This is a mathematical appendix to my post "Why you must maximize expected utility", giving precise statements and proofs of some results about von Neumann-Morgenstern utility theory without the Axiom of Continuity. I wish I had the time to make this post more easily readable, giving more intuition; the ideas are rather straight-forward and I hope they won't get lost in the line noise!

The work here is my own (though closely based on the standard proof of the VNM theorem), but I don't expect the results to be new.

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I represent preference relations as total preorders on a simplex ; define , , and in the obvious ways (e.g., iff both and , and iff but *not* ). Write for the 'th unit vector in .

In the following, I will always assume that *satisfies the independence axiom*: that is, for all and , we have if and only if . Note that the analogous statement with weak preferences follows from this: holds iff , which by independence is equivalent to , which is just .

**Lemma 1 **(more of a good thing is always better)**.** *If ** and , **then .*

*Proof. *Let . Then, and . Thus, the result follows from independence applied to , , , and .

**Lemma 2.** *If and , then there** is a unique such that for and for .*

*Proof. *Let be the supremum of all such that (note that by assumption, this condition holds for ). Suppose that . Then there is an such that . By Lemma 1, we have , and the first assertion follows.

Suppose now that . Then by definition of , we do * not* have , which means that we have , which was the second assertion.

Finally, uniqueness is obvious, because if both and satisfied the condition, we would have .

**Definition 3.** is *much better* than , notation or , if there are neighbourhoods of and of (in the relative topology of ) such that we have for all and . (In other words, the graph of is the interior of the graph of .) Write or when ( is *not much better* than ), and ( is *about as good* as ) when both and .

**Theorem 4** (existence of a utility function). *There is a ** such that for all *,

*Unless for all and , there are such that .*

*Proof. *Let be a worst and a best outcome, i.e. let be such that for all . If , then for all , and by repeated applications of independence we get for all , and therefore again for all , and we can simply choose .

Thus, suppose that . In this case, let be such that for every , equals the unique provided by Lemma 2 applied to and . Because of Lemma 1, . Let .

We first show that implies . For every , we either have , in which case by Lemma 2 we have for arbitrarily small , or we have , in which case we set and find . Set . Now, by independence applied times, we have ; analogously, we obtain for arbitrarily small . Thus, using and Lemma 1, and therefore as claimed. Now note that if , then this continues to hold for and in a sufficiently small neighbourhood of and , and therefore we have .

Now suppose that . Since we have and , we can find points and arbitrarily close to and such that the inequality becomes strict (either the left-hand side is smaller than one and we can increase it, or the right-hand side is greater than zero and we can decrease it, or else the inequality is already strict). Then, by the preceding paragraph. But this implies that , which completes the proof.

**Corollary 5.** * is a preference relation (i.e., a total preorder) that satisfies independence and the von Neumann-Morgenstern continuity axiom.*

*Proof. *It is well-known (and straightforward to check) that this follows from the assertion of the theorem.

**Corollary 6.** * is unique up to affine transformations.*

*Proof. *Since is a VNM utility function for , this follows from the analogous result for that case.

**Corollary 7.** *Unless for all , for all the set has lower dimension than (i.e., it is the intersection of with a lower-dimensional subspace of ).*

*Proof. *First, note that the assumption implies that . Let be given by , , and note that is the intersection of the hyperplane with the closed positive orthant . By the theorem, is not parallel to , so the hyperplane is not parallel to . It follows that has dimension , and therefore can have at most this dimension. (It can have smaller dimension or be the empty set if only touches or lies entirely outside the positive orthant.)