# Does a closed set with a non-empty “half-core” necessarily have an interior?

Background:

Suppose $$X$$ is a Banach space. Denote by $$[x, y]$$ the line segment between $$x$$ and $$y$$. Given $$C \subseteq X$$, we define $$\operatorname{core} C = \{c \in C : \forall x \in X, \exists\lambda > 0 : [c, c + \lambda x] \subseteq C \}.$$ In other words, $$c \in \operatorname{core} C$$ if and only if there is a small, non-trivial line segment contained in $$C$$, in any direction from $$c$$.

Clearly any element of the interior of $$C$$ lies in the core, however the converse is not necessarily true. However, there's a nice little folklore result about closed sets with non-empty cores:

Suppose $$C \subseteq X$$ is closed and $$\operatorname{core} C \neq \emptyset$$. Then $$\operatorname{int} C \neq \emptyset$$.

The proof uses Baire Category Theorem. We can assume without loss of generality that $$0 \in \operatorname{core} C$$. By the definition of $$\operatorname{core} C$$, we know that, for every $$x \in X$$, there exists some $$n$$ such that $$x \in nC$$. Thus, $$X = \bigcup_{n=1}^\infty nC.$$ Note that $$nC$$ is closed for all $$n$$, so by the Baire Category Theorem, at least one of the $$nC$$ sets must have a non-empty interior. But, since they are all just scales of each other, they all have non-empty interior. In particular, $$\operatorname{int} C \neq \emptyset$$.

My Question:

I'm wondering if this result can be generalised to something I've decided to call the half-core:

$$\operatorname{half-core} C = \{c \in C : \forall x \in X, \exists \lambda > 0 : [c, c + \lambda x] \subseteq C \text{ or } [c, c - \lambda x] \subseteq C\}.$$

That is, given a line passing through $$c$$, there is a line segment contained in $$C$$ along that line, in one or the other direction. My question is:

If $$\operatorname{half-core} C \neq \emptyset$$, is it true that $$\operatorname{int} C \neq \emptyset$$?

I haven't had many useful thoughts so far, and I cannot seem to decide whether I think it's true or false. My current intuition (which I do not trust very much at all) is that it might be true in separable spaces, but not true in inseparable spaces, but it's difficult for me to articulate why.

Any thoughts are welcome!

If $$C$$ is closed, then $$\text{half-core}(C)\neq\emptyset$$ implies that $$C$$ has a nonempty interior, and a similar proof as in the case you described works. Without loss of generality, if $$0\in\text{half-core}(C)$$, then $$X=\bigcup_{n=-\infty}^{\infty}nC,$$ and then use Baire's theorem.
No. Let C = {(0,0)}, a subset of R$$^2$$.