# $x\sim y$ implies $x+a\sim y+a$ for any $a\geq0$ and $x,y\in\mathbb R^n$, then the preference is linear?

$$x,y,a$$ 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 $$x,y,a$$.

• Isn't the proof just restating the definition? Mar 21 '20 at 21:24
• @HerrK.: In the definition of linearity the indifference has to hold for all $a$, by assumption we only know that it holds for $a\ge 0$. Mar 23 '20 at 15:54

It is not true. Let us consider $$\mathbb{R}^2$$ so bundles are $$x = (x_1,x_2)$$.
(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.