# Continuous rational and monotone preference relation implies $x\succsim0$?

I updated my proof to a general version as follows: please share your thoughts & 2cent. Thanks

Show a monotone continuous complete preorder on $\mathbb{R^L_+}$ has $y\geq x\rightarrow y\succsim x$.

Point of Clarification
$X=\mathbb{R^L_+}$
Preorder means the usual reflexivity and transitivity.
Complete means for any $x,y\in X$, have $x\succsim y$ or $y\succsim x$
Continuous means the relation is preserved under limits.
Monotone means for any $x,y\in X$, if $y\gg x$, then $y\succ x$.
$\succ$ and $\sim$ are respectively asymmetic and symmetric parts of $\succsim$

Outline of Proof
Go through two cases: (1) $y\gg x$. Easily get the result by definition. (2) Some components are equal while else y is strictly greater x. Use continuity where you add a sequence of small positive $\epsilon$ to y, making it a sequence $y^n_\epsilon$ where for every n, $y^n_\epsilon\gg x$ $\forall n$.

Proof
Suppose $\succsim$ is a monotone, continuous, complete preorder on $X=\mathbb{R^L_L}$.
Case (1) $y\gg x$ (i.e. $y_i>x_i$ $\forall i\in B=\{1,\dots,L\}$).
By definition, $y\succ x$, which implies $y\succsim x$.

Case (2) $y_j=x_j$ for some $j\in B$. For $\forall k\not=j,k\in B, y_k>x_k.$
For some $\epsilon>0$, let the sequence $\epsilon^n\in\mathbb{R^L_+}$ such that $\epsilon_j=\epsilon$, $\epsilon_k=0$.
Denote the sequence $y^n_\epsilon=y+\epsilon^n$.
Then, for any $\epsilon>0$ and $\forall n$, $y^n_\epsilon\succ x$, hence $y^n_\epsilon\succsim x$.
By continuity of $\succsim$, $$\lim_{\epsilon \to 0} {y^n_\epsilon} = y$$
Hence, $y\succsim x$. $\blacksquare$

OLDER VERSION

My question is what is the valid reasoning behind that continuous rational and monotone preference relation implies $x\succsim0$. I have put a proof below and would appreciate if you share your 2 cent on the validity/rigor of the proof. Thanks!

Suppose $x\in\mathbb{R^+_L}=\{x\in\mathbb{R^L}:x_l\geq0$ $\forall l=1,\dots,L\}$.

Claim: For every $x\in\mathbb{R^+_L}$, monotonicity implies $x\succsim0$.

Proof:

(1) Suppose $x=(0,\dots,0)$. Then, $x\sim0$ is possible.

(2) Suppose $x\gg y$. Then, by Definition of monotone preference, $x\succ y$ is possible.

(3) Suppose $\exists$ some $j$ such that $x_j>0$ and $1\leq j\leq L$.Then, I have the following process of elimination:

• $x\succsim0$ is possible.
• $x\succsim0$ and $0\succsim x$ $\iff x\sim0$ is possible.
• $x\succsim0$ but not $0\succsim x$ $\iff x\succ0$ is possible.
• $0\succsim x$ but not $x\succsim 0$ $\iff 0\succ x$ is impossible.

The single preference relation that is common in all three scenarios is $x\succsim0$. Hence, for every $x\in\mathbb{R^+_L}$, monotonicity implies $x\succsim0$. Q.E.D

• 1.) Where does rationality or continuity of preferences come in? 2.) In your third case, you say $x_j > 0$, so how can $0 \succsim x$ in that case? Jun 18 '16 at 3:12
• @KitsuneCavalry (1) Never. LOL (2) I reference my reasoning back in Chapter 3 where the authors state that "if $\succsim$ is monotone, we may have indifference with respect to an increase in the amount of some but not all commodities." An intuitive example I thought of was a set of complementary goods, so maybe a gaming console and its joystick. Suppose only one person can play per console, so one joystick used per console. It doesn't really make much difference if you have 1 Playstation and 0 joystick versus 0 playstation and 0 joystick. Either way, you can't operate the game console. Jun 18 '16 at 7:35

If we take the definition of monotonicity to be if $x\geqq y$ then $x \succeq y$, you can simplify the proof (though it looks right).

Note $\mathbf{0}\leq x$ for all $x\in \mathbb{R}_+^l$. So by the definition of monotonicity (essentially replacing $y$ with $\mathbf{0}$ above), $x\succeq \mathbf{0}$. I don't think continuity is required (check lexicographic preferences to see it is not necessary for the stated result).

• Because if your first sentence is true, then it trivially follows that $x\succsim0$. But because MWG states only "by monotonicity", I was compelled to work out like above case by case. The IFFY part pointed out by Kitsune is, is $0\succsim x$ ever possible? Because MWG mentions "we may have indifference with respect to an increase in the amount of some but not all commodities". From this, I inferred $x\sim0$ is feasible. Share your thoughts! Jun 23 '16 at 19:34
• Pburg, thanks for the response. The problem I was facing is exactly your first sentence. To have that $x\geq y\rightarrow x\succsim y$, I came to the conclusion that this requires actually both monotonicity AND Continuity of preference relation. My other attempt is Here Jun 23 '16 at 19:39
• I see how you're using continuity in the other attempt. That would be necessary if you only knew that $x>>y$ implies $x\succ y$. But I understand monotonicity to give us more, namely that case 2 in your other attempt is already covered by the definition. I don't have MWG in front of me and maybe books differ in how they include qualifiers like "weak," but I would guess monotonicity to cover $x\geqq 0 \implies x\succeq 0$. Because I think it's a matter of weakly greater vectors being weakly preferred. Jun 23 '16 at 20:02
• And I think $x\sim 0$ is feasible. By the definition I am using, $u(x)=0$ for all $x$ would be monotonic. Or we could assume perfect complements. Jun 23 '16 at 20:04
• Thanks for the comments. This is excellent. Yeah, I just followed the definition of monotonicity as written by MWG and it doesn't spell out the weakly greater vectors being weakly preferred. Of course, to some perhaps many, it is obvious... I had to go through the proof I linked above to see that clearly. On intuitive level, it definitely makes sense. As for $x\sim0$, perfect complements works exactly for the example I gave to Kitsune, which was xbox one and controller. If I had 0 xbox and 0 controller, it wouldn't make much difference if I had 1 xbox and 0 controller. LOL Jun 23 '16 at 20:20

I did not understand the last part when you said $0≻x$ is impossible. I think it's necessary to use continuity + monotonicity in (3).

For instance, let's take the $\Bbb R^2$ case. Assume we have two elements $x_1=(0,1)$ and $x_2=(1,0)$. Then, any combination is greater than $0$: $\alpha*x_1+(1-\alpha)*x_2 >>0$ if $\alpha \in (0,1)$. Re-definining $\alpha=1/n$. Then, we have built a sequence $x_n=(1/n)*x_1+(1-1/n)*x_2$ such that $x_n >> 0$ for any $n=2,3,...$. By monotonicity, $x_n≿0$ for all $n=2,...$. Finally, by continuity of preferences, $\lim_{n\to\infty}x_n≿0$. That is, $x_2≿0$ (the same for $x_1≿0$).

This can be done in $\Bbb R^n$ and for any vector that have $x_j=0$ for some $j$. Hence, adding this with the proof you have made for the other cases ($x=0$ and $x>>0$) we have $x≿0$. The (1) case is implied by completeness, which imply reflexivity.

• Alvaro, I agree with you, and that is what I did in my second attempt. The above actually does not constitute a proof...yeah, because I thought about it for a bit and realized I was roped into MWG's statement that "we may have indifference with respect to an increase in the amount of some but not all commodities". Jun 24 '16 at 3:37
• So from that statement, now I figure that the authors probably meant that perfect complements like video game console and controller or computer and keyboard or mouse type of pairs actually make the indifference $x\sim0$ possible. Jun 24 '16 at 3:39
• I have updated my post with a new proof so please take a look and share thoughts. Jun 24 '16 at 3:39
• I think the proof is fine! It seems that without continuity we can have some x s.t $0≻x$. Jun 24 '16 at 17:46
• I checked the definition of monotonic in MWG, and I agree that per that definition, you do need continuity. It seems permissible that $(0,0)$ could be preferred to all points $(a,b)$ with $a=0$ or $b=0$ since they aren't ordered by $>>$. Imagine it like perfect complements without free disposal. You've got a left shoe and no right shoe, now the left shoe is just taking up space and you'd rather throw it away than deal with the clutter. Jun 24 '16 at 19:14