<p>Let <span class="math-container">$\succsim$</span> be a lexicographic preference represented with <span class="math-container">$(u_1,u_2)$</span>. <span class="math-container">$x\succsim y$</span> if <span class="math-container">$u_1(x)&gt;u_1(y)$</span> OR <span class="math-container">$u_1(x)=u_1(y)$</span> and <span class="math-container">$u_2(x)\geq u_2(y)$</span>.</p>
<p>Is it obvious that <span class="math-container">$\succsim$</span> must be both complete and transitive? What is a  proof for this?</p>

CC BY-SA 4.0
2024-04-16T12:39:55.023Z
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.
BakerStreet
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2024-04-29T10:36:43.203Z
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
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2024-05-01T21:39:35.820Z
@201p I though if u_1 and u_2 are functions then they are naturally complete and trans?
dodo
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2024-05-01T23:24:08.080Z
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).
Brennan
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2024-05-01T23:25:04.387Z
Do you understand why the "usual" lexicographic ordering on $\mathbb{R}^2_+$ is complete and transitive?
Michael Greinecker
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2024-05-02T11:58:48.320Z
@dodo It is little more than an immediate consequence of the transitivity and *total ordering* axioms for $\mathbb {R}$  (*completeness* for $\mathbb{R}$  has a different meaning as *complete* for preferences), just we have pairs of real numbers and we have to compare them examining the possible various cases.
BakerStreet
CC BY-SA 4.0
2024-05-12T23:14:39.040Z
@MichaelGreinecker I won't say I fully understand but I heard about it.
dodo
<microeconomics><preferences><decision-theory><theory>
Prove that any lexicographic preference $(u_1,u_2)$ must be complete and transitive
dodo
# Prove that any lexicographic preference $(u_1,u_2)$ must be complete and transitive

(Asked by: username_0 on 2024-04-16)

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?

**Tags:** microeconomics, preferences, decision-theory, theory
> **Comments:**
> **username_1**: 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.  
> **username_2**: 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.  
> **username_0**: @201p I though if u_1 and u_2 are functions then they are naturally complete and trans?  
> **username_3**: 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).  
> **username_4**: Do you understand why the "usual" lexicographic ordering on $\mathbb{R}^2_+$ is complete and transitive?  
> **username_1**: @dodo It is little more than an immediate consequence of the transitivity and *total ordering* axioms for $\mathbb {R}$  (*completeness* for $\mathbb{R}$  has a different meaning as *complete* for preferences), just we have pairs of real numbers and we have to compare them examining the possible various cases.  
> **username_0**: @MichaelGreinecker I won't say I fully understand but I heard about it.  

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