The star of a simplex $\sigma$ is defined to be the union of the interiors of the simplices that have $\sigma$ as a face. I need to show that the star of $\sigma$ is the intersection of all the star of its vertices, that is
$st(\sigma)=\bigcap\limits_{i=0}^{n}st(v_{i})$
Below are some of my ideas. I intend to prove that $LHS\subseteq RHS$ and $RHS\subseteq LHS$.
Since $st(\sigma)\subseteq st(v_{i})$ $\forall i=0,1,...,n$ (Is this obvious??? Do I have to prove this?)
$\therefore st(\sigma)\subseteq\bigcap\limits_{i=0}^{n}st(v_{i})$
Now suppose that $x$ is in $\bigcap\limits_{i=0}^{n}st(v_{i})$, then $x=t_{0}v_{0}+t_{1}v_{1}+...+t_{n}v_{n}+...$ where $t_{i}>0$ for all $i=0,1,...n$. Thus $x$ must be in one of the interior of the simplex which contains $\sigma$ as a face and this implies that $x$ is in $st(\sigma)$.
$\therefore st(\sigma)\supseteq\bigcap\limits_{i=0}^{n}st(v_{i})$
Hence, $st(\sigma)=\bigcap\limits_{i=0}^{n}st(v_{i})$.
Is my argument above clear? Any suggestion?