Is the set of doubly stochastic matrices of positive measure in all symmetric matrices Let $n\geq 2$ and  $a=(a_{ij})$ satsifies  
$a_{ij}\geq 0$ for all $i,j\in \{1,...n\}$,  and 
$\sum_{j=1}^n a_{ij}=1$ for all $i=1,...,n$, and
$\sum_{i=1}^n a_{ij}=1$ for all $j=1,...,n$.
Let $A$ be the set of all such $a$, or the set of  all doubly stochastic matrices. 
Is $A$ an open set and have positive  measure in $B=\{a\in \mathbb R^{n\times n}: a_{ij}=a_{ji}, \forall i,j=1,...,n\}$?
Thanks.
 A: We will assimilate the set of $n \times n$ matrices with $\mathbb R^{n \times n}$.
Let us call SDS the set of symmetrical doubly stochastic matrices.
SDS is defined as a finite intersection of closed sets:


*

*(1) the $2n$ equations $\sum_{i}a_{ij}=1$, and $\sum_{j}a_{ij}=1$, 

*(2) the $n^2$ inequations $a_{ij}\geq 0$,

*(3) the $n(n-1)/2$ equations $a_{ij}-a_{ji}=0$.
They are closed because all of them are reciprocal images of closed sets either $\{1\}$, $[0,+\infty)$or $0$ by continuous functions.
Besides, due to (3), space SDS is included in a subspace of $\mathbb R^{n \times n}$ with dimension $n^2 - n(n-1)/2$. This subspace has global measure $O$ (Think to $R^2$ whose measure is $0$ in $\mathbb R^3$). Thus SDS has measure $0$. 
This would be different if we consider SDS in the ambient space of symmetrical matrices due to the fact that the set of doubly stochastic matrices is the convex hull of permutation matrices (Birkhoff-Von Neumann theorem see the interesting reference (https://www.duo.uio.no/bitstream/handle/10852/37698/MariaMehlumThesis.pdf?)
A: And no, it has Lebesgue measure zero.  Any subset of $\mathbb R^{n \times n}$ contained in a hyperplane has measure zero.
