6
$\begingroup$

Given a totally ordered set $I$.

It is well known that $I$ becomes a normal Hausdorff space when endowed with the order topology.

What can be said about finite products $I \times \cdots \times I$ ? When are they normal? Are there sufficient and necessary conditions on the order of $I$ for all finite powers to be normal?

$\endgroup$
4
  • $\begingroup$ Do you have an example of when it's not normal? I'm having trouble thinking of one. $\endgroup$
    – Will Brian
    Commented Dec 10, 2016 at 12:07
  • $\begingroup$ I haven't read the paper below, but it seems that it contains relevant information in its Theorem 1. The author (R.A. Conover) apparently only looks at locally compact linearly ordered topological spaces, but provides some necessary and sufficient conditions for a finite product of such spaces to be normal. sciencedirect.com/science/article/pii/0016660X72900086 $\endgroup$ Commented Dec 10, 2016 at 19:53
  • $\begingroup$ @MathieuBallif: Thanks a lot for this paper. It contains all the information I needed! If you write this links as an answer, I would be happy to accept it! $\endgroup$
    – Tom
    Commented Dec 11, 2016 at 16:57
  • $\begingroup$ Glad you found it helpful ! $\endgroup$ Commented Dec 12, 2016 at 22:11

1 Answer 1

4
$\begingroup$

To answer Will Brian's question, notice that the product space $(\omega_1 + 1) \times \omega_1$ is not normal. (This fact is well-known; the Pressing Down Lemma implies that the diagonal and the right edge cannot be completely separated.) Let $I$ be the ordinal space $\omega_1 + \omega_1$. Since $I$ contains each of the spaces $\omega_1 + 1$ and $\omega_1$ as a closed subspace, the product $I \times I$ is not normal.

$\endgroup$
1
  • $\begingroup$ Yes, this is a very nice example! :-) $\endgroup$
    – Tom
    Commented Dec 11, 2016 at 16:59

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .