Is the convex hull always open? Let $M\subseteq\mathbb{R}^m$ and the convex hull is defined by $\text{conv}(M)=\{\sum_{i=1}^{n}\lambda_ix_i:x_i\in M, \lambda_i>0,\sum_{i=1}^{n}\lambda_i=1,n\in \mathbb{N}\}$. 
I am wondering if $\text{conv}(M)$ is always open? Care has to be taken that $\lambda_i\neq0$.
 A: The definition you give is correct, but rather confusingly written.  What it means is the set of values $\sum_{i=1}^n \lambda_i x_i$ where $\lambda_1,\ldots,\lambda_n$ are positive real numbers with sum $1$ and $x_1,\ldots,x_n$ are taken among $M$.  Now in fact we can replace "positive" by "nonnegative" in this definition, because we can always remove those $x_i$ for which the corresponding $\lambda_i$ is zero.  Also, we can indifferently allow several $x_i$ to be equal or require that they all be distinct, because we can always replace $\lambda_i x_i + \lambda_{i'} x_{i'}$ by $(\lambda_i+\lambda_{i'}) x_i$ if $x_i = x_{i'}$.
That being said, the convex hull is generally not open (although it can be open, as when $M = \mathbb{R}^m$).  For example, when $M$ is a finite nonempty set (say $\{x_1,\ldots,x_n\}$), its convex hull is nonempty and compact (because it is the image of the compact standard simplex $\{(\lambda_1,\ldots,\lambda_n) : \lambda_i\geq 0, \sum_{i=1}^n\lambda_i = 1\}$ by a continuous map), so it is never open.
A: Take $m=2$ and two points in $\mathbb{R}^2$. Their convex hull is open line segment: $\left\{\lambda x + (1- \lambda)y \hspace{2 pt} | \hspace{2 pt} 0<\lambda<1 , x,y \in \mathbb{R}^2 \right\}$. But this set is not open w.r.t. $\mathbb{R}^2$.
