# [FOM] Axiom of Choice and separation

K. P. Hart K.P.Hart at tudelft.nl
Thu Aug 16 07:06:13 EDT 2007

On Tue, Aug 14, 2007 at 08:56:30AM +0530, Saurav Bhaumik wrote:
> Dear Experts,
>
> 1. It is evident that if we assume Axiom of Choice, any linearly ordered
> space is Normal, not only that, it is monotonically normal. But in
> absence of Axiom of choice, however, a non-normal orderable space can be
> constructed.
>
> Consider the hierarchy: T_0<T_1<T_2<T_(2 1/2)<T_3<T_(3 1/2)<T_4 (or any
> denser separation hierarchy).
>
> Within ZF, how far can a general linear order might go?
> It is evidently T_2.
>
> Is it Completely Hausdorff? Is it regular?
>
> What if we assume that the order is dense? What if it is order complete?
> What if it is connected?

After some further searching:

In
\bib{MR0143705}{article}{
author={L{\"a}uchli, H.},
title={Auswahlaxiom in der Algebra},
language={German},
journal={Comment. Math. Helv.},
volume={37},
date={1962/1963},
pages={1--18},
issn={0010-2571},
review={\MR{0143705 (26 \#1258)}},
}
(or http://www.ams.org/mathscinet-getitem?mr=143705 )
one finds a permutation model in which the rationals in in the interval [0,1]
form an ordered continuum that satisfies the T_4-axiom, yet every continuous
real-valued function defined on it is constant.
Therefore without sufficient choice Urysohn's Lemma need not hold and
linearly ordered topological spaces need not be completely regular.

Connectedness is the only stumbling block:
if no interval in the space is connected then there are sufficiently
many 0-1-valued functions to separate points as well as points and
closed sets so those linearly ordered spaces are completely Hausdorff
and completely regular.

KP Hart

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