Interior points in a convex set can be represented as convex combination of different points from the set

Question

Can we assume that any interior point $z$ in a convex set $S\subseteq R^n $ be represented by $2$ points $x \in S$ and $y \in S$ such that $z = \lambda x +(1-\lambda)y $, where $x\neq y \neq z$ , and $\lambda \in (0,1)$?

What I understand is that every point $\in S$ can be represented as a convex combination of 2 points in $S$, and that any exterior point (say $v$) can't be constructed by 2 different points $x$ and $y$ ($x \neq y \neq v$) such that $v = \lambda x + (1-\lambda)y$ where $\lambda \in (0,1)$.

Intuitively, I think for any interior point it should be true unless the set only has 1 element.

How should I go about proving this seemingly trivial-looking result (if it is correct) or can I simply state that?

Answer 1

As you already figured out, $B(z,\epsilon) \subset S$ for some $\epsilon > 0$. Then $$ z \pm r v \in S $$ for $0 < r < \epsilon$ and an arbitrary unit vector $v$, and $$ z = \frac 12 (z-rv) + \frac 12 (z+rv) \, . $$

With respect to

... it should be true unless the set only has 1 element.

note that a set with only one element does not have interior points.

Answer 2

Pick any $u \neq z$. Let $x=z+\epsilon (u-z)$ and $y=z+\epsilon (z-u)$. Then $z=\frac 1 2 x+\frac 1 2 y$. Can you check that $x,y \in S$ if $\epsilon$ is sufficiently small? This is where you use the fact that $z$ is an interior point. [$x \to z$ and $y \to z$ as $\epsilon \to 0$. To be explicit it is enough to take $0<\epsilon <\frac r {\|u-z\|}$ where $r$ is such that $B(z,r) \subset S$]. Geometrically, you just draw a line in some abitrary direction passing through $z$ and take two points on the line close to $z$. Then the chosen points belong to $S$ and $z$ is a convex combination of those points.