Cone generated by a $w^*-$compact set is $w^*-$closed Let $X$ be normed and $X^*$ it's dual. Consider a $w^*-$compact set $G^*\subseteq X^*$ such that $0^*\notin G^*.$ I may have proved that $$cone (G^*)=\{t x^*:t\geq 0,\; x^*\in G^*\}$$ is $w^*-$closed. The motivation came from the finite dimensional setting, from which the result is well known. Does this version in infinite dimensions already exists? If it does, I would appreciate any reference. Thanks in advance
 A: This is true in a Hausdorff topological vector space: Let $G$ be compact such that $0\ne G$. Then 
$$
cone(G) =\{ tx: t\ge0, x\in G\}
$$
is closed.
Let $x\not\in cone(G)$. Then for each $\lambda g \in cone(G)$ with $\lambda>0$ and $g\in G$ there are open sets $U_{\lambda,g}$ and $V_{\lambda,g}$ such that $x\in U_{\lambda,g}$, $\lambda g\in V_{\lambda,g}$, and $U_{\lambda,g} \cap V_{\lambda,g}=\emptyset$.
In addition there are open sets $U_0$, $V_0$ such that $x\in U_0$, $0\in V_0$, and $U_0\cap V_0=\emptyset$.
The open sets $\lambda^{-1}V_{\lambda,g}$, $\lambda>0$, $g\in G$, cover $G$. Hence we can take a finite selection still covering $G$, $G\subset\bigcup_{i=1}^n \lambda_i^{-1}V_{\lambda_i,g_i}$. This implies
$$
cone(G)\subset V_0 \cup \bigcup_{i=1}^n cone^+(V_{\lambda_i,g_i}),
$$
where $cone^+(B)=\{tb: t>0,b\in B\}$. Then 
$$
U:=U_0 \cap \bigcap_{i=1}^n cone^+(U_{\lambda,g})
$$
is an open neighborhood of $x$ that is disjoint with $cone(G)$. Hence, $x\not\in cl\ cone(G)$.
And $cone(G)$ is closed.
A: Let $\{t_{i} x_i\} \subseteq cone(G) $ be a convergent net, say  $t_{i} x_i \to y.$ with $t_i \in [0, +\infty)$, and $ x_i \in G $. WLOG we may assume $x_i \to x\in  G \setminus\{0\}$ this shows $\{ t_{i} \}$ has to be a bounded net, so Again WLOG assume $t_{i} \to t \in [0, + \infty )$ that means $y = tx \in cone(G) .$   DONE
