# Existence of Linear Representation

Consider a preference relation $\succeq$ on $X=\mathbb{R^2_{+}}$. If $\succeq$ satisifies: \begin{align} &1.\mbox{ }(a_1,a_2)\succeq (b_1,b_2)\implies(a_1+t,a_2+s)\succeq (b_1+t,b_2+s),\forall t,s\\ &2.\mbox{ }a_1\geq b_1 \mbox{ and } a_2\geq b_2 \implies (a_1,a_2)\succeq (b_1,b_2)\mbox{ (and the analogous for }\succ\mbox{)}\\ &3.\mbox{ Continuity } \end{align} Then: exists a linear representation for $\succeq$.

Could anyone give me some hints on how to prove this?

Thanks for helping! :D

• You can look at Koopman's very famous Econometrica paper to have some insight on these kind of stuff. – optimal control Aug 31 '15 at 0:10
• I remember having seen this proof somewhere before, I tried finding it on Mas-colell but without success. Maybe your suggestion can help me somehow. Thanks! :D BTW, which paper are you talking about? – Guilherme Salomé Aug 31 '15 at 0:11
• Here it is ; mikael.cozic.free.fr/koopmans60.pdf – optimal control Aug 31 '15 at 10:28
• You could make this into an answer. – BKay Aug 31 '15 at 22:53
• Who proposed this independence axiom? I cannot find it on Koopmans 1960 – High GPA Jun 5 '19 at 22:05

How about this: For each vector $(x,y)$, there is a unique $z\in \mathbb R$ such that $(x,y) \sim (z,z)$. WLOG assume $x \geq y$. Then to see this claim, first notice by A2 that $(x,x) \succeq (x,y) \succeq (y,x)$. Then traveling along the $45^\circ$ from $(y,y)$ to $(x,x)$, A3 ensures the existence of our $z$. (Strict) Monotonicity assures uniqueness in the obvious way. Let $u: (x,y) \mapsto z$ where $z$ is defined in this way.
Now let $(x,y) \sim (z,z)$ and $(x',y') \sim (z',z')$. Then by A1 we have \begin{align} (x + x',y+y') &\sim (z + x',z + y') \\ (x' + z,y' +z) &\sim (z'+z,z'+z) \end{align} so by transitivity, $(x+x',y+y') \sim (z+z',z+z')$. Verifying that $u$ is continuous is trivial.
So $u$ is additive and continuous. As this Math SE post explains if $\mathbb R^2 \to \mathbb{R}$ is additive and continuous then it is linear.