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Re: Countable space
Posted:
Nov 28, 2011 11:30 AM
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[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, 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.
Lee Rudolph
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