
Re: Nhood Space
Posted:
Jul 1, 2013 10:57 AM


On Sun, 30 Jun 2013 19:03:00 0700, William Elliot <marsh@panix.com> wrote:
>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.
Ah, that's different.
> >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? >

