<p>I do not have a full answer, but here are my notes when I studied it that hopefully someone can extend to a full answer.</p>
<p>Sketch of Proof:</p>
<ol>
<li><p>Consider the linear space with basis <span class="math-container">$\cup_{i =1}^N X_i$</span>, and we can identify any <span class="math-container">$x \in X$</span> by <span class="math-container">$\sum_i x_i$</span>.</p>
</li>
<li><p>Define the convex cone <span class="math-container">$D = \{\lambda(x-y): x\succeq y;\lambda &gt; 0\}$</span></p>
</li>
<li><p>Let <span class="math-container">$D^{-}$</span> be the convex hull of <span class="math-container">$\{x - y:x \prec y\}$</span></p>
</li>
<li><p>An additive utility representation is a hyperplane through the origin that strictly separates <span class="math-container">$D$</span> from <span class="math-container">$D^{-}$</span>.</p>
</li>
<li><p>These sets must be disjoint, as if they were not, we could multiply the weights on convex combinations large enough that at least one coordinate is strictly preferred.</p>
</li>
</ol>
<p>This seems very Mixture Space theoremy so I suspect the proof is not too difficult for those familiar with it.</p>

CC BY-SA 4.0
2021-04-15T01:43:29.713Z
Thanks for the response, but we aren't looking for an additive representation- simply an ordinal one. Singleton separability is not enough to get additivity.
Andrew McMillan
CC BY-SA 4.0
2021-04-15T00:59:02.070Z
2021-04-15T00:59:02.070Z
Walrasian Auctioneer
<p>Here is the sketch of a proof. All we need is that every continuous weak order on each <span class="math-container">$X_i$</span> admits a continuous utility representation. One sufficient condition is that each <span class="math-container">$X_i$</span>  is a connected separable topological space by a theorem of Eilenberg.  A proof of Eilenberg's theorem is given in Debreu's book <a href="https://cowles.yale.edu/sites/default/files/files/pub/mon/m17-all.pdf" rel="noreferrer">Theory of Value</a>. Debreu assumes the domain there to be Euclidean, but the proof easily generalizes to the present setting. One can also use the assumption that each <span class="math-container">$X_i$</span> is a second-countable topological space (for metrizable spaces implied by separability). Debreu showed that continuous utility representations exist in that setting too, but the proof is considerably harder.</p>
<p>Define <span class="math-container">$\succeq_i$</span> on <span class="math-container">$X_i$</span> by <span class="math-container">$x_i\succeq_i y_i$</span> if <span class="math-container">$x_iz\succeq y_iz$</span> for some <span class="math-container">$z$</span>. Clearly, this defines a continuous weak order and the <span class="math-container">$z$</span> used does not matter by singleton separability. There exists a continuous utility function <span class="math-container">$u_i:X_i\to\mathbb{R}$</span>.</p>
<p>Next, we can show that <span class="math-container">$x_i~\sim_iy_i$</span> for <span class="math-container">$i=1,\ldots,N$</span> implies <span class="math-container">$x\sim y$</span>. By repeated applications of singleton separability we get <span class="math-container">$(x_1,y_2,y_3,\ldots,y_N)\succeq (y_1,y_2,y_3,\ldots,y_N)$</span>, <span class="math-container">$(x_1,x_2,y_3,\ldots,y_N)\succeq (x_1,y_2,y_3,\ldots,y_N)$</span>, and so on, until we have <span class="math-container">$(x_1,x_2,\ldots,x_{N-1},x_N)\succeq(x_1,x_2,\ldots,x_{N-1},y_N)$</span>. By transitivity, <span class="math-container">$x\succeq y$</span>. Similarly, <span class="math-container">$y\succeq x$</span>, and, therefore <span class="math-container">$x\sim y$</span>.</p>
<p>Let <span class="math-container">$U=\prod_{i=1}^N u_i(X_i)$</span>, a product of (possibly unbounded) intervals. Define <span class="math-container">$\succeq^*$</span> on <span class="math-container">$U$</span> by <span class="math-container">$u\succeq^* u'$</span> if for some <span class="math-container">$x,y\in X$</span> one has <span class="math-container">$u=\big(u_1(x_1),\dots,u_N(x_N)\big)$</span>, <span class="math-container">$u'=\big(u_1(y_1),\dots,u_N(y_N)\big)$</span>, and <span class="math-container">$x\succeq y$</span>. By what we have just shown, <span class="math-container">$\succeq^*$</span> contains, together with <span class="math-container">$u_1,\ldots, u_N$</span>, the same information as <span class="math-container">$\succeq$</span>. It is also obvious that <span class="math-container">$\succeq^*$</span> is strictly monotone on <span class="math-container">$U$</span>. If a utility representation <span class="math-container">$W:U\to\mathbb{R}$</span> of <span class="math-container">$\succeq^*$</span> exists, it must be strictly increasing. We are done if we can show a continuous such representation exists. Using any of the mentioned utility representation theorems, it suffices to show that <span class="math-container">$\succeq^*$</span> is continuous.</p>
<p>So let <span class="math-container">$u\succ^* u'.$</span> There exist <span class="math-container">$x,y\in X$</span> such that <span class="math-container">$u=\big(u_1(x_1),\dots,u_N(x_N)\big)$</span>, <span class="math-container">$u'=\big(u_1(y_1),\dots,u_N(y_N)\big)$</span>, and <span class="math-container">$x\succ y$</span>. Since <span class="math-container">$\succeq$</span> is continuous and by the definition of the product topology, there exist open neighborhoods <span class="math-container">$V_i\ni x_i$</span> and <span class="math-container">$W_i\ni y_i$</span> such that <span class="math-container">$x'\in\prod_{i=1}^N V_i$</span> and <span class="math-container">$y'\in\prod_{i=1}^N W_i$</span> implies that <span class="math-container">$x'\succ y'$</span>.  Let <span class="math-container">$V$</span> be the interior of <span class="math-container">$\prod_{i=1}^N u_i(V_i)$</span> and <span class="math-container">$W$</span> the interior of <span class="math-container">$\prod_{i=1}^N u_i(W_i)$</span>. Then for <span class="math-container">$u''\in V$</span> and <span class="math-container">$u'''\in W$</span> one has <span class="math-container">$u''\succ^* u'''$</span>, so <span class="math-container">$\succeq^*$</span> is, indeed, continuous.</p>

CC BY-SA 4.0
2023-10-19T08:08:55.193Z
Nice proof! Do we have any reference for this method/proof in literature?
High GPA
CC BY-SA 4.0
2023-10-19T10:11:04.310Z
@HighGPA I'm not sure, I might have picked up some relevant ideas when looking at Bergson–Samuelson social welfare functions, which tend to have such a structure.
Michael Greinecker
CC BY-SA 4.0
2023-11-02T13:45:29.740Z
Many thanks! I wonder if you happen to have a good text that introduce the Bergson–Samuelson social welfare functions with rigorous mathematical definitions?
High GPA
CC BY-SA 4.0
2021-04-15T04:58:26.777Z
2021-04-15T04:58:26.777Z
Michael Greinecker
<p>Let <span class="math-container">$X_i$</span> be a separable, compact, Banach space.</p>
<p><strong>Definition:</strong> A weak order <span class="math-container">$\succeq$</span> on <span class="math-container">$X=\prod_{i=1}^NX_i$</span> has an <em>ordinally separable representation</em> if there exists <span class="math-container">$u_i: X_i\rightarrow \mathbb{R}$</span> and <span class="math-container">$W:\prod_{i=1}^Nu_i(X_i)\rightarrow \mathbb{R}$</span> strictly increasing in each coordinate such that <span class="math-container">\begin{equation}
U(x_1,...,x_N)=W(u_1(x_1),...,u_N(x_N))
\end{equation}</span>
represents <span class="math-container">$\succeq$</span>.</p>
<p><strong>Definition:</strong> Given a weak order <span class="math-container">$\succeq$</span>, an index <span class="math-container">$i\in I$</span> is <em>separable</em> if for all <span class="math-container">$x,y,z,z'\in X$</span>, <span class="math-container">$x_iz\succeq y_iz\iff x_iz'\succeq y_iz'$</span>, where <span class="math-container">$x_iz$</span> means <span class="math-container">$x_i$</span> in the i-coordinate of a vector <span class="math-container">$x\in X$</span> and z otherwise. A weak order satisfies <em>singleton separability</em> if all <span class="math-container">$i\in I$</span> are separable.</p>
<p>In <em>Decision Theory</em> by Tomasz Strzalecki chapter 3,, there is the following theorem without proof:</p>
<p><strong>Theorem:</strong> Suppose that <span class="math-container">$\succeq$</span> is a weak order on <span class="math-container">$X=\prod_{i=1}^NX_i$</span>. The ordering <span class="math-container">$\succeq$</span> satisfies singleton separability and continuity if and only if it is has an ordinally separable representation with continuous functions <span class="math-container">$u_i$</span> and <span class="math-container">$W$</span>.</p>
<p>He claims that the proof is quite hard. Does anyone know how to prove this or have a reference?</p>
<p><strong>Remark:</strong> The assumption that <span class="math-container">$X_i$</span> is separable, compact and Banach is not essential here, but let's just make the spaces <span class="math-container">$X_i$</span> as nice as possible for the time being.</p>

CC BY-SA 4.0
2021-04-14T18:54:18.947Z
Some points are nor clear: what is the dimension of $x_i$? What does $x_iz$ mean?
Bertrand
CC BY-SA 4.0
2021-04-14T19:29:14.863Z
What is "Wakker's Decision Theory text?" He wrote more than one book on decision theory.
Michael Greinecker
CC BY-SA 4.0
2021-04-14T19:33:19.203Z
I had the wrong reference. The text is "Decision Theory" by Strzalecki, but he references Wakker's "Additive Representations of Preferences- A New Foundation of Decision Analysis"
Andrew McMillan
CC BY-SA 4.0
2021-04-14T21:38:44.493Z
If $x_i$ is "the i-coordinate of a vector $x\in X$" then it is always possible to write $U(x)=W(u_1(x_1),...,u_N(x_N))$ for any increasing functions $u_i$... Or I probably completely miss the point.
Bertrand
CC BY-SA 4.0
2021-04-14T21:56:11.577Z
@Bertrand You need to construct the functions $W$ and $u_i$...
Andrew McMillan
<mathematical-economics><reference-request><utility><decision-theory>
Ordinally Separable Utility Representation
Andrew McMillan
# Ordinally Separable Utility Representation

(Asked by: username_0 on 2021-04-14)

Let $X\_i$ be a separable, compact, Banach space.

**Definition:** A weak order $\succeq$ on $X=\prod\_{i=1}^NX\_i$ has an *ordinally separable representation* if there exists $u\_i: X\_i\rightarrow \mathbb{R}$ and $W:\prod\_{i=1}^Nu\_i(X\_i)\rightarrow \mathbb{R}$ strictly increasing in each coordinate such that \begin{equation}
U(x\_1,...,x\_N)=W(u\_1(x\_1),...,u\_N(x\_N))
\end{equation}
represents $\succeq$.

**Definition:** Given a weak order $\succeq$, an index $i\in I$ is *separable* if for all $x,y,z,z'\in X$, $x\_iz\succeq y\_iz\iff x\_iz'\succeq y\_iz'$, where $x\_iz$ means $x\_i$ in the i-coordinate of a vector $x\in X$ and z otherwise. A weak order satisfies *singleton separability* if all $i\in I$ are separable.

In *Decision Theory* by Tomasz Strzalecki chapter 3,, there is the following theorem without proof:

**Theorem:** Suppose that $\succeq$ is a weak order on $X=\prod\_{i=1}^NX\_i$. The ordering $\succeq$ satisfies singleton separability and continuity if and only if it is has an ordinally separable representation with continuous functions $u\_i$ and $W$.

He claims that the proof is quite hard. Does anyone know how to prove this or have a reference?

**Remark:** The assumption that $X\_i$ is separable, compact and Banach is not essential here, but let's just make the spaces $X\_i$ as nice as possible for the time being.

**Tags:** mathematical-economics, reference-request, utility, decision-theory

> **Comments:**
> **username_1**: Some points are nor clear: what is the dimension of $x_i$? What does $x_iz$ mean?  
> **username_2**: What is "Wakker's Decision Theory text?" He wrote more than one book on decision theory.  
> **username_0**: I had the wrong reference. The text is "Decision Theory" by Strzalecki, but he references Wakker's "Additive Representations of Preferences- A New Foundation of Decision Analysis"  
> **username_1**: If $x_i$ is "the i-coordinate of a vector $x\in X$" then it is always possible to write $U(x)=W(u_1(x_1),...,u_N(x_N))$ for any increasing functions $u_i$... Or I probably completely miss the point.  
> **username_0**: @username_1 You need to construct the functions $W$ and $u_i$...  

---

## 2 answers

### Answer by username_2 on 2021-04-15. Score: 7 | Accepted answer

Here is the sketch of a proof. All we need is that every continuous weak order on each $X\_i$ admits a continuous utility representation. One sufficient condition is that each $X\_i$ is a connected separable topological space by a theorem of Eilenberg. A proof of Eilenberg's theorem is given in Debreu's book [Theory of Value](https://cowles.yale.edu/sites/default/files/files/pub/mon/m17-all.pdf). Debreu assumes the domain there to be Euclidean, but the proof easily generalizes to the present setting. One can also use the assumption that each $X\_i$ is a second-countable topological space (for metrizable spaces implied by separability). Debreu showed that continuous utility representations exist in that setting too, but the proof is considerably harder.

Define $\succeq\_i$ on $X\_i$ by $x\_i\succeq\_i y\_i$ if $x\_iz\succeq y\_iz$ for some $z$. Clearly, this defines a continuous weak order and the $z$ used does not matter by singleton separability. There exists a continuous utility function $u\_i:X\_i\to\mathbb{R}$.

Next, we can show that $x\_i~\sim\_iy\_i$ for $i=1,\ldots,N$ implies $x\sim y$. By repeated applications of singleton separability we get $(x\_1,y\_2,y\_3,\ldots,y\_N)\succeq (y\_1,y\_2,y\_3,\ldots,y\_N)$, $(x\_1,x\_2,y\_3,\ldots,y\_N)\succeq (x\_1,y\_2,y\_3,\ldots,y\_N)$, and so on, until we have $(x\_1,x\_2,\ldots,x\_{N-1},x\_N)\succeq(x\_1,x\_2,\ldots,x\_{N-1},y\_N)$. By transitivity, $x\succeq y$. Similarly, $y\succeq x$, and, therefore $x\sim y$.

Let $U=\prod\_{i=1}^N u\_i(X\_i)$, a product of (possibly unbounded) intervals. Define $\succeq^\*$ on $U$ by $u\succeq^\* u'$ if for some $x,y\in X$ one has $u=\big(u\_1(x\_1),\dots,u\_N(x\_N)\big)$, $u'=\big(u\_1(y\_1),\dots,u\_N(y\_N)\big)$, and $x\succeq y$. By what we have just shown, $\succeq^\*$ contains, together with $u\_1,\ldots, u\_N$, the same information as $\succeq$. It is also obvious that $\succeq^\*$ is strictly monotone on $U$. If a utility representation $W:U\to\mathbb{R}$ of $\succeq^\*$ exists, it must be strictly increasing. We are done if we can show a continuous such representation exists. Using any of the mentioned utility representation theorems, it suffices to show that $\succeq^\*$ is continuous.

So let $u\succ^\* u'.$ There exist $x,y\in X$ such that $u=\big(u\_1(x\_1),\dots,u\_N(x\_N)\big)$, $u'=\big(u\_1(y\_1),\dots,u\_N(y\_N)\big)$, and $x\succ y$. Since $\succeq$ is continuous and by the definition of the product topology, there exist open neighborhoods $V\_i\ni x\_i$ and $W\_i\ni y\_i$ such that $x'\in\prod\_{i=1}^N V\_i$ and $y'\in\prod\_{i=1}^N W\_i$ implies that $x'\succ y'$. Let $V$ be the interior of $\prod\_{i=1}^N u\_i(V\_i)$ and $W$ the interior of $\prod\_{i=1}^N u\_i(W\_i)$. Then for $u''\in V$ and $u'''\in W$ one has $u''\succ^\* u'''$, so $\succeq^\*$ is, indeed, continuous.

> **Comments:**
> **username_4**: Nice proof! Do we have any reference for this method/proof in literature?  
> **username_2**: @HighGPA I'm not sure, I might have picked up some relevant ideas when looking at Bergson–Samuelson social welfare functions, which tend to have such a structure.  
> **username_4**: Many thanks! I wonder if you happen to have a good text that introduce the Bergson–Samuelson social welfare functions with rigorous mathematical definitions?  

---

### Answer by username_3 on 2021-04-15. Score: 2

I do not have a full answer, but here are my notes when I studied it that hopefully someone can extend to a full answer.

Sketch of Proof:

1. Consider the linear space with basis $\cup\_{i =1}^N X\_i$, and we can identify any $x \in X$ by $\sum\_i x\_i$.
2. Define the convex cone $D = \{\lambda(x-y): x\succeq y;\lambda > 0\}$
3. Let $D^{-}$ be the convex hull of $\{x - y:x \prec y\}$
4. An additive utility representation is a hyperplane through the origin that strictly separates $D$ from $D^{-}$.
5. These sets must be disjoint, as if they were not, we could multiply the weights on convex combinations large enough that at least one coordinate is strictly preferred.

This seems very Mixture Space theoremy so I suspect the proof is not too difficult for those familiar with it.

> **Comments:**
> **username_0**: Thanks for the response, but we aren't looking for an additive representation- simply an ordinal one. Singleton separability is not enough to get additivity.  

---

