
Egyptian topology on Q
Posted:
May 21, 2004 5:22 PM


Let S be the sequence {1,1/2,1/3,...,0}, a compact topological space. Let X be the disjoint union of all the spaces S^n, n in N. Then X is locally compact, Lindelof, normal, nice space. Actually, even metric, but that is not important. Map X to Q by addition: (u_1,u_2,...,u_n) > u_1 + u_2 + ... + u_n. This induces a quotient topology which, for obvious reasons I call the Egyptian topology. I have an argument, not entirely solid, but I think it right, that suggests this is strictly finer than the standard topology. I can prove that it is a normal Hausdorff space. Has anyone seen this before and can anyone give me more information on it?
Incidentally, with a slight modification (replace S^n, but T_n(S) which consists of all sequences (0,...,0,u_i,...,u_n) for which 0 < u_i < ... < u_n, i = 1,...,n+1, you get what really should be called the Egyptian topology, but I am pretty well convinced it is the same. If not, the same question for it.)
The topology can be described as follows. Let the length of a rational r be the number of terms in the shortest possible decomposition of r as a sum of unit fractions. Then the topology is that of convergence of all sequences of bounded length. In other words a function f on Q is continuous iff whenever r_1,r_2,... > r is a sequence whose terms all have length =< l (for some l), then f(r_1), f(r_2_,... > f(r).

