A problem about convex domains Let $\Omega\subseteq\mathbb{R}^{n}$ be an open convex domain and let $A$
be a compact subset in $\Omega$. Prove that, there is $c>0$ such that 
$$
x+\left(1-c\left|x-y\right|\right)\left(z-y\right)\in\Omega,\quad\forall z\in\Omega,\forall x,y\in A,   \left|x-y\right|<\dfrac{1}{c}.
$$
 A: I think that the statement is incorrect.
Example: $\Omega = (0,4)$, $A = [\epsilon, 2]$, with $\epsilon >0$ and $\epsilon \ll 1$.
Let $x = \epsilon$, $y = 2\epsilon$. Then the constraint implies that $\forall z \in \Omega$:
$$\epsilon + (1 - c \epsilon)(z-2\epsilon) > 0$$
Taking $z \to 0$, we must have
$$\epsilon + (1-c\epsilon)(-2\epsilon) \geq 0 \Leftrightarrow \epsilon(2c\epsilon -1) \geq 0 \Rightarrow c \geq \frac{1}{2\epsilon}.$$
Now take $x=1$, $y =2$, $z=3$. Then 
$$x + (1 - c|x-y|)(z-y) = 1 + (1 - c) = 2 - c \leq 2 - \frac{1}{2\epsilon},$$
which is negative for $\epsilon$ small enough.
Edit 1: When I wrote this, the OP had forgotten to mention that the condition must be satisfied only for $|x-y|\leq 1/c$.
Edit 2: Here is a proof of the weaker result.
Rearranging, we have
$$x + (1 - c|x-y|)(z-y) = (1 - c|x-y|) z + c|x-y|\,\underbrace{\left(\frac{1}{c|x-y|}\, x + \left(1 - \frac{1}{c|x-y|}\right)y\right)}_{:=w}.$$
Since $\Omega$ is convex, it suffices to show that there exists $c$ such that $w \in \Omega$ for all $x,y \in A$. For this, let us compute
$$|w-x| = \left(\frac{1}{c|x-y|} - 1 \right)|x-y| = \frac{1}{c}(1 - c|x-y|) \leq \frac{1}{c}.$$
Since $A$ is compact the distance from $A$ to $\mathbb R^n \backslash \Omega$ is strictly positive; let us call it $d(A,\mathbb R^n \backslash \Omega)$. By choosing $c > 1/d(A,\mathbb R^n \backslash \Omega)$, $|w - x| \leq d(A,\mathbb R^n \backslash \Omega)$, so $w\in \Omega$ and the result is proved.
