# Prove that any lexicographic preference $(u_1,u_2)$ must be complete and transitive

Let $$\succsim$$ be a lexicographic preference represented with $$(u_1,u_2)$$. $$x\succsim y$$ if $$u_1(x)>u_1(y)$$ OR $$u_1(x)=u_1(y)$$ and $$u_2(x)\geq u_2(y)$$.

Is it obvious that $$\succsim$$ must be both complete and transitive? What is a proof for this?

• I think that the question is: where and to whom? In a lesson to beginning students it can be appropriate, or it could be appropriate to say 'Verify it for exercise, ('Prove' maybe is too high-sounding), in other contexts, as in a book of microeconomics or in a paper or speaking to a more advanced audience it is inappropriate, at times even ridiculous. If you mean ‘we’ as ‘human kind’, yes, we need, but it is not worth spending but a little time about it. Books of mathematics are full of expressions as ‘Trivial’, ‘It can be easily shown that…, or ‘Properties 1 and 2 are obvious…’, and so on. Commented Apr 16 at 12:39
• It is worth noting also that this is only true if $u_1$ and $u_2$ are themselves complete any transitive. In other words, that the co-domain of the $u$'s is a total order.
– 201p
Commented Apr 29 at 10:36
• @201p I though if u_1 and u_2 are functions then they are naturally complete and trans?
– dodo
Commented May 1 at 21:39
• Yes. If you use $u(x) > u(y)$ to define your preference relation then it is transitive. (e.g. 3 > 2 > 1 => 3 > 1). Completeness here follows from the fact that any two alternatives have technically been `ranked' in the sense that the functions are such that I can compare any two values of $x$ and $y$ (e.g. $u(5,000,000)$ with $u(1)$). This argument readily extends to comparing two different utility functions, which is somewhat what lexicographic preferences capture. (I think of it like, I compare them based on one quality and if it doesn't satisfy that, I look at the second). Commented May 1 at 23:24
• Do you understand why the "usual" lexicographic ordering on $\mathbb{R}^2_+$ is complete and transitive? Commented May 1 at 23:25