<p>It is not true. Let us consider <span class="math-container">$\mathbb{R}^2$</span> so bundles are <span class="math-container">$x = (x_1,x_2)$</span>.</p>

<p>Consider the preference:</p>

<p>(i) If <span class="math-container">$x_1 \leq 0$</span>, preferences are lexicographic, i.e. 
<span class="math-container">$$
x \succ y \Leftrightarrow 
\begin{cases}
x_1 &gt; y_1 \\
\text{ or } \\
x_1 = y_1 \text{ and } x_2 &gt; y_2
\end{cases}
$$</span> 
(ii) If <span class="math-container">$x_1 \geq 0$</span>, <span class="math-container">$u(x_1,x_2)=x_1+x_2$</span>.</p>

<p>Notice that no indifference occurs on <span class="math-container">$\mathbb{R}_{&lt;0} \times \mathbb{R}$</span>, and the required condition holds on <span class="math-container">$\mathbb{R}_{\geq 0}\times \mathbb{R}$</span> since it is linear there.</p>

<p>However, <span class="math-container">$(0,4) \sim (2,2)$</span>, choose <span class="math-container">$a = (-1,-1)$</span>, we get 
<span class="math-container">$$(2-a,2-a) = (1,1) \sim (0,2) \succ (-1,3) = (0-a,4-a)
$$</span>
hence the second condition fails.</p>

CC BY-SA 4.0
2020-03-21T09:33:57.020Z
Hi Walrasian I am sorry that I forgot to mention the preference has to be continuous. I will open another question! Thank you for your smart counterexample by the way!
High GPA
CC BY-SA 4.0
2020-03-21T07:14:21.780Z
2020-03-21T07:14:21.780Z
Walrasian Auctioneer
<p><span class="math-container">$x,y,a$</span> are vectors in <span class="math-container">$\mathbb R^n$</span></p>

<p>We say <span class="math-container">$a\geq0$</span> if all directions of the vector <span class="math-container">$a$</span> is greater or equal to zero.</p>

<p>We want to prove (or disprove by counterexample) that: </p>

<p>Suppose <span class="math-container">$x\sim y$</span> implies <span class="math-container">$x+a\sim y+a$</span> for any <span class="math-container">$a\geq0$</span> and <span class="math-container">$x,y\in\mathbb R^n$</span>, </p>

<p>Then the preference is linear. </p>

<hr>

<p>One definition of linear preference is that <span class="math-container">$x\sim y$</span> implies <span class="math-container">$x+a\sim y+a$</span> for any <span class="math-container">$x,y,a$</span>.</p>

CC BY-SA 4.0
2020-03-21T21:24:53.443Z
Isn't the proof just restating the definition?
Herr K.
CC BY-SA 4.0
2020-03-23T15:54:47.597Z
@HerrK.: In the definition of linearity the indifference has to hold for all $a$, by assumption we only know that it holds for $a\ge 0$.
VARulle
<microeconomics><game-theory><preferences><linear-algebra>
$x\sim y$ implies $x+a\sim y+a$ for any $a\geq0$ and $x,y\in\mathbb R^n$, then the preference is linear?
High GPA
# $x\sim y$ implies $x+a\sim y+a$ for any $a\geq0$ and $x,y\in\mathbb R^n$, then the preference is linear?

(Asked by: username_0 on 2020-03-21)

$x,y,a$ are vectors in $\mathbb R^n$

We say $a\geq0$ if all directions of the vector $a$ is greater or equal to zero.

We want to prove (or disprove by counterexample) that:

Suppose $x\sim y$ implies $x+a\sim y+a$ for any $a\geq0$ and $x,y\in\mathbb R^n$,

Then the preference is linear.

---

One definition of linear preference is that $x\sim y$ implies $x+a\sim y+a$ for any $x,y,a$.

**Tags:** microeconomics, game-theory, preferences, linear-algebra

> **Comments:**
> **username_1**: Isn't the proof just restating the definition?  
> **username_2**: @HerrK.: In the definition of linearity the indifference has to hold for all $a$, by assumption we only know that it holds for $a\ge 0$.  

---

## 1 answers

### Answer by username_3 on 2020-03-21. Score: 1 | Accepted answer

It is not true. Let us consider $\mathbb{R}^2$ so bundles are $x = (x\_1,x\_2)$.

Consider the preference:

(i) If $x\_1 \leq 0$, preferences are lexicographic, i.e.
$$
x \succ y \Leftrightarrow
\begin{cases}
x\_1 > y\_1 \\
\text{ or } \\
x\_1 = y\_1 \text{ and } x\_2 > y\_2
\end{cases}
$$
(ii) If $x\_1 \geq 0$, $u(x\_1,x\_2)=x\_1+x\_2$.

Notice that no indifference occurs on $\mathbb{R}\_{<0} \times \mathbb{R}$, and the required condition holds on $\mathbb{R}\_{\geq 0}\times \mathbb{R}$ since it is linear there.

However, $(0,4) \sim (2,2)$, choose $a = (-1,-1)$, we get
$$(2-a,2-a) = (1,1) \sim (0,2) \succ (-1,3) = (0-a,4-a)
$$
hence the second condition fails.

> **Comments:**
> **username_0**: Hi username_3 I am sorry that I forgot to mention the preference has to be continuous. I will open another question! Thank you for your smart counterexample by the way!  

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