For every pair of points $x,y\in S\subset\mathbb{R}^2,\exists z\in S$ s.t. $z$ lies on the straight line formed by $x,y.$ Can $S$ be finite? Actual question: If $S\subset\mathbb{R}^2$ contains at least $3$ noncollinear points and for every pair of points $x_1,x_2$ of $S$, there exists a third point $x_3$ which lies on the (extended) straight line formed by $x_1$ and $x_2$, then can $S$ be finite?
I have more questions on this topic (in particular, what can we say about $S$ if $S$ is bounded - e.g. does $S$ have to be uncountable, or connected, or open...?), but it's one question per question, so leave it as this for now.
My attempt so far (I suspect the answer is no): Suppose, by way of contradiction that the answer is yes. Then $S$ is finite and is therefore bounded. Consider the convex hull, $H$, of $S.$ The boundary of $H$ has $n\geq3$ vertices. On each straight line joining adjacent vertices of the boundary of $H$, there is at least one other point of $S$. Now what?
 A: No such finite set exists. The Sylvester-Gallai theorem asserts that for any finite set of points $S$ in the plane, there is either a line passing through exactly two of them, or all points are collinear. Since we want $S$ to have at least 3 non-collinear points, if $S$ is finite then there is a line passing through exactly two points, i.e. there are two points $P,Q \in S$ for which there isn't another point in the set lying on the line $PQ$.
Now it's not necessary for a bounded set $S$ with the presented property to be uncountable. Let $a,b,c$ be the complex numbers corresponding to the vertices of a triangle in the plane, and let
$$S = \{\alpha a + \beta b + \gamma c \mid \alpha + \beta + \gamma = 1, \alpha , \beta, \gamma \in \mathbb{Q}_{\geq 0} \}$$
This set is clearly bounded and countable, since the number of triples of $(\alpha, \beta,\gamma)$ is smaller than $\mathbb{Q}^3,$ which is countable. To see that $S$ has the required property, let $x, y \in S$ be two points in $S$, so that $x = \alpha_1 a + \beta_1 b + \gamma_1 c$, and $y = \alpha_2 a + \beta_2 b + \gamma_2 c$. Then the midpoint of the segment connecting these two points, aka the point
$$\frac{\alpha_1 + \alpha_2}{2} a + \frac{\beta_1 + \beta_2}{2} b +  \frac{\gamma_1 + \gamma_2}{2} c $$
is also clearly in $S$, so $S$ has the required property.
Note that this set is dense in the region bounded by the triangle; After some scribbling it seems to me rather clear (although I haven't showed it rigorously) that any bounded such $S$ has at least one limit point (as opposed to, say, an infinite square lattice, which also satisfies the conditions but does not have any limit points).
