
Re: Nhood Space, correction
Posted:
Jul 1, 2013 4:50 AM


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? 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.

