<p>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).</p>

<p>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 $0$: $\alpha*x_1+(1-\alpha)*x_2 &gt;&gt;0$ 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 $x_n &gt;&gt; 0$ 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 $x_1≿0$).</p>

<p>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 $x&gt;&gt;0$) we have $x≿0$. The (1) case is implied by completeness, which imply reflexivity.</p>

CC BY-SA 3.0
2016-06-24T03:37:49.453Z
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".
Frank Swanton
CC BY-SA 3.0
2016-06-24T03:39:05.787Z
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.
Frank Swanton
CC BY-SA 3.0
2016-06-24T03:39:31.837Z
I have updated my post with a new proof so please take a look and share thoughts.
Frank Swanton
CC BY-SA 3.0
2016-06-24T17:46:58.447Z
I think the proof is fine! It seems that without continuity we can have some x s.t $0≻x$.
Belisario
CC BY-SA 3.0
2016-06-24T19:14:57.677Z
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.
Pburg
CC BY-SA 3.0
2016-06-27T12:46:56.973Z
@Pburg, Alvaro, If MWG meant what they really meant, then it should've said "By continuity and monotonicity". If they wanted to exclude continuity assumption, you would have to work case by case as I did in the first attempt. But then, that's not really proof, because the contentious case of showing the example of $0\succsim$ or $\succ x$ simply says it is "possible".
Frank Swanton
CC BY-SA 3.0
2016-06-23T23:02:38.703Z
2016-06-23T23:02:38.703Z
Belisario
<p>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). </p>

<p>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). </p>

CC BY-SA 3.0
2016-06-23T19:34:45.737Z
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!
Frank Swanton
CC BY-SA 3.0
2016-06-23T19:39:49.947Z
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](http://math.stackexchange.com/questions/1837246/order-relation-monotone-continuous-complete-preorder-on-mathbbrl-has)
Frank Swanton
CC BY-SA 3.0
2016-06-23T20:02:37.663Z
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.
Pburg
CC BY-SA 3.0
2016-06-23T20:04:04.147Z
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.
Pburg
CC BY-SA 3.0
2016-06-23T20:20:01.310Z
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
Frank Swanton
CC BY-SA 3.0
2016-06-23T20:38:34.210Z
From your comments, regarding $u(.)$, would it be fair to say that monotonicity implies the utility function is monotone increasing, too? Because, we do have increasing $u(.)$, $u(x)>u(y)$ if $x\gg y$. Is the chain of logic valid: $x\geq y\rightarrow x\succsim y$, hence $u(x)\geq u(y)$?
Frank Swanton
CC BY-SA 3.0
2016-06-23T20:53:41.553Z
@FrankSwanton, yes the chain of logic is valid for the utility function (when a utility function exists--it might not, as in the case of lexicographic preferences).
Pburg
CC BY-SA 3.0
2016-06-23T21:40:12.687Z
Okay. I would have to look into little bit though for any real valued function defined on a vector space over R whether a inc function implies it is also a monotone increasing. Basic graphing on $\mathbb{R^2_+}$ says yes in my head. Would you consider the original attempt case by case constitute a formal proof or no given MWG's monotonicity defines only for $y\gg x$?
Frank Swanton
CC BY-SA 4.0
2024-04-06T03:52:38.983Z
According to MWG  Exercise 3.B.2, the preference relation $\succsim$ defined on the consumption set $X=\mathbb{R}^L_+$ is said to be *weakly monotone* if and only if $x\geq y$ implies that $x\succsim y$. This exercise, in fact, asked us to show that if $\succsim$ is transitive, locally nonsatisted, and weakly monotone, then it is monotone.
Beerus
CC BY-SA 4.0
2024-04-06T03:54:29.477Z
Moreover, based on MWG's definition of *monotonicity* and *weakly monotonicity*, I think NEITHER ***monotonicity implies weakly monotonicity*** NOR ***weakly monotonicity implies monotonicity***.
Beerus
CC BY-SA 3.0
2016-06-23T19:19:45.100Z
2016-06-23T19:19:45.100Z
Pburg
<p><em><strong>I updated my proof to a general version as follows: please share your thoughts &amp; 2cent. Thanks</strong></em></p>
<p>Show a monotone continuous complete preorder on <span class="math-container">$\mathbb{R}^L_+$</span> has <span class="math-container">$y\geq x\rightarrow y\succsim x$</span>.</p>
<blockquote>
<p><em><strong>Point of Clarification</strong></em>
<br/> <span class="math-container">$X=\mathbb{R}^L_+$</span>
<br/> Preorder means the usual reflexivity and transitivity.
<br/> Complete means for any <span class="math-container">$x,y\in X$</span>, have <span class="math-container">$x\succsim y$</span> or <span class="math-container">$y\succsim x$</span>
<br/> Continuous means the relation is preserved under limits.
<br/> Monotone means for any <span class="math-container">$x,y\in X$</span>, if <span class="math-container">$y\gg x$</span>, then <span class="math-container">$y\succ x$</span>.
<br/> <span class="math-container">$\succ$</span> and <span class="math-container">$\sim$</span> are respectively asymmetic and symmetric parts of <span class="math-container">$\succsim$</span></p>
</blockquote>
<p><em><strong>Outline of Proof</strong></em>
<br/> Go through two cases: (1) <span class="math-container">$y\gg x$</span>. 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 <span class="math-container">$\epsilon$</span> to y, making it a sequence <span class="math-container">$y^n_\epsilon$</span> where for every n, <span class="math-container">$y^n_\epsilon\gg x$</span> <span class="math-container">$\forall n$</span>.</p>
<blockquote>
<p><em><strong>Proof</strong></em>
<br/> Suppose <span class="math-container">$\succsim$</span> is a monotone, continuous, complete preorder on <span class="math-container">$X=\mathbb{R}^L_+$</span>.
<br/> Case (1) <span class="math-container">$y\gg x$</span> (i.e. <span class="math-container">$y_i&gt;x_i$</span> <span class="math-container">$\forall i\in B=\{1,\dots,L\}$</span>).
<br/> By definition, <span class="math-container">$y\succ x$</span>, which implies <span class="math-container">$y\succsim x$</span>.
<br/> <br/> Case (2) <span class="math-container">$y_j=x_j$</span> for some <span class="math-container">$j\in B$</span>. For <span class="math-container">$\forall k\not=j,k\in B, y_k&gt;x_k.$</span>
<br/> For some <span class="math-container">$\epsilon&gt;0$</span>, let the sequence <span class="math-container">$\epsilon^n\in\mathbb{R}^L_+$</span> such that <span class="math-container">$\epsilon_j=\epsilon$</span>, <span class="math-container">$\epsilon_k=0$</span>.
<br/> Denote the sequence <span class="math-container">$y^n_\epsilon=y+\epsilon^n$</span>.
<br/> Then, for any <span class="math-container">$\epsilon&gt;0$</span> and <span class="math-container">$\forall n$</span>, <span class="math-container">$y^n_\epsilon\succ x$</span>, hence <span class="math-container">$y^n_\epsilon\succsim x$</span>.
<br/> By continuity of <span class="math-container">$\succsim$</span>, <span class="math-container">$$\lim_{\epsilon \to 0} {y^n_\epsilon} = y$$</span>
<br/> Hence, <span class="math-container">$y\succsim x$</span>. <span class="math-container">$\blacksquare$</span></p>
</blockquote>
<p><br/> <em><strong>OLDER VERSION</strong></em></p>
<p><em>My question is what is the valid reasoning behind that continuous rational and monotone preference relation implies <span class="math-container">$x\succsim0$</span>. I have put a proof below and would appreciate if you share your 2 cent on the validity/rigor of the proof. Thanks!</em></p>
<p>Suppose <span class="math-container">$x\in\mathbb{R}^L_+=\{x\in\mathbb{R}^L:x_l\geq0$</span> <span class="math-container">$\forall l=1,\dots,L\}$</span>.</p>
<p><strong>Claim</strong>: For every <span class="math-container">$x\in\mathbb{R}^L_+$</span>, monotonicity implies <span class="math-container">$x\succsim0$</span>.</p>
<p><strong>Proof</strong>:</p>
<p>(1) Suppose <span class="math-container">$x=(0,\dots,0)$</span>.
Then, <span class="math-container">$x\sim0$</span> is possible.</p>
<p>(2) Suppose <span class="math-container">$x\gg y$</span>. Then, by Definition of monotone preference, <span class="math-container">$x\succ y$</span> is possible.</p>
<p>(3) Suppose <span class="math-container">$\exists$</span> some <span class="math-container">$j$</span> such that <span class="math-container">$x_j&gt;0$</span> and <span class="math-container">$1\leq j\leq L$</span>.Then, I have the following process of elimination:</p>
<ul>
<li><span class="math-container">$x\succsim0$</span> is possible.</li>
<li><span class="math-container">$x\succsim0$</span> and <span class="math-container">$0\succsim x$</span> <span class="math-container">$\iff x\sim0$</span> is possible.</li>
<li><span class="math-container">$x\succsim0$</span> but not <span class="math-container">$0\succsim x$</span> <span class="math-container">$\iff x\succ0$</span> is possible.</li>
<li><span class="math-container">$0\succsim x$</span> but not <span class="math-container">$x\succsim 0$</span> <span class="math-container">$\iff 0\succ x$</span> is impossible.</li>
</ul>
<p>The single preference relation that is common in all three scenarios is <span class="math-container">$x\succsim0$</span>.
Hence, for every <span class="math-container">$x\in\mathbb{R}^L_+$</span>, monotonicity implies <span class="math-container">$x\succsim0$</span>. Q.E.D</p>
<hr />
<p><strong>Reference</strong>: Proposition 3.C.1 in <em>Microeconomic Theory</em> by  Andreu Mas-Colell, Michael Whinston, and Jerry Green.</p>

CC BY-SA 3.0
2016-06-18T03:12:54.247Z
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?
Kitsune Cavalry
CC BY-SA 3.0
2016-06-18T07:35:35.373Z
@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.
Frank Swanton
<microeconomics><consumer-theory><preferences><choice-theory>
Continuous rational and monotone preference relation implies $x\succsim0$?
Frank Swanton
# Continuous rational and monotone preference relation implies $x\succsim0$?

(Asked by: username_0 on 2016-06-17)

***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 $X=\mathbb{R}^L\_+$.
>   
>  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

---

**Reference**: Proposition 3.C.1 in *Microeconomic Theory* by Andreu Mas-Colell, Michael Whinston, and Jerry Green.

**Tags:** microeconomics, consumer-theory, preferences, choice-theory
> **Comments:**
> **username_1**: 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?  
> **username_0**: @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.  

---

## 2 answers

### Answer by username_3 on 2016-06-23. Score: 5 | Accepted answer

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).

> **Comments:**
> **username_0**: 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!  
> **username_0**: 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](http://math.stackexchange.com/questions/1837246/order-relation-monotone-continuous-complete-preorder-on-mathbbrl-has)  
> **username_3**: 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.  
> **username_3**: 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.  
> **username_0**: 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  
> **username_0**: From your comments, regarding $u(.)$, would it be fair to say that monotonicity implies the utility function is monotone increasing, too? Because, we do have increasing $u(.)$, $u(x)>u(y)$ if $x\gg y$. Is the chain of logic valid: $x\geq y\rightarrow x\succsim y$, hence $u(x)\geq u(y)$?  
> **username_3**: @FrankSwanton, yes the chain of logic is valid for the utility function (when a utility function exists--it might not, as in the case of lexicographic preferences).  
> **username_0**: Okay. I would have to look into little bit though for any real valued function defined on a vector space over R whether a inc function implies it is also a monotone increasing. Basic graphing on $\mathbb{R^2_+}$ says yes in my head. Would you consider the original attempt case by case constitute a formal proof or no given MWG's monotonicity defines only for $y\gg x$?  
> **username_4**: According to MWG  Exercise 3.B.2, the preference relation $\succsim$ defined on the consumption set $X=\mathbb{R}^L_+$ is said to be *weakly monotone* if and only if $x\geq y$ implies that $x\succsim y$. This exercise, in fact, asked us to show that if $\succsim$ is transitive, locally nonsatisted, and weakly monotone, then it is monotone.  
> **username_4**: Moreover, based on MWG's definition of *monotonicity* and *weakly monotonicity*, I think NEITHER ***monotonicity implies weakly monotonicity*** NOR ***weakly monotonicity implies monotonicity***.  

---

### Answer by username_2 on 2016-06-23. Score: 2

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.

> **Comments:**
> **username_0**: 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".  
> **username_0**: 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.  
> **username_0**: I have updated my post with a new proof so please take a look and share thoughts.  
> **username_2**: I think the proof is fine! It seems that without continuity we can have some x s.t $0≻x$.  
> **username_3**: 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.  
> **username_0**: @Pburg, Alvaro, If MWG meant what they really meant, then it should've said "By continuity and monotonicity". If they wanted to exclude continuity assumption, you would have to work case by case as I did in the first attempt. But then, that's not really proof, because the contentious case of showing the example of $0\succsim$ or $\succ x$ simply says it is "possible".  

---

