
Re: Subspaces of limit ordinals
Posted:
Jun 30, 2014 11:40 PM


On Mon, 30 Jun 2014, Herman Rubin wrote: > On 20140630, William Elliot <marsh@panix.com> wrote:
> > Let eta be a limit ordinal and A a subspace of eta. > > > Is the following correct? > > A is homeomorphic to eta iff A is an unbounded, closed subset. > > Here is a counterexample. Let eta = omega^2, omega being the order > type of the rationals. Then the sequence omega*n, n \in omega, is > unbounded and closed, but is homeomorphic to omega, not eta.
As neither omega nor eta is an ordinal, how is that a counter example
Here's a counter example. Let eta = omega_0 + omega_0 and A = { omega_0 + n  n in omega_0 } Unbounded, closed A homeomorphic to omega_0.
> The theorem is true if eta is a cardinal, not the sum of a smaller > number of cardinals.
Here's a counter example. Let eta = aleph_(omega_0), A = { aleph_n  n in omega_0 } Unbounded, closed A homeomorphic to aleph_0.
Is the theorem true only for regular cardinals?
On the other hand, is there a counter example to the proposition that if A (subset eta) is homeomorphic to (limit ordinal) eta then A is an unbounded, closed subset.
Yes, eta = omega_1, A = eta  {omega_0} which isn't closed.
Thus what is the theroem? Is this it? If kappa is a regular cardinal and A an unbounded, closed subset of kappa, then A is homeomorphic to kappa.

