
Re: Nhood Space
Posted:
Jul 1, 2013 12:13 AM


On Sun, 30 Jun 2013, William Elliot wrote: > On Sun, 30 Jun 2013, Peter Percival 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 > > > > Is this the same as neighbourhood space defined as follows. > > > > (S, N), S a set, N a map S > PPS (P for power set) and > > i) x in S => N(x) =/= 0 > > ii) x in S, M in N(x) => x in M > > iii) x in S, M in N(x) => (L superset M => L in N(x) > > iv) x in S, L, M in N(x) => L intersect M in N(x) > > v) x in S, M in N(x) => exists L in N(x) s.t. > > L subset M and, forall y in L, L in N(y)
Let's see. Define N(x) = { A  {x} << A } 1. For all x, {x} << S. 2. A << B implies A subset B 3. A << B and B subset C imply A << C 4. A << B/\C iff A << B, A << C 5. {x} << S and for all y in S, {y} << S. (Seems trivial.)
Conversely given N(x), how is A << B to be defined? {x} << A when A in N(x) ? A << B when for all a in A, {a} << B ?
Thus my second question is poignant. If for all a in A, {a} << B, does A << B?
If we take A << B to be cl A subset int B and let A = { 1/n  n in N } and B = [0,1], then not A << B.
So that rejects my second question leaving me wondering how to define A << B from the N(x)'s.
I think your nhood space is more general that my proximal nhood space. How is an open set defined using N(x)'s? U open when for all x in U, some V in N(x) with V subset U? Thus empty set is open.
It seems forthwith, that the collection of open sets defines a topology and every topology gives a nhood space, which if derived from N(x)'s, will give the same nhood space in return.
It seems proximal nhood spaces don't generate every topology and accordingly less general. They were, in fact, not intended to describe topological spaces but merely to generalize uniform spaces.

