# Continuity of preferences

Let $$\succsim$$ be a transitive and reflexive relation on a metric space $$X$$ with closed upper and lower contour sets. If $$\succsim$$ is not complete, does it hold that: for all converging sequences with $$x_n\succsim y_n$$ for each $$n\geq 1$$ and $$x_n\to x$$, $$y_n\to y$$, we have $$x\succsim y$$? I think that without the completeness of $$\succsim$$ this cannot hold, but I failed to provide an example.

• If $X$ is not complete that means there are $x$ and $y$ such that neither $x \succeq y$ nor $x \preceq y$ hold. You can easily fabricate an example where $x_n \to x$ and $y_n \to y$ while for all $n$ you have $x_n \preceq y_n$. Then all you have to show is that the contour sets are indeed closed, right? Mar 24 '20 at 11:56
• Yes, that's what I am struggling to show. I can't find a relation with closed contour sets, which is not "continuous" in terms of sequences. Mar 24 '20 at 12:42
• A result by Schmeidler (1971) tells us if $X$ is a connected topological space with a transitive, continuous (i.e. upper contour sets are closed and strict upper contour sets are open) and non-trivial (i.e. not every element is indifferent to one another), it is a complete preference. You must break one of these requirements for a counterexample. (e.g. using the discrete topology as per the answer). Mar 24 '20 at 23:40

Giskard is right lexicographic preferences would make the job being dicsontinuous, but the problem is to find relations with closed contour sets. Maybe an idea would be to start from constructing a preference relation for which the only possible upper and lower contour sets are $$X$$ and $$\emptyset$$. In that case, you take the indiscrete topology, and all nets (or generalized sequences) converge to all the points in $$X$$. Anyway, I have not been able to construct such a relation yet.