From: Fred Galvin <galvin@math.ukans.edu>
Subject: Re: normal linear orders
Date: Fri, 20 Jul 2001 02:07:56 -0500
Newsgroups: sci.math
Summary: Linearly Ordered Topological Spaces are completely normal

On Thu, 19 Jul 2001, William Elliot wrote:

> Let S be linear or total ordered set with the open interval topology.
> It's not so hard to show S is regular, but how prove S is normal...
> [...]
> Is this a viable approach to this problem or is there an easier way?

I don't know; I'm too lazy to study your approach. However, here is a
proof of that theorem from some ancient lecture notes of mine. My
notes are kind of sketchy, so you will still have some work to do
filling in the details.

DEFINITION. A topological space is "hereditarily normal" (or
"completely normal") if every subspace is normal. Equivalently, X is
hereditarily normal if, whenever A and B are subsets of X such that
each is disjoint from the closure of the other, there are disjoint
open sets U and V such that A is a subset of U and B is a subset of V.

THEOREM. Every linearly ordered topological space (LOTS) is
hereditarily normal.

PROOF. Let X be a LOTS, and suppose A and B are subsets of X, each
disjoint from the closure of the other. We may assume, without loss of
generality, that no point of A or B is an endpoint of X. [That is
mainly for notational convenience.] For each point a in A choose p_a
and q_a in X so that:

(1) p_a < a < q_a;

(2) (p_a, q_a) is disjoint from B;

(3) either (i) q_a is in A,
    or else (ii) (a, q_a) is empty,
    or else (iii) q_a is not in B, and (a, q_a) is disjoint from A;

(4) either (i) p_a is in A,
    or else (ii) (p_a, a) is empty,
    or else (iii) p_a is not in B, and (p_a, a) is disjoint from B.

Let U be the union of the intervals (p_a, q_a) over all a in A; then B
is disjoint from the closure of U, so let V be the complement of the
closure of U.
==============================================================================

From: Fred Galvin <galvin@math.ukans.edu>
Subject: Re: normal linear orders
Date: Fri, 20 Jul 2001 02:28:09 -0500
Newsgroups: sci.math

On Fri, 20 Jul 2001, Fred Galvin wrote:

> (4) either (i) p_a is in A,
>     or else (ii) (p_a, a) is empty,
>     or else (iii) p_a is not in B, and (p_a, a) is disjoint from B.
                                                     ^^^^^^^^^^^^^^^
OOPS, sorry! That should read "disjoint from A".

[quotes of previous message trimmed --djr]