completeness, we present a unified and simplified proof of these characterizations.

[**(characterizations of relative interior for convex sets in $\R^n$).**]{}\[pts2\] Let $\Omega$ be a nonempty convex set in $\R^n$ and let $\ox\in \R^n$. The following properties are equivalent:\
[(a)]{} $\ox\in\ri(\Omega)$.\
[(b)]{} $\ox\in\Omega$ and for every $x\in\Omega$ with $x\ne\ox$ there exists $u\in\Omega$ such that $\ox\in(x,u)$, where $(x,u):=\{tx+(1-t)u\in\R^n\;|\;t\in(0,1)\}$ signifies the open interval between the points $x,u$.\
[(c)]{} $\ox\in\Omega$ and $\cone(\Omega-\ox)$ is a linear subspace of $\R^n$.\
[(d)]{} $\ox\in\Omega$ and $\overline{\cone}(\Omega-\ox)$ is a linear subspace of $\R^n$.\
[(e)]{} $\ox\in\Omega$ and the normal cone $N(\ox;\Omega)$ is a subspace of $\R^n$.

[**Proof.**]{} (a)$\Longrightarrow$ (b): Suppose that $\ox\in\ri(\Omega)$ and fix $x\in\Omega$ with $x\ne\ox$. It follows from that $\ox\in\Omega$ and there exists $\delta>0$ such that $$\label{affc}
\B(\ox;\delta)\cap\aff(\Omega)\subset\Omega.$$ Let us choose $0<t<1$ so small that $u:=\ox+t(\ox-x)\in\B(\ox;\delta)$. Hence $u\in\aff(\Omega)$, and we get from that $u\in\Omega$. This tells us that $$\ox=\frac{t}{1+t}x+\frac{1}{1+t}u\in(x,u),$$ which therefore verifies the conclusions in (b).

(b)$\Longrightarrow$ (c): It suffices to show that for every $a\in K:=\cone(\Omega-\ox)$ we get $-a\in K$. Fix any $a\in K$ and find by definition such $t\ge 0$ and $w\in\Omega$ that $a=t(w-\ox)$. If $w=a$, we have $a=0$ and hence $-a=0\in K$. In the case where $w\ne\ox$, take $u\in\Omega$ with $\ox\in(w,u)$ and then find $\gamma>0$ 