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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 $u_i: X\rightarrow \mathbb{R}$ 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.

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    $\begingroup$ Some points are nor clear: what is the dimension of $x_i$? What does $x_iz$ mean? $\endgroup$ – Bertrand Apr 14 at 18:54
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    $\begingroup$ What is "Wakker's Decision Theory text?" He wrote more than one book on decision theory. $\endgroup$ – Michael Greinecker Apr 14 at 19:29
  • $\begingroup$ 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" $\endgroup$ – Andrew McMillan Apr 14 at 19:33
  • $\begingroup$ 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. $\endgroup$ – Bertrand Apr 14 at 21:38
  • $\begingroup$ @Bertrand You need to construct the functions $W$ and $u_i$... $\endgroup$ – Andrew McMillan Apr 14 at 21:56
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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. 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.

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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.

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  • $\begingroup$ 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. $\endgroup$ – Andrew McMillan Apr 15 at 1:43

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