>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, S-X is homeomorphic to Q, and therefore also to the subset Q+ of positive rationals; let f be a homeomorphism from S-X 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 S-X 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.