Countable space

[Maarten Bergvelt]

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?

[Lee Rudolph]

[The quoted message is from a post submitted by William Elliot<ma...@rdrop.com>,
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

[William Elliot]

On Mon, 28 Nov 2011, Lee Rudolph wrote:
>
>> 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+.

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?

[Waldek Hebisch]

Lee Rudolph <lrud...@panix.com> wrote:
> [The quoted message is from a post submitted by William Elliot<ma...@rdrop.com>,
> not by Maarten Bergvelt; something went wrong in the moderating
> process. Sorry. MB]
>
> be...@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;

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.

--
Waldek Hebisch
heb...@math.uni.wroc.pl

[William Elliot]

On Wed, 30 Nov 2011, Waldek Hebisch wrote:
> Lee Rudolph <lrud...@panix.com> wrote:
>>
>>> 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.
>>>

> 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.

> Consequently, S as a subset is homeomorphic to subset of Q.
>
Nice proof. Another suggestion in a similar vein, was to
use the second theorem to prove SxQ is homeomorphic to Q.

> OTOH, both theorems have similar proofs, with the second

> beeing harder (more details to get right), so I feel that

> the second should be considered extension of the first.

----