29th Putnam 1968

A compact set of real numbers is closed and bounded. Show that we cannot find compact sets A₁, A₂, A₃, ... such that (1) all elements of A_n are rational and (2) given any compact set K whose members are all rationals, K ⊆ some A_n.

Solution

Use a diagonalisation argument.

Suppose we can find such A_n. Let S_n be the interval [1/2²ⁿ, 1/2^2n-1], and let T_n be the set of rational points in S_n. Then the closure of T_n is S_n, which contains irrational points, so there must be points of T_n which are not in A_n. Let x_n be one such. Now consider {x_n}. Its only limit point is 0, so K, the union of {x_n} and {0}, is a compact set of rationals. But K is not contained in any of the A_n, because it has a member x_n not in A_n for each n. Contradiction.

29th Putnam 1968

© John Scholes
jscholes@kalva.demon.co.uk
14 Jan 2002