Uncountable subsets of omega_1
William Elliot
Has anybody anything to add to this?
Let A be an uncountable subset of omega_1.
Then A is well ordered and being uncountable
has order type omega_1. The order type is at least omega_1
because omega_1 is the smallest uncountable ordinal and it
can not be greater than omega_1 because A can have no element
a with uncountably elements less than a.
Thus every uncountable subset of omega_1 is order isomorphic to
omega_1, hence homeomorphic to omega_1.
Similar can be claimed of all initial ordinals.
William Elliot
This all seems correct. Now however let's ask if the order topology
of A is homeomorphic to the subspace topology. I think not.
Is the following correct and how would it be shown?
The subspace topology of A is homeomorphic to omega_1
iff A is closed.
Dave L. Renfro
Well, you need the subset to be uncountable, which you
said in your earlier post but left out here, but unless
I'm missing something, this is all you need. In particular,
you don't need to assume the subset is closed in omega_1.
Every uncountable well-ordered set, such that each of its
proper initial segments is countable, is order-isomorphic
to omega_1.
This is fairly immediate from the fact that, given two
well-ordered sets (B, <) and (B', <'), exactly one of the
following holds:
(1) (B, <) is order-isomorphic to (B', <').
(2) (B, <) is order-isomorphic to a proper initial
segment of (B', <').
(3) (B', <') is order-isomorphic to a proper initial
segment of (B, <).
Among other things, this implies that every uncountable
subset (closed or not) of omega_1 is order-isomorphic
to omega_1. Since the topological type of linearly
ordered sets does not change for any two linearly ordered
sets belonging to the same order type, it follows that
every uncountable subset of omega_1, with the subspace
topology it inherits from omega_1, is homeomorphic to
omega_1.
Dave L. Renfro
G. A. Edgar
> William Elliot wrote:
>
> > Is the following correct and how would it be shown?
> > The subspace topology of A is homeomorphic to omega_1
> > iff A is closed.
>
L. Renfro <renf...@cmich.edu> wrote:
> Well, you need the subset to be uncountable, which you
> said in your earlier post but left out here, but unless
> I'm missing something, this is all you need. In particular,
> you don't need to assume the subset is closed in omega_1.
>
> Every uncountable well-ordered set, such that each of its
> proper initial segments is countable, is order-isomorphic
> to omega_1.
>
But that may not be the subspace topology.
For example, A is the set of all non-limit
ordinals. A is uncountable and discrete in
the subspace topology, not homeomorphic
to omega_1 .
--
G. A. Edgar http://www.math.ohio-state.edu/~edgar/
Dave L. Renfro
>> [...] but unless I'm missing something, this is all
>> you need. In particular, you don't need to assume the
>> subset is closed in omega_1.
>>
>> Every uncountable well-ordered set, such that each of its
>> proper initial segments is countable, is order-isomorphic
>> to omega_1.
G. A. Edgar wrote:
> But that may not be the subspace topology.
> For example, A is the set of all non-limit
> ordinals. A is uncountable and discrete in
> the subspace topology, not homeomorphic
> to omega_1 .
I guess I _was_ missing something. Although the order type
determines the topological type for *linearly ordered sets*
by the usual method of assigning a topology to a linearly
ordered set, after choosing a subspace of omega_1 the object
we have is a *linearly ordered topological space*. That is,
a topology is already in place, and there's no a prior reason
why this topology should be the same as the one we'd get by
the usual assignment of topologies to linearly ordered sets.
And, in fact, it's _not_ always the same topology, as you
pointed out.
Dave L. Renfro
William Elliot
Newsgroups: sci.math
Subject: Re: Uncountable subsets of omega_1
> > William Elliot wrote:
> > > Is the following correct and how would it be shown?
> > > The subspace topology of A is homeomorphic to omega_1
> > > iff A is closed.
> Dave L. Renfro <renf...@cmich.edu> wrote:
> > Well, you need the subset to be uncountable, which you
> > said in your earlier post but left out here, but unless
> > I'm missing something, this is all you need. In particular,
> > you don't need to assume the subset is closed in omega_1.
>
> > Every uncountable well-ordered set, such that each of its
> > proper initial segments is countable, is order-isomorphic
> > to omega_1.
> But that may not be the subspace topology.
> For example, A is the set of all non-limit ordinals.
> A is uncountable and discrete in the subspace topology,
> not homeomorphic to omega_1.
Or for example, in the subspace
omega_1\omega_0 = { xi | xi < omega_1 }\omega_0
the set
omega_0 = { xi | xi < omega_0 }
is closed and open while in the intrinstic linear order topology it is
open and not closed.
I think the following may be useful. Does
it answer the question in the affrimative?
--
Let (X,<) be a linear order and A be a subset of X.
DEFINITION
A point p in X is an UPPER LIMIT POINT of A if p is not the first
element of X and for all x in X, x < p implies there is a in A such
that x < a < p. Similarly define LOWER LIMIT POINT
A point is a TWO-SIDED LIMIT POINT of A if it is both
an upper limit point and lower limit point of A.
Let X be a linear order with the order topology.
If Y is a subset of X, there are two natural topologies on Y:
(1) The relative topology on Y as a topological subspace of X,
(2) the order topology on Y as a suborder of X.
FACT
The order topology (2) on Y is weaker than the subspace topology (1).
Example: Let X = (R,<) and Y = [0,1) \/ [2,3].
Then the order topology on Y is strictly weaker than the subspace
topology on Y; Y with the order topology is homeomorphic to [0,1],
but Y with the subspace topology is neither connected nor compact.
Under certain conditions the order topology on a subset coincides
with the subspace topology:
FACT
Let X be a linear order with the order topology. For a subset Y of X,
if every point of Y is a two-sided limit point of Y, then the order
topology on Y coincides with the subspace topology.
----
Rupert
Yes, this is correct.
> Similar can be claimed of all initial ordinals.
What generalization do you have in mind here? I can think of at least
two possible generalizations, and only one of them is correct.
William Elliot
then order type A = eta. In addition if eta is closed subset,
then the subspace topology of A is homeomorphic to eta.
This latter and the converse are yet to be established. See my
other post which describes instances when subspace topology and
intrinsic linear order topology of the subset coincide. I think
that is key but have yet to wade through the details or find insight
to make it easy to manage. On the other hand, perhaps there's a less
complex way.