1991 IMO Problems/Problem 6
An infinite sequence of real numbers is said to be bounded if there is a constant such that for every .
Given any real number , construct a bounded infinite sequence such that for every pair of distinct nonnegative integers .
Since , the series is convergent; let be the sum of this convergent series. Let be the interval (or any bounded subset of measure ).
Suppose that we have chosen points satisfying
for all distinct . We show that we can choose such that holds for all distinct . The only new cases are when one number (WLOG ) is equal to , so we must guarantee that for all .
Let be the interval , of length . The points that are valid choices for are precisely the points of , so we must show that this set is nonempty. The total length is at most the sum of the lengths . This is .
Therefore the total measure of is , so has positive measure and thus is nonempty. Choosing any and continuing by induction constructs the desired sequence.
The argument above would not work for , since only converges for . But Osmun Nal argues in this video that satisfies the stronger inequality for all distinct ; in other words, this sequence simultaneously solves the problem for all simultaneously.