Countable space

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?

 

[The quoted message is from a post submitted by William Elliot<>,
not by Maarten Bergvelt; something went wrong in the moderating
process. Sorry. MB]

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

 

No, removing infinitely many isolated points can create more isolated
points. For example, S = { m + 1/n | m,n in N }

The set of non-isolated points of S is N.
So the set of isolated points of S is S\N
and S - (S\N) = N is all isolated points.

Note, S is homeomorphic to (omega_0)^(omega_0).

Let S^(1) = S with it's isolated points removed.
S^(eta + 1) S^(eta) with it's isolated points removed
S^(eta) = /\{ S^(xi) | xi < eta } when eta is a limit ordinal.

There will be a countable ordinal beta with S^(beta + 1) = S^(beta).
S^(beta), called the perfect kernel, will not have any isolated points.

An interesting illustration of this is the countable ordinal
epsilon_0 = (omega_0)^(omega_0)^(omega_0)^...

Now starting with the perfect kernel of S embedded in Q,
is it possible to back tract to S, embedding each of
a transfinite series of increasing subspaces?

 

Wrong: S-X may have isolated points. One can deduce the first
theorem from the second, but for this one needs to _add_
points: replace each element of x of X by an interval i_x
from Q and then identify x with 0 in i_x. The extended
space is still countable and has no isolated points, so
is homeomorphic to Q. Conseqently, S as a subset is
homeomorphic to subset of Q.

OTOH, both theorems have similar proofs, with the second
beeing harder (more datails to get right), so I feel that
the second should be considered extension of the first.

 

Nice proof. Another suggestion in a similar vein, was to
use the second theorem to prove SxQ is homeomorphic to Q.

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