Showing a set has nonempty interior Let $A$ be a finite set.  Let $M : A \times A \to {\bf R}$ be a symmetric function, which is positive-semidefinite when regarded as an $A \times A$ matrix.
Let $P(A)$ be the set of vectors $p = (p_a)_{a \in A} \in {\bf R}^A$ such that for all $a \in A$, $p_a \ge 0$, and $\sum_{a \in A} p_a = 1$.
Then let $S(p) = \sum_{(a,b) \in A \times A} M(a,b) p_a p_b$, and $U_t = \{p \in P(a) : S(p) \leq r\}$.
I want to prove that if $U_t$ is nonempty, it also has nonempty interior under the norm $||p - q || = \sum_{a \in A} |p_a - q_a|$.
I have used Cauchy-Schwarz to show that the $U_t$ are convex.

This is not true.
If $S^{-1}((-\infty, r])$ is non-empty, then it contains each of the open sets $S^{-1}((-\infty, s])$ for $0\leq s<r$ and is thus of non-empty interior unless all of those sets are empty, in other words when $r$ is the global minimum of $S$ on $P(A)$.
But the global minimum of $S$ might well be $r>0$. For example if $M$ is positive definite, since $S(p)=p^TMp$, the minimum of $S$ on the compact set $P(A)$ must be attained, but cannot be zero.
