how many unit balls are needed to cover a unit sphere (1-dense set on a unit sphere) There is an exercise in a geometry textbook to prove that "any $1$-dense set in the unit sphere $S^{n-1}$ has at least $\frac{1}{2}e^{n/8}$ points". It is supposed to be easy.
A set $T$ is $\epsilon$-dense on the sphere if $\left(\forall x\in S^{n-1}\right)\left(\exists y\in T\right)\ d(x, y)\leq\epsilon$.
It will probably use the measure contcentration theorem for the sphere. That is, for your convinience: Let $A\subseteq S^{n-1}$ be a measurable set with $\mu(A)\geq\frac12$, let $A_t$ denote the $t$-neighbourhood of $A_t=\left\{x\in S^{n-1} : d(x,A)\leq t \right\}$. Then $1-\mu(A_t)\leq 2e^{-t^2n/2}$.
The spheric measure $\mu(X)$ for $X\subseteq S^{n-1}$ is defined as the fraction of the volume of the unit ball occupied by the points connecting $X$ to $0$, that is $$\mu(X)=\frac{\lambda\left(\{tx : x\in X, 0\leq t\leq1 \}\right)}{\lambda(B^n)}$$
Thank you for any assistance.
 A: Let's have $p$ the north pole ($p_n=1$, $p_1,\ldots,p_{n-1}=0$) and $A$ the southern hemisphere $A=\{x\in S^{n-1}:x_n\leq0\}$. Then in euclidean metric $d(A_{1/2},p)>1$ thus we know that the pole covers at most the area of $1-\mu(A_{1/2})$. By the MCT we have $1-\mu(A_{1/2})\leq 2e^{-n/8}$ - upper bound on the area covered by one point. It follows that we need at least $\frac{1}{2e^{-n/8}}=\frac 1 2e^{n/8}$ points.
Thanks @how-about-a-nice-big-cup-of for the tip.

Lemma: $d(A_{1/2},p)>1$
Proof: Let's have a point $x\in S^{n-1}$ such that $x\in A_{1/2}$. We will show that $d(x,p)\ > 1$.
W.l.o.g. $x=(x_1,0,\ldots,0,x_n)$ (from the symmetry). And $x$ is on the northern hemisphere ($x_n\geq0$).
From $x\in S^{n-1}$ we have $x_1^2+x_n^2=1$.
From $x\in A_{1/2}$ we know that the distance of $x$ from the equator ($x_n$=0) is at most $\frac 1 2$. The nearest point on the equator is $y=(1,0,\ldots,0)$, then $$\left(\frac 1 2\right)^2 \geq (d(y,x))^2=(x_1-1)^2+x_n^2=(1-x_1)^2+(1-x_1^2)=2-2x_1$$
so we have $x_1\geq \frac 7 8$. And equivalently $x_n=\sqrt{1-x_1^2}\leq\sqrt{1-\left(\frac 7 8\right)^2}=\frac{\sqrt{15}}{8}$.
At last we compute the distance to the pole:
$$(d(x,p))^2=x_1^2+(x_n-1)^2=(1-x_n^2)+(x_n-1)^2=2-2x_n\geq2-2\frac{\sqrt{15}}{8}>1$$ $\square$
