This is a follow-up question from [my previous question](https://mathoverflow.net/questions/439653/maximum-number-of-vectors-with-bounds-on-inner-products/440643?noredirect=1#comment1136719_440643).


Suppose there are (2n+1) vectors $\{m\_1,m\_2,...,m\_n\}$, $\{p\_1,p\_2,...,p\_n\}$ and $p^\*$ in $R^{k+1}$. $m\_i$ are weakly positive vectors. $p\_i$ and $p^\*$ are probability vectors (whose coordinates are weakly positive and sum to 1).


The vectors satisfy the following properties:
$$ m\_i p\_j \geq m\_i p^\* = 1 \quad \forall i\neq j,$$
$$ m\_i p\_i = 0 \quad \forall i.$$


The question is: what is the maximum number of $n$?



Maximum number of vectors with bounds on inner products (follow up question)
This is not an answer to the question, but here are some upper/lower bounds. Firstly, if we let $A\_i\subseteq\{0,1,\dots,k+1\}$ be the non zero coordinates of $m\_i$, then we can't have $A\_i\subseteq A\_j$ for $i\neq j$, because then we would have $m\_ip\_j=0$ (you already mentioned this in the other question I think). This means that $(A\_i)\_i$ are all different and form an antichain of $\{1,\dots,k+1\}$, thus $n\leq\binom{k+1}{\left\lfloor\frac{k+1}{2}\right\rfloor}$ by [Sperner's theorem](https://en.wikipedia.org/wiki/Sperner%27s_theorem).


For a superpolynomial lower bound, suppose that $\sqrt{k+1}$ is an even integer for simplicity. Let $p^\*=(\frac{1}{k+1},\dots,\frac{1}{k+1})$, and for each $A\subseteq\{1,\dots,\sqrt{k+1}\}$ of size $\frac{\sqrt{k+1}}{2}$, let $p\_A$ have coordinates $(p\_i)\_j=\frac{2}{\sqrt{k+1}}$ when $j\in A$ and $0$ if not, and let $m\_A$ have coordinates $2\sqrt{k+1}$ in $\{1,\dots,\sqrt{k+1}\}\setminus A$ and $0$ else.


The number of vectors $p\_A$, $m\_A$ is then $\binom{\sqrt{k+1}}{\frac{\sqrt{k+1}}{2}}$, which grows faster than polynomially.


