Set of marginals is convex Let $[n] = \{1,2,\cdot,n\}$ and $[m] = \{1,2,\cdot,m\}$. Let $Z_{1,2}$ denote the set of all probability distribution on the Cartesian product $[m]\times [n]$. Let 
$S_{1,2}$ denote a convex closed subset of $Z_{1,2}$. Let $S_1$ denote the set of probability distributions on the set $[n]$ obtained my marginalizing every probability in the set $S_{1,2}$. Is $S_1$ a convex and closed set.    

Think of this as a linear map from $\phi: Z_{1,2} \subset \mathbb{R}^{mn} \to \mathbb{R}^n$ where $Z_{1,2}$ is your compact set and the map is the marginalizing map. Since $\phi$ is linear, it is continuous and since and $S_{1,2}$ is compact, the image $\phi (S_{1,2})$ is compact (and therefore closed).
On the other hand, since $\phi$ is linear, it preserves convex combinations. Thus the image is convex.   
Note: The marginalizing map is the map that sends the joint probability vector to its marginal on $[n]$.
