Re: Super densely ordered sets

On Fri, 15 Feb 2008 marcjhg@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?

Obviously nulset and {a} are super dense linear orders. ;-)

Q, R are not superdense.
Let X = { -1/n | n in N }, Y = { 0, 1/n | n in N }
or X = { 0 | n in N }, Y = { 1/n | n in N }

Let S be a complete linear order.
Let X = { aj | j in N } be an increasing sequence
a1 < a2 < a3 <
X and Y = { sup X } shows S is not superdense.

In fact for a superdense linear order to exist, it's necessary
it's a dense order,
every increasing sequence does not have a supremum,
every decreasing sequence does not have an infinum.

This implies, it's at least totally disconnected.

I'd suppect it's extremely disconnected.

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

I think that there are no super densely ordered denumerable sets
because:

Let assume that S is denumerable and w is an element of S such that
there are x and y in S with x < w < y. Then the sets X = {a : a < w}
and Y = {b : w <= b} will satisfy |X| <= Alpeh_0, |Y| <= Aleph_0 and a
< b for all a in X, all b in Y. However according to the super density
property in this case there must exist c in S such that a < c < b for
all a in X, all b in Y. So c must lie outside of X and Y, but X \/ Y =
S. Therefore c is in S and is not in S, which is a contradiction, so
the assumption is false.

Is this reasoning correct?

Thanks in advance,
Marc

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