Does a "full enough" convex set have an internal point?

Question

Let $C \subseteq V$ be a convex subset of a vector space (not necessarily finite dimensional). Suppose that for every $v \in V$, there are some $a ∈ \mathbb{R}^+$ and $x, y ∈ C$ such that $v = a(x - y)$. Intuitively, $C$ has some positive "thickness" in every direction, somewhere. Does it follow that $C$ has an internal point?

Here's a bit more detail. For a vector $v ∈ V$, call a point $x ∈ C$ $v$-internal iff for some $α > 0$, for every $0 < β ≤ α$, $x + βv$ is in $C$. Given that $C$ is convex, this is equivalent to just saying that $x + αv ∈ C$ for some $α > 0$. A point $x ∈ C$ is internal iff it is $v$-internal for every $v ∈ V$.

My question amounts to this: given that, for every $v ∈ V$, there is some $v$-internal point $x ∈ C$, does it follow that there is a point $x ∈ C$ which is $v$-internal for every $v ∈ V$?

In the finite dimensional case this checks out. Given a basis $e₁, …, e_n$, and some points $x₁, …, x_n, y₁, …, y_n ∈ C$ which are $e₁$-internal, …, $e_n$-internal, $-e₁$-internal, …, and $-e_n$-internal, respectively, the mean of all of these points must be an internal point.

But I don't know how things would go in an infinite dimensional vector space.

Answer 1

Sorry, misread your question. Here is an example at least showing that your example does not generalize:

Let $V=\ell_2(\mathbb{R})$ and let $C = \ell_2^+$. Then for every standard unit vector $e_r$, there exists a point in $C$ such that it is $\pm e_r$ internal, namely $\tfrac{1}{2}e_r$. However, the core of $C$ is empty. (In fact, even the quasi relative interior is empty.)