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Re: Countable space
Posted:
Nov 28, 2011 11:30 AM


[The quoted message is from a post submitted by William Elliot<marsh@rdrop.com>, not by Maarten Bergvelt; something went wrong in the moderating process. Sorry. MB]
bergv@u19.math.uiuc.edu (Maarten Bergvelt) writes:
>Let S be a countable, 2nd countable, regular T0 space >(equivalently, countable metrizable space). How to >show that S embeds in the rationals? > >Compare with the theorem: if S is a countable, 2nd countable, >regular T0 space without isolated points (equivalently, >perfect countable metrizable space) then S is homeomorphic to Q. > >Are the proofs of these two theorems similar? >Is the latter proof an extension or corollary >of the former?
It seems offhand (and the moderators will surely correct me if I'm wrong...) as if the (implicit) first theorem is a pretty immediate corollary of the (explicit) second theorem. Let S be a countable, 2nd countable, regular T0 space, and X its subset of isolated points. By the second theorem, SX is homeomorphic to Q, and therefore also to the subset Q+ of positive rationals; let f be a homeomorphism from SX onto Q+. On the other hand, I think (but could be wrong) that X is a finite or countable discrete space; if so, let g be a homeomorphism from X into the negative integers. Then h from S to Q, defined to be f on SX and g on X, is an embedding of S in Q.
Note that, even if correct, this doesn't answer the last two questions you asked; but it does (even if incorrect) answer the first one.
Lee Rudolph



