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A preference relation on lotteries is complete when every pair is comparable, allowing an indifference relation. Completeness does not require strict ranking of distinct lotteries. Together with a transitive relation, it gives the ordering assumption in the Von Neumann-Morgenstern utility theorem.

For a preorder, the indifference relation is an equivalence relation. Passing to its equivalence classes removes ties and produces a partial order.

First pass from the preorder $R$ to a partial order. Define the indifference relation $y\sim z$ by $yRz$ and $zRy$. Reflexivity and transitivity make this an equivalence relation. On the quotient set $C={\mathbb{R}}^L/\sim$, putTransitivity makes this independent of the representatives, and the quotient relation is antisymmetric, hence a partial order.

We need its order-extension property, including freedom to order incomparable classes. Here is the argument. Order all partial order extensions of $⊵$ by inclusion. The union of a chain of extensions is again a partial order: any finite list of comparisons needed to check an axiom occurs together in one member of the chain. Zorn's lemma therefore supplies a maximal extension. If two classes $A,B$ remain incomparable, adjoin $A⊵B$ and take the transitive closure of a relation. The only possible new comparisons have the form $X⊵A⊵B⊵Y$. A cycle would require a previously existing path $B⊵A$, contradicting incomparability. The enlarged relation is still a partial order, contradicting maximality. Consequently the maximal extension is a total order. This also proves the Szpilrajn extension theorem used here.

Lift such a total order to the original bundles by comparing their equivalence classes. The lifted preference relation $R^{\prime }$ is complete and transitive, contains $R$, and preserves its strict preference: if $yPz$, their distinct classes satisfy $[y]⊵[z]$, and antisymmetry of the extended order prevents the reverse comparison. In particular, the class $\mathcal{P}$ is nonempty and $R\subseteq {\bigcap }_{R^{\prime }\in \mathcal{P}}{R}^{\prime }$.

For the reverse inclusion, take a pair with $\neg (yRz)$. If $zRy$, then $zPy$, so every member of $\mathcal{P}$ strictly ranks $z$ above $y$ and excludes $y{R}^{\prime }z$. If neither comparison holds, their classes are incomparable. First adjoin $[z]⊵[y]$, which creates no cycle by the preceding argument, and then extend to a total order. Its lift belongs to $\mathcal{P}$ and again excludes $y{R}^{\prime }z$. Thus every pair outside $R$ is outside at least one member of $\mathcal{P}$, provingThe intersection proof is an order-theoretic fact about any preorder. The congruence axiom additionally ensures exact rationalization of the observed demand correspondence: every chosen bundle weakly outranks every affordable bundle, while an affordable unchosen bundle is strictly worse by part (b). Hence each completion's budget-maximizing set is exactly the observed chosen set.

No: uniform convergence of utilities on bounded sets does not imply closed convergence of preferences. It does imply the upper condition. Indeed, suppose $x_j\to x$, $y_j\to y$ and $u_{n_j}(x_j)\ge u_{n_j}(y_j)$. These bundles all lie in one bounded ball. Uniform convergence there, together with continuity of $u$, givesso $x⪰y$.

For a strict limiting comparison $u(x)>u(y)$, the fixed pair itself satisfies $u_n(x)>u_n(y)$ eventually. The obstruction is the lower condition at an indifference relation, where the strict approximating rankings need not disappear.

For a counterexample valid in every dimension $L\ge 1$, takeOn $\Vert x\Vert \le k$, $|u_n(x)-u(x)|\le k/n\to 0$, exactly as required. Every $u_n$ induces $x{⪰}_{n}y$ if and only if $x_1\ge y_1$, while $u$ makes every pair indifferent. In particular, $0⪰e_1$ holds in the limit. Any sequences $x_n\to 0$, $y_n\to e_1$ eventually have $(x_n{)}_1<(y_n{)}_1$, so they cannot satisfy $x_n{⪰}_n{y}_n$. The lower graph-limit condition fails.

Equivalently, an open neighborhood of $(0,e_1)$ sufficiently small to keep the first coordinate of its first bundle below that of its second bundle meets the limiting graph and none of the approximating graphs. The corresponding hit-open neighborhood in the Fell topology proves failure of convergence directly. HenceThe uniform utility convergence need not preserve closed preference convergence phenomenon reflects the loss of ranking information when the limiting utility has a flat indifference region.

${\ell }_2\succ {\ell }_1$. The distinction is small but strict: their normalized expected utilities are $18/25$ and $7/10$, respectively.

By the Von Neumann-Morgenstern utility theorem, use a representing utility function and apply a positive affine function to set $u({\omega }_1)=1$ and $u({\omega }_4)=0$. This is possible because the given strict preference implies $u({\omega }_1)>u({\omega }_4)$. The two specified indifference relations then giveTaking the expected utility of each lottery yieldsTheir difference is $1/50>0$, establishing the stated strict preference. These numbers measure utility, not monetary reward or an assumed ordering of the subscripts.

Here is the underlying argument directly in terms of the axioms. The independence axiom for lottery preferences permits substitution of indifferent lotteries inside a common mixture. Replacing ${\omega }_3$ by its given mixture of ${\omega }_1,{\omega }_4$, and then replacing ${\omega }_2$, followed by reduction of compound lotteries, givesand consequentlyThe transitive relation axiom combines the successive substitutions. Finally, mixture monotonicity says that among mixtures of a strictly better and a strictly worse outcome, the greater probability of the better outcome is strictly preferred; this follows from the independence axiom for lottery preferences and transitivity. Since $18/25>7/10$ and ${\omega }_1\succ {\omega }_4$, the comparison follows. Complete lottery preferences and continuity of lottery preferences are part of the representation theorem's hypotheses; the direct substitution calculation uses the already supplied indifferences and does not need to invoke continuity of lottery preferences again.^[ref]

Under the expected utility hypothesis, use positive affine uniqueness of affine preference representations to normalize the extreme outcomes to $u({\omega }_1)=1$ and $u({\omega }_4)=0$. The stated indifference relations then giveThe two lotteries have expected utilitiesHence ${\ell }_2\succ {\ell }_1$, with normalized expected utility difference $3/100$. Equivalently, substitution reduces them to lotteries with probabilities $0.45$ and $0.48$ of the better extreme outcome. The inference uses the independence axiom for lottery preferences; the three displayed outcome comparisons alone would not determine preferences over all compound lotteries without the expected utility hypothesis.

A weak preference relation records that one alternative is at least as desirable as another. Rational preferences are complete and transitive. Their strict preference and indifference relation are the asymmetric and symmetric parts. On a consumption space, continuity is expressed by a closed set of weakly ranked pairs.

Apply the independence axiom for lottery preferences in both weak directions of the indifference relation. At $p=0$ the two mixtures are identical. Repeating this substitution and using transitivity lets one replace any finite collection of indifferent component lotteries, after which reduction of compound lotteries combines their outcome probabilities.

Every partial order has a total order extension. Apply Zorn's lemma to extensions ordered by inclusion: the union of a chain is a partial order. If a maximal extension leaves two elements incomparable, adjoining one comparison and taking its transitive closure of a relation creates no cycle, contradicting maximality. Quotienting a preorder by its indifference relation and lifting a total order gives a complete weak-order extension preserving its strict part.