[FOM] Axiom of Choice and separation
Saurav Bhaumik
saurav1b at gmail.com
Thu Aug 16 01:05:01 EDT 2007
K. P. Hart wrote:
> 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).
>
Yes, my hierarchy was not linear; instead, the one T_2<T_3<T_4, which is
linear, may be the hierarchy I should talk about.
> 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).
>
I have, forgive me, doubt whether this absolutely excludes choice; when
you say "pick an interval" there is a full course choice, and when you
say "shrink I to J whose..." there is again a matter of choice. If
choice was to be taken, much more than regularity can be asserted.
Can you, if I am not wrong in the above, please give a proof or
counterexample that a linear order (perhaps dense) is necessarily regular?
> 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.
>
Great proof! Thanks a lot.
Is this a well known point or you devised this proof after I had asked it?
Saurav.
> KP Hart
>
>
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