Convexity and Strict Preferences

Question

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.

Answer 1

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.