar{v}|} quad Bigg| quad bar{v} = sum_{v in S} v, quad forall S subseteq P(V)proper}

$$

That’s, every subset $S subseteq P(V)$ contributes to a vector in $V_{sum}$ fashioned as a sum of all of the vectors within the subset $S$ after which taking the unit vector in that path.

Be aware that there may very well be duplicates. For instance, for $d = 3$, the vector $(frac{1}{sqrt{3}},frac{1}{sqrt{3}},frac{1}{sqrt{3}})$ might be fashioned as a sum of vectors of any of the next subsets $$S_1 = {(1,0,0),(0,1,0),(0,0,1)}, S_2 = {(1,1,0

),(1,0,1),(0,1,1)}, S_3 = {(1,1,1)}.$$

and plenty of extra potentialities.

Now I wish to discover the utmost of Euclidean distance between any vector in $V_{sum}$ to its closest vector in $V_{sum}$. Is there a simple option to higher sure this max distance?

In different phrases, if I think about $V_{sum}$ to be an $epsilon$-set to the floor of the unit ball in $d$-dimensions, then I wish to discover an higher sure on $epsilon$. Any weak higher sure on $epsilon$ ought to suffice. The objective is to indicate that $V_{sum}$ types a greater $epsilon$-net than the unit vectors fashioned from the vectors in $V$.