[FOM] Axiom of Choice and separation

[K. P. Hart]

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?
>   
If by T_(2.5) and Completely Hausdorff you mean separation of points by 
continuous
functions then it and regularity are incomparable, i.e., your hierarchy 
is not linear
(there are regular spaces on which every real-valued continuous function 
is constant).
Also: the implication T_4 -> T_(3.5) needs a bit of choice via Urysohn's 
Lemma
(Dependent Choice suffices).

Linearly ordered spaces are regular: if x in O then first pick an 
interval I between x and O
and then shrink I to an interval J whose closure is contained in I (you 
have to consider
cases depending on whether the order is dense in either piece of I).
The question about Complete Hausdorffness is equivalent to asking about 
Complete
Regularity and can be proven using Dependent Choice only but I haven't 
gotten rid of
Choice completely.
> What if we assume that the order is dense? What if it is order complete?
> What if it is connected?
>   
The problem is mainly with density: you need to build a copy of the 
rationals in the
Dedekind completion and that is where some Choice seems necessary.

> 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?
>   
No: given a point x and an open set G define A(x,G) to be the following 
subset of
the field: {a>0: for all n the ball B(x,na) is a subset of G}; this set 
is nonempty by
non-Archimedeanness.
Now let U(x,G) be the union of the B(x,a) with a in A(x,G).
If x is not in H and y is not in G then d(x,y) is infinitely larger than 
all elements of A(x,G)
and of A(y,H) and so, by the triangle inequality, U(x,G) and U(y,H) are 
disjoint.

KP Hart

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