```Date: Jul 1, 2013 4:50 AM
Author: William Elliot
Subject: Re: Nhood Space, correction

On Mon, 1 Jul 2013, Peter Percival wrote:> William Elliot 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.)This will not do.  Define int U = { x | {x} << U }.Now again is the need to prove int int U = int U.With than, if {x} << U, {x} << int U subset U,for all y in int U, {y} << int U will prove the fifth axiomIf {x} << U:  x in int U = int int U;  {x} << int UIf y in int U:  y in int int U;  {y} << int UAny notions how to prove int int A = A?> > 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?> > A set is open if it is a neighbourhood of each of its points.That's simpler and equivalent to what I surmised below.Simularly an open set in a p-nhood space would be a set U withfor all x in U, {x} << U.  The showing int int U = int U wouldbe equivalent to showing int U is open.> > U open when for all x in U, some V in N(x) with V subset U?> > Thus empty set is open.Hoe would the interior of a set A, be defined?As the largest open set contined in A?> > 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.
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