[FOM] Axiom of Choice and separation
Saurav Bhaumik
saurav1b at gmail.com
Mon Aug 13 23:26:30 EDT 2007
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?
2. GENERAL metric space:
A general metric on a nonempty set X may be considered a function
d: XxX --> F, with the commutativity, triangle inequality etc , where F
is an arbitrary ordered field.
It is evident that a general metric space is monotonically normal if
Axiom of Choice is assumed, or if the field is Archimedean:
For then, if x in G,G open, we may take μ(x,G): = B(x,1 / 2n(x,G)),
where n(x,G) = min{n in N: B(x,1/n) \subset G}, and the trick is done
without choice.
But if the field is NOT Archimedean, is there an example without choice
that a general metric may not be normal?
Thanks in advance,
Saurav Bhaumik
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