[FOM] Axiom of Choice and separation

[K. P. Hart]

On Thu, Aug 16, 2007 at 10:35:01AM +0530, Saurav Bhaumik wrote:
> K. P. Hart wrote:
> > Saurav Bhaumik wrote:

> > 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 `pick an interval' involves choosing one element from one nonempty
set, to wit {I : I is an interval and x is in I and I is a subset of O}.
This set is nonempty by the definition of the order topology and this one
choice is permissible by logic alone.
Once we have I we proceed as follows:
if x has a direct successor or x is the maximum of the set put J_1={y:y<=x},
this is an open interval, otherwise choose two points s and t in I such that
x<s<t and set J_1={y:y<s}.
if x has a direct predecessor or x is the minimum of the set put J_2={y:y>=x},
this is an open interval, otherwise choose two points p and q in I such that
p<q<x and set J_2={y:y>qs}.
Then J = J_1 intersection J_2 is an open interval around x whose closure
is contained in I and hence in O.

The above involves (at most) five choices from nonempty sets and first-order
logic allows that.

> > 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?

I'm not sure if this particular proof is well-known; I thought of it after
you posed the question.

Best,
KP Hart

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