Countable space

Discussion in 'Math Research' started by Maarten Bergvelt, Nov 27, 2011.

  1. 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?
    Maarten Bergvelt, Nov 27, 2011
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  2. Maarten Bergvelt

    Lee Rudolph Guest

    [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
    Lee Rudolph, Nov 28, 2011
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  3. 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?
    William Elliot, Nov 30, 2011
  4. 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, Nov 30, 2011
  5. Nice proof. Another suggestion in a similar vein, was to
    use the second theorem to prove SxQ is homeomorphic to Q.
    William Elliot, Dec 1, 2011
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