"algebraic openness" of the epigraph of a sublinear function Let $X$ be a real linear space, $p:X \rightarrow \mathbb{R}$ a sub-linear function, i.e:
$$\forall x,y\in X\;p(x)+p(y) \geq p(x+y)$$
$$\forall \lambda \geq 0,x \in X \; p(\lambda x)=\lambda p(x)$$
Consider the strict upper epigraph:
$$\Gamma^+(p)=\{(x,y) \,\vert\, x \in X,y > f(x)\} \subseteq X\times\mathbb{R}$$
One sees that $p$ is convex and thus $\Gamma^+(p)$ is convex.
My question is about it's "algebraic openness", formally define that $v \in \Gamma^+(p)$ is an internal point  if for all $u \in X\times \mathbb{R}$ there exists $\delta_0 >0$ so that for all $\vert \delta \vert \leq \delta_0$ we have $v+\delta u \in \Gamma^+(p)$. My Question is whether it is true that every point in $\Gamma^+(p)$ is an internal point. For those who are curious this is for an application of the Geometric version of the Hahn Banach Theorem.
Any help is appreciated. Thanks
Note: an affirmative answer is tempting since for $X=\mathbb{R^n}$, $p$ is continues and thus it's strict epigraph is (topologically and thus "algebraically") open
 A: Turns out this questions isn't so hard, I solved it after having another go at it.
As I mentioned the topology here is a starting pont. But since we are only talking internal points, we can use the topology on one dimension at a time.
Let $(x,y)\in \Gamma^+(p),\; y>p(x)$. Let $(w,r) \in X \times \mathbb{R}$ be some vector. Note that $p(x+tw)$ is a continues function of $t$ at $0$ since:
$$p(x+tw) \leq p(x) + p(tw) = p(x) + \vert t \vert p(sign(t)x) \underset{t \rightarrow 0}{\rightarrow} p(x)$$
Similarly,
$$p(x+tw) \geq p(x) - p(tw) = p(x) - \vert t \vert p(sign(t)x) \underset{t \rightarrow 0}{\rightarrow} p(x)$$
and so $p(x+tw) \underset{t \rightarrow 0}{\rightarrow} p(x)$. Since $y > p(x)$ we can 
take $\delta_0$ small enough so that for all $\vert \delta \vert < \delta_0$ we have $p(x+\delta w) < y$. Now let's check the definition of $(x,y)$ being internal: we want:
$$p(x+ \delta w) \underset{?}{<} y + \delta r$$
But for $\delta_0$ even smaller we can guarantee it, since $p(x+\delta w) < y$ and $r$ is constant. $\blacksquare$
