
Re: Nhood Space
Posted:
Jun 30, 2013 10:03 PM


On Sun, 30 Jun 2013, dullrich@sprynet.com wrote: > On Sat, 29 Jun 2013 21:39:59 0700, William Elliot <marsh@panix.com> > wrote: > > >(S,<<) is a nhood space when << is a binary relation for P(S) and > >for all A,B,C subset S > > empty set << A << S > > A << B implies A subset B > > A << B implies S\B << S\A > > A << B/\C iff A << B and A << C > > (i) this is an actual definition you saw somewhere
Yes, you'll find it included with proximity spaces in Wikipedia, Willard and other places.
> _If_ (ii), then it seems likely that this is an > attempt to axiomatize the relation "A is a > nbd of B" in a topological space. But it > definitely does not do that. If we start > with a topological space and then define > << by saying A << B if B is a nbd of A > then the axiom
It's a generalization of uniform spaces.
> A << B implies S\B << S\A If A << B is interpeted to mean cl A subset int B, then cl S\B subset int S\A.
As you've shown, A << B can't be interpeted to mean A subset int B.
> does not hold. (For example, if A is any > open set then A << A. But S\A = C can > be any closed set, and a closed set C need > not be a nbd of itself.) This indicates why the topology inheritent in nhood, or proximity spaces are general topological spaces. If fact, their Tychonov.
> >A << B is taken to mean B is a nhood of A. > >Thus {x} << A would mean A is a nhood of x. > > > >Additional axioms are separation > > for all x,y, if x /= y, then {x} << S\y > >and normality > > for all A,B, if A << B, then there's some K with > > A << K << B > > > >Useful theorems > > The statements below cannot be theorems, because > there are no hypotheses!
Read it in the consted of nhood spaces.
> Of course it's easy to guess what you intend the hypotheses to be  the > problem is that there are at least two natural guesses. Presumably you're > assuming we have a Nhood space. _Are_ you assuming the "additional axioms"? > Include them as or if needed. Separated nhood spaces are T1 and separated normal nhood spaces are Hausdorff.
Often normal is included as part of the definition.
> > are > >A << B, B subset C implies A << C > >A subset C, B << C implies A << C > > That must be a typo for > > A subset B, B << C implies A << C.
Correct.
> >Define the interior of a set A, int A = { x  {x} << A }. > >Easy theorems are > > int empty set = empty set; int S = S > > int A/\B = int A /\ int B; int A subset A > > A subset B implies int A subset int B. > > > >How would one prove int int A = int A? > > I don't know. I don't have a counterexample, > but I tend to doubt that it's true. (I also tend > to suspect I'm correct in my speculations above, > meaning that you simply don't have the definition > of "Nhood space" right...) > Nhood spaces are the DeMorgan like duals of proximity spaces. Williard gives a closure operator cl, for a proximity space and leaves it as an exercise to show cl is a closure operator. The part I'm having trouble with is proving cl cl A = cl A, which in the dual nhood space is int int A = int A.
> >Since int A subset A, int int A subset int A. > >So the question actually is how to prove > > int A subset int int A? > > > >Another question. > >If for all x in A, {x} << B, is A << B provable? > > > >From the axiom > > A << B/\C iff A << B and A << C > > > > S\B \/ S\C << S\A iff S\B << S\A and S\C << S\A > > > >Thus for all A,B,C > > B \/ C << A iff B << A, C << A > > > >Hence if A if finite and for all x in A, {x} << B, then A << B. > >What if A isn't finite?

