A *compact* set of real numbers is closed and bounded. Show that we cannot find compact sets A_{1}, A_{2}, A_{3}, ... 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 [1/2^{2n}, 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.

© John Scholes

jscholes@kalva.demon.co.uk

14 Jan 2002