A relation $\succeq $ over a vector space $X$ is rational if it is transitive and complete.
We say $x\succ y$ iff $x\succeq y$ and NOT $y \succeq x$
Moreover $x\sim y$ iff $x\succeq y$ and $y \succeq x$
We say it is convex iff $\ \ \ \forall x,y,a \in X, \ \ \ \ \ $
$ x \succeq a, y \succeq a \Rightarrow tx+(1-t)y \succeq a \ \ \forall t\in (0,1) $
Similarly it is strictly convex iff $\ \ \ \forall x,y,a \in X, \ \ \ \ $
$ x \succeq a, y \succeq a \Rightarrow tx+(1-t)y \succ a \ \ \ \forall t\in (0,1) $
An indifference set is $\{x\in X : x\sim a \}$ for some $a$
Does there exists a rational preference on the positive quadrant of $\mathbb{R}^2$ which is convex but NOT strictly convex and such that every indifference set is a singleton?
Lexicographic preferences are rational and satisfy the singleton indifference set requirement but they are strictly convex...
UPDATE: I am quite certain this should not exist, simply because strict convexity and convexity seem to me to be equivalent in the case of singleton indifference sets, it is very hard to me to formalize this though
Thanks.
I think it is possible by exploiting the boundary of the set, but I hope you will double check me because I could be mistaken. Start with any lexicographic ordering $\succeq$.
Let's consider the set of points $A_0=\{(0,x):x\in\mathbb{R}\}$. These points are strictly ordered according to the $x$ value (assume increasing). Then modify the ordering to where $(0,0)\succ (0,1)$. Now, using your notation, let $a=(0,1)$. This new ordering fails strict convexity with $x=(0,0), y=(0,2)$ and $t=0.5$. But it is still transitive and complete with singleton indifference sets. We just let $(0,1)$ and $(0,0)$ trade identities so to speak.
Alternatively, points in $A_0$ could be ordered according to their distance to $(0,1)$ with an additional lexicographic tie breaking criterion $(0,1+c)\succ(0,1-c)$. This is convex so long as closer is better.