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Topic: Egyptian topology on Q
Replies: 9   Last Post: May 24, 2004 10:19 PM

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Michael Barr

Posts: 748
Registered: 12/6/04
Egyptian topology on Q
Posted: May 21, 2004 5:22 PM
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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).

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