
Re: Nhood Space
Posted:
Jul 2, 2013 1:19 AM


On Mon, 1 Jul 2013, fom wrote: > On 6/30/2013 9:03 PM, William Elliot wrote:
> > 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. > > Look at the example in Willard concerning > the diagonal uniformities. And then look > at the definition for neighborhoods in a > uniformity in the previous sections.
Arrg. It's hard to follow what your thinking and how it apllies to the problem at hand. From what you've consider, can you directly show cl cl A = A where cl A = { x  {x}nA }, from the axioms (separable or normal included if needed) for a proximity space?
n proximity for S when n binary relation for P(S), for all x in S, A,B,C subset S not nulset n S (nullitive) {x}n{x} (reflective) AnB ==> BnA (symmetric) A n B\/C iff AnB or AnC (additive) B proximity nhood, nnhood A, A << B when not AnS\B
(S,n) proximity space when n proximity for S (S,n) separated when for all x,y, ({x}n{y} ==> x = y) (S,n) normal when for all A,B, not AnB ==> some disjoint U,V with A << U, B << V
> Then for cl(A) one has > > x near A implies that for every > entourage (surrounding) D there exists > some t in S such that > > (x,t) implies t in D[{x}] > and > (y,t) implies t in D[A] > > So, D[A] fixes the values of t > that determine whether any given > x is near. > > So, this should mean that > D[A] = D[cl(A)] > > so > cl(A) = cl(cl(A)) > > Anyway, its late. Maybe I screwed this up. > > But, it should give you a different way to > think about the problem.

