
Re: Nhood Space, correction
Posted:
Jul 1, 2013 5:13 AM


William Elliot wrote: > 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 axiom > > If {x} << U: x in int U = int int U; {x} << int U > If y in int U: y in int int U; {y} << int U > > Any 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 pnhood space would be a set U with > for all x in U, {x} << U. The showing int int U = int U would > be 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?
x is in the interior of A if A is a neighbourhood of x.
> As the largest open set contined in A?
Equivalent.
>>> 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. >
 I think I am an Elephant, Behind another Elephant Behind /another/ Elephant who isn't really there.... A.A. Milne

