Re: Super densely ordered sets

On Feb 15, 3:09 pm, "Mariano Suárez-Alvarez"
<mariano.suarezalva...@xxxxxxxxx> wrote:
On Feb 15, 12:27 pm, marc...@xxxxxxxxx wrote:

Let S be infinite set and > be a strict linear order on S. Let say
that S is super densely ordered by > iff:

For all subsets X and Y of S with |X| <= Aleph_0 and |Y| <= Aleph_0
such that a < b for all a in X, all b in Y, there exists c in S such
that a < c < b for all a in X, all b in Y.

Do super densely ordered sets exist?

If k is a cardinal, let us say that a total order S
is k-super dense if whenever X and Y are
non-empty subsets of S of cardinality at most k,
and such that for all x in X and all y in Y we have
x < y, there exists an element z in S such that
for all x in X and all y in Y we have x < z < y.

Pick any linear order (S, <) and pick a cardinal k.

Consider the set P(S) of all pairs (X, Y) of
non-empty subsets of S such that

- |X| <= k, |Y| <= k;

- for all x in X, for all y in Y, x < y; and

- there does not exist an element z in S such
that for all x in X and for all y in Y,
x <= z <= y

Let S' = S union P(S), and define a relation << on
S' such that

- if x, y are in S, then x << y if and only if
x < y;

- if z is in S and (X, Y) is in P(S), we put
z << (X, Y) iff there exists x in X
such that z < x;

- if z is in S and (X, Y) is in P(S), we put
(X, Y) << z iff there exists y in Y
such that y < z;

- if (X, Y) and (U, V) are in P(S), then
(X, Y) << (U, V) iff there exist y in Y
and u in U such that y < u.

The relation << is anti-symmetric and transitive.
Therefore its reflexive clausure is an order on
S'. It is not necessarily a total order, but
it is well know that there is another relation
<<< on S' extending << which is total.

Notice that whenever (X, Y) is in P(S), we have

x <<< (X, Y) <<< y

for all x in X and all y in Y.

Thus S' is `closer' than S from being k-super dense,
since we've added all the missing points. Of course,
since we have added new points, we cannot claim that
S' itself is k-super dense.

But we can do transfinite induction so as to
repeat the process many times. That is, let us we put
S_0 = S, and define inductively for any ordinal
alpha

S_{alpha+1} = (S_alpha)'

and, when alpha has no immediate predecessor,

S_alpha = union_{beta < alpha} S_beta

with the obvious order.

Then I claim that there exists an ordinal eta such
that S_eta is k-super dense. I don't have space
in this margin, though. (The proof should be along
very standard lines)

If eta is the least ordinal such that S_eta
is k-super dense, call S_eta the k-super dense
closure of S and call eta the k-wholiness of S

If so, do we get smaller collection of such sets if we replace Aleph_0
with a greater cardinal in the above definition?

I have not thought this through, but I would expect that
if k and k' are two cardinals such that k < k', and we start
with a totally ordered set S of cardinal strictly less than k,
then the k-wholiness of S is strictly less that the k'-wholiness
of S, and that the cardinal of the k-super dense closure of S
to be strictly less than the cardinal of the k'-super dense
closure of S. This in particular would entail that
some k-super dense sets are not k'-super dense.

-- m

FWIW, the above construction is not really correct.
One needs a slightly weaker order on the set S', for
otherwise things do not work as well as implied above.

I wrote up the whole thing, hopefully corrected, and
the result is available at <http://mate.dm.uba.ar/~aldoc9/
Publicaciones/Notas/dense.pdf>
Any comments welcome, of course.

-- m
.

Relevant Pages

  • Re: Super densely ordered sets
    ... non-empty subsets of S of cardinality at most k, ... iff there exists x in X ... Thus S' is `closer' than S from being k-super dense, ... and, when alpha has no immediate predecessor, ...
    (sci.math)
  • Re: Super densely ordered sets
    ... let us say that a total order S ... non-empty subsets of S of cardinality at most k, ... iff there exists x in X ... S' itself is k-super dense. ...
    (sci.math)
  • Re: Super densely ordered sets
    ... non-empty subsets of S of cardinality at most k, ... iff there exists x in X ... Thus S' is `closer' than S from being k-super dense, ... Then I claim that there exists an ordinal eta such ...
    (sci.math)
  • Re: Super densely ordered sets
    ... Do super densely ordered sets exist? ... non-empty subsets of S of cardinality at most k, ... iff there exists x in X ... Thus S' is `closer' than S from being k-super dense, ...
    (sci.math)
  • Re: Cardinality of the surreals
    ... But a class is a proper class iff there is a bijective correspondence with the class of all sets: ... Levy's book on set theory gives all the excruciating details of how ... I have no idea how far we can talk about cardinality of proper classes in this theory. ...
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