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

  • $\begingroup$ Isn't the proof just restating the definition? $\endgroup$
    – Herr K.
    Mar 21, 2020 at 21:24
  • 1
    $\begingroup$ @HerrK.: In the definition of linearity the indifference has to hold for all $a$, by assumption we only know that it holds for $a\ge 0$. $\endgroup$
    – VARulle
    Mar 23, 2020 at 15:54

1 Answer 1


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.

  • $\begingroup$ 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! $\endgroup$
    – High GPA
    Mar 21, 2020 at 9:33

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