<p>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.</p>

<p>Now let $(x,y) \sim (z,z)$ and $(x',y') \sim (z',z')$. Then by A1 we have
\begin{align}
(x + x',y+y') &amp;\sim (z + x',z + y') \\
(x' + z,y' +z) &amp;\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.</p>

<p>So $u$ is additive and continuous. As <a href="https://math.stackexchange.com/questions/116223/proof-that-a-continuous-additive-homomorphism-mathbbrn-to-mathbbrm-is">this Math SE post</a> explains if $\mathbb R^2 \to \mathbb{R}$ is additive and continuous then it is linear.</p>

CC BY-SA 4.0
2018-07-09T20:58:49.703Z
2018-07-09T20:58:49.703Z
201p
<p>Consider a preference relation $\succeq$ on $X=\mathbb{R^2_{+}}$. If $\succeq$ satisifies:
$$
\begin{align}
&amp;1.\mbox{ }(a_1,a_2)\succeq (b_1,b_2)\implies(a_1+t,a_2+s)\succeq (b_1+t,b_2+s),\forall t,s\\
&amp;2.\mbox{ }a_1\geq b_1 \mbox{ and } a_2\geq b_2 \implies (a_1,a_2)\succeq (b_1,b_2)\mbox{ (and the analogous for }\succ\mbox{)}\\
&amp;3.\mbox{ Continuity }
\end{align}
$$
Then: exists a linear representation for $\succeq$.</p>

<p>Could anyone give me some hints on how to prove this?</p>

<p>Thanks for helping! :D</p>

CC BY-SA 3.0
2015-08-31T00:10:14.433Z
You can look at Koopman's very famous Econometrica paper to have some insight on these kind of stuff.
optimal control
CC BY-SA 3.0
2015-08-31T00:11:27.017Z
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 Salome
CC BY-SA 3.0
2015-08-31T10:28:50.877Z
Here it is ;

http://mikael.cozic.free.fr/koopmans60.pdf
optimal control
CC BY-SA 3.0
2015-08-31T22:53:15.843Z
You could make this into an answer.
BKay
CC BY-SA 4.0
2019-06-05T22:05:52.780Z
Who proposed this independence axiom? I cannot find it on Koopmans 1960
High GPA
<utility><preferences>
Existence of Linear Representation
Guilherme Salome
# Existence of Linear Representation

(Asked by: username_0 on 2015-08-30)

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

**Tags:** utility, preferences

> **Comments:**
> **username_1**: You can look at Koopman's very famous Econometrica paper to have some insight on these kind of stuff.  
> **username_0**: 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?  
> **username_1**: Here it is ;  http://mikael.cozic.free.fr/koopmans60.pdf  
> **username_2**: You could make this into an answer.  
> **username_3**: Who proposed this independence axiom? I cannot find it on Koopmans 1960  

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## 1 answers

### Answer by username_4 on 2018-07-09. Score: 2

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](https://math.stackexchange.com/questions/116223/proof-that-a-continuous-additive-homomorphism-mathbbrn-to-mathbbrm-is) explains if $\mathbb R^2 \to \mathbb{R}$ is additive and continuous then it is linear.


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