<p>Assumptions <span class="math-container">$1$</span> to <span class="math-container">$3$</span> are sufficient to obtain a linear representation when <span class="math-container">$X$</span> is open and convex. We proceed in two steps.</p>

<hr>

<p><strong>Step <span class="math-container">$1$</span>:</strong></p>

<p>We will repeatedely use the following consequence of continuity and <span class="math-container">$A1$</span>:
<em>If <span class="math-container">$x \sim x^{\prime}$</span>, then <span class="math-container">$x \sim x + \lambda (x^{\prime} - x)$</span> for every <span class="math-container">$\lambda \in \mathbb{R}$</span> such that <span class="math-container">$x + \lambda (x^{\prime} - x) \in X$</span>.</em></p>

<p>First, <span class="math-container">$x + 0.5 (x^{\prime} - x)$</span> is contained in <span class="math-container">$X$</span> as we assumed <span class="math-container">$X$</span> to be convex. Second, by completeness of the relation, <span class="math-container">$x \succeq x + 0.5 (x^{\prime} - x)$</span> or <span class="math-container">$x \preceq x + 0.5 (x^{\prime} - x)$</span>. Assume the latter (the other case is treated similarly). By <span class="math-container">$A1$</span>, 
<span class="math-container">$$x \preceq x + 0.5 (x^{\prime} - x) \quad \Leftrightarrow \quad x + 0.5 (x^{\prime} - x) \preceq x + 0.5 (x^{\prime} - x) + 0.5 (x^{\prime} - x).$$</span> 
The latter comparison is equivalent to  <span class="math-container">$x + 0.5 (x^{\prime} - x) \preceq x^{\prime}$</span>. This shows <span class="math-container">$$x \preceq x + 0.5 (x^{\prime} - x) \preceq x^{\prime} \sim x.$$</span></p>

<p>Repeating this argument, we conclude that for any <span class="math-container">$n, k \in\mathbb{N}$</span> with <span class="math-container">$k \leq 2^{n}$</span> the comparison
<span class="math-container">$$
x \sim x + \frac{k}{2^{n}}(x^{\prime} - x)
$$</span>
holds.
For any number <span class="math-container">$\lambda \in [0, 1]$</span> there is a sequence of fractions of the form <span class="math-container">$k / 2^{n}$</span> converging to <span class="math-container">$\lambda$</span>. Since the relation is continuous, this establishes that <span class="math-container">$x \sim x + \lambda (x^{\prime} - x)$</span> for any <span class="math-container">$\lambda \in [0, 1]$</span>.</p>

<p>To get indifference for arbitrary <span class="math-container">$\lambda \geq 0$</span>, find <span class="math-container">$n\in \mathbb{N}$</span> such that <span class="math-container">$n \leq \lambda \leq n + 1$</span>.
Note that <span class="math-container">$x \sim x^{\prime}$</span> iff <span class="math-container">$x^{\prime} \sim x^{\prime} + (x^{\prime} - x)$</span> iff <span class="math-container">$x^{\prime} + (x^{\prime} - x) \sim x^{\prime} + 2(x^{\prime} - x)$</span>, etc. Thus, <span class="math-container">$x \sim x + n (x^{\prime} - x ) \sim x + (n+1) (x^{\prime} - x )$</span>.
Since <span class="math-container">$\lambda$</span> may be written as a convex combination of <span class="math-container">$n$</span> and <span class="math-container">$n+1$</span>, earlier arguments now imply that <span class="math-container">$x\sim x + n (x^{\prime} - x ) \sim x + \lambda (x^{\prime} - x )$</span>.
Lastly, when <span class="math-container">$\lambda  \leq 0$</span>, go through the same steps but consider <span class="math-container">$x - n (x^{\prime} -x)$</span>, etc.</p>

<hr>

<p><strong>Step <span class="math-container">$2$</span>:</strong></p>

<p>Let <span class="math-container">$x$</span> be an arbitrary element of <span class="math-container">$X$</span>.
By continuity and monotonicity we may find a point <span class="math-container">$x^{\prime}$</span> unequal to <span class="math-container">$x$</span> such that <span class="math-container">$x\sim x^{\prime}$</span>. 
The argument is as follows: 
For <span class="math-container">$\varepsilon$</span> sufficiently small, the points <span class="math-container">$x - \varepsilon (1, 1)$</span> and <span class="math-container">$x + \varepsilon (1, 1)$</span> are contained in <span class="math-container">$X$</span> as <span class="math-container">$X$</span> is open. 
By monotonicity, <span class="math-container">$$x - \varepsilon (1, 1) \prec x \prec x + \varepsilon (1, 1).$$</span>
Then for <span class="math-container">$\varepsilon^{\prime}$</span> sufficiently small, 
<span class="math-container">$$x - \varepsilon (1, 1) + \varepsilon^{\prime}(-1, 1) \prec x \prec x + \varepsilon (1, 1) + \varepsilon^{\prime}(-1, 1)$$</span>
 by continuity (and for small <span class="math-container">$\varepsilon^{\prime}$</span> these points are once again contained in <span class="math-container">$X$</span>). Now consider the line segment between <span class="math-container">$x - \varepsilon (1, 1) + \varepsilon^{\prime}(-1, 1) $</span> and <span class="math-container">$ x + \varepsilon (1, 1) + \varepsilon^{\prime}(-1, 1)$</span> and use continuity and the fact that <span class="math-container">$X$</span> is convex.</p>

<p>Now suppose <span class="math-container">$y$</span> and <span class="math-container">$y^{\prime}$</span> are two other distinct points in <span class="math-container">$X$</span> such that <span class="math-container">$y \sim y^{\prime}$</span>. 
Without loss, let's label the points such that <span class="math-container">$x_{1} &lt; x_{1}^{\prime}$</span> and <span class="math-container">$y_{1} &lt; y_{1}^{\prime}$</span>.
We will show that <span class="math-container">$x^{\prime} - x$</span> and <span class="math-container">$y^{\prime} - y$</span> are parallel.
If they are not, then we may assume that <span class="math-container">$(x^{\prime}_{2} - x_{2}) / (x_{1}^{\prime} - x_{1}) &gt; (y^{\prime}_{2} - y_{2}) / (y_{1}^{\prime} - y_{1}) $</span> is satisfied (the argument is analogous for the other case).</p>

<p>By <span class="math-container">$A1$</span>, 
<span class="math-container">$$x \sim x + \lambda_{x} (x^{\prime} - x) \Rightarrow y \sim y + \lambda_{x} (x^{\prime} - x)$$</span> for all <span class="math-container">$\lambda \in [0, 1]$</span>.
Also, <span class="math-container">$y \sim y + \lambda_{y} (y^{\prime} - y)$</span> for all <span class="math-container">$\lambda_{y} \in [0, 1]$</span> since <span class="math-container">$y\sim y^{\prime}$</span>.
All of these comparisons are well-defined as <span class="math-container">$X$</span> is convex.
Choosing <span class="math-container">$\varepsilon &gt; 0$</span> sufficiently small, these comparisons hold, in particular, for <span class="math-container">$\lambda_{x} = \varepsilon / (x_{1}^{\prime} - x_{1})$</span> and <span class="math-container">$\lambda_{y} = \varepsilon / (y_{1}^{\prime} - y_{1})$</span>. (Choosing <span class="math-container">$\varepsilon$</span> sufficiently small is necessary to guarantee <span class="math-container">$\lambda_{x}, \lambda_{y} \in (0, 1)$</span>.)
Now we note that 
<span class="math-container">\begin{align*}
    \lambda_{x} (x^{\prime} - x) &amp;= 
    \varepsilon
    \begin{pmatrix}
     1 \\
     \frac{x_{2}^{\prime} - x_{2}}{x_{1}^{\prime}- x_{1}}
    \end{pmatrix},
    \\
    \text{and}\quad 
        \lambda_{y} (y^{\prime} - y) &amp;= 
    \varepsilon
    \begin{pmatrix}
     1 \\
     \frac{y_{2}^{\prime} - y_{2}}{y_{1}^{\prime}- y_{1}}
    \end{pmatrix}.
\end{align*}</span>
From our assumptions we then conclude that <span class="math-container">$\lambda_{x} (x^{\prime} - x) &gt; \lambda_{y} (y^{\prime} - y)$</span>. 
Given monotonicity, this contradicts the fact that <span class="math-container">$y\sim y + \lambda_{x} (x^{\prime} - x)$</span> and  <span class="math-container">$y \sim y + \lambda_{y} (y^{\prime} - y)$</span>.</p>

<p>So far we have thus shown that there is a vector <span class="math-container">$r$</span> such that <span class="math-container">$z \sim z^{\prime}$</span> only if <span class="math-container">$r \cdot z = r \cdot z^{\prime}$</span>. 
To prove the converse, consider our initial points <span class="math-container">$x, x^{\prime}$</span> which we know to be equivalent and satisfy <span class="math-container">$r \cdot x = r \cdot x^{\prime}$</span>. Now, if <span class="math-container">$r \cdot z = r \cdot z^{\prime}$</span>, then travelling from <span class="math-container">$z$</span> to <span class="math-container">$z^{\prime}$</span> involves travelling in a line parallel to <span class="math-container">$x - x^{\prime}$</span>. Indifference <span class="math-container">$z \sim z^{\prime}$</span> then follows from <span class="math-container">$A1$</span> and continuity.</p>

<hr>

<hr>

<p>One way to get away from a linear representation is to drop <span class="math-container">$A1$</span>. 
For instance, if you replaced it with <span class="math-container">$x \sim y \Leftrightarrow x \sim x + \lambda (x - y)$</span> for all <span class="math-container">$\lambda \in \mathbb{R}$</span> (such that <span class="math-container">$x + \lambda (x - y) \in X$</span>), then all utility representations may fail to be linear. 
Loosely speaking, if you only impose that each indifference curve is a hyperplane in <span class="math-container">$\mathbb{R}^{2}$</span>, then restricting <span class="math-container">$X$</span> appropriately allows you to arrange indifference curves in a way such that they are not parallel and do not intersect. Of course, not restricting <span class="math-container">$X$</span> wouldn't work since nonparallel hyperplanes always intersect somewhere in <span class="math-container">$\mathbb{R}^{2}$</span>.</p>

<p>Here is one such example:
Let <span class="math-container">$X = (0, 1] \times [-1, 0]$</span> and let <span class="math-container">$u(x_{1}, x_{2}) = x_{2} / x_{1}$</span>.
In other words, the utility assigned to the point <span class="math-container">$x$</span> is the slope of the line segment that connects it to the origin.
This utility function is strictly increasing in both arguments (since <span class="math-container">$x_{2} \leq 0 &lt; x_{1}$</span> on <span class="math-container">$X$</span>) and continuous. You can easily check that
<span class="math-container">$$
\frac{x_{2}}{x_{1}} = \frac{y_{2}}{y_{1}} \Leftrightarrow
\frac{x_{2}}{x_{1} } = \frac{x_{2} + \lambda (y_{2} - x_{2})}{x_{1}+ \lambda (y_{1} - x_{1})} 
\quad \forall \lambda
$$</span>
is satisfied, showing that indifference curves are hyperplanes.
These hyperplanes are not parallel (by construction) and would only intersect at <span class="math-container">$(0, 0)$</span>, but we removed the origin from <span class="math-container">$X$</span>.</p>

<p>To see that this example violates <span class="math-container">$A1$</span> and linearity, consider the points <span class="math-container">$(0.5, -0.5)$</span> and <span class="math-container">$(-1, 1)$</span>. Clearly, <span class="math-container">$- 0.5 / 0.5 = - 1 / 1$</span>, i.e. <span class="math-container">$(0.5, -0.5) \sim (-1, 1)$</span>.
However, <span class="math-container">$$(0.5, -0.5 + 0.25) \succ (1, -1 + 0.25)$$</span> since <span class="math-container">$$(- 0.5 + 0.25) / 0.5 = - 0.5 &gt; - 0.75 = (-1 + 0.25) / 1.$$</span></p>

CC BY-SA 4.0
2019-11-15T22:26:31.730Z
Sorry but your third paragraph using A1 does not seem right in my limited intuition. Your third paragraph basically says that $x\sim 0.5x+0.5x'$; I don't know how to prove this. I guess you might try to prove it from $0.5x+0.5x\sim 0.5x+0.5x'$; however, first we have to notice that $0.5x$ is not necessarily ind to $0.5x'$. Additionally, $0.5x$ might not be in $X$.
High GPA
CC BY-SA 4.0
2019-11-16T02:38:25.477Z
Sorry for being a little vague there; I added a longer explanation. The basic intuition is as follows: Imagine travelling from $x$ to $x^{\prime}$ via a finite number of equidistant steps in the direction $x^{\prime} - x$. Since the endpoints of that path are equivalent, $A1$ and transitivity are sufficient to conclude that all points along the path are equivalent. By continuity, choosing the steps of such a path sufficiently small implies that any points inbetween $x$ and $x^{\prime}$ must be equivalent.
induction_is_a_laddah
CC BY-SA 4.0
2019-11-16T14:48:55.560Z
Thank you so much for your detailed answer! You are a very good teacher to me! Last question: would semicontinuity be enough to generate the proof in your Step 1?
High GPA
CC BY-SA 4.0
2019-11-16T14:54:32.360Z
By the way, A2 is "monotonicity" and I think you used in step 2; but I am not saying that you are wrong: I agree that A2 can be dropped without affecting the result.
High GPA
CC BY-SA 4.0
2019-11-15T05:23:53.077Z
2019-11-17T14:21:16.497Z
2019-11-17T14:21:16.497Z
induction_is_a_laddah
<p>Consider a preference relation <span class="math-container">$\succeq$</span> on <span class="math-container">$X\subseteq\mathbb R^2$</span>. If <span class="math-container">$\succeq$</span> satisifies:
<span class="math-container">$$
\begin{align}
&amp;1.\mbox{ }(a_1,a_2)\succeq (b_1,b_2)\implies(a_1+t,a_2+s)\succeq (b_1+t,b_2+s),\forall t,s\\
&amp;2.\mbox{ }a_1\geq b_1 \mbox{ and } a_2\geq b_2 \implies (a_1,a_2)\succeq (b_1,b_2)\mbox{ (and the analogous for }\succ\mbox{)}\\
&amp;3.\mbox{ Continuity }
\end{align}
$$</span>
Then would there exists a linear representation for <span class="math-container">$\succeq$</span>?</p>

<hr>

<p>When <span class="math-container">$X=\mathbb R^2$</span>, the old proof is copied here:</p>

<p>Step1: For each vector <span class="math-container">$(x,y)$</span>, there is a unique <span class="math-container">$z\in \mathbb R$</span> such that <span class="math-container">$(x,y) \sim (z,z)$</span>. WLOG assume <span class="math-container">$x \geq y$</span>. Then to see this claim, first notice by A2 that <span class="math-container">$(x,x) \succeq (x,y) \succeq (y,x)$</span>. Then traveling along the <span class="math-container">$45^\circ$</span> from <span class="math-container">$(y,y)$</span> to <span class="math-container">$(x,x)$</span>, A3 ensures the existence of our <span class="math-container">$z$</span>. (Strict) Monotonicity assures uniqueness in the obvious way. Let <span class="math-container">$u: (x,y) \mapsto z$</span> where <span class="math-container">$z$</span> is defined in this way.</p>

<p>Step2: Now let <span class="math-container">$(x,y) \sim (z,z)$</span> and <span class="math-container">$(x',y') \sim (z',z')$</span>. Then by A1 and transitivity we have <span class="math-container">$(x+x',y+y') \sim (z+z',z+z')$</span>. Additivity+transitivity implies linearity.</p>

<hr>

<p>However, in our case of <span class="math-container">$X\subseteq \mathbb R^2$</span>, for example let's set <span class="math-container">$X=[2,3]\times [2,3]$</span>, then step 2 does not work anymore: because if <span class="math-container">$x,x'\in [2,3]$</span>, then <span class="math-container">$x+x'\not\in [2,3]$</span>.</p>

<p>Therefore, I hypothesize that the preference is not necessarily linear. It can be a power function like <span class="math-container">$u(x,y)=ax^b+cy^d$</span> where <span class="math-container">$a,b,c,d$</span> can be positive or negative. Also, <span class="math-container">$u$</span> must be analytic.</p>

<p>For 3+ dimensions, the <span class="math-container">$u$</span> must be separable.</p>

<microeconomics><mathematical-economics><utility><preferences>
Linear Utility?
High GPA
# Linear Utility?

(Asked by: username_0 on 2019-11-14)

Consider a preference relation $\succeq$ on $X\subseteq\mathbb R^2$. 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 would there exists a linear representation for $\succeq$?

---

When $X=\mathbb R^2$, the old proof is copied here:

Step1: 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.

Step2: Now let $(x,y) \sim (z,z)$ and $(x',y') \sim (z',z')$. Then by A1 and transitivity we have $(x+x',y+y') \sim (z+z',z+z')$. Additivity+transitivity implies linearity.

---

However, in our case of $X\subseteq \mathbb R^2$, for example let's set $X=[2,3]\times [2,3]$, then step 2 does not work anymore: because if $x,x'\in [2,3]$, then $x+x'\not\in [2,3]$.

Therefore, I hypothesize that the preference is not necessarily linear. It can be a power function like $u(x,y)=ax^b+cy^d$ where $a,b,c,d$ can be positive or negative. Also, $u$ must be analytic.

For 3+ dimensions, the $u$ must be separable.

**Tags:** microeconomics, mathematical-economics, utility, preferences


---

## 1 answers

### Answer by username_1 on 2019-11-15. Score: 2 | Accepted answer

Assumptions $1$ to $3$ are sufficient to obtain a linear representation when $X$ is open and convex. We proceed in two steps.

---

**Step $1$:**

We will repeatedely use the following consequence of continuity and $A1$:
*If $x \sim x^{\prime}$, then $x \sim x + \lambda (x^{\prime} - x)$ for every $\lambda \in \mathbb{R}$ such that $x + \lambda (x^{\prime} - x) \in X$.*

First, $x + 0.5 (x^{\prime} - x)$ is contained in $X$ as we assumed $X$ to be convex. Second, by completeness of the relation, $x \succeq x + 0.5 (x^{\prime} - x)$ or $x \preceq x + 0.5 (x^{\prime} - x)$. Assume the latter (the other case is treated similarly). By $A1$,
$$x \preceq x + 0.5 (x^{\prime} - x) \quad \Leftrightarrow \quad x + 0.5 (x^{\prime} - x) \preceq x + 0.5 (x^{\prime} - x) + 0.5 (x^{\prime} - x).$$
The latter comparison is equivalent to $x + 0.5 (x^{\prime} - x) \preceq x^{\prime}$. This shows $$x \preceq x + 0.5 (x^{\prime} - x) \preceq x^{\prime} \sim x.$$

Repeating this argument, we conclude that for any $n, k \in\mathbb{N}$ with $k \leq 2^{n}$ the comparison
$$
x \sim x + \frac{k}{2^{n}}(x^{\prime} - x)
$$
holds.
For any number $\lambda \in [0, 1]$ there is a sequence of fractions of the form $k / 2^{n}$ converging to $\lambda$. Since the relation is continuous, this establishes that $x \sim x + \lambda (x^{\prime} - x)$ for any $\lambda \in [0, 1]$.

To get indifference for arbitrary $\lambda \geq 0$, find $n\in \mathbb{N}$ such that $n \leq \lambda \leq n + 1$.
Note that $x \sim x^{\prime}$ iff $x^{\prime} \sim x^{\prime} + (x^{\prime} - x)$ iff $x^{\prime} + (x^{\prime} - x) \sim x^{\prime} + 2(x^{\prime} - x)$, etc. Thus, $x \sim x + n (x^{\prime} - x ) \sim x + (n+1) (x^{\prime} - x )$.
Since $\lambda$ may be written as a convex combination of $n$ and $n+1$, earlier arguments now imply that $x\sim x + n (x^{\prime} - x ) \sim x + \lambda (x^{\prime} - x )$.
Lastly, when $\lambda \leq 0$, go through the same steps but consider $x - n (x^{\prime} -x)$, etc.

---

**Step $2$:**

Let $x$ be an arbitrary element of $X$.
By continuity and monotonicity we may find a point $x^{\prime}$ unequal to $x$ such that $x\sim x^{\prime}$.
The argument is as follows:
For $\varepsilon$ sufficiently small, the points $x - \varepsilon (1, 1)$ and $x + \varepsilon (1, 1)$ are contained in $X$ as $X$ is open.
By monotonicity, $$x - \varepsilon (1, 1) \prec x \prec x + \varepsilon (1, 1).$$
Then for $\varepsilon^{\prime}$ sufficiently small,
$$x - \varepsilon (1, 1) + \varepsilon^{\prime}(-1, 1) \prec x \prec x + \varepsilon (1, 1) + \varepsilon^{\prime}(-1, 1)$$
by continuity (and for small $\varepsilon^{\prime}$ these points are once again contained in $X$). Now consider the line segment between $x - \varepsilon (1, 1) + \varepsilon^{\prime}(-1, 1) $ and $ x + \varepsilon (1, 1) + \varepsilon^{\prime}(-1, 1)$ and use continuity and the fact that $X$ is convex.

Now suppose $y$ and $y^{\prime}$ are two other distinct points in $X$ such that $y \sim y^{\prime}$.
Without loss, let's label the points such that $x\_{1} < x\_{1}^{\prime}$ and $y\_{1} < y\_{1}^{\prime}$.
We will show that $x^{\prime} - x$ and $y^{\prime} - y$ are parallel.
If they are not, then we may assume that $(x^{\prime}\_{2} - x\_{2}) / (x\_{1}^{\prime} - x\_{1}) > (y^{\prime}\_{2} - y\_{2}) / (y\_{1}^{\prime} - y\_{1}) $ is satisfied (the argument is analogous for the other case).

By $A1$,
$$x \sim x + \lambda\_{x} (x^{\prime} - x) \Rightarrow y \sim y + \lambda\_{x} (x^{\prime} - x)$$ for all $\lambda \in [0, 1]$.
Also, $y \sim y + \lambda\_{y} (y^{\prime} - y)$ for all $\lambda\_{y} \in [0, 1]$ since $y\sim y^{\prime}$.
All of these comparisons are well-defined as $X$ is convex.
Choosing $\varepsilon > 0$ sufficiently small, these comparisons hold, in particular, for $\lambda\_{x} = \varepsilon / (x\_{1}^{\prime} - x\_{1})$ and $\lambda\_{y} = \varepsilon / (y\_{1}^{\prime} - y\_{1})$. (Choosing $\varepsilon$ sufficiently small is necessary to guarantee $\lambda\_{x}, \lambda\_{y} \in (0, 1)$.)
Now we note that
\begin{align\*}
\lambda\_{x} (x^{\prime} - x) &=
\varepsilon
\begin{pmatrix}
1 \\
\frac{x\_{2}^{\prime} - x\_{2}}{x\_{1}^{\prime}- x\_{1}}
\end{pmatrix},
\\
\text{and}\quad
\lambda\_{y} (y^{\prime} - y) &=
\varepsilon
\begin{pmatrix}
1 \\
\frac{y\_{2}^{\prime} - y\_{2}}{y\_{1}^{\prime}- y\_{1}}
\end{pmatrix}.
\end{align\*}
From our assumptions we then conclude that $\lambda\_{x} (x^{\prime} - x) > \lambda\_{y} (y^{\prime} - y)$.
Given monotonicity, this contradicts the fact that $y\sim y + \lambda\_{x} (x^{\prime} - x)$ and $y \sim y + \lambda\_{y} (y^{\prime} - y)$.

So far we have thus shown that there is a vector $r$ such that $z \sim z^{\prime}$ only if $r \cdot z = r \cdot z^{\prime}$.
To prove the converse, consider our initial points $x, x^{\prime}$ which we know to be equivalent and satisfy $r \cdot x = r \cdot x^{\prime}$. Now, if $r \cdot z = r \cdot z^{\prime}$, then travelling from $z$ to $z^{\prime}$ involves travelling in a line parallel to $x - x^{\prime}$. Indifference $z \sim z^{\prime}$ then follows from $A1$ and continuity.

---

---

One way to get away from a linear representation is to drop $A1$.
For instance, if you replaced it with $x \sim y \Leftrightarrow x \sim x + \lambda (x - y)$ for all $\lambda \in \mathbb{R}$ (such that $x + \lambda (x - y) \in X$), then all utility representations may fail to be linear.
Loosely speaking, if you only impose that each indifference curve is a hyperplane in $\mathbb{R}^{2}$, then restricting $X$ appropriately allows you to arrange indifference curves in a way such that they are not parallel and do not intersect. Of course, not restricting $X$ wouldn't work since nonparallel hyperplanes always intersect somewhere in $\mathbb{R}^{2}$.

Here is one such example:
Let $X = (0, 1] \times [-1, 0]$ and let $u(x\_{1}, x\_{2}) = x\_{2} / x\_{1}$.
In other words, the utility assigned to the point $x$ is the slope of the line segment that connects it to the origin.
This utility function is strictly increasing in both arguments (since $x\_{2} \leq 0 < x\_{1}$ on $X$) and continuous. You can easily check that
$$
\frac{x\_{2}}{x\_{1}} = \frac{y\_{2}}{y\_{1}} \Leftrightarrow
\frac{x\_{2}}{x\_{1} } = \frac{x\_{2} + \lambda (y\_{2} - x\_{2})}{x\_{1}+ \lambda (y\_{1} - x\_{1})}
\quad \forall \lambda
$$
is satisfied, showing that indifference curves are hyperplanes.
These hyperplanes are not parallel (by construction) and would only intersect at $(0, 0)$, but we removed the origin from $X$.

To see that this example violates $A1$ and linearity, consider the points $(0.5, -0.5)$ and $(-1, 1)$. Clearly, $- 0.5 / 0.5 = - 1 / 1$, i.e. $(0.5, -0.5) \sim (-1, 1)$.
However, $$(0.5, -0.5 + 0.25) \succ (1, -1 + 0.25)$$ since $$(- 0.5 + 0.25) / 0.5 = - 0.5 > - 0.75 = (-1 + 0.25) / 1.$$

> **Comments:**
> **username_0**: Sorry but your third paragraph using A1 does not seem right in my limited intuition. Your third paragraph basically says that $x\sim 0.5x+0.5x'$; I don't know how to prove this. I guess you might try to prove it from $0.5x+0.5x\sim 0.5x+0.5x'$; however, first we have to notice that $0.5x$ is not necessarily ind to $0.5x'$. Additionally, $0.5x$ might not be in $X$.  
> **username_1**: Sorry for being a little vague there; I added a longer explanation. The basic intuition is as follows: Imagine travelling from $x$ to $x^{\prime}$ via a finite number of equidistant steps in the direction $x^{\prime} - x$. Since the endpoints of that path are equivalent, $A1$ and transitivity are sufficient to conclude that all points along the path are equivalent. By continuity, choosing the steps of such a path sufficiently small implies that any points inbetween $x$ and $x^{\prime}$ must be equivalent.  
> **username_0**: Thank you so much for your detailed answer! You are a very good teacher to me! Last question: would semicontinuity be enough to generate the proof in your Step 1?  
> **username_0**: By the way, A2 is "monotonicity" and I think you used in step 2; but I am not saying that you are wrong: I agree that A2 can be dropped without affecting the result.  

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