for which $w=\ox+\gamma(\ox-u)$. It follows therefore that $$-a=t(\ox-w)=-t\gamma(\ox-u)=t\gamma(u-\ox)\in K,$$ which implies that $K$ is a linear subspace of $\R^n$.

(c)$\Longrightarrow$ (d): This implication is obvious since every linear subspace is closed in $\R^n$.

(d)$\Longrightarrow$ (e): Fix any $v\in N(\ox;\Omega)$ and show that $-v\in N(\ox;\Omega)$. We have from that $$\la v,x-\ox\ra\le 0\;\mbox{\rm for all }\;x\in\Omega.$$ Using this and denoting $K:=\overline{\cone}(\Omega-\ox)$ tells us that $\la v,z\ra\le 0$ for all $z\in K$. Since $K$ is a subspace, for any $x\in \Omega$ we get that $\ox-x\in K$, and hence $$\ox-x=\lim_{k\to\infty}t_k(w_k-\ox),$$ where $t_k\ge 0$ and $w_k\in\Omega$ for every $k\in\N$. It follows therefore that $$\la-v,x-\ox\ra =\lim_{k\to\infty}t_k\la v,w_k-\ox\ra\le 0,$$ which yields $-v\in N(\ox;\Omega)$. Thus $N(\ox;\Omega)$ is a linear subspace in $\R^n$.

(e)$\Longrightarrow$ (a): Assuming that (e) is satisfied and arguing by contradiction, suppose that $\ox\notin\ri(\Omega)$. It follows from Theorem \[pst1\] that there exists $v\in\R^n$ such that $$\label{cd1}
\la v,x\ra\le\la v,\ox\ra\;\mbox{\rm for all }\;x\in\Omega,$$ and furthermore there exists $\tilde{x}\in\Omega$ for which $\la v,\tilde{x}\ra<\la v,\ox\ra$. Then implies that $v\in N(\ox;\Omega)$, and so $-v\in N(\ox;\Omega)$. This yields $$\la-v,\tilde{x}-\ox\ra=\la v,\ox\ra-\la v,\tilde{x}\ra\le 0,$$ which gives us a contradiction, which shows that (a) holds. $\h$

The obtained finite-dimensional characterizations of relative interior motivate the major extensions of this notion to infinite dimensions, which 