Let's say you have a hyper-cube of $d$-dimensions $A$. How can you prove that there is no $d$-hyper-cube $B$, that encloses all the vertices of $A$ except for one.
I was thinking of a proof by contradiction but it seems incomplete/not rigurous.
WLOG, we can set our hyper-cube $A$ to be the hyper-cube made by permuting the $2^d$ coordinates $( \pm 1, \pm1,...)$.
Since we do not want to enclose ONLY one vertex, choose a random vertex $V_{out}$ which we don't want covered by $B$ and is located in some arbitrary $d-1$-face $F$.
Since we don't want just one vertex, we want all the other vertices inside $B$. Therefore, all of the vertices on the opposite $d-1$-face $F_{op}$ (or is it called $d-1$-cell?), should be inside $B$.
Then, for $B$ to cover those vertices on $F$, since it is the opposite $d-1$-face, it has to grow the distance from $F$ to $F_{op}$. Since all vertices on $F$ are separated by the same distance to those in $F_{op}$, $B$ has to grow enough to cover all of them.
Therefore, we must cover all of the points in $F$, so singling one out is impossible.
I'm trying to do the same for excluding $2^{d-1} - d$ vertices.