Proposition: Suppose $S \subset \mathbb{R}^2$ contains at least $3$ noncollinear points. Suppose further that for every noncollinear $3$ points of $S$, the triangle formed by those $3$ points contains inside of it - and not on it's boundary - another point of $S$.
Is $S$ a dense subset of some convex hull $H \subset \mathbb{R}^2$?
The converse statment:
$S$ is a dense subset of some convex hull $\implies$ every noncollinear $3$ points of $S$ contains inside it's triangle - and not on the boundary - another point of $S$,
is obviously true due to the definition of "dense subset of $\mathbb{R}^2:$ inside every triangle there will be a point of $S.$
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I think the proposition is true: if not, there exists a neighbourhood $N_x(r)$ (circle) around some point $x \in (\mathbb{R}^2 - S)$ inside the convex hull of $S$ such that (at least) $3$ points of $S$ lie on the circumference of this circle, but no member of $S$ lies inside $N_x(r)$, and this would contradict the property in the question. However, I'm not 100% certain this is correct and I'm not sure how to formalise this.
Thanks in advance.
It's false. Take, for example, take two points $u, v \in \Bbb{R}$ with distance $3$ between them, and let $$S = B(u; 1) \cup B(v; 1),$$ i.e. the union of the two open unit balls with centres $u$ and $v$. The closure is just the union of the closed unit balls: $$B[u; 1] \cup B[v; 1],$$ which is clearly not convex.
But, $S$ has the desired property. If we take non-colinear $x, y, z \in S$, then two of the points must lie in the same open ball, which means the line segment between them lies in the convex open subset of $S$, which we can perturb slightly to get something in the interior of the triangle $\operatorname{conv}\{x, y, z\}$.