Proof Hypercube 1

Theorem: Let $v\in \mathbb {S}$.

$N(v)\subseteq \mathbb {S} \Rightarrow vol(P_{v})=1/2^{n}$

Proof: Let $e_{i}=0\ldots 1\ldots 0$ where 1 appears at ith position.

$N(v)=\{v\oplus e_{i}\}_{i=1}^{n}$

There will be some technicalities in the proof. One way to get rid of them is to consider $v=0^{n}$.

Consider a hyperplane $H_{i}$ halving the middle point between v and $v\oplus e_{i}$ The hyperplane is defined by $(x-(v+v\oplus e_{i})/2))\cdot e_{i}=0$. Working out the calculation,

$x_{i}-1/2=0$

So, P_v is defined by the intersection of halfspaces $x_{i}\leq 1/2$ for $i=1,\ldots ,n$. It is obvious that the volume of this intersection is $1/2^{n}$ (If you don't believe, you can do Reimann integration of this set).

Since the volume can't be any smaller, we conclude that, in this case, $Vol(P_{v})=1/2^{n}$