$a\geq 0$, $x\succsim y$ implies $x+a\succsim y+a$ so the preference is linear?

$$\succsim$$ is a continuous and local non-satiate weak order.

$$x,y,a$$ 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 $$x\succsim y$$ implies $$x+a\succsim y+a$$ for any $$a\geq0$$ and $$x,y\in\mathbb R^n$$ (condition 1),

Then the preference is linear.

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

Proof by contradiction. Suppose $$\succsim$$ is not linear then there exists $$x\succsim y$$ but $$x+a\prec y+a$$.

By non-satiation and continuity, there exists $$x+\epsilon\succ y$$ and $$x+a+\epsilon \prec y+a$$

Denote $$x'=x+\epsilon$$

Here if $$a_i\geq 0$$ or $$a_i\leq 0$$ for all index $$i\in\{1,..,n\}$$ then the proof is done.

Now suppose that $$a_i\geq0$$ for some indexes but $$a_j\leq 0$$ for some other indexes.

Let $$c_i=\min\{0,a_i\}$$

$$v:=x'+c$$ is a point such that $$v\leq x'$$ and $$v\leq x'+a$$

$$w:=y+c$$ is a point such that $$w\leq y$$ and $$w\leq y+a$$

If $$v\succsim w$$, then by condition (1) we must have $$x'\succsim y$$ and $$x'+a\succsim y+a$$, contradition!

If $$v\precsim w$$, then by condition (1) we must have $$x'\precsim y$$ and $$x'+a\precsim y+a$$, contradition!

Is the proof sounds rigorous?

• That's maybe not the most elegant way to write it down, but yes, I think it's correct. Mar 30 '20 at 12:40