Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 36 / 5

Put $X={\mathbb{R}}_{+}^L$ and identify a continuous preference relation $R$ with its closed graph $G_R=\{(x,y)\in X\times X:xRy\}$. The closed convergence of preference relations is the topology induced by the Fell topology on these closed subsets of $X\times X$.

Concretely, a subbasis consists of the setsFinite intersections of these requirements form a basis, restricted to the specified complete, transitive, continuous preference relations. Thus a neighborhood may require some comparisons in prescribed open regions and forbid any comparison in prescribed compact regions. The use of compact sets in the avoidance condition is what makes this the Fell topology, rather than a stronger hyperspace topology on an unbounded consumption space.

There are two equivalent graph-limit requirements. Both are necessary. The upper condition is that, for every subsequence $n_j$ and convergent pairs $x_j\to x$, $y_j\to y$,The lower condition is that every limiting comparison can be approximated: for every $x⪰y$, there exist $x_n\to x$ and $y_n\to y$ withThe upper condition excludes spurious limiting comparisons; the lower condition prevents genuine limiting comparisons from disappearing. The subsequence quantifier in the upper condition is important: a violation need occur along only some of the indices. On the locally compact separable metric space $X\times X$, these two conditions are exactly closed convergence of preference relations.

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.