Convexity and Strict Preferences

I was looking into convexity while reading Varian's intermediate micro textbook, and an article by Richter and Rubinstein said, "The canonical definition of convex preferences requires that if a is preferred to b, then any convex combination of a and b is also preferred to b." (I interpret "preferred" in the strong sense in this statement, although that might be wrong.) I looked at Kreps' textbook, and he defined it slightly differently: a preference relation is convex if for all $$x$$ and $$y$$, if $$x\succeq y$$ and $$z$$ is a convex combination of $$x$$ and $$y$$, then $$z\succeq y$$. I am trying to prove that Kreps' definition implies Richter and Rubinstein's. (As I think about this, though, I'm not sure whether this is true given my understanding of the latter.)

For the forwards direction, let $$x\succ y$$ and $$z$$ be a convex combination of $$x$$ and $$y$$. Because $$x\succeq y$$ and $$\succeq$$ is convex, $$z\succeq y$$. Now, I need to prove that the statement $$y\succeq z$$ is false, which is where I am stuck. I tried a proof by contradiction by trying to get $$x$$ as a convex combination of $$y$$ and $$z$$, but that didn't get me anywhere.

Let $$X$$ be a convex subset of $$\mathbb{R}^n$$. We define

C1: $$\forall x, y, z\in X$$, if $$x\succeq y$$ and $$z$$ is a convex combination of $$x$$ and $$y$$, then $$z\succeq y$$.

C2: $$\forall x, y, z\in X$$, if $$x\succ y$$ and $$z\neq y$$ is a convex combination of $$x$$ and $$y$$, then $$z\succ y$$.

Proposition. C1 does not imply C2.

Proof. Consider $$X=[0,2)$$, define $$\succeq$$ on $$X$$ as follows $$x\succeq y$$ if and only if $$\lfloor x\rfloor\geq \lfloor y\rfloor$$. In other words, $$\succeq$$ is represented by the utility function $$u:[0,2)\rightarrow\mathbb{R}$$ defined as $$u(x)=\lfloor x\rfloor = \begin{cases} 0 & \text{if } x \in [0,1) \\ 1 & \text{if } x \in [1,2) \end{cases}$$. Clearly, $$\succeq$$ satisfy C1 but not C2. To see that it does not satisfy C2, consider $$x=1.5$$ and $$y=0$$ such that $$x\succ y$$, now $$z=0.5$$ is a convex combination of $$x$$ and $$y$$, but it is not the case that $$z \succ y$$. Therefore, C2 is not satisfied.

• Thank you so much!
– RHyp
Commented Aug 22 at 21:30